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296,088 |
Thanks to the fibrations \begin{align*}
SO(n) \to SO(n+1) &\to S^n\\
SU(n) \to SU(n+1) &\to S^{2n+1}\\
Sp(n) \to Sp(n+1) &\to S^{4n+3}
\end{align*} we know that \begin{align*}
\pi_i(SO(n)) \cong \pi_i(SO(n+1)) \cong \pi_i(SO), \quad i &\leq n-2\\
\pi_i(SU(n)) \cong \pi_i(SU(n+1)) \cong \pi_i(SU), \quad i &\leq 2n - 1 = (2n+1) - 2\\
\pi_i(Sp(n)) \cong \pi_i(Sp(n+1)) \cong \pi_i(Sp), \quad i &\leq 4n+1 = (4n + 3) - 2.
\end{align*} These values of $i$ are known as the stable range. So the first unstable groups are $\pi_{n-1}(SO(n))$, $\pi_{2n}(SU(n))$, and $\pi_{4n+2}(Sp(n))$ respectively. I was able to find $\pi_{n-1}(SO(n))$ for $1 \leq n \leq 16$ by combining the tables on the nLab page for the orthogonal group and appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics . The groups are $$0, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2\oplus\mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}.$$ There doesn't seem to be any pattern here, so I guess that there is no general result for $\pi_{n-1}(SO(n))$. ( Feel free to correct me if I'm wrong .) I just noticed that every second term contains a copy of $\mathbb{Z}$, while every fourth term contains two copies. The case of $SU(n)$ is completely different: in The space of loops on a Lie group , Bott proved, among other things, that $\pi_{2n}(SU(n)) \cong \mathbb{Z}_{n!}$, see Theorem 5. Again consulting the Encyclopedic Dictionary of Mathematics, I was able to find $\pi_{4n+2}(Sp(n))$ for $n = 1, 2, 3$. The groups are $\mathbb{Z}_{12}$, $\mathbb{Z}_{120}$, and $\mathbb{Z}_{10080}$. This seems to suggest that this case is more similar to $SU(n)$ than $SO(n)$, so one might hope there is a Bott-type result. Is there an analogue of Bott's result for $Sp(n)$? That is, is there some increasing function $f : \mathbb{N} \to \mathbb{N}$ such that $\pi_{4n+2}(Sp(n)) \cong \mathbb{Z}_{f(n)}$? OEIS has no sequences beginning $12, 120, 10080$, so I have no guess what $f(n)$ could be. It is interesting to note that $12 \mid 120$ and $120 \mid 10080$ which is another similarity with the $SU(n)$ case. Of course, three groups is not much to go on, so this may be a completely misguided guess. Some questions that would be nice to answer before seriously hoping for such a result are: Is $\pi_{4n+2}(Sp(n))$ always cyclic? Is $\pi_{4n+2}(Sp(n))$ always finite? Is $|\pi_{4n+2}(Sp(n))|$ increasing in $n$? Any information regarding these three questions would also be interesting to know. Falling short of answering any of these questions, have any more of these groups (namely $\pi_{18}(Sp(4)), \pi_{22}(Sp(5)), \dots$) been computed? Update: I added the sequence $|\pi_{4n+2}(Sp(n))|$ to the OEIS: A301898 . Also, the answer to the question I asked was also in the Encyclopedic Dictionary of Mathematics on page 1746.
|
The answer appears to be in the paper Homotopy groups of symplectic groups by Mimura and Toda. They claim the calculation was already in a paper of Harris , but that was stated in terms of a symmetric space and it's not immediately obvious to me how to translate into information about the groups. They state that the group is $\mathbb Z_{(2n+1)!}$ if $n$ is even and $\mathbb Z_{(2n+1)! \cdot 2}$ if $n$ is odd, which agrees with your data.
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|
296,162 |
Let $X$ be a compact complex smooth manifold with holomorphically trivial canonical class. It is true that any (sufficiently small?) deformation of the complex structure of $X$ also has holomorphically trivial canonical class?
|
The answer in general is no. Nakamura has constructed here (pp.90, 96-99, solvmanifolds of type III-(3b)) an example of a compact complex (non-Kähler) manifold $M$ with $TM$ holomorphically trivial (so in particular $K_M$ is holomorphically trivial) which has arbitrarily small deformations $M_t$ with negative Kodaira dimension. On the other hand, if $M$ is compact Kähler with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ are Kähler (Kodaira-Spencer) and still have $K_{M_t}$ holomorphically trivial. Indeed, we have $c_1(K_{M_t})=0$ in $H^2(M_t,\mathbb{Z})$ (a topological condition), so by the Calabi-Yau theorem they admit Ricci-flat Kähler metrics $g_t$. On the other hand $\dim H^0(M_t,K_{M_t})=h^{n,0}(M_t)$ is locally constant hence equal to $1$, so you have a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. A Bochner formula gives $\Delta_{g_t}|\Omega_t|^2_{g_t}=|\nabla \Omega_t|^2_{g_t},$ which can be integrated on $M_t$ to see that $\Omega_t$ is parallel with respect to $g_t$, hence it must be nowhere vanishing. This gives you a holomorphic trivialization of $K_{M_t}$. Prompted by Piotr Achinger's comment below, let me also note the following generalization. If $M$ is compact complex with Hodge-de Rham (aka Frölicher) spectral sequence degenerating at $E_1$ (this is true for all compact Kähler manifolds), and with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ also have spectral sequence degenerating at $E_1$ (again by Kodaira-Spencer), and $K_{M_t}$ holomorphically trivial. Indeed, it is well-known that the degeneration at $E_1$ is equivalent to the equality
$$b_k(M)=\sum_{p+q=k}h^{p,q}(M),$$
for all $k$, so in particular the Hodge numbers $h^{p,q}(M_t)$ are locally constant. As above, you get a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. Now we don't have Ricci-flat Kähler metrics anymore, but by Proposition 1.6 and 1.1 of this paper , we can find Hermitian metrics $g_t$ on $M_t$ whose first Chern form (i.e. Chern-Ricci curvature) vanishes, and then the Bochner formula goes through essentially as before ($\nabla$ now being the Chern connection of $g_t$), see Lemma 2.1 in that paper.
|
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|
296,192 |
Let $L/K$ be a finite separable field extension and let $\theta$ be a primitive element for $L/K$ with minimal polynomial $\mu(t) \equiv \mu_{\theta/K}(t) = \sum_{k=0}^n c_k t^k$. I am trying to compute the powers $\theta^n,\dots,\theta^{2n-2}$ in terms of $1,\theta,\dots,\theta^{n-1}$. In other words, I am trying to compute the structure constants of the basis $1,\theta,\dots,\theta^{n-1}$ in terms of the coefficients of $\mu(t)$, but the expressions quickly become unmanageable. I can hardly imagine that I am the first one to attempt such a calculation, so my question is Are there perhaps some nice general formulas or manageable expressions/patterns that capture the structure constants of such a basis in terms of the coefficients of $\mu(t)$? Is there some general context where these appear? Or is it hopeless? Notice that using the dual basis with respect to the trace essentially seems to run into the same issues. I have looked into J.S.Milne's notes on Fields and Galois Theory and on Algebraic Number Theory as well as in Neukirch's Algebraic Number Theory, but in this regard they don't seem to go beyond calculating the discriminant.
|
The answer in general is no. Nakamura has constructed here (pp.90, 96-99, solvmanifolds of type III-(3b)) an example of a compact complex (non-Kähler) manifold $M$ with $TM$ holomorphically trivial (so in particular $K_M$ is holomorphically trivial) which has arbitrarily small deformations $M_t$ with negative Kodaira dimension. On the other hand, if $M$ is compact Kähler with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ are Kähler (Kodaira-Spencer) and still have $K_{M_t}$ holomorphically trivial. Indeed, we have $c_1(K_{M_t})=0$ in $H^2(M_t,\mathbb{Z})$ (a topological condition), so by the Calabi-Yau theorem they admit Ricci-flat Kähler metrics $g_t$. On the other hand $\dim H^0(M_t,K_{M_t})=h^{n,0}(M_t)$ is locally constant hence equal to $1$, so you have a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. A Bochner formula gives $\Delta_{g_t}|\Omega_t|^2_{g_t}=|\nabla \Omega_t|^2_{g_t},$ which can be integrated on $M_t$ to see that $\Omega_t$ is parallel with respect to $g_t$, hence it must be nowhere vanishing. This gives you a holomorphic trivialization of $K_{M_t}$. Prompted by Piotr Achinger's comment below, let me also note the following generalization. If $M$ is compact complex with Hodge-de Rham (aka Frölicher) spectral sequence degenerating at $E_1$ (this is true for all compact Kähler manifolds), and with $K_M$ holomorphically trivial, then its sufficiently small deformations $M_t$ also have spectral sequence degenerating at $E_1$ (again by Kodaira-Spencer), and $K_{M_t}$ holomorphically trivial. Indeed, it is well-known that the degeneration at $E_1$ is equivalent to the equality
$$b_k(M)=\sum_{p+q=k}h^{p,q}(M),$$
for all $k$, so in particular the Hodge numbers $h^{p,q}(M_t)$ are locally constant. As above, you get a nontrivial holomorphic section $\Omega_t$ of $K_{M_t}$. Now we don't have Ricci-flat Kähler metrics anymore, but by Proposition 1.6 and 1.1 of this paper , we can find Hermitian metrics $g_t$ on $M_t$ whose first Chern form (i.e. Chern-Ricci curvature) vanishes, and then the Bochner formula goes through essentially as before ($\nabla$ now being the Chern connection of $g_t$), see Lemma 2.1 in that paper.
|
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|
296,312 |
Do you know of any very important theorems that remain unknown? I mean results that could easily make into textbooks or research monographs, but almost
nobody knows about them. If you provide an answer, please: State only one theorem per answer. When people will vote on your answer they will vote on a particular theorem. Provide a careful statement and all necessary definitions so that a well educated graduate student
working in a related area would understand it. Provide references to the original paper. Provide references to more recent and related work. Just make your answer useful so other people in the mathematical community can use it right away. Add comments: how you discovered it, why it is important etc. Please, make sure that your answer is written at least as carefully as mine. I did invest quite a lot of time writing my answers. As an example I will provide three answers to this question. I discovered these results while searching for papers related to the questions I was working on.
|
Monotony is a superfluous hypothesis in the Monotone convergence theorem for Lebesgue integral .
In fact the following is true. Theorem - Let $(X, \tau, \mu)$ be a measurable space, $f_n : X \rightarrow [0,\infty]$ a sequence of measurable functions converging
almost everywhere to a function $f$ so that $f_n \leq f$ for all $n$.
Then $$\lim_{n\rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu.$$ Proof :
$$
\int_X f d\mu = \int_X \underline{\lim} \, f_n d\mu \leq \underline{\lim} \int_X f_n d\mu \leq \overline{\lim} \int_X f_n d\mu \leq \int_X f d\mu.
$$ I learnt this result from an article by J.F. Feinstein in the American Mathematical Monthly, but I never saw it in any textbook. Since the Monotone convergence theorem is important, I wish to argue that this is also an important theorem. Here is an illustration. Let $(X, \tau, \mu)$ be a measurable space, $f : X \rightarrow [0,\infty]$ a measurable function. Then
$$
\int_X f d\mu = \lim_{r \rightarrow 1, r>1} \sum_{n\in {\bf Z}} r^n \mu\Bigl( f^{-1}([r^n, r^{n+1}))\Bigr).
$$
Neither the dominated nor the monotone convergence theorem apply here. Note that this is a way to define the Lebesgue integral of nonnegative functions. Computing integrals by geometrically dividing the $x$ axis is due to Fermat.
|
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|
296,549 |
The Knight's Tour is a well-known mathematical chess problem. There is an extensive amount of research concerning this question in two/higher dimensional finite boards. Here, I would like to tackle this problem on the unbounded infinite board, $\mathbb{Z}\times\mathbb{Z}$ . The first issue is how to construct a knight's tour on the infinite plane. An intuitive way is to build it locally by mimicking the king's spiral tour in $\mathbb{Z}\times\mathbb{Z}$ board as follows: Roughly speaking, a knight may follow the king's pattern by completing a knight's tour on an $n\times n$ block and then moving to the next block in the spiral order. For instance, in the below picture each square represents a certain $n\times n$ part of the $\mathbb{Z}\times\mathbb{Z}$ plane and the blue path is a knight tour performed in that particular block. It is connected to a point in the neighbor blocks with a knight move. Of course, the existence of such a nice natural number $n$ should be checked. However, it sounds reasonable to expect that for a sufficiently large $n$ (which there are so many open knight tours with various beginning and ending points) there are some tours which can serve as the building blocks of the knight's king-like spiral. (Actually only a few straight and right -oriented $n\times n$ tours with appropriate beginning and ending points are needed. The other blocks will be redundant up to isomorphism). Here, the first question arises: Question 1. Is there (a concrete example of) a knight's tour on the infinite plane? In the special case, does the already described strategy for building a knight's tour locally actually work? If yes, what is the minimum required $n$ ? Assuming an affirmative answer to the above question, the next step is to speculate on the general structure of all possible knight tours on the infinite plane: Question 2. Do all infinite knight's tours on the infinite plane arise as a combination of local finite knight's tours on a grid of blocks of the same size (whose local knight's tours are attached to their neighbor counterparts in a king's move fashion)? Precisely, is it true that for any given knight's tour $t$ on the infinite plane $\mathbb{Z}\times\mathbb{Z}$ , there is a (probably large) natural number $n$ and a partition $p$ of the plane into $n\times n$ blocks in a grid such that the restriction of $t$ into any block in $p$ is an $n\times n$ knight's tour in that particular block itself? We call the partition $p$ a localization of the knight's tour $t$ into a grid of $n\times n$ blocks. If no, what is an example of an infinite knight's tour which can't be seen as a combination of local knight's tours in a grid of finite blocks of the same size (no matter what the resolution of the gird is)? If yes, what is: $$min\{n_t|~t~\text{is a knight's tour of $\mathbb{Z}\times\mathbb{Z}$ & there is a localization of}~t~\text{into a grid of}~n\times n~\text{blocks}\}$$ In better words, what is the finest possible resolution of a grid for a knight's tour of the infinite plane, $\mathbb{Z}\times\mathbb{Z}$ ? As a remark it is worth mentioning that not all partitions of the infinite plane into $n\times n$ blocks are in the form of a perfect grid. For example, see the left picture in the following diagram which is a partition of the plane via $n\times n$ blocks different from a grid of $n\times n$ blocks (right): So, for the sake of completeness, one may reduce the grid partition condition in question 2 to merely a partition of the infinite plane into $n\times n$ blocks: Question 3. What is the answer to question 2 if we merely consider partitions of the infinite plane into $n\times n$ blocks, rather than those partitions which arrange blocks in a perfect grid? The question could be asked in a more general sense as well: Question 4. Is the knight's tour on the infinite plane always decomposable into finite knight's tours? Precisely, is it true that for every knight's tour $t$ on the infinite plane there is a partition $p$ of the plane into finite squares (not necessarily consisting of blocks in a grid or of the same size) such that the restriction of $t$ to any block in $p$ is a finite knight's tour as well? Update. Thanks to Joel's answer and Eric's counterexample , the questions 1 and 2-4 have been answered positively and negatively respectively. Thus, we know that there are open knight's tours of the unbounded infinite plane, $\mathbb{Z}\times\mathbb{Z}$ , of different nature , namely those which are formed of local knight's tours of finite planes and those total tours which don't arise this way. In this sense the answer to the question in the title is: "Sometimes but not always!" Remark. As an additional piece of information, it is worth mentioning that there are some simple infinite sub-spaces of $\mathbb{Z}\times\mathbb{Z}$ -plane which don't admit a knight's tour while some others do. For example, it is easy to see that one can obtain a knight's tour for $\mathbb{N}\times\mathbb{N}$ sub-plane through a local block by block construction via Joel's $5\times 5$ blocks introduced in his answer. However, it is intuitively clear that the infinite strip (i.e. $[-n,+n]\times\mathbb{Z}$ for some $n$ ) doesn't admit a knight's tour because the knight in such a sub-space must cover both infinite sides of the strip in a back and forth movement and so it should pass the central part of the strip infinitely many times which is impossible simply because after finitely many passages there will be no empty room left in the central region. Anyway, it sounds an interesting question to classify all infinite sub-spaces of $\mathbb{Z}\times\mathbb{Z}$ -plane which admit a knight's tour. The same has already been done for all finite rectangle-shape sub-spaces of $\mathbb{Z}\times\mathbb{Z}$ -plane, namely $m\times n$ -boards.
|
Consider the following open knight's tour on a $5\times 5$ board, starting at position $1$ and then touring the $5\times 5$ board in the indicated move order. The final position is $25$, from which the knight can exit at either $A$ or $B$. A
5 22 17 12 7 B
16 11 6 25 18
21 4 23 8 13
10 15 2 19 24
C 3 20 9 14 1
D So the tour starts at a corner, and can exit in either direction at an adjacent corner. By a suitable reflection of the same tour, we could also arrange to exit at C or D, if desired. Therefore, with this pattern and its copies, we can proceed to tile any $5\times 5$ board and then arrange so as to enter any of the four adjacent $5\times 5$ boards on a corner as desired. This allows us to tile any sequence of $5\times 5$ boards that are each connected to the next by a common edge. In particular, we can carry out your spiral pattern, since in that pattern each block has a common edge with the next block. Thus, the answer is affirmative, there is a knight's tour of the infinite $\mathbb{Z}\times\mathbb{Z}$ chessboard. Indeed, since this method allows one to follow any tiling of the plane by $5\times 5$ blocks, where each block is attached to a next block by an edge, it follows that there are continuum such knight's tours.
|
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|
296,771 |
I’m working on a paper that makes heavy use of colorful diagrams to supplement the text. For most of these it would probably not be possible to create grayscale versions that convey the same information as effectively. I’m a bit worried about this because (1) I imagine that some people like to print out papers to read them but these people might not want/be able to print in color, and (2) some readers may be colorblind. What are the expectations on an author in my situation? Will it be considered rude to leave as-is, so long as the diagrams are not technically necessary to verify the arguments? Am I expected to include a description in the captions (“this region is red, this region is blue...”)? Or most stringently, would I be expected to post a “colorblind version” somewhere that tries to recreate the diagrams in grayscale as best as I can? I’m interested in all opinions, but especially those of people who would have trouble with color for whatever reason.
|
The concern with color figures for grey scale printing is less of an issue these days, when pretty much all displays are in color, so the reader can always check which color is which even if the document is printed in grayscale. (Many journals no longer insist that the figures should display well when printed in grayscale.) The issue with color blind readers is more substantial. Best practice ("etiquette") is to Avoid ambiguous color combinations: green & brown, blue & purple, green & blue, light green & yellow, blue & grey, green & grey, green & black; Use a high enough contrast, which most color blind people can still distinguish. Use textures instead of / in addition to colors; Add a label or graphical element to distinguish different colors. Here is a color-blind safe palette:
|
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|
297,281 |
$A$ and $B$ take turns to pick integers: $A$ picks one integer and then $B$ picks $k > 1$ integers ($k$ being fixed). A player cannot pick a number that his opponent has picked. If $A$ has $5$ integers that form an arithmetic sequence, then $A$ wins. $B$'s goal is to prevent that from happening. Who has a winning strategy (after finite number of turns)? Does it depend on $k$?
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I claim that Player A has a winning strategy in your game, and furthermore, it is a winning strategy for her simply to play the smallest available number. Let me consider the game along with several natural variations. Player A wins the original arithmetic progression game. To be a little more specific about the game, let us consider what I should like to propose we call the arithmetic progression game $G_k^n$ for positive integers $k$ and $n$, a two-player game of perfect information. On each turn, player A plays a new integer and player B plays up to $k$ integers, with all played numbers distinct; player A wins if there is an arithmetic progression of length $n$ amongst her numbers, and player B is aiming to prevent this. In this notation, your original game is $G_k^5$. These are all open games for player A, meaning that any win for player A, if it occurs, will occur at a finite stage of play. It follows by the Gale-Stewart theorem that one of the players will have a winning strategy. But in fact, we can explicitly describe a winning strategy for player A: let her simply play the smallest available positive integer. I find it interesting that the same strategy works uniformly in $n$ and $k$, and player A doesn't even have to know these game parameters in advance. Further, the strategy doesn't depend on the history of play, but only on the current sets of already-played numbers. To see that this is a winning strategy, consider any infinite play of the game played according to it. For such a play, the key observation to make is that this strategy ensures that the set of numbers played by A will have proportion at least $1/(k+1)$ in any initial segment of the positive integers. In particular, the set played by A will have positive asymptotic density. It follows by Szemerédi's theorem that the set played by A must contain arbitrarily long arithmetic progressions. So player A will have already won at some stage. So it is a winning strategy. Let us consider several natural variations of the game. Player A wins the arbitrarily long finite progressions game. Consider an infinite version of the game, denoted $G_k^{<\omega}$, where play continues through all finite stages, and player A wins such an infinite play if the set of her numbers contains arithmetic progressions of every finite length. The argument above using Szemerédi's theorem shows that the play-the-smallest-available-number strategy is still a winning strategy for player A in this modified game, since she can ensure that her set has positive density and therefore contains arithmetic progressions of any desired length. Player B wins the infinite arithmetic progression game. If we modify the winning condition of the game, however, to the game $G_k^\omega$, where player A wins only when her set contains an infinite arithmetic progression, then I claim that player B has a winning strategy. The reason is that there are only countably many infinite arithmetic progressions in the integers, since each is determined by its starting point and the difference. Since there are only finitely many numbers played by any given finite stage of play, it follows that player B can block the $n^{th}$ infinite arithmetic progression with a single number played on move $n$. So even when $k=1$, player B has a winning strategy to block an infinite arithmetic progression in A's set. Player B wins the variant where $k$ is not fixed. Lastly, consider the version of the game where $k$ is not fixed, so that player B is free to play an arbitrary finite set on each move. We may denote it by $G_{<\omega}^n$. For this variant, player B has a winning strategy to block all arithmetic progressions even of length $n=3$ for A. The strategy is simply to fill in all the numbers up to double the current largest number played. In this way, no three numbers played by A can have their differences in an arithmetic progression, since each next number played by A has a larger difference than any two previous numbers. So this is a winning strategy to prevent A from forming an arithmetic progression of length $3$. Further remarks and questions. In the original $G_k^n$ game, the play-the-smallest-available-number strategy is a winning strategy, but I am not sure how long this takes to win, or whether this strategy is efficient in any sense in winning. Perhaps the finite combinatorics experts can place bounds on how long the game will take. Presumably there is a number $N$, depending on $n$ and $k$, such that player A can force a win in at most $N$ moves, but I don't know how big $N$ will be or even whether there is such an $N$. The existence of such a number $N$ is equivalent to the assertion that this game has a finite game value (but note that not all open games have a finite game value). Question 1. How quickly can player A win the game $G_k^n$? Does the play-the-smallest-available-number strategy achieve the best bound? Question 2. For fixed $n$ and $k$, is there a uniform bound $N_k^n$ for the length of play, not depending on the moves of player B? In the game where $k$ is not fixed, one can imagine the game $G_f^n$, where the function $f:\mathbb{N}\to\mathbb{N}$ is used to specify the size $f(i)$ of the set played by B on move $i$. The argument I gave shows that if $f$ is sufficiently exponential, then player B wins. Meanwhile, constant functions $f(i)=k$ correspond to the original arithmetic progression game. Question 3. For which functions $f$ does player A still have a winning strategy? For example, if $f$ is linear or even polynomial, does player A have a winning strategy? What if $f=o(2^i)$?
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297,872 |
Let's say players take turns placing numbers 1-9 on a sudoku board. They must not create an invalid position (meaning that you can not have the same number in within a row, column, or box region). The first player who can't move loses, and the other player wins. Given a partially filled sudoku board, what is a way to evaluate the winner (or even better, the nimber) of the position (besides brute force)? Additionally, has this perhaps been analyzed before? (A natural generalization is to allow certain players to only play certain digits, in which case each position can be assigned a CGT game value.)
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Update. I made a blog post about Infinite Sudoku and the Sudoku game , following up on ideas in this post and the comments below. I claim that the second player wins the even-sized empty Sudoku boards and the first player wins for odd-sized empty Sudoku boards, including the main $9\times 9$ case. (The odd-case solution uses a key idea of user orlp in the comments.) Consider first the even-sized board case, which is a little easier. For example, perhaps we have a board of size $16\times 16$, divided into subsquares of size $4\times 4$. The second player can win in this case by the mirror-play strategy. (This argument was pointed out to me by my daughter, 11 years old.) That is, given any move by the other player, let the second player play the mirror image of that move through the origin. The new play cannot violate the Sudoku condition if the previous play did not, since the new violation would reflect to an earlier violation. The point is that on an even-sized board, the reflection of any row, column or subsquare will be a totally different row, column or subsquare, and so by maintaining symmetry, the second player can ensure that any violation of the Sudoku conditions will arise with the first player. This copying strategy breaks down on the odd-sized board, however, including the main $9\times 9$ case, since there is a central row and column and a central subsquare and copying a move there would immediately violate the Sudoku conditions. Nevertheless, user orlp explained in the comments how to adapt the mirroring strategy to the odd case. Namely, in the main $9\times 9$ case, let's have the first player play a $5$ in the center square, and thereafter play the ten's complement mirror image of the opponent's moves. That is, if opponent played $x$, the first player should play $10-x$ in the mirror location. In this way, the first player can ensure that after her moves, the board is ten's complement symmetric through the origin. This implies that any violation of the Sudoku requirement will reduce by reflection to an earlier violation in the reflected moves, and so it is a winning strategy. More generally, in the general odd case $k^2\times k^2$ for $k^2=2n-1$, player one will play $n$ in the middle square, and then proceed to play the $2n$'s complement mirroring move of the opponent. In this way, the first player ensures that after her play, the board remains $2n$'s complement symmetric, and this implies that she will not be the first to violate the Sudoku conditions. So it is a winning strategy. Notice that in the even case, the second player could also have won by playing the complement mirror strategy, rather than the mirror strategy, since again any violation of the Sudoku condition would reflect to an earlier but complementary violation. Finally, see my blog post for the winning strategy in the case of the Infinite Sudoku game , which came up in the comments.
|
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|
297,880 |
In the book Discriminants, Resultants, and Multidimensional Determinants of Andrei Zelevinsky,M.M. Kapranov and Izrail' Moiseevič Gel'fand, the authors give the following definition of degree of a hypersurface in a Grassmannian. As they say, in generale a hypersurfaces in a projective variety is not given by the vanishing of a polynomial in its coordinate ring, but for Grassmannians this is true, since its coordinate ring is a UFD, therefore every height-one prime is principal by Krull Theorem. However I'm stuck on the definition of degree of a hypersurface in a Grassmannian. To be more precise...I wuold prove that this definition is well posed, as in the case of projective hypersurfaces: The set of pencils $P_{NM}$ is parametrized by the flag variety $\mathcal{Fl}(n;k-1,k+1):=\lbrace N\subset M \: : \: \dim N=k-1, \dim M=k+1 \rbrace$. I would check that there exists a nonempty open subset $U$ of $\mathcal{Fl}(n;k-1,k+1)$ such that the intersection number of $Z$ with any pencil $P_{NM}$ in $U$ is equal to the maximal number of intersection points of $Z$ with a pencil $P\not\subset Z$. Any help?
|
Update. I made a blog post about Infinite Sudoku and the Sudoku game , following up on ideas in this post and the comments below. I claim that the second player wins the even-sized empty Sudoku boards and the first player wins for odd-sized empty Sudoku boards, including the main $9\times 9$ case. (The odd-case solution uses a key idea of user orlp in the comments.) Consider first the even-sized board case, which is a little easier. For example, perhaps we have a board of size $16\times 16$, divided into subsquares of size $4\times 4$. The second player can win in this case by the mirror-play strategy. (This argument was pointed out to me by my daughter, 11 years old.) That is, given any move by the other player, let the second player play the mirror image of that move through the origin. The new play cannot violate the Sudoku condition if the previous play did not, since the new violation would reflect to an earlier violation. The point is that on an even-sized board, the reflection of any row, column or subsquare will be a totally different row, column or subsquare, and so by maintaining symmetry, the second player can ensure that any violation of the Sudoku conditions will arise with the first player. This copying strategy breaks down on the odd-sized board, however, including the main $9\times 9$ case, since there is a central row and column and a central subsquare and copying a move there would immediately violate the Sudoku conditions. Nevertheless, user orlp explained in the comments how to adapt the mirroring strategy to the odd case. Namely, in the main $9\times 9$ case, let's have the first player play a $5$ in the center square, and thereafter play the ten's complement mirror image of the opponent's moves. That is, if opponent played $x$, the first player should play $10-x$ in the mirror location. In this way, the first player can ensure that after her moves, the board is ten's complement symmetric through the origin. This implies that any violation of the Sudoku requirement will reduce by reflection to an earlier violation in the reflected moves, and so it is a winning strategy. More generally, in the general odd case $k^2\times k^2$ for $k^2=2n-1$, player one will play $n$ in the middle square, and then proceed to play the $2n$'s complement mirroring move of the opponent. In this way, the first player ensures that after her play, the board remains $2n$'s complement symmetric, and this implies that she will not be the first to violate the Sudoku conditions. So it is a winning strategy. Notice that in the even case, the second player could also have won by playing the complement mirror strategy, rather than the mirror strategy, since again any violation of the Sudoku condition would reflect to an earlier but complementary violation. Finally, see my blog post for the winning strategy in the case of the Infinite Sudoku game , which came up in the comments.
|
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|
297,942 |
Things like the first-order completeness theorem and the Löwenheim-Skolem theorem are considered foundational in mathematical logic. The modern approach seems to be, usually, to interpret a "model" specifically as a set in some other (typically first-order) "set metatheory." So when we talk about a model of PA, for instance, we typically mean that we are formalizing it as a subtheory of something like first-order ZFC. Models of ZFC can be formalized in stronger set theories, such as those obtained by adding large cardinals, etc. But when Godel proved that a first-order sentence has a finite proof if and only if it holds in every "model" -- what was he talking about? Likewise, how can we understand the Löwenheim-Skolem theorem if models didn't even exist at the time? It is clear that these researchers were not talking about using first-order ZFC as a metatheory, as that theory didn't even gain popularity until after Cohen's work on forcing in the 60s. Likewise, NBG set theory had not yet been formalized. And yet, they were obviously talking about something . Did they have a different notion of semantics than the modern set-theoretic one? In closing, two questions: In general, how did early researchers (let's say pre-Cohen) formalize semantic concepts such as these? Have any of these original works been translated into English, just to see directly how they treated semantics?
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I don’t know the history well enough for a full answer, but here is a partial answer, on the mathematical aspects. When you write: It is clear that these researchers were not talking about using first-order ZFC as a metatheory […] And yet they were obviously talking about something. Did they have a different notion of semantics than the modern set-theoretic one? and The modern approach seems to be, usually, to interpret a "model" specifically as a set in some other (typically first-order) "set metatheory." you seem to be following a somewhat common misconception: that one can’t do set-based semantics without having some set theory in mind as a metatheory. But this isn’t the case! The fundamental definition of a (Tarskian) model is just as a set with certain extra structure — just like a group, or a ring, or similar. Not “a set in ZFC”, or “a set in NBG”, but just a set, which we can then reason about using whatever techniques and principles we use for mathematical reasoning in general. Of course, in that reasoning, we’re likely to follow some established principles, like those justified by ZFC or NBG or some other specific theory. (Historically, such foundational theories were developed exactly to try to codify/justify the principles generally used and accepted.) And logicians are, for a variety of reasons, more likely than other mathematicians to be explicit about what principles they’re following in a particular piece of work. But fundamentally, you don’t need an explicit set-theoretic metatheory to study set-based semantics, any more than you need one to study groups or rings or Riemann surfaces. As I said, I’m not especially well-read historically, but from the papers I’ve read from that period, my impression is mostly that most researchers in the period were using the modern (Tarskian) notion of semantics, and that some authors wrote explicitly about what sort of metatheory they were using, while others didn’t. But the lack of an explicit metatheory is not any failure of rigour or clarity in their notion of models — it’s normal mathematical practice, certainly of the time and at least arguably of today as well.
|
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297,971 |
Suppose, $G = \mathbb{Z} \ast H$, where $H$ is an arbitrary group. Suppose, $g \in G$ and $g \notin \langle\langle H \rangle \rangle $.
Is $\langle\langle g \rangle \rangle \cap H$ always trivial?
($\ast$ stands for free product, and $\langle \langle \dots \rangle \rangle$ stands for normal closure) Yesterday, I have asked this question on math.stackexchange.com and was advised to re-ask it there: Is the intersection of two subgroups, defined below, always trivial? v2.0 Any help will be appreciated.
|
Your question is related to a famous conjecture: Kervaire Conjecture: Given a non-trivial group $H$ and an element $g \in H \ast \mathbb{Z}$, the quotient $(H \ast \mathbb{Z} ) / \langle \!\langle g \rangle\!\rangle$ is non trivial. In fact, a positive answer to your question turns out to be equivalent to the strong Kervaire conjecture: Strong Kervaire Conjecture: Given a non-trivial group $H$ and an element $g \in H \ast \mathbb{Z}$, the canonical map $$H \to (H \ast \mathbb{Z}) / \langle\!\langle g \rangle\!\rangle$$ is injective if $\exp(g) \neq 0$. In the statement, $\exp(g)$ refers to the sum of exponents $\sum\limits_{i=1}^r \epsilon_i$ if you write $g$ as a reduced word $h_1 t^{\epsilon_1} \cdots h_r t^{\epsilon_r} h_{r+1}$ where $h_1, \ldots, h_r \in H$ and $t$ is a fixed generator of $\mathbb{Z}$. Notice that the condition $\exp(g) = 0$ is equivalent to $g \in \langle \langle H \rangle \rangle$, and that the map above is injective precisely when $\langle\! \langle g \rangle\!\rangle \cap H$ is trivial. You can find relevant references in this document: Presentation of the Kervaire conjecture (notes by A. Ould Houcine) .
|
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|
298,600 |
I search for a big list of open problems which have been solved in a PhD thesis by the Author of the thesis (or with collaboration of her/his supervisor). In my question I search for every possible open problem but I prefer (but not limited) to receive answers about those open problems which had been unsolved for at least (about) 25 years and before the appearance of the ultimate solution, there had been significant attentions and efforts for solving it. I mean that the problem was not a forgotten problem. If the Gauss proof of the fundamental theorem of algebra did not had a
gap, then his proof could be an important example of such dissertations. I ask the moderators to consider this question as a wiki question.
|
I find George Dantzig's story particularly impressive and inspiring. While he was a graduate student at UC Berkeley, near the beginning of
a class for which Dantzig was late, professor Jerzy Neyman wrote two
examples of famously unsolved statistics problems on the blackboard.
When Dantzig arrived, he assumed that the two problems were a homework
assignment and wrote them down. According to Dantzig, the problems
"seemed to be a little harder than usual", but a few days later he
handed in completed solutions for the two problems, still believing
that they were an assignment that was overdue. Six weeks later, Dantzig received a visit from an excited professor
Neyman, who was eager to tell him that the homework problems he had
solved were two of the most famous unsolved problems in statistics.
Neyman told Dantzig to wrap the two problems in a binder and he would
accept them as a Ph.D. thesis. The two problems that Dantzig solved were eventually published in: On the Non-Existence of Tests of "Student's" Hypothesis Having Power Functions Independent of σ (1940) and in On the Fundamental Lemma of Neyman and Pearson (1951).
|
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|
298,609 |
On a smooth maniflod $M$ of dimension $n$, a current of degree $n-p$ is a functional on the space of compactly supported differential $p$-forms which is continuos. We denote the space of currents of degree $n-p$ by $D^{'n-p}(M)$. If we consider the functionals on $D^{'n-p}(M)$ with an approperiate comapctness and continuity assumptions, then what are these functionals? Are they just differential $p$-forms, or could be more then that? I am asking this question because I want to understand why there is no definition for the pullback of current $T$ in general (as far as I know). Let $f: M_1 \to M_2$ be a map between manifolds. Then the pullback $f^*T$ should be defined as (formally) $$\langle f^*T, u\rangle = \langle T, f_*u \rangle.$$ Here $u$ is a differential $p$-form, in particular, $f_*u$ is well-defined as a current under some compactness assumption. Thus the problem is to make sense of $\langle T, f_*u \rangle$, i.e. can a current be a functional on the space of current? (Of course, it is enough to have $T$ be a functional on the space $\{f_*u\}$, that is why pullback of current is well-defined for submersion maps).
|
I find George Dantzig's story particularly impressive and inspiring. While he was a graduate student at UC Berkeley, near the beginning of
a class for which Dantzig was late, professor Jerzy Neyman wrote two
examples of famously unsolved statistics problems on the blackboard.
When Dantzig arrived, he assumed that the two problems were a homework
assignment and wrote them down. According to Dantzig, the problems
"seemed to be a little harder than usual", but a few days later he
handed in completed solutions for the two problems, still believing
that they were an assignment that was overdue. Six weeks later, Dantzig received a visit from an excited professor
Neyman, who was eager to tell him that the homework problems he had
solved were two of the most famous unsolved problems in statistics.
Neyman told Dantzig to wrap the two problems in a binder and he would
accept them as a Ph.D. thesis. The two problems that Dantzig solved were eventually published in: On the Non-Existence of Tests of "Student's" Hypothesis Having Power Functions Independent of σ (1940) and in On the Fundamental Lemma of Neyman and Pearson (1951).
|
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|
298,749 |
Starting to write an introduction to the philosophy of mathematics, I find tons of positions that are of historical interest. Which philosophical positions are explicitly considered these days, say in the last ten years?
|
Let me mention a few current issues on which I have been involved in the philosophy of
mathematics. Of course there are also many other issues on which people are working. Debate on pluralism. First, there is currently a lively or indeed raging
debate on the issue of pluralism in the philosophy of set theory.
If one takes set theory as a foundational theory, in the sense that
essentially every mathematical argument or construction can be
viewed as taking place or modeled within set theory (whether or not
it could also be represented in other foundational theories), then
the question arises whether set-theoretic questions have
determinate answers. On the singularist or universist view, every
set-theoretic question has a final, determinate truth value in the
one true set-theoretic universe, the Platonic realm of set theory.
On the pluralist or multiverse views, we have different conceptions
of set giving rise to different set-theoretic truths. Both views
are a form of realism, and so the debate breaks apart the question
of realism or Platonism from the question of the uniqueness of the
intended interpretation. I discuss these issues at length in my paper Hamkins, Joel David, The set-theoretic multiverse , Rev. Symb. Log. 5, No. 3, 416-449 (2012). DOI:10.1017/S1755020311000359 , ZBL1260.03103 . There is a growing literature discussing these issues, part of which you can find here . See also the multiverse tag on my blog. Potentialism. Another currently active topic of research is the issue of potentialism. This topic arises classically in the idea of potential versus
actual infinity, with quite a long history, but current work is
looking into various aspects of the classical debate, particularly
with respect to considering potentialism as a modal theory. That is, one separates the potentialist idea from the issue of
infinity, and looks upon the potentialism as concerned with the
idea of having a family of partial universes, or universe
fragments, which can be extended to one another as the universe
unfolds. Øystein Linnebo has emphasized this modal nature to potentialism,
and he and I undertook to analyze the precise modal commitments of
various kinds of set-theoretic potentialism in our paper: Joel David Hamkins and Øystein Linnebo, The modal logic of set-theoretic potentialism and the potentialist maximality principles , arXiv:1708.01644 . See also the slides for the tutorial lectures series I gave recently on Set-theoretic potentialism, Winter School in Abstract Analysis 2018 . For an example, I believe that the use of Grothendieck-Zermelo
universes in category theory exemplifies the potentialist outlook,
since one works inside a given universe until a need arises for a
larger universe concept, in which case one freely moves to the
larger universe concept. My work on arithmetic potentialism in Joel David Hamkins, The modal logic of arithmetic potentialism and the universal algorithm , arxiv:1801.04599 . provides a way to understand the philosophy of ultrafinitism,
viewing it ultimately as a form a potentialism.
|
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|
298,807 |
Eilenberg and Mac Lane formally defined categories in their 1945 paper General Theory of Natural Equivalences . Their definition of a category starts as follows: " A category {A,a} is an aggregate of abstract elements A (for example, groups), called the objects of the category and etc. " When they consider their first example of categories on P. 239, namely the category of all sets, they immediately remark: " This category obviously leads to paradoxes of set theory. " Obviously, all sets collected together don't form a set. However, their definition does not require the objects of a category to form a set. They use the term "aggregate" in their definition, perhaps to avoid this particular issue. That is, a set can be an aggregate, a class (in the sense of NBG) can also be an aggregate. Since the "aggregate" of all sets is a class, I do not see the "paradox" that they say will arise from considering the category of all sets. Unless by "aggregate" they really meant set. So my question is: what is the paradox that they were referring to? Please note, my question is specific to Eilenberg and Mac Lane's comment about the category of all sets. Obviously, there are other paradoxes caused by their definition of categories, and allowing the notion of a class doesn't eliminate all set-theoretic paradoxes from category theory. But I am not asking about these topics. On the other hand, as pointed out on page 245 of R. Kromer's book (Tool and Object): "Eilenberg and Mac Lane use the term 'set' in the combination the set of all objects of [a] category , on page 238 of their paper." Is this an evidence that by "aggregate" they really meant "set"?
|
I interpret the question as not being about the problems of size in category theory in general and how to deal with them (which are fairly well-understood and the subject of other questions on this site), but about what Eilenberg and MacLane actually meant in their original paper. The phrasing of that particular footnote is sloppy, but I think section 6 of their paper ("Foundations") suggests that what they meant is that "this category would lead to paradoxes if we required the objects of a category to form a set rather than something like a proper class". My guess is that they used the word "aggregate" in the definition in section 2 as a nod to the fact that to be formal, one may want to take these to be proper classes (or something related), but assumed that the average mathematician reading the paper would interpret "aggregate" as "set" at least until they got to section 6. So they added a footnote pointing out that they were aware of the issue, but deferred a fuller discussion of it (and an explanation of what "aggregate" can formally be defined to mean, or other ways one can deal with the problem while still interpreting "aggregate" as "set") to the later section. For instance, in section 6 they wrote "we have chosen to adopt the intuitive standpoint, leaving the reader free to insert whatever type of logical foundation (or absence thereof) he may prefer".
|
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299,304 |
I had posted the following problem on stack exchange before. Suppose $\lambda$ is a real number in $\left( 0,1\right)$, and let $n$ be a positive integer. Prove that all the roots of the polynomial $$
f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}
$$ have modulus $1$. I do not seem to know how to do it, to show if the roots of the polynomial have modulus one. Putnam 2014 B4 Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers.
This problem is very similar to the Putnam problem.
|
This is a special case of the Lee-Yang theorem. Let $G$ be a finite graph with vertex set $V$ and edge set $E$. Let $\beta \in (0,1)$ be a real number. Let $\sigma$ denote a ``spin function" $\sigma: V \to \{ -1, 1\}$. Put $m(\sigma)$ to be the number of vertices with positive spin, and $d(\sigma)$ to be the number of edges with vertices of opposite spin. Form the polynomial
$$
Z_\beta(x) = \sum_{\sigma} \beta^{d(\sigma)} x^{m(\sigma)},
$$
where the sum is over all possible choices of the spins $\sigma$. The Lee-Yang theorem then states that all the zeros of $Z_{\beta}(x)$ lie on the unit circle. Amazing! Apply the Lee-Yang theorem to the complete graph on $n$ vertices. The polynomial $Z_{\beta}(x)$ in this case is exactly $\sum_{k=0}^{n} \binom{n}{k} \beta^{k(n-k)} x^k$, and so this polynomial has all its roots on the unit circle. For a discussion of the Lee-Yang theorem see these notes from a course by Nikhil Srivastava. The original Lee-Yang paper can be found here (behind a paywall). For more work on the Lee-Yang theorem (among other things), see Borcea and Branden . In particular, Theorem 8.4 there states the following more general form of the Lee-Yang theorem: Let $A=(a_{ij})$ be a Hermitian $n\times n$ matrix with all entries inside the closed unit disc. Then the polynomial
$$
f(z) = \sum_{S\subset [n]} z^{|S|} \prod_{\substack{i\in S \\ j\not\in S}} a_{ij}
$$
has all its zeros on the unit circle. (Here $S$ runs over all subsets of $[n]=\{1, \ldots, n\}$.)
|
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299,572 |
$\def\Cl{\mathcal C\ell}
\def\CL{\boldsymbol{\mathscr{C\kern-.1eml}}(\mathbb R)}$
I'm not an expert in neither of the fields I'm touching, so don't be too rude with me :-) here's my question. A well known definition of Clifford algebras is the following: Fix a quadratic form $q$ on a vector space $V$[¹] and consider the quotient of the tensor algebra $T(V)$ under the ideal $\mathfrak i_q = \langle v\otimes v - q(v){\bf 1}\mid v\in V\rangle$; Remark : as such quotient enjoys a suitable universal property, it is easy to see that $(V,q)\mapsto \Cl(V,q)$ determines a functor from the category of quadratic spaces to the category of algebras. Remark: It's easy to see that if $q=0$ the Clifford algebra $\Cl(V,0)$ coincides with the exterior algebra $\bigwedge^*\! V$ (this is clear, $\mathfrak i_0=\langle v\otimes v\mid v\in V\rangle$!). Now. I would like to know to which extent it is possible to regard Clifford algebras as "deformations of exterior algebras": I believe that fixing a suitable real number $0<\epsilon \ll 1$, and a map $t\mapsto q_t$ defined for $t\in]-\epsilon,\epsilon[$ with codomain the space of quadratic forms, a useful interpretation of this construction regards $\Cl(V,q_\epsilon)$ as a suitably small deformation of $\bigwedge^*\! V$.[²] I can try to be more precise, but this pushes my moderate knowledge of differential geometry to its limit. I'm pretty sure there is a topology on the don't say moduli, don't say moduli --"space" of Clifford algebras $\CL$ so that I can differentiate those curves $\gamma : I \to \CL$ for which $\gamma(0)=\bigwedge^*\!V$. If this is true, what is the linear term of a "series expansion"
$$\textstyle
\gamma(t) \approx \bigwedge^*\!V + t(\text{higher order terms})?
$$ Is there any hope to prove that the functor $\Cl$ is "analytic" in the sense that there are suitable vector-space valued coefficients $a_n(V,q)$ for which
$$\textstyle
\Cl(V,q) \cong \bigoplus_{n\ge 0} a_n(V,q)\otimes V^{\otimes n}
$$
or even less prudently: $\Cl(V,q) \cong \bigoplus_{n\ge 0} a_n(V,q)\otimes (\bigwedge^*V)^{\otimes n}$ (I'd call these "semi-analytic", or something)? Regarding quadratic forms as symmetric matrices, one can wonder what are the properties of Clifford algebras induced by the fact that $\|q\|\ll 1$. For example, an exterior algebra $\Cl(V,0)$ decomposes as a direct sum of homogeneous components
$$\textstyle
\Cl(V,0) \cong \bigoplus_{p=0}^{\dim V} \bigwedge^p\!V
$$
is there a similar decomposition of $\Cl(V,q_\epsilon)$ for a sufficiently small $\epsilon$[³]? I can't help but admit I wasn't able to come up with a good idea for this question. :-) especially because well, I'm not a geometer. [¹] I will assume that vector spaces are finite dimensional, over a field of characteristic zero, sorry if you like arithmetic geometry! Even better, if you want, vector spaces are real. [²] I guess it is possible to say simply that I'm given a "Clifford bundle" $p : E \to \mathbb R$, and the preimage of a small neighbourhood of zero is a neighbourhood of $(V, 0)$ -the zero quadratic space over the typical fiber of $p$. [³] I must admit this might be a trivial question: I remember that $\Cl(V,q)$ has an even-odd decomposition in a natural way, but I can't find a reference for a complete decomposition in irreducibles. And, well, I don't have a book on this topic :-)
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As Igor mentions in the comments, this is really a question about deformations of the multiplication map of the exterior algebra in the space of associative multiplications. Since this is a pretty general story, let me try to sketch how it works in this case. Let's fix some field $k$ and a $k$ -algebra $(A,\mu)$ . We want to understand deformations of $(A,\mu)$ , and to start only infinitesimal deformations. For this we fix an Artinian augmented $k$ -algebra $R\to k$ and build a groupoid $\operatorname{Def}_R$ : objects are flat $R$ -algebras $B$ with an isomorphism $\rho: B\otimes_R k\cong A$ , and morphisms from $(B,\rho)$ to $(B',\rho')$ are isomorphisms of $R$ -algebras over $k$ . $R$ being Artinian just means that $R \cong k\oplus\mathfrak m$ , with $\mathfrak m$ being nilpotent and finite-dimensional. In fact, one does not lose all that much by restricting to $R_n = k[\hbar]/(\hbar^{n+1})$ . By flatness, any $R_n$ -deformation is of the form $B\cong R_n\otimes_k A\cong A[\hbar]/(\hbar^{n+1})$ , but with an interesting multiplication $\mu_{B}$ . It is determined by $\mu_B(a,a') = \sum_{i=0}^n\hbar^i\mu_i(a,a')$ for some maps $\mu_i:A\otimes_k A\to A$ . Similarly, any automorphism $\lambda$ of $B$ is determined by $\lambda(a) = \sum_{i=0}^n\hbar^n\lambda_i(a)$ for some maps $\lambda_i:A\to A$ . Then the condition that $\mu_B$ is associative leads to the following constraints: $\mu_0 = \mu$ by checking compatibility with $\rho$ . From now on, I'll write $\cdot = \mu$ . Similarly, $\lambda_0 = \operatorname{id}$ . $\mu_1(f,g)h + \mu_1(fg,h) = f\mu_1(g,h) + \mu_1(f,gh)$ , and $\lambda$ is a morphism from $\mu_1$ to $\mu_2$ iff $\mu_1(f,g) - \mu_2(f,g) = \lambda_1(fg) - f\lambda_1(g) - \lambda_1(f)g$ . The first condition is that $\mu_1$ is a closed $1$ -cocycle in the Hochschild chain complex $C^n(A) = \operatorname{Hom}(A^{\otimes(n+1)},A)$ with differential $$
\mathrm d\mu(a_0,\dots,a_{n+1}) = a_0\mu(a_1,\dots,a_n) + \sum_{i=0}^{n-1} (-1)^{i+1}\mu(a_0,\dots,a_ia_{i+1},\dots,a_n) + (-1)^{n+1}\mu(a_0,\dots,a_{n-1})a_n\ .
$$ The multiplications of two such cocycles are isomorphic iff the cocycles are cohomologous. In this case the isomorphisms are a torsor over the set of closed $0$ -cycles. All in all, the category $\operatorname{Def}_{k[x]/x^2}$ has objects in bijection with $HH^1(A)$ , and each object has automorphism group $HH^0(A)$ , where $HH^i(A) = H^i(C^*(A))$ . $f\mu_2(g,h) - \mu_2(fg,h) + \mu_2(f,gh) - \mu_2(f,g)h = \mu_1(f,\mu_1(g,h)) - \mu_1(\mu_1(f,g),h)$ . Thus given a first-order deformation $\mu_1$ , one may form the element $\mu_1(-,\mu_1(-,-)) - \mu_1(\mu_1(-,-),-)$ , which one can check to be a closed $2$ -cocycle. The existence of $\mu_2$ is equivalent to the corresponding cohomology class vanishing. Thus for every first-order deformation there is an obstruction lying in $HH^2(A)$ which vanishes iff this first-order deformation extends to a second-order deformation. In this case the extensions are a torsor over $HH^1(A)$ , and similarly for the automorphisms. This story continues: If we have found $\mu_1,\dots,\mu_n$ , the existence of $\mu_{n+1}$ is equivalent to some obstruction class in $HH^2(A)$ vanishing, in which case the posible extensions are a torsor over $HH^1(A)$ . Note however that the obstruction depends on all previous choices. In fact, essentially all of these formal deformation problems arise in the following way (at least if $k$ is characteristic $0$ ): One has a dgla $(\mathfrak g,[-,-])$ and a natural isomorphism $\operatorname{Def}_{k\oplus\mathfrak m} \cong \{x\in \mathfrak m\otimes \mathfrak g^1\mid \mathrm dx + \frac{1}{2}[x,x] = 0\}//\mathfrak m\otimes \mathfrak g^0$ , where the nilpotent Lie algebra $\mathfrak m\otimes \mathfrak g^0$ acts on this set of "Maurer-Cartan elements" by the Lie bracket and one takes the action groupoid of the action of the corresponding nilpotent Lie group. In our case the dgla is $C^*(A)$ equipped with the Gerstenhaber bracket which takes two cocycles and takes the sum over all ways to plug one into the other (with some signs). In general, the deformation functor only depends on the dgla up to quasiisomorphism. This has one important consequence: If $\mathfrak g$ is formal, i.e. quasiisomorphic to its cohomology, then a Maurer-Cartan element of $\mathfrak g$ is "the same" as an element $x\in H^1(\mathfrak g)$ satisfying $[x,x] = 0$ , i.e. a first-order deformation which extends to a second-order deformation. This then gives rise to a Maurer-Cartan element for any $R$ , i.e. a lift to arbitrarily high order in $\hbar$ , without having to check whether any more obstructions vanish! One can then put all of these lifts together to obtain an associative multiplication $\mu_\hbar = \mu + \cdots$ on $A[[\hbar]]$ . For a differential geometer, the most interesting case is that $A = C^\infty(M)$ . In this case one should restrict to the subcomplex of the Hochschild complex of maps $C^\infty(M)^{\otimes(n+1)}\to C^\infty(M)$ which are differential operators in each variable. The Hochschild-Kostant-Rosenberg theorem computes the homology of this complex as polyvector fields $\oplus_{i\ge 1} \Gamma(M,\Lambda^i TM)$ . The bracket on the homology groups is the Schouten bracket (extension of the Lie bracket as a biderivation). Amazingly, as proven by Kontsevich this dgla is actually formal! By the previous discussion this means that a first-order deformation, i.e. a biderivation $\pi\in\Gamma(M,\Lambda^2 TM)$ , extends to all orders iff it extends to a second-order deformation, i.e. iff $[\pi,\pi] = 0$ , iff $f\otimes g\mapsto \{f,g\} = \pi(\mathrm df,\mathrm dg)$ is a Poisson structure on $M$ . This is called deformation quantization since quantization should associate to each function $f$ on $M$ an operator $\widehat f$ on some Hilbert space such that $[\widehat f,\widehat g] = \hbar \widehat{\{f,g\}} + O(\hbar^2)$ . Deformation quantization does away with the Hilbert space and formally expands $\widehat f \cdot \widehat g = \widehat{f\cdot g} + \sum_{i\ge 1}\hbar^i\widehat{\mu_i(f,g)}$ . What does any of this have to do with your question about Clifford algebras? One can think of the graded commutative algebra $\Lambda^*V$ as functions on the supermanifold $\Pi V^\vee$ (" $V$ dual in odd degree"). The metric is a graded anti symmetric bilinear form on $V^\vee$ and hence induces a symplectic, and by extension Poisson, structure on this supermanifold, which by the previous discussion has an essentially unique deformation quantization. In fact, one can give an explicit formula for it by putting the right "fermion signs" into the Moyal product . This is a very special deformation quantization since only finitely many $\mu_i$ are nonzero. Thus for any value of $\hbar$ the series $\sum_{i\ge 0}\hbar^i \mu_i$ converges, and one may set $\hbar = 1$ to recover the Clifford algebra.
|
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|
300,013 |
In mathematics and in physics, people use the terms "modular..." and "moduli space" very often. I was puzzled by the etymology, the origins and the similarity/equivalence/differences for these usages/concepts behind for some time. Could I seek for expert's explanations on the contrast/comparison of the "modular..." and "moduli space" in the following contexts: "Modular" in the modular tensor category. (Also the "pre-modular" category. ) "Modular" in the modular form. "Modular" in the topological modular form (tmf). "Modular" in the modular invariance (e.g. in Conformal Field Theory ). "Moduli space" in algebraic geometry "Moduli space" in Non-supersymetric Quantum Field Theories or gauge theories "Moduli space" in Supersymetric gauge theories "Module" in the "mod" or modular arithmetic. "Modular" in modular representation theory of (finite) groups/algebraic groups. [As André Henriques suggested] "Module": $G$-module in the group (co)homology and topological $G$-module for the topological group. [Likely related to 9.] "Modulus" of convergence and "Modulus" of continuity in analysis. Thank you in advance for solving/illuminating this rather puzzling issue.
|
The word modulus (moduli in plural, cf. radius and radii, focus and foci, locus and loci) comes from Latin as a word meaning "small measure" or "unit of measure". This is why the absolute value of a complex number z is sometimes called the modulus of z and why the word is used in physics for Young's modulus. In 1800 Gauss introduced the congruence relation $a \equiv b \bmod m$ with m being called the modulus because this equivalence relation on integers a and b was being "measured" according to the integer m . The later term "modular representation" for representations in characteristic p comes from the simplest source of fields with characteristic p being the integers mod p (and finite extensions of it). The term modulus, from its meaning as a standard of measure, drifted into a more general usage as "parameter". This meaning led to the terms modular function and modular form by the end of the 19th century. In the study of elliptic integrals like $u(x) = \int_0^x \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ the parameter $k$ that shows up has traditionally been called the elliptic modulus of the elliptic integral. It pins down which elliptic integral you are speaking about, since the integral varies with $k$ . This name for k goes back to Jacobi's work on elliptic integrals in the 1820s. Inverting the relation between u and x , $x(u)$ extends from a neighborhood of $u = 0$ to all of $\mathbf C$ as a doubly periodic meromorphic function, and if $\tau$ is a ratio of two carefully chosen periods (taken in the upper half-plane) then Gauss found a formula for k in terms of some theta-functions evaluated at $\tau$ . These theta-functions are invariant under a finite-index subgroup of $\text{SL}_2(\mathbf Z)$ (specifically, invariant under $\Gamma(2)$ ), although such work of Gauss was unpublished until it appeared after his death (1855) in his collected works. Earlier, less than 10 years after Jacobi's work, in volume 18 (1838) of Crelle's Journal, Gudermann wrote a long paper Theorie der Modular-Functionen und der Modular-Integrale (pp. 1-54 and 220-258). Jacobi's modulus k and the Jacobi elliptic functions sn, cn, dn appear prominently in it; while I can't read this German very well, other sources that refer to this paper say Gudermann's use of "modular function" refers to what we would call elliptic functions (and, correspondingly, what he calls "modular integrals" are elliptic integrals). I mention Gudermann's paper just to point out that the term "modular function" was in use before 1840 (hence preceding Riemann's work), even if its meaning over time would change. Paul Garrett says in an answer to a math stackexchange question here that in the 1870s Dedekind introduced in his work on algebraic number theory the term Modul for what we'd nowadays call a lattice (in Euclidean space) or finite-free $\mathbf Z$ -module, and it is suggested in one of the answers here that this term might have been chosen because it was a general kind of structure you could "mod" out by. I think that is what the word eventually came to mean (related to the algebraic notion of a module), but I took a look at Dedekind's famous Chapter XI of Dirichlet and Dedekind's Vorlesungen über Zahlentheorie, where he first introduces the term Modul, in section 168, and he doesn't actually use Modul to refer to lattices at all. Dedekind is thinking of a number field as a subset of $\mathbf C$ (there were no abstract fields in those days, except for perhaps finite fields) and he defines a Modul $\mathfrak a$ as any set (he writes "System") of real and complex numbers closed under subtraction. From any elements $\alpha$ and $\beta$ in $\mathfrak a$ you get $\alpha-\alpha=0$ and then $0 - \alpha = -\alpha$ and then $\beta - (-\alpha) = \alpha+\beta$ , so basically a Modul is just an additive subgroup of $\mathbf C$ (not of a general $\mathbf R^n$ ). Dedekind is most interested in the case when $\mathfrak a$ is a finite Modul, meaning it has a finite basis, such as $\mathbf Z + \mathbf Z\sqrt{2}$ and $\mathbf Z +
\mathbf Z\sqrt[3]{2} + \mathbf Z\sqrt[3]{4}$ in $\mathbf R$ . Neither of these are lattices, since we are not putting all the real and complex embeddings of a number field together to make such groups discrete in a larger Euclidean space. (In section 177 Dedekind defines an ideal to be a special kind of Modul in a number field.) Perhaps over time the concept of a Modul did turn into a lattice, and at least inside the integers of imaginary quadratic fields, which are so important for complex multiplication, they are the same thing. Lattices in $\mathbf C$ , up to real scaling, can be written as $\mathbf Z + \mathbf Z\tau$ with $\tau$ in the upper half-plane being determined by that lattice up to the action of $\text{SL}_2(\mathbf Z)$ . This group had been used by Gauss in the early 19th century in his work on binary quadratic forms and the arithmetic-geometric mean, but the study of this group and especially its subgroups really took off in the late 19th century in work of people like Dedekind, Fricke, and Klein. On p. 364 here Klein says the term "elliptic modular function" comes from Dedekind's 1877 article here . What about modular forms? Eisenstein series were introduced in the mid-19th century and they were viewed later as certain homogeneous functions on lattices. The word "form" is a concise term for homogeneous function (like "linear form" and "quadratic form"), so these homogeneous functions on lattices got the name "modular form". According to this article the term "modular form" was introduced by Klein on pp. 143-144 here , and even if you don't know any German you can recognize the German for "homogeneous function" on p. 144. See also pp. 127-128. (From a different direction, in 1909 Dickson wrote a paper where he used "modular form" to mean "homogeneous polynomial over a finite field"!) Riemann introduced the term "moduli" (Moduln) in his 1857 paper on abelian functions (see bottom of p. 33 here ) to refer to a count of $3p-3$ parameters or coordinates, which is the source of the later idea of a moduli space or parameter space for geometric objects of a common type. In summary, there are several sources of all this "modular" terminology: Gauss's modulus in 1800, Jacobi's elliptic modulus $k$ in the 1820s, Riemann's moduli in the 1850s, and Dedekind's module in the 1870s. I don't see how Riemann's moduli (leading to moduli spaces) is related to the development of the other mod-type terms in the 19th century.
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|
300,134 |
What is the value of this triple integral $$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz?$$ It has to do with some Schwarz lemma.
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Comment This is related to some papers (e.g. this ) by P.M. Borwein et. al. on short random walks in the plane. Let $X_1, X_2, \dots$ be i.i.d. random variables, uniformly distributed on the unit circle $|z|=1$ in the complex plane. Then
$$
X_1+X_2+X_3
$$
is a random variable in the plane, and your integral is
$$
(2\pi)^3\;\mathbb{E}\Big[\big|\mathrm{Re}(X_1+X_2+X_3)\big|\Big]
$$ Borwein and collaborators have information on moments
$$
W_3(s) := \mathbb{E}\Big[\big|X_1+X_2+X_3\big|^s\Big]
$$
including "closed form" in terms of hypergeometric functions when $s \in \mathbb N$. In particular
$$
W_3(1) = \frac{3}{16}\;\frac{2^{1/3}}{\pi^4}\Gamma({\textstyle \frac{1}{3}})^6
+
\frac{27}{4}\;\frac{2^{2/3}}{\pi^4}\Gamma({\textstyle \frac{2}{3}})^6 .
$$ Now, if the distribution of
$$
Y = X_1+X_2+X_3
$$
is rotationally symmetric in the complex plane, and we have the exact value of $\mathbb E[|Y|]$, can we find $\mathbb E[|\mathrm{Re}\;Y|]$ ??
|
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|
300,407 |
Let $(M,g)$ be a (say closed) Riemannian manifold. One can try to understand the geometry/topology of $(M,g)$ by studying the eigenvalues of the Laplacian (this I guess has two versions: when considering the Laplacian on functions only, or on differential forms. feel free to answer about whichever). Sometimes (for example for nice bounded planar domains) the topology of the manifold is completely determined by the spectrum of the Laplacian (on functions). Also, by the Hodge theorem, for an arbitrary closed Riemannian manifold all of the Betti numbers can be read off the spectrum of the Laplacian (on forms).
Sadly, it is well known that there are examples of isospectral but not isometric Riemannian manifolds. Q1: Are there examples of pairs of Riemannian manifolds which are isospectral but not diffeomorphic/homeomorphic/homotopy equivalent? If so, what is the simplest one? assuming the answer to Q1 is "Yes" or "It is not known": Q2: What are the best positive results in this direction? How much can I know about the topology of $M$ only from my knowledge of the spectrum of the Laplacian? I'm sorry if the question is too vague, but still any information will be nice.
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There are examples due to Ikeda of isospectral Lens spaces which are not homotopy equivalent . Likeliest the simplest examples are the compact connected 3-dimensional flat manifolds which are a tetracosm and didicosm . These are isospectral but not homotopy equivalent. I'm not sure if they are known to have the same spectrum for the Laplace-de Rham operator (i.e. on forms). Presumably this could be computed explicitly. There are many other examples of non-isometric manifolds which are strongly isospectral (having the same spectrum on forms) using Sunada's method, and arithmetic techniques (originally due to Vignéras ) for locally symmetric spaces. Coupled with the Mostow rigidity theorem for locally symmetric spaces, this implies that these manifolds are not homotopy equivalent. In 1-dimension, the spectrum determines the manifold. For connected two-dimensional surfaces, the Laplace-de Rham spectrum determines the betti numbers, and hence the topological type (the surface $S$ is orientable iff $b_2(S)=1$, and then $b_1(S)$ determines the topological type if it is orientable or non-orientable). (For surfaces with boundary, an example of Bérard-Webb shows that the Neumann spectrum does not determine orientability.) Thus, the 3-dimensional examples coming from Sunada's construction or Vignéras' are minimal dimensional examples which are strongly isospectral. I don't know if the Laplace spectrum determines the topological type of closed surfaces. See section 5 of the survey paper by Carolyn Gordon for more information on how topology cannot be detected by the spectrum and references. Note also that Lubotzky-Samuels-Vishne showed that there are isospectral arithmetic lattices which are not commensurable (so not obtained by Sunada's method or Vigneras' criterion).
|
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|
300,466 |
Is the Riemann zeta function surjective or does it miss one value?
|
The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an essential singularity at $z=0$, hence $\zeta(s)$ misses at most one value. If $\zeta(s)=w$ then $\zeta(\overline{s})=\overline{w}$, hence $\zeta(s)$ can only miss a real value. However, $\zeta(s)$ maps the real interval $(1,\infty)$ onto $(1,\infty)$, and the real interval $(-2,1)$ onto $(-\infty,0)$. Also $\zeta(-19)>1$ and $\zeta(-18)=0$, hence $\zeta(s)$ maps $[-19,-18]$ onto a real interval that contains $[0,1]$. So $\zeta(s)$ does not miss any real value, and hence it does not miss any complex value.
|
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|
300,570 |
First note that there exists a natural measure $\mu$ on $P(\omega \times \omega)$, inherited from the Lebesgue measure on the reals (by identifying the reals with $P(\omega)$ and $\omega$ with $\omega \times \omega$ in the natural ways) Let us consider models of $ZFC$ of the form $(\omega, E),$ where $E \subseteq \omega \times \omega$ is interpreted as the $\in$-relation. Given a theory $T \supseteq ZFC,$ let us consider the set
$M_T=\{E \subseteq \omega \times \omega:(\omega, E)$ is a model of $T\}$ Question 1. Is the set $M_{ZFC}$ measurable? If so, what is its measure? In the case that the answer to the above question is yes, and the measure is positive, then one may ask questions like the following which can measure the truth or falsity of statements like $CH$. Question 2 What is the measure of $M_{ZFC+CH}?$ What about $M_{ZFC+\neg CH}?$ If the answer to question 1 is negative, then one may ask the following: Question 3. What about questions 1 and 2 if we replace the measure with the outer measure of the required sets? Edit: By the answer given by Wojowu, and his suggestion, I would also like to ask the following: Question 4. What happens if we replace $ZFC$ with $ZFC -$foundation?
|
For any $E$ modelling ZFC and for each $n\in\omega$, we must have $(n,n)\not\in E$. Therefore $M_{ZFC}$ is contained in the cylinder set defined by $(0,0)\not\in E,\dots,(n,n)\not\in E$ which has measure $2^{-n-1}$. Therefore $M_{ZFC}$ has outer measure zero, thus is measurable and has measure zero. Had we excluded the diagonal from $\omega\times\omega$, we can't have both $(n,m)\in E,(m,n)\in E$ for $n\neq m$, so we get infinitely many independent conditions each of measure $3/4$, so again we get outer measure zero. Excluding foundation from ZFC makes this method not work, so it can make for a more interesting question, but perhaps we can use a similar trick. Edit: Without foundation, let $M_n$ be the set of those models of ZFC-Foundation for which $n$ represents the empty set. We get infinitely many conditions $(m,n)\not\in E$, so $M_n$ has outer measure zero. Since $M_{ZFC-Foundation}$ is a countable union of $M_n$, it has measure zero too. Since I don't know of a reasonable variant of ZFC which doesn't prove existence of the empty set, I think that closes the case.
|
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|
300,768 |
Let $X$ be a projective variety (so, with some ( edit: fixed nondegenerate closed) embedding) with the following curious property: for every hyperplane section $H$, we have that $X-H \cong \mathbb{A}^n$. Then is $X$ necessarily isomorphic to $\mathbb{P}^n$? Assuming we are over the complex numbers for simplicity, one can show that $X$ has the same Hodge diamond as $\mathbb{CP}^n$ using purely topological arguments (such as long exact sequence of a pair, duality as in the proof of Lefschetz hyperplane, etc.) but this does not rule out these potential fake projective spaces. This question is motivated by the fact that spheres are characterized by a similar property; i.e., a closed oriented manifold that is contractible upon removal of any point will be homeomorphic to a sphere.
|
The hypotheses above are very strong, and they impose strong hypotheses on the cohomology of the complement $U$ of the universal hyperplane section. Using Leray spectral sequences for both projections, this quickly gives the result. Denote the dimension of $X$ by $n$, and denote by $m$ the dimension of the ambient projective space $\mathbb{CP}^m$ in which $X$ is embedded as a linearly nondegenerate variety. Proposition. Assume that for every hyperplane $L$ in $\mathbb{CP}^m$, the intersection $H=L\cap X$ is a Cartier divisor in $X$ such that the open complement $X\setminus H$ is isomorphic to $\mathbb{C}^n$. Then $n$ equals $m$, and $X$ equals all of $\mathbb{CP}^m$. Proof. Denote by $\widehat{\mathbb{CP}}^m$ the dual projective space of hyperplanes $L$ in $\mathbb{CP}^m$. Consider the following closed subset $Y$ of $X\times \widehat{\mathbb{CP}}^m$, $$Y :=\{(x,[L])\in X\times \widehat{\mathbb{CP}}^m : x\in L\}.$$ By hypothesis, the open complement $U$ of $Y$ in $X\times \widehat{\mathbb{CP}}^m$ is an affine space bundle over $\widehat{\mathbb{CP}}^m$. By the Leray spectral sequence for cohomology applied to the second projection, $$\text{pr}_2:X\times \widehat{\mathbb{CP}}^m \to \widehat{\mathbb{CP}}^m,$$ the cohomology of $U$ equals the cohomology of $\widehat{\mathbb{CP}}^m$. In particular, the cohomological dimension of $U$ equals $2m$. On the other hand, we also have the first projection, $$\text{pr}_1:X\times \widehat{\mathbb{CP}}^m \to X.$$ The restriction of $\text{pr}_1$ to $U$ is again an affine space bundle, but now of relative dimension $m$. Thus, the cohomology of $U$ also equals the cohomology of $X$. In particular, the cohomological dimension of $U$ equals $2n$. Since the cohomological dimension of $U$ equals both $2n$ and $2m$, the dimension $n$ of $X$ equals the dimension $m$ of the ambient projective space $\mathbb{P}^m$. Therefore $X$ equals $\mathbb{P}^m$. QED However, this is an interesting question, and variants often arise (with weaker hypotheses, of course). So here are some further observations.
One of the coarsest topological invariants is the Euler characteristic of cohomology with compact support. For a complex algebraic variety $X$ considered as a set of $\mathbb{C}$-rational points with the classical / Euclidean / analytic topology, for a Zariski closed subset $Z\subseteq X$ with its induced topology, and for the open complement $X\setminus Z$ with its induced topology, there is a long exact sequence of compactly supported cohomology, $$\dots \to H^r_c(X\setminus Z;\mathbb{Z}) \to H^r_c(X;\mathbb{Z}) \to H^r_c(Z;\mathbb{Z}) \xrightarrow{\delta} H^{r+1}_c(X\setminus Z;\mathbb{Z})\to \dots $$ This gives rise to an equality of compactly supported Euler characteristics, $$\chi_c(X) = \chi_c(X\setminus Z) + \chi_c(Z).$$ Thus, for a linearly nondegenerate, irreducible, Zariski closed subset $X$ of $\mathbb{P}^n$ that is smooth, if for every hyperplane section $H$ the open complement $X\setminus H$ has equal Betti numbers, then also the Euler characteristic $\chi_c(H)$ is independent of the choice of hyperplane section. On the other hand, for a field $k$ of characteristic zero , such as $k=\mathbb{C}$, for an integral closed subscheme $X$ of $\mathbb{P}^n_k$ that is smooth, a general pencil $(H_t)_{t\in \Pi}$ of hyperplane sections $H_t$ will be a Lefschetz pencil . There are many references for this; one reference is Corollary 2.10, p. 46 of the following. Voisin, Claire , Hodge theory and complex algebraic geometry. II. Transl. from the French by Leila Schneps, Cambridge Studies in Advanced Mathematics. 77. ZBL1032.14002 . In positive characteristic, this can fail, although it is often still true. The standard reference is the second volume of SGA 7. In characteristic $0$, there will be only finitely many elements $t$ of the pencil, say $t\in\{t_1,\dots,t_\delta\}$, such that $H_t$ is singular, and each singular member $H_t$ will have a single ordinary double point. The finite set $\{t_1,\dots,t_\delta\}$ is the discriminant locus . For each such $t$, the Betti numbers of $H_t$ and of the nearby fibers $H_s$ will differ in precisely one degree, coming from a vanishing cycle, so that the difference of Betti numbers is $\pm 1$ (depending on the parity of the cohomological degree of the vanishing cycle). Thus, $\chi_c(H_t)-\chi_c(H_s)$ equals $+1$ or $-1$ for every $t$ in the discriminant locus. The precise cardinality of the discriminant locus is computed in my answer to a previous MathOverflow question: Bounding the number of critical points in a Lefschetz pencil If this cardinality is nonzero, then there are hyperplane sections with varying Euler characteristics, and thus there open complements also have varying Euler characteristic. When the cardinality of the discriminant set equals $0$, then $X$ has defective dual variety , sometimes also called "defective discriminant variety", "degenerate dual variety", "degenerate discriminant variety", etc. Starting with Griffiths-Harris, then Ein, and then many others, varieties with defective dual variety have been classified in low dimensions. One general result proved by Beltrametti-Fania-Sommese is that every such variety admits a Fano fibration (possibly with base equal to a point) whose general fiber is a Fano manifold that also has defective dual variety. However, for a general hyperplane section $H$, since $H$ is irreducible (I am assuming that $\text{dim}(X)\geq 2$ since the result is trivial in dimension one), the exact sequence of Picard groups gives $$\mathbb{Z}\cdot [H] \to \text{Pic}(X) \to \text{Pic}(X\setminus H) \to 0.$$ If $X\setminus H$ is affine space for a general hyperplane section $H$, then $\text{Pic}(X\setminus H)$ is zero. Thus, $\text{Pic}(X)$ equals $\mathbb{Z}\cdot [H]$. If the target of the fibration has positive dimension, then the pullback of an ample divisor class from the target contradicts that $\text{Pic}(X)$ equals $\mathbb{Z}\cdot [H]$. Therefore,
if $X$ is a variety with defective dual variety such that $X\setminus H$ is an affine space for a general hyperplane section $H$, then $X$ is a Fano manifold (hence simply connected) with Picard rank $1$. By a theorem of Fujita and Libgober-Wood, the only fake projective spaces of dimension $\leq 6$ that are simply connected are honest projective spaces. In conclusion, for a smooth projective variety $X$ of dimension $n\leq 6$, if $X$ has dual defective variety and if $X\setminus H$ is isomorphic to affine space for a general hyperplane section $H$ of $X$, then $X$ is isomorphic to projective space, and $H$ is a general linear hyperplane in that projective space.
|
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|
300,785 |
I have been struggling with this equation for some time and I do not seem to find any conclusive answer (it's from my research, not a homework). It has to do with the real solutions $x$ to the following equation $$ x + x f(x) = 1 + f(1),$$
where
$$ f(x) = 2\sum_{n=1}^\infty \mathrm{e}^{(-ax^2-b) n^2} $$
with $a$ and $b$ strictly positive. I know that $x=1$ solves the equation trivially; from simulations, I cannot find a contradiction to the fact that it should be the only solution. However, I cannot prove nor disprove that $x=1$ is the only solution. I have tried using the upper bound
$$ f(x) \leq \frac{ax^2+b+1}{ax^2+b}\mathrm{e}^{-ax^2 -b},$$
and i have tried relating $f(x)$ with the Elliptic theta function
$$ f(x) = -1 + \theta_3(0,\mathrm{e}^{-ax^2 -b});$$
I have also tried to prove that $x = -x f(x) + 1 +f(1)$ is a contractive mapping; however, I have only found (quite restrictive) sufficient conditions on $a$ and $b$ for it to be true. If someone manages to solve it or help me find a counterexample, I will gladly acknowledge their contribution in the paper i am writing. EDIT: $x = -x f(x) +1 +f(1)$ is not a contractive mapping in general. I have found counterexamples where it is not (but the equation still only has one solution $x=1$).
|
The hypotheses above are very strong, and they impose strong hypotheses on the cohomology of the complement $U$ of the universal hyperplane section. Using Leray spectral sequences for both projections, this quickly gives the result. Denote the dimension of $X$ by $n$, and denote by $m$ the dimension of the ambient projective space $\mathbb{CP}^m$ in which $X$ is embedded as a linearly nondegenerate variety. Proposition. Assume that for every hyperplane $L$ in $\mathbb{CP}^m$, the intersection $H=L\cap X$ is a Cartier divisor in $X$ such that the open complement $X\setminus H$ is isomorphic to $\mathbb{C}^n$. Then $n$ equals $m$, and $X$ equals all of $\mathbb{CP}^m$. Proof. Denote by $\widehat{\mathbb{CP}}^m$ the dual projective space of hyperplanes $L$ in $\mathbb{CP}^m$. Consider the following closed subset $Y$ of $X\times \widehat{\mathbb{CP}}^m$, $$Y :=\{(x,[L])\in X\times \widehat{\mathbb{CP}}^m : x\in L\}.$$ By hypothesis, the open complement $U$ of $Y$ in $X\times \widehat{\mathbb{CP}}^m$ is an affine space bundle over $\widehat{\mathbb{CP}}^m$. By the Leray spectral sequence for cohomology applied to the second projection, $$\text{pr}_2:X\times \widehat{\mathbb{CP}}^m \to \widehat{\mathbb{CP}}^m,$$ the cohomology of $U$ equals the cohomology of $\widehat{\mathbb{CP}}^m$. In particular, the cohomological dimension of $U$ equals $2m$. On the other hand, we also have the first projection, $$\text{pr}_1:X\times \widehat{\mathbb{CP}}^m \to X.$$ The restriction of $\text{pr}_1$ to $U$ is again an affine space bundle, but now of relative dimension $m$. Thus, the cohomology of $U$ also equals the cohomology of $X$. In particular, the cohomological dimension of $U$ equals $2n$. Since the cohomological dimension of $U$ equals both $2n$ and $2m$, the dimension $n$ of $X$ equals the dimension $m$ of the ambient projective space $\mathbb{P}^m$. Therefore $X$ equals $\mathbb{P}^m$. QED However, this is an interesting question, and variants often arise (with weaker hypotheses, of course). So here are some further observations.
One of the coarsest topological invariants is the Euler characteristic of cohomology with compact support. For a complex algebraic variety $X$ considered as a set of $\mathbb{C}$-rational points with the classical / Euclidean / analytic topology, for a Zariski closed subset $Z\subseteq X$ with its induced topology, and for the open complement $X\setminus Z$ with its induced topology, there is a long exact sequence of compactly supported cohomology, $$\dots \to H^r_c(X\setminus Z;\mathbb{Z}) \to H^r_c(X;\mathbb{Z}) \to H^r_c(Z;\mathbb{Z}) \xrightarrow{\delta} H^{r+1}_c(X\setminus Z;\mathbb{Z})\to \dots $$ This gives rise to an equality of compactly supported Euler characteristics, $$\chi_c(X) = \chi_c(X\setminus Z) + \chi_c(Z).$$ Thus, for a linearly nondegenerate, irreducible, Zariski closed subset $X$ of $\mathbb{P}^n$ that is smooth, if for every hyperplane section $H$ the open complement $X\setminus H$ has equal Betti numbers, then also the Euler characteristic $\chi_c(H)$ is independent of the choice of hyperplane section. On the other hand, for a field $k$ of characteristic zero , such as $k=\mathbb{C}$, for an integral closed subscheme $X$ of $\mathbb{P}^n_k$ that is smooth, a general pencil $(H_t)_{t\in \Pi}$ of hyperplane sections $H_t$ will be a Lefschetz pencil . There are many references for this; one reference is Corollary 2.10, p. 46 of the following. Voisin, Claire , Hodge theory and complex algebraic geometry. II. Transl. from the French by Leila Schneps, Cambridge Studies in Advanced Mathematics. 77. ZBL1032.14002 . In positive characteristic, this can fail, although it is often still true. The standard reference is the second volume of SGA 7. In characteristic $0$, there will be only finitely many elements $t$ of the pencil, say $t\in\{t_1,\dots,t_\delta\}$, such that $H_t$ is singular, and each singular member $H_t$ will have a single ordinary double point. The finite set $\{t_1,\dots,t_\delta\}$ is the discriminant locus . For each such $t$, the Betti numbers of $H_t$ and of the nearby fibers $H_s$ will differ in precisely one degree, coming from a vanishing cycle, so that the difference of Betti numbers is $\pm 1$ (depending on the parity of the cohomological degree of the vanishing cycle). Thus, $\chi_c(H_t)-\chi_c(H_s)$ equals $+1$ or $-1$ for every $t$ in the discriminant locus. The precise cardinality of the discriminant locus is computed in my answer to a previous MathOverflow question: Bounding the number of critical points in a Lefschetz pencil If this cardinality is nonzero, then there are hyperplane sections with varying Euler characteristics, and thus there open complements also have varying Euler characteristic. When the cardinality of the discriminant set equals $0$, then $X$ has defective dual variety , sometimes also called "defective discriminant variety", "degenerate dual variety", "degenerate discriminant variety", etc. Starting with Griffiths-Harris, then Ein, and then many others, varieties with defective dual variety have been classified in low dimensions. One general result proved by Beltrametti-Fania-Sommese is that every such variety admits a Fano fibration (possibly with base equal to a point) whose general fiber is a Fano manifold that also has defective dual variety. However, for a general hyperplane section $H$, since $H$ is irreducible (I am assuming that $\text{dim}(X)\geq 2$ since the result is trivial in dimension one), the exact sequence of Picard groups gives $$\mathbb{Z}\cdot [H] \to \text{Pic}(X) \to \text{Pic}(X\setminus H) \to 0.$$ If $X\setminus H$ is affine space for a general hyperplane section $H$, then $\text{Pic}(X\setminus H)$ is zero. Thus, $\text{Pic}(X)$ equals $\mathbb{Z}\cdot [H]$. If the target of the fibration has positive dimension, then the pullback of an ample divisor class from the target contradicts that $\text{Pic}(X)$ equals $\mathbb{Z}\cdot [H]$. Therefore,
if $X$ is a variety with defective dual variety such that $X\setminus H$ is an affine space for a general hyperplane section $H$, then $X$ is a Fano manifold (hence simply connected) with Picard rank $1$. By a theorem of Fujita and Libgober-Wood, the only fake projective spaces of dimension $\leq 6$ that are simply connected are honest projective spaces. In conclusion, for a smooth projective variety $X$ of dimension $n\leq 6$, if $X$ has dual defective variety and if $X\setminus H$ is isomorphic to affine space for a general hyperplane section $H$ of $X$, then $X$ is isomorphic to projective space, and $H$ is a general linear hyperplane in that projective space.
|
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|
300,885 |
Suppose that $X$ is a metric space. Is the family of all real-valued uniformly continuous functions on $X$ dense in the space of all continuous functions with respect to the topology of uniform convergence on compact sets?
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Yes, and even more is true. The argument is as follows: let $f\colon X \to \mathbb R$ be a continuous function and let $K\subset X$ be a compact set. Then $f|_K$ is uniformly continuous; let $\omega$ be its nondecreasing subadditive modulus of continuity. By McShane-Whitney's extension formula $f|_K$ admits a uniformly continuous extension to $X$ with the same modulus - more concretely, $$F(x)=\inf\{f(k)+\omega(d(k,x))\colon k\in K\}$$ is this extension which is uniformly continuous on the whole $X$.
|
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"https://mathoverflow.net/questions/300885",
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|
300,961 |
A knot is an embedding of a circle $S^{1}$ in $3$-dimensional Euclidean space, $\mathbb{R}^3$. Knots are considered equivalent under ambient isotopy . There are two different types of knots, tame and wild . A tame knot is any knot equivalent to a polygonal knot, that is a knot whose image in $\mathbb{R}^3$ is the union of a finite set of line segments. A wild knot is a knot that is not tame. In contrary to the tame knots, wild knots are not well-behaved creatures. They often provide counterexamples to the general version of those theorems which hold for the tame knots. There are several invariants associated with the knots. For example, the crossing number of a knot is a knot invariant defined as the smallest number of crossings of any diagram of the knot. Wild knots (such as the following picture) may have infinite crossing numbers anywhere between $\aleph_0$ and $2^{\aleph_0}$. Remark 1. Some cautiousness is needed in the previous statement. According to what Joel has mentioned in his comment , the last statement can be made more accurate. In fact, the set of crossing points of a wild knot seems to be projective which implies that assuming the existence of infinitely many Woodin cardinals, it is either countable or of size continuum because the axiom of Projective Continuum Hypothesis holds under this plausible large cardinal assumption. Furthermore, based on Andreas' comment , one may remove the large cardinal assumption in the last argument too. While there has been an extensive amount of research dedicated to the classification of tame knots (particularly the iconic prime ones ), up to ambient isotopy, it turned out that the classification of wild knots is generally much more complicated. (In this direction see Kulikov's paper: A non-classification result for wild knots ). My question simply is how to make the chaos caused by wild knots even worse by adding new ones to the universe through forcing. Question 1. Assuming failure of Continuum Hypothesis, let $\aleph_0<\kappa\leq 2^{\aleph_0}$ be an uncountable cardinal. Is there a cardinal preserving forcing notion $\mathbb{P}$ of the universe $V$ such that in $V^{\mathbb{P}}$ there exists a wild knot $K$ of crossing number $\kappa$ which is different from all wild knots of crossing number $\kappa$ in the ground model up to ambient isotopy (definable in $V^{\mathbb{P}}$)? Remark 2. Potentially, one may find some questions, related to the possible number/behavior of wild knots of the certain knot invariant (such as bridge , unknotting , stick numbers, etc.) in a forcing extension, interesting. However, for the sake of this question, I preferred to stick to a very specific one concerning crossing number. Please feel free to share your thoughts about other possibly interesting problems along these lines in the comments below. Update. According to Ian's remark in the comments section, it turned out that the notion of crossing number for wild knots retains some ambiguity, as (in contrary to the case for tame knots) there is no straightforward theorem guaranteeing the existence of a regular projection for wild knots, which is somehow essential for counting the number of crossing points in a wild knot. Roughly speaking, if you have a wild knot, you can't even be sure how much knotted it is! Another indication of why these crazy creatures deserve the name "wild"! Anyway, the soul of the original question remains valid even without any specification to the case of tame/wild knots of any particular characteristics in terms of knot invariants. The main problem is whether it is possible to add a morphologically new knot to the mathematical universe via forcing. Precisely: Question 2. Starting from a model $V$ of $ZFC$, is there a (cardinal preserving) forcing notion $\mathbb{P}$ in $V$ such that in $V^{\mathbb{P}}$ there exists a (possibly wild) knot $K$ which is different from all knots of the ground model up to ambient isotopy (definable in $V^{\mathbb{P}}$)?
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Yes, forcing can add fundamentally new knots, not equivalent to any ground model knot. Indeed, whenever you extend the set-theoretic universe to add new reals, then you must also have added fundamentally new knots. Theorem. If $V\subset W$ are two models of set theory with the same ordinals and different reals (for example, any forcing extension with new reals), then there are new knots in $W$ not equivalent in $W$ to any ground model knot from $V$. (Note that indeed it makes sense to consider the ground model knots $k$ as knots in any extension; simply view them as a Borel set, to be re-interpreted as the corresponding Borel set in the extension, constructed according to the same process; the assertion that a given Borel code represents a knot is a $\Sigma^1_1$ property and therefore absolute. One should not think of a ground model knot merely as a point set, since this exact set will no longer be a knot in a forcing extension with new real numbers.) To prove the theorem, we use the work of Vadim Kulikov, who has already done the heavy knot-tying for this question. Specifically, in his answer to the question on the Descriptive set-theoretic complexity of knot equivalence , Vadim explains that the relation of knot equivalence is strictly above the isomorphism relation of countable structures in the hierarchy of Borel reducibility. This means that there is a Borel map $\pi$ from the space of countable structures to knots, such that structures $A$ and $B$ are isomorphic if and only if knots $\pi(A)$ and $\pi(B)$ are equivalent (but no such map in the other direction). One should think of the map $\pi$ as a means to code an arbitrary countable structure into a knot. Using $\pi$, one can code a countable graph, group or partial order $A$ into the knot $\pi(A)$, in a way that respects knot equivalence. In this way, Vadim's result fulfills the suggestion I had made in the comments on the question that what we want to do is code arbitrary binary sequence into knots. Given the countable structure $A$, the knot $\pi(A)$ codes $A$ in that we can recover $A$ up to isomorphism from any knot that is equivalent to $\pi(A)$. The recovery process is sufficiently explicit that it is absolute between a model of set theory and its forcing extensions. This analysis of the complexity of not equivalence is evidently the main result of his paper: Kulikov, Vadim , A non-classification result for wild knots , Trans. Amer. Math. Soc. 369 (2017), 5829-5853, ZBL06722528 . ( arxiv:1504.02714 ) Kulikov states that it follows from his theorem that, "wild knots cannot be completely classified by real numbers considered up to any Borel equivalence relation." So there can be no classification of the wild knots along the same lines as the tame knot classification. My point here is that it follows from the positive part of the reduction — that isomorphism of countable structures reduces to knot equivalence — that in any extension $W$ with new real numbers, there will also be new knots, not equivalent to any ground model knot. To see this, suppose that $z$ is a real in an extension $W$, but not in $V$. Let $A_z$ be a countable structure, say, a graph, coding $z$ in some fundamental way, making it definable from the theory of this structure. Let $k=\pi(A_z)$ be the associated knot in $W$. I claim that $k$ is not equivalent to any ground model knot. If $k$ were equivalent to some $k'$ in the ground model $V$, then the assertion $\exists y\ k'\sim\pi(A_y)$ is true about $k'$ in $W$. This is a $\Sigma^1_1$ statement about $k'$, which by Lévy absolutelenss must already be true in the ground model. So there is some ground model real $y$ such that $k'$ is equivalent to $\pi(A_y)$ in $V$. Thus, $\pi(A_y)$ and $\pi(A_z)$ are equivalent in $W$, and so $A_y$ must be isomorphic to $A_z$, contradicting that $z$ is not in $V$ and therefore not definable in $A_y$. So $k$ is a totally new knot, as desired. This phenomenon is not possible with tame knots. Since there are only countably many tame knots up to equivalence, there is a list of countably many tame knots, and the assertion that every tame knot is equivalent to one of them is a $\Pi^1_2$ assertion about that list. By absoluteness, this remains true in all forcing extensions. So you cannot add fundamentally new tame knots. (This answer addresses what I take to be the main part of your questions, whether forcing adds new knots, and not specifically about the crossing number issue.)
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|
301,385 |
I have now some problems about my research Career, I would like to tell my stories. I am a Chinese guy, but now a Ph.D. candidate in Germany, in the field of so called 'Geometric Analysis', but I do not feel happy when I work in such a field. Somebody said that a wrong choice of study field or supervisor means several years' Frustration, it describes my situation very well. I choose this geometric analysis only because I want to study geometry with some analytical methods, but gradually I found that the most researcher in this field only have very narrow knowledge about PDE and Riemannian geometry, I think this field is lack of real beautiful idea, but full of papers, I am not judging that this is not good, but this is not my taste. Compare to the normal researchers in this field, I have relative comprehensive mathematical knowledge, I am not only familiar with second order elliptic PDE and differential geometry (Riemannian geometry, geometry of fiber bundles, Chern-Weil theory) but also with algebraic topology (homology, cohomology, characteristic class and spectral sequence), complex geometry(the whole book of Demailly'Complex differential and analytic geometry' and some complex Hodge theory in the first book of C.Voisin). Several years ago, I have also studied abstract algebraic geometry and Index theorem in courses, but now I am not familiar with such kind of material. In China I have already known, for my future, I should not only have one math-tool. At that time, a professor in algebraic geometry has suggested me that as a young guy, algebraic geometry may be a better choice for the future and invited me to be his student, but I hesitated and refused, that was maybe a stupid decision. During my research in Germany, I was always very depressive because what I was working were only some trivial and unnature PDE estimates. Now my career is a little hopeless, somebody suggested that I could try to contact some experts who work in Symplectic geometry with analytic methods, like members of Hofer's school in Germany. Yes, I actually want to study some deep topics in (complex-) algebraic geometry or symplectic geometry, but the problem is, I will soon have a Ph.D. degree, how can I change my research area after my graduation? As a normal young guy the experts in another field do not know me at all, maybe it's very hard to get a postdoc position from them, so should I do another Ph.D. in math? Is that worthy? I am frustrated and asking for help, maybe some guys have similar experience. If I cannot research what I like, I must try to find a Job in Industry.
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Finish this degree, then switch to whatever interests you. Many (most?) mathematicians change fields at some point in their research careers. Bob Solovay told me once that the most important research was the first new thing you did after you got your degree. His thesis was A Functorial Form of the Differentiable Riemann–Roch theorem . He finished it in a hurry so he could move on to mathematical logic, where he's famous. https://en.wikipedia.org/wiki/Robert_M._Solovay
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|
301,512 |
Lately, I have been constructing finite involution monoids that generate varieties with $2^{\aleph_0}$ subvarieties. One construction requires groups that violate the identity ${ [x,y]^2 \approx 1 }$, where ${ [x,y] = x^{-1} y^{-1} xy }$. Is there a name for groups satisfying the identity ${ [x,y]^2 \approx 1 }$? Has there been any work done on these groups?
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This variety of groups has indeed been considered in the literature. It is known that the following conditions hold for every group $G$ satisfying the identity $[x,y]^2=1$: $[[x,y_1,\ldots,y_m],[x,z_1,\ldots,z_n]]=1$ for all
$x,y_1,\ldots,y_m,z_1,\ldots,z_n\in G$ (see [1]). $[[x_1,x_2],[x_3,x_4]]=[[x_{\pi(1)},x_{\pi(2)}],[x_{\pi(3)},x_{\pi(4)}]]$ for all $x_1,\ldots,x_4\in G$ and $\pi\in S_4$ (see [2]). $[[x,y],[z,w]]=[x,y,z,w][x,y,w,z]$ (see [1]). $[\gamma_2(G),\gamma_3(G)]=1$ (see [2]). $[G'',G]=1$ (see [2]). $G'^4=1$ (see [2]). The group $G$ need not be metabelian (see [3]). If $G$ is finite, then $G=P\rtimes H$, where $P$ is a normal Sylow $2$-subgroup of $G$, $H$ is abelian of odd order, and $[P,H]$ is an elementary abelian $2$-group (see [1]). References: M. Farrokhi D. G., On groups satisfying a symmetric Engel word , Ric. Mat. 65 (2016), 15–20. I. D. Macdonald, On certain varieties of groups , Math. Z. 76 , (1961) 270–282. B. H. Neumann, On a conjecture of Hanna Neumann , Proc. Glasgow Math. Assoc. 3 (1956), 13–17.
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{
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"https://mathoverflow.net/questions/301512",
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|
301,533 |
For any undirected simple graph $G=(V,E)$ we define for $v\in V$ the set $N(v) = \{w\in V: \{v,w\}\in E\}$. Suppose $A, B$ are finite, disjoint sets, and $G = (A\cup B, E)$ is a bipartite graph with bipartition $(A,B)$ -- that is, for every $e\in E$ we have $e\cap A \neq \emptyset \neq e\cap B$. Informally speaking, I am trying to prove that if every $b\in B$ has "enough neighbors" in $A$, then $2$ members of $A$ "cover" $B$. More precisely, I would like to know whether the following statement is true: If $N(b) > |A|/2$ for all $b\in B$, then there are $a_1, a_2\in A$ such that $N(a_1) \cup N(a_2) = B$.
|
This variety of groups has indeed been considered in the literature. It is known that the following conditions hold for every group $G$ satisfying the identity $[x,y]^2=1$: $[[x,y_1,\ldots,y_m],[x,z_1,\ldots,z_n]]=1$ for all
$x,y_1,\ldots,y_m,z_1,\ldots,z_n\in G$ (see [1]). $[[x_1,x_2],[x_3,x_4]]=[[x_{\pi(1)},x_{\pi(2)}],[x_{\pi(3)},x_{\pi(4)}]]$ for all $x_1,\ldots,x_4\in G$ and $\pi\in S_4$ (see [2]). $[[x,y],[z,w]]=[x,y,z,w][x,y,w,z]$ (see [1]). $[\gamma_2(G),\gamma_3(G)]=1$ (see [2]). $[G'',G]=1$ (see [2]). $G'^4=1$ (see [2]). The group $G$ need not be metabelian (see [3]). If $G$ is finite, then $G=P\rtimes H$, where $P$ is a normal Sylow $2$-subgroup of $G$, $H$ is abelian of odd order, and $[P,H]$ is an elementary abelian $2$-group (see [1]). References: M. Farrokhi D. G., On groups satisfying a symmetric Engel word , Ric. Mat. 65 (2016), 15–20. I. D. Macdonald, On certain varieties of groups , Math. Z. 76 , (1961) 270–282. B. H. Neumann, On a conjecture of Hanna Neumann , Proc. Glasgow Math. Assoc. 3 (1956), 13–17.
|
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"https://mathoverflow.net/questions/301533",
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|
301,630 |
Forcing construction in set theory leads to a new understanding of the mathematical (multi)universe by providing a machinery through which one can construct new models of the universe from the existing ones in a fairly controlled and comprehensible way and connect them to one another through forcing extensions. Since the very beginning, the true nature of this rather bizarre construction method has been a matter of debate. A widely accepted (?) analogy emphasizes on a degree of similarity between forcing method in set theory and field extensions in Galois theory where one constructs new fields using the existing ones by adding new elements. Besides some papers and textbooks, a good description of this analogy could be found in this MathOverflow good oldie: Forcing as a new chapter of Galois Theory? . However, as Dorais mentioned in his answer , despite the undeniable similarity, this analogy is not as perfect as what it seems at the first glance. So, forcing is not a one hundred percent field extension-like construction anyway. This fact causes some confusion concerning the possible translation of Galois theory machinery into the language of forcing and set theory. It is often not immediately clear what the forcing analogy of a given notion in the field extensions would be. Also, sometimes there is more than one approach towards defining a corresponding Galois theoretic notion in forcing so that one needs to decide which the natural one is. One such central concept in Galois theory is the notion of the degree of a field extension, that is the dimension of the extended field while viewed as a vector space over the ground field. The question that arises here is whether there is any corresponding well-defined, well-behaved and natural similar notion in set-theoretic forcing. One may think if a forcing extension could be viewed as a kind of vector space -like structure over the ground model. In this sense, there might be a kind of basis associated with any such pair of models which itself may satisfy some uniqueness properties that give rise to the existence of a well-defined notion of (relative) dimension of a generic forcing extension with respect to its ground model. Our axiomatic expectations of the possible behavior of any such notion of basis/dimension is also an interesting topic to explore even before defining any such notion. Of course, we need them to behave as natural as possible and resemble their Galois theoretic counterparts fairly closely. Question. What are examples of defining a notion of dimension or basis for forcing generic extensions in the set-theoretic literature (possibly close to the same fashion that exists for the field extensions and vector spaces)? I couldn't find much along these lines in the literature except a short unpublished note of Golshani in which he takes a Galois theoretic approach towards dimension in forcing by dealing with mutually generic sequences and the chain of forcing extensions. It also shares many features with Hamkins' set-theoretic geology project. However, this notion of forcing dimension seems not to be complete enough to cover all aspects of a vector space view towards generic extensions but could be a really good starting point anyway. Remark. As Peter stated in his comment , absoluteness would be an issue while defining a notion of (relative) dimension in the set-theoretic multiverse. One may ask, from whose perspective are you trying to calculate the relative dimension of two set-theoretic universes and why? In fact, due to the highly contradictory views of different models of $\sf ZFC$ towards anything beyond the narrow scope of universally absolute properties, there is not more than a little hope to obtain an absolute notion of forcing dimension after all. However, one may argue that absoluteness is not a crucial condition in the long list of our expectations from a possible natural definition of dimension for generic models. The same situation happens in the Hamkins' Well-foundedness Mirage Principle stating that " Every universe $V$ is non-well-founded from the perspective of another universe " and so there is no standard model of $\sf ZFC$. Update. According to the Joel and Mohammad 's answers, it turned out that there is more than one approach towards developing a dimension theory for forcing extensions; each with their own characteristics. While Joel's definition requires all forcing dimensions to be infinite, Mohammad's approach allows finite dimensions as well as infinite ones. Also, as Monroe pointed out both definitions fail to satisfy the so-called downwards closure property . In general, Joel's approach sounds a little bit vector space-like to me while Mohammad's reminds me the Krull's dimension in commutative algebra. There is also a chance that these different dimensions be related or coincide with each other under certain circumstances. The point is that in the absence of a general common sense about the expected behavior of a nice forcing dimension , it is not immediately clear whether the mentioned features are flaws or advantages of the presented definitions. Maybe one needs to fix some abstract list of required properties for any such dimension operator and then search for its existence in the forcing extensions. Anyway, if there is any suggestion for such an axiomatic approach towards forcing dimension, I will be so happy to hear about it.
|
My co-authors and I introduced a notion of dimension for forcing
extensions in the following paper: Hamkins, Joel David; Leibman, George; Löwe, Benedikt , Structural connections between a forcing class and its modal logic , Isr. J. Math. 207, Part 2, 617-651 (2015). ZBL1367.03095 , arxiv:1207.5841 , blog post . Specifically, for any forcing extension $V\subset V[G]$, we defined
the essential size of the extension to be the smallest
cardinality in $V$ of a complete Boolean algebra $\mathbb{B}\in V$
such that $V[G]$ is realized as a forcing extension of $V$ using
$\mathbb{B}$, so that $V[G]=V[H]$ for some $V$-generic
$H\subset\mathbb{B}$. More recently, I have been inclined to call
this the forcing dimension of $V[G]$ over $V$. This can indeed be seen as a dimension, in light of the following (essentially lemma 23 of the paper above): Theorem. If $V\subset V[G]$ has essential size $\delta$, then
the essential size of any further extension $V[G][H]$ over $V$ is
at least $\delta$. Proof. By combining the forcing into an iteration, we may view
$V[G][H]$ as a single-step forcing extension of $V$, and so it has
some essential size over $V$. Since we have an intermediate model
$V\subset V[G]\subset V[G][H]$, it follows by the intermediate
model theorem that $V[G]$ can be realized as an extension by a
complete subalgebra of that forcing notion. So the smallest size of
a complete Boolean algebra realizing $V[G]$ is not larger than the
smallest size of a compete Boolean algebra realizing $V[G][H]$ over
$V$. $\Box$ We had used the fact that there is a definable dimension, in the
forcing extensions over $L$, to show in general circumstances that
the modal logic of $\Gamma$-forcing over $L$ is contained in S4.3,
for a wide collection of forcing classes $\Gamma$. Monroe Eskew points out in the comments that, contrary to my
initial thoughts about this, the size of the smallest partial order
giving rise to the extension will also serve as a forcing
dimension. The reason is that the density of a complete subalgebra of a complete Boolean algebra is at most the density of the whole algebra, simply by projecting any dense set of the larger algebra to the subalgebra. It follows by the same argument as in the theorem above that the poset-based forcing dimension of any intermediate model in a forcing extension is bounded by the minimal size of a partial order giving rise to the whole extension. I propose that we officially adopt the poset-based notion as the forcing dimension of a forcing extension $V\subset V[G]$, denoting this dimension by $\left[V[G]\mathrel{:}\strut V\right]$. We may now observe the following attractive identity, confirming the suggestion of Will Brian. Theorem. For any successive forcing extensions $V\subset
V[G]\subset V[G][H]$, we have
$$\left[V[G][H]\mathrel{:}\strut V\right]=\left[V[G]\mathrel{:}\strut V\right]\cdot\left[V[G][H]\mathrel{:}\strut V[G]\right].$$ Proof. Suppose that $G\subset\mathbb{P}\in V$ and
$H\subset\mathbb{Q}\in V[G]$, where these are the minimal-size partial
orders realizing the extensions. Suppose that $\mathbb{Q}$ has size
$\kappa$ in $V[G]$. So without loss there is $\mathbb{P}$-name for a relation $\dot\leq$
on $\kappa$, such that
$\mathbb{Q}=\langle\kappa,{\dot\leq}_G\rangle$ in $V[G]$. We can
now use the partial order $\{(p,\check\alpha)\mid
p\in\mathbb{P},\alpha<\kappa\}$, which is dense inside
$\mathbb{P}*\dot{\mathbb{Q}}$, to realize $V\subset V[G][H]$. This shows
$\leq$ of the desired identity. Conversely, if we can realize $V[G][H]$ as a forcing extension of
$V$ by some partial order $\mathbb{R}$, then $V[G]$ arises as a
subforcing notion, and $V[G][H]\supset V[G]$ arises as quotient forcing.
The quotient forcing $\mathbb{R}/G$ can be thought of as the
conditions in $\mathbb{R}$ that are compatible with every element
of (the image of) $G$ in (the Boolean completion of) $\mathbb{R}$.
So the smallest partial order giving rise to $V[G][H]$ over $V[G]$
is at most the size of the smallest partial order giving rise to
$V[G][H]$ over $V$, as desired. $\Box$
|
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|
301,645 |
Let $a_1,\dots,a_n$ and $b_1,\dots,b_n$ be two sequences of non negative numbers such that for every positive integer $k$,
$$ a_1^k+\cdots+a_n^k \leq b_1^k+\cdots+b_n^k,$$
and
$$a_1+\cdots+a_n = b_1+\cdots+b_n.$$
Can we conclude $$\sqrt{a_1}+\cdots+\sqrt{a_n}\geq \sqrt{b_1}+\cdots+\sqrt{b_n}$$
|
As a counter-example with $n=3$ $$a_1=6, a_2=42,a_3=52$$ $$b_1=12, b_2=22, b_3=66$$ Then $a_1+a_2+a_3 = 100 = b_1+b_2+b_3$ $a_1^p+a_2^p+a_3^p \lt b_1^p+b_2^p+b_3^p$ for $p \gt 1$ $\sqrt{a_1}+\sqrt{a_2}+\sqrt{a_3} \lt 16.2 \lt \sqrt{b_1}+\sqrt{b_2}+\sqrt{b_3}$ though notice that $a_1^p+a_2^p+a_3^p \gt b_1^p+b_2^p+b_3^p$ for $0.851 \le p \lt 1$, and I suspect all such counterexamples with $n=3$ reverse the inequality in a small interval below $1$ Added Perhaps a more interesting counter-example is $$a_1=1, a_2=4,a_3\approx 5.3931524748543$$ $$b_1=2, b_2=2, b_3\approx 6.3931524748543$$ where $ 6.3931524748543$ is an approximation to the solution of $x^x=16 (x-1)^{x-1}$, so $\sum a_i = \sum b_i$ and $\prod a_i^{a_i} = \prod {b_i}^{b_i}$ This has $$a_1^p+a_2^p+a_3^p \le b_1^p+b_2^p+b_3^p$$ for all non-negative real $p$ (integer or not), and equality only when $p=0$ or $1$
|
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|
301,844 |
Suppose $\mathbf{v},\mathbf{w} \in \mathbb{R}^n$ (and if it helps, you can assume they each have non-negative entries), and let $\mathbf{v}^2,\mathbf{w}^2$ denote the vectors whose entries are the squares of the entries of $\mathbf{v}$ and $\mathbf{w}$. My question is how to prove that
\begin{align*}
\|\mathbf{v}^2\|\|\mathbf{w}^2\| - \langle \mathbf{v}^2,\mathbf{w}^2\rangle \leq \|\mathbf{v}\|^2\|\mathbf{w}\|^2 - \langle \mathbf{v},\mathbf{w}\rangle^2.
\end{align*} Some notes are in order: The Cauchy-Schwarz inequality tells us that both sides of this inequality are non-negative. Thus the proposed inequality is a strengthening of Cauchy-Schwarz that gives a non-zero bound on the RHS. I know that this inequality is true, but my method of proving it is extremely long and roundabout. It seems like it should have a straightforward-ish proof, or should follow from another well-known inequality, and that's what I'm looking for.
|
Here is a proof for every $n$. Using the notation $\mathbf{v}=(v_1,\dots,v_n)$ and $\mathbf{w}=(w_1,\dots,w_n)$, the inequality reads
$$\left(\sum_i v_i^4\right)^{1/2}\left(\sum_i w_i^4\right)^{1/2}-\sum_i v_i^2 w_i^2\leq
\left(\sum_i v_i^2\right)\left(\sum_i w_i^2\right)-\left(\sum_i v_i w_i\right)^2.$$
Rewriting the right hand side in a familiar way, and then rearranging and squaring, we obtain the equivalent form
$$\left(\sum_i v_i^4\right)\left(\sum_i w_i^4\right)\leq\left(\sum_i v_i^2 w_i^2+\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$
Rewriting the left hand side in a familiar way, we obtain the equivalent form
$$\left(\sum_i v_i^2w_i^2\right)^2+\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq\left(\sum_i v_i^2 w_i^2+\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$
Equivalently,
$$\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq 2\left(\sum_k v_k^2w_k^2\right)\sum_{i<j}(v_iw_j-v_j w_i)^2+\left(\sum_{i<j}(v_iw_j-v_j w_i)^2\right)^2.$$
It will be clear in a moment why we renamed the variable $i$ to $k$ in the first sum on the right hand side.
Namely, we claim that the following stronger inequality holds:
$$\sum_{i<j}(v_i^2w_j^2-v_j^2w_i^2)^2\leq 2\sum_{i<j}(v_i^2w_i^2+v_j^2w_j^2)(v_iw_j-v_j w_i)^2+\sum_{i<j}(v_iw_j-v_j w_i)^4.$$
Indeed, this inequality can be rearranged to
$$0\leq 2\sum_{i<j}(v_iw_i-v_jw_j)^2(v_iw_j-v_jw_i)^2,$$
and we are done.
|
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|
302,492 |
Quantum computing is a very active and rapidly expanding field of research. Many companies and research institutes are spending a lot on this futuristic and potentially game-changing technology. Some even built toy models for a quantum computer in the lab. For instance, see IBM's 50-qubit quantum computer . However, some scientists are not that optimistic when it comes to the predicted potential advantages of quantum computers in comparison with the classical ones. They believe there are theoretical obstacles and fundamental limitations that significantly reduce the efficiency of quantum computing. One mathematical argument against quantum computing (and the only one that I am aware of) is based on the Gil Kalai's idea concerning the sensitivity of the quantum computation process to noise, which he believes may essentially affect the computational efficiency of quantum computers. Question. I look for some references on similar theoretical (rather than practical) mathematical arguments against quantum computing — if there are any. Papers and lectures on potential theoretical flaws of quantum computing as a concept are welcome. Remark. The theoretical arguments against quantum computing may remind the so-called Goedelian arguments against the artificial intelligence, particularly the famous Lucas-Penrose's idea based on the Goedel's incompleteness theorems. Maybe there could be some connections (and common flaws) between these two subjects, particularly when one considers the recent innovations in QAI such as the Quantum Artificial Intelligence Lab .
|
Scott Aaronson has this list of Eleven Objections, involving both mathematics and physics arguments. What I did is to write out every skeptical argument against the
possibility of quantum computing that I could think of. We'll just go
through them, and make commentary along the way. Let me just start by
saying that my point of view has always been rather simple: it's
entirely conceivable that quantum computing is impossible for some
fundamental reason. If so, then that's by far the most exciting thing
that could happen for us. That would be much more interesting than if
quantum computing were possible, because it changes our understanding
of physics. To have a quantum computer capable of factoring
10000-digit integers is the relatively boring outcome -- the outcome
that we'd expect based on the theories we already have. So, on to the Eleven Objections: Works on paper, not in practice. Violates Extended Church-Turing Thesis. Perhaps the most precise objection from a mathematical point of view. Here is an extended discussion . Not enough "real physics." Small amplitudes are unphysical. Leonid Levin's objection Exponentially large states are unphysical. Paul Davies' objection Quantum computers are just souped-up analog computers. Robert Laughlin's objection Quantum computers aren't like anything we've ever seen before. Dyakonov's objection Quantum mechanics is just an approximation to some deeper theory. Decoherence will always be worse than the fault-tolerance threshold. Gil Kalai's objection We don't need fault-tolerance for classical computers. Errors aren't independent. This is (part of) Gil Kalai's objection UPDATE (September 2019): objection 2, the violation of the extended Church-Turing thesis, may now have been removed, as discussed by Scott Aaronson.
|
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"https://mathoverflow.net/questions/302492",
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|
302,876 |
When modular forms are usually introduced, it is by: "We have the standard action of $SL(2,\mathbb Z)$ on the upper half-plane, so let us study functions which are (almost) invariant under such transformations". But modular forms were discovered in the 19th century, before group theory was available. Therefore, I'd like to ask how were modular forms originally discovered? In his book The 1-2-3 of modular forms D. Zagier writes that this was the reason: Proposition 21 Let $f(z)$ be a modular form of weight $k$ on some group $\Gamma$ and $t(z)$ a modular function with respect to $\Gamma$. Express $f(z)$ locally as $\Phi(t(z))$. Then the function $\Phi(t)$ satisfies a linear differential equation of order $k+1$ with algebraic coefficients After the proof (actually, 3 of them) there is an example: The function $\sqrt[4] E_4$ is formally a mod. form of weight one, which means: $\sqrt[4] E_4(z)=F(\frac 1 {12}, \frac 5 {12}, 1; 1728/j(z))=1+\frac {60} {j(z)}+...$ He states that this is a classical identity found in the works of Fricke and Klein. This, however, sparks a few questions: 1) I know that hypergeometric series were investigated in that time, but how did they find this one and similar ones? 2) Did they find something special about the resulting functions, like the $SL(2,\mathbb Z)$ transformation law?
|
You don't need the language of group theory to talk about some aspects of groups. For example, number theorists going back to Fermat were studying the group of units mod $m$ (including things like the order of a unit mod $m$) and geometers were studying groups of motions in space long before anyone defined a "group". The first modular forms (of level $4$, not level $1$) were found by Gauss in his work on the arithmetic-geometric mean around 1800, and it would take until the end of the 19th century for the term "modular form" to be introduced, in 1890. I provided links for this near the end of my answer here , but I don't want to discuss this further. Instead I want to show how one reason for interest in modular forms in the 19th century was the construction of nonconstant meromorphic functions on Riemann surfaces of higher genus. This will explain why someone might care about functions satisfying a transformation rule like $f((a\tau+b)/(c\tau+d)) = (c\tau+d)^kf(\tau)$. Compact Riemann surfaces of genus 1 arise in the form $\mathbf C/L$ where $L$ is a discrete subgroup of $\mathbf C$. Meromorphic functions on $\mathbf C/L$ are the elliptic functions, which in various forms were studied throughout much of the 19th century (by Jacobi, Weierstrass, et al.). We'd like to find a similar story for compact Riemann surfaces of genus greater than $1$. By letting a discrete group $Γ \subset {\rm SL}_2(\mathbf R)$ act on the upper half-plane $\mathfrak h$ by linear fractional transformations, the coset space $\Gamma\setminus\mathfrak h$ will be a compact Riemann surface (with finitely many points missing) and the higher-genus Riemann surfaces arise in this way.
A natural collection of discrete subgroups of ${\rm SL}_2(\mathbf R)$ is ${\rm SL}_2(\mathbf Z)$ and its finite-index subgroups, such as the congruence subgroups. When I write $\Gamma$ below, you could consider it to be one of these groups of integral matrices. How can we create nonconstant meromorphic functions on $\Gamma\setminus\mathfrak h$? That is the same thing as creating nonconstant meromorphic functions $F : \mathfrak h \rightarrow \mathbf C$ that are $\Gamma$-invariant: $F(\gamma\tau) = F(\tau)$ for all $\gamma \in \Gamma$ and $\tau \in \mathfrak h$. If we aren't clever enough to be able to write down $\Gamma$-invariant functions directly, we can still make progress by finding some non-invariant functions as long as they are non-invariant in the same way: if
$$
f\left(\frac{aτ + b}{cτ + d}\right) = (cτ + d)^kf(τ)
$$
and
$$
g\left(\frac{aτ + b}{cτ + d}\right) = (cτ + d)^kg(τ)
$$
for some "weight" $k \in \mathbf Z$ and all $(\begin{smallmatrix} a& b\\c &d\end{smallmatrix}) ∈ \Gamma$ and $\tau ∈ \mathfrak h$, then the ratio $f(τ)/g(τ)$ is $\Gamma$-invariant:
$$
\frac{f((a\tau+b)/(c\tau+d))}{g((a\tau+b)/(c\tau+d))} = \frac{(cτ + d)^kf(τ)}{(cτ + d)^kg(τ)} = \frac{f(\tau)}{g(\tau)}.
$$
As long as $f$ and $g$ are not constant multiples of each other (essentially, the space of modular forms $f$ and $g$ live in is more than one-dimensional), the ratio $f/g$ will be a nonconstant $\Gamma$-invariant function. But why should we use fudge factors of the form $(cτ + d)^k$? Suppose for a function $f : \mathfrak h \rightarrow \mathbf C$ that $f(γτ)$ and $f(τ)$ are always related by a nonvanishing factor determined by $γ ∈ Γ$ and $τ ∈ \mathfrak h$ alone, not by $f$:
$$
f(γτ) = j(γ, τ)f(τ)
$$
for some function $j : \Gamma × \mathfrak h \rightarrow \mathbf C^\times$.
If also $g(γτ) = j(γ, τ)g(τ)$ then $f(γτ)/g(γτ) = f(\tau)/g(\tau)$, so $f/g$ is $\Gamma$-invariant. We want to figure out how anyone might imagine using the fudge factor $j(\gamma,\tau) = (c\tau+d)^k$, where $γ = (\begin{smallmatrix} a& b\\ c& d\end{smallmatrix})$. Since $(γ_1γ_2)τ = γ_1(γ_2τ)$, we have $f((γ_1γ_2)τ) = f(γ_1(γ_2τ))$. This turns
the above displayed equation into
$$
j(γ_1γ_2,τ)f(τ) = j(γ_1,γ_2τ)f(γ_2τ).
$$
Since $f(γ_2τ) = j(γ_2,τ)f(τ)$, the above equation holds if and only if the "cocycle condition"
$$j(γ_1γ_2, τ) = j(γ_1,γ_2τ)j(γ_2,τ)$$
holds, which looks a lot like the chain rule $(f_1 ◦ f_2)'(x) = f_1'(f_2(x))f_2'(x)$. This suggests a natural example of the function $j(\gamma,\tau)$ using differentiation: when
$γ = (\begin{smallmatrix} a& b\\ c& d\end{smallmatrix})$ and $\tau \in \mathfrak h$, set
$$
j(γ, τ ) := \left(\frac{aτ + b}{cτ + d}\right)' =
\frac{a(cτ + d) − c(aτ + b)}{(cτ + d)^2} =\frac{ad − bc}{(cτ + d)^2},
$$
and for $γ ∈ {\rm SL}_2(\mathbf R)$ this says $j(γ, τ) = 1/(cτ + d)^2$.
When $j(γ,τ)$ satisfies the cocycle condition, so does $j(γ,τ)^m$
for each $m \in \mathbf Z$, which motivates the consideration of the transformation rule for modular forms with factors $(cτ + d)^k$, at least for even $k$ (use $k = -2m$). The top answer at https://math.stackexchange.com/questions/312515/what-is-the-intuition-between-1-cocycles-group-cohomology shows how the same cocycle condition arises when trying to write down line bundles on an elliptic curve over $\mathbf C$.
|
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|
302,933 |
We call an integer $k\geq 1$ good if for all $q\in\mathbb{Q}$ there are $a_1,\ldots, a_k\in \mathbb{Q}$ such that $$q = \prod_{i=1}^k a_i \cdot\big(\sum_{i=1}^k a_i\big).$$
Euler showed that $k=3$ is good. Is the set of good positive integers infinite?
|
I suspect that $k = 4$ is good, but am not sure how to prove it. However, every positive integer $k \geq 5$ is good. This follows from the fact (see the proof of Theorem 1 from this preprint ) that for any rational number $x$, there are rational numbers $a$, $b$, $c$, $d$ so that $a+b+c+d = 0$ and $abcd = x$. In particular, one can take
$$
a(x) = \frac{2(1-4x)^{2}}{3(1+8x)}, b(x) = \frac{-(1+8x)}{6}, c(x) = \frac{-(1+8x)}{2(1-4x)}, d(x) = \frac{18x}{(1-4x)(1+8x)},
$$
as long as $x \not\in \{1/4, -1/8\}$. (For $x = 1/4$ one can take $(a,b,c,d) = (-1/2,1/2,-1,1)$ and for $x = -1/8$ one can take $(a,b,c,d) = (-2/3,25/12,-1/15,-27/20)$.) Now, fix $k \geq 5$, let $q \in \mathbb{Q}$ and take $a_{1} = a(q/(k-4))$, $a_{2} = b(q/(k-4))$, $a_{3} = c(q/(k-4))$, $a_{4} = d(q/(k-4))$ and $a_{5} = a_{6} = \cdots = a_{k} = 1$. We have that
$$
a_{1} + a_{2} + a_{3} + a_{4} + \cdots + a_{k} = 0 + a_{5} + \cdots + a_{k} = k-4
$$
and $a_{1} a_{2} a_{3} a_{4} \cdots = \frac{q}{k-4} \cdot 1 \cdot 1 \cdots 1 = \frac{q}{k-4}$. Thus
$$
\left(\prod_{i=1}^{k} a_{i}\right) \left(\sum_{i=1}^{k} a_{i}\right) = \frac{q}{k-4} \cdot (k-4) = q.
$$
|
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|
303,149 |
What is the probability that three pairs $(a,b) $ , $(c,d) $ and $(e,f) $ of integers generate $\mathbb Z^2$? As usual the probability is the limit as $n\to \infty$ of the same probability for the $n\times n$ square. It is well known that for $\mathbb Z $ the probability of two numbers to generate is $6/\pi^2$.
|
According to Proposition 1 in the paper G. Maze, Gérard, J. Rosenthal, U. Wagner: Natural density of rectangular unimodular integer matrices , Linear Algebra Appl. 434 , No. 5 (2011), 1319-1324, ZBL1211.15044 , the probability that $n$ random vectors generate $\mathbb{Z}^{n-1}$ is $$p_n = \prod_{j=2}^n \zeta(j)^{-1}.$$ For $n=2$ this gives $p_2=\zeta(2)^{-1}=6/\pi^2$, whereas for $n=3$ we obtain $$p_3= \zeta(2)^{-1} \zeta(3)^{-1} \simeq 0.505739038$$
|
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303,793 |
Here is a naive outsiders perspective on set theory: A typical set-theoretical result involves constructing new models of set theory from given ones (typically with different theories for the original model and the resulting model). Typically, from a meta-perspective we are allowed (encouraged, or even required) to assume that the models are countable. To the extent that this view is correct, set-theoretic constructions correspond to partial, multivalued operations $T : \subseteq \{0,1\}^\mathbb{N} \rightrightarrows \{0,1\}^\mathbb{N}$ which are defined on sequences coding a model of the original theory, and are outputting a sequence coding a model of the desired theory. These are multivalued, because the constructions may involve something like "pick a generic filter for this forcing notion". Question: Are these operations typically computable, and if not, how non-computable are they? The Weihrauch degrees ( https://arxiv.org/abs/1707.03202 ) provide a framework for classifying non-computability of operations of these types. An answer, however, could also take forms like "For arguments like this, a resulting model is typically computable in the join of the original model and a 1-generic."
|
The question is extremely interesting, and I have looked into this kind of thing with various colleagues (including Russell Miller and Kameryn Williams), although our investigation has not yet resulted in any paper. One important realization to which we had come was that the question is highly sensitive to whether the input data includes just the set-membership information, or instead the diagram (the truths) of the original model. With an oracle for the elementary diagram of a countable model $\langle M,\in^M\rangle$ of a model of set theory, that is, an oracle listing all the true statements of the model, with names for every element, then for any forcing notion $\mathbb{P}$ in $M$, one can compute an $M$-generic filter $G$ and the elementary diagram of the corresponding forcing extension $M[G]$. This can be seen as a strong positive answer to your question. The reason you can find such a $G$ is that the oracle for the diagram of $M$ allows you to find a canonical enumeration of the dense sets of $M$, and so by the usual diagonalization, one can compute a filter getting into all those dense sets, and this filter will therefore be generic. (In this sense, your remarks about multi-valued seem to be unnecessary — one can compute a specific generic filter.) Using this filter $G$, one can evaluate the diagram of the extension $V[G]$ in terms of names, using the forcing relation, which is expressed in the diagram of the original model. Thus, from the diagram of a model $M$, you can compute the diagram of the forcing extension $M[G]$. In particular, this shows that from the diagram of any model of ZFC, you can compute the diagram of a model of ZFC+CH or of ZFC+$\neg$CH or ZFC+MA and so on, for all the various results obtained by forcing. One can also compute the elementary diagram of many other kinds of set-theoretic constructions, such as inner models $L$ and $\text{HOD}$, and ultrapowers $\text{Ult}(M,\mu)$ by an ultrafilter $\mu$ in $M$, or an extender ultrapower. You don't need the whole diagram to compute the generic filter $G$, and in fact, even just the atomic diagram, which contains information only about the membership relation $n\in m$ for specific objects, is sufficient to compute a generic filter $G$. One simply fixes the object whose elements are the dense subsets of $\mathbb{P}$ and an object for the order relation, and then you can carry out the diagonalization. Nevertheless, and this is an important point, if all you have is the atomic diagram of the original model $M$, then it turns out that you cannot compute much about the model or its forcing extensions at all. For example, you cannot even reliably identify from the atomic diagram which is the empty set of $M$, since for any computable algorithm, there is a model $M$ that will fool it. And there are many further such examples. I take these examples to illustrate that one shouldn't think of a model of set theory merely as its atomic diagram, but as a bit more.
|
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|
303,861 |
It has been recently mentioned by a speaker (his talk is completely not relevant to random matrix theory/RMT though) that modern statistics, especially random matrices theory, will help solving some number theoretic problems. I was quite intrigued and ask for more explanation but the speaker himself said he did not understand the philosophy well enough but point me to a few keywords to search for, one among which is "Katz-Sarnak philosophy". After a few searches, I figured out that all sources seem to point to [KS]. The major results in [KS] seems like saying that the class of general linear (compact) groups have the same n-level correlations. While some researchers [Kowalski][Miller] do mention that they applied KS philosophy, but what they have done seems very different...So for number theorists and experts on elliptic curves: (1) When you refer to "Katz-Sarnak philosophy", what kind of
thinking/technique do you actually mean? (2) Is there a formalism/explanation of this philosophy in language of
RMT? I asked this because it might be helpful to understand it in this
perspective (at least to a probabilist). Any inputs are highly appreciated. Reference [KS]Katz, Nicholas M., and Peter Sarnak, eds. Random matrices, Frobenius eigenvalues, and monodromy. Vol. 45. American Mathematical Soc., 1999. Google books [Kowalski] http://blogs.ethz.ch/kowalski/2008/07/30/finding-life-beyond-the-central-limit-theorem/ [Miller]Miller, Steven J. "One-and two-level densities for rational families of elliptic curves: evidence for the underlying group symmetries." Compositio Mathematica 140.4 (2004): 952-992.
|
The "Katz-Sarnak philosophy" is just the idea that statistics of various kinds for $L$-functions should, in the large scale limit, match statistics for large random matrices from some particular classical compact group. First you need to decide what kinds of zeros to look at: the high zeros of an individual $L$-function, the low zeros (near the real axis) in some family of related $L$-functions (e.g., low zeros in all Dirichlet $L$-functions, maybe with a restriction on the parity of the character), and so on. The choice of what sort of zeros you look at can have an effect, e.g., looking at statistics of the low zeros in a family can reveal a "symmetry" (similar statistics to a specific classical compact group) that is not evident when looking at high zeros of an individual $L$-function (universality, not distinguishing between different $L$-functions). Next you need to normalize the (nontrivial) zeros. For example, zeros of an individual $L$-function tend to get close to each other high up the critical line, so you count the number of zeros up to height $T$ and then rescale them so they get average spacing 1. It's those rescaled zeros that you work with when computing your various statistics. Next you need to work with a suitable class of test functions and be able to actually carry out statistical calculations to reveal similarity with some class of random matrices. Part of the motivation for people to look at these questions is the hope that it could suggest a spectral interpretation of the nontrivial zeros. Look up the Hilbert-Polya conjecture (Hilbert had nothing to do with it). You mentioned the Katz-Sarnak book in your question. Keep in mind that this book is entirely about $L$-functions for varieties over finite fields, not the ones like $\zeta(s)$ that are associated to number fields. The blog post and paper you mention by Kowalski and Miller are about $L$-functions over number fields, and that is perhaps why they look very different to you from the Katz-Sarnak book. Another paper where the number field case is treated head-on is Rudnick, Sarnak, "Zeros of principal $L$-functions and random matrix theory," Duke Math. J. 81 (1996), 269–322.
If you want to see a textbook's treatment of the relation between $L$-functions over number fields (like $\zeta(s)$ and Dirichlet $L$-functions) and random matrix theory, consider the book "An Invitation to Modern Number Theory" by Steve Miller and Ramin Takloo-Bighash. The random matrix theory is in Part 5.
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|
304,486 |
The invention of the Jones polynomial led to hundreds of papers and a Fields medal. However, as far as I can tell it had few consequences in topology. After all, after Thurston’s work we already had algorithms to completely classify knots, so by itself a new invariant seems to be of limited value. Given all the excitement, however, I suspect that I might be missing something. Why was this so important?
|
Your question presupposes that people were excited about the Jones polynomial because it would help them to classify/distinguish knots. In fact, I suspect the interest came from the fact that this knot invariant was originally defined using operator algebras (rather than in the more combinatorial way people usually define it now). Operator algebras had grown out of an attempt to formalise quantum mechanics/QFT, and it was surprising that a knot invariant should appear naturally in a completely different subject. At around the same time, other manifold invariants (Donaldson invariants, instanton Floer homology) appeared that were also inspired by constructions in physics. These other manifold invariants had very definite topological consequences, solving huge open questions in 4-dimensional topology. Witten showed that all of these invariants (including the Jones polynomial) can be obtained formally by performing path integrals. For example, roughly speaking*, the Jones polynomial can be obtained by looking at all possible connections A on a suitable bundle on your 3-manifold, taking the trace of the holonomy of A around your knot, multiplying by e^(iCS(A)) where CS is the Chern-Simons invariant of A, and then integrating the result (over the infinite-dimensional space of all connections, which is a kind of path integral). Whether or not you care about physics, that is a pretty cool way to define an invariant.** This led mathematicians to study topological quantum field theories, and the associated manifold invariants. In particular, Khovanov was led to discover his homological refinement of the Jones polynomial, which is extraordinarily useful as a knot invariant: it is functorial under cobordisms of knots, so can be used to study surfaces in 4-manifolds, for example giving lower bounds on slice genus of knots. So if you're looking for topological results in pure knot theory that wouldn't exist if it weren't for the Jones polynomial, you should check out Khovanov homology and its applications. (*) I corrected what I had written earlier (it's been a long time since I looked at Witten's paper). (**) The reason I think this is a cool way to define an invariant is that, since the thing you're integrating is covariant under diffeomorphisms, provided your path integral measure is covariant under diffeomorphisms, it would be obvious that you get an invariant. Of course, even if you knew how to define these measures, the devil is in proving that they are invariant (often, classical symmetries of a Lagrangian do not descend to a symmetry of the quantum theory, and the theory is called anomalous).
|
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|
304,526 |
For natural numbers $m, r$, consider the ratio of the number of subsets of size $m$ taken from a set of size $2(m+r)$ to the number of subsets of the same size taken from a set of size $m+r$: $$R(m,r)=\frac{\binom{2(m+r)}{m}}{\binom{m+r}{m}}$$ For $r=0$ we have the central binomial coefficients, which of course are all integers: $$R(m,0)=\binom{2m}{m}$$ For $r=1$ we have the Catalan numbers, which again are integers: $$R(m,1)=\frac{\binom{2(m+1)}{m}}{m+1}=\frac{(2(m+1))!}{m!(m+2)!(m+1)}=\frac{(2(m+1))!}{(m+2)!(m+1)!}=C_{m+1}$$ However, for any fixed $r\ge 2$, while $R(m,r)$ seems to be mostly integral, it is not exclusively so. For example, with $m$ ranging from 0 to 20000, the number of times $R(m,r)$ is an integer for $r=2,3,4,5$ are 19583, 19485, 18566, and 18312 respectively. I am seeking general criteria for $R(m,r)$ to be an integer. Edited to add: We can write: $$R(m,r) = \prod_{k=1}^m{\frac{m+2r+k}{r+k}}$$ So the denominator is the product of $m$ consecutive numbers $r+1, \ldots, m+r$, while the numerator is the product of $m$ consecutive numbers $m+2r+1,\ldots,2m+2r$. So there is a gap of $r$ between the last of the numbers in the denominator and the first of the numbers in the numerator.
|
Put $n=m+r$, and then we can write $R(m,r)$ more conveniently as
$$
R(m,r) = \frac{(2n)!}{m! (n+r)!} \frac{m! r!}{n!} = \frac{\binom{2n}{n} }{\binom{n+r}{r}}.
$$
So the question essentially becomes one about which numbers $n+k$ for $k=1$, $\ldots$, $r$ divide the middle binomial coefficient $\binom{2n}{n}$. Obviously when $k=1$, $n+1$ always divides the middle binomial coefficient, but what about other values of $k$? This is treated in a lovely Monthly article of Pomerance . Pomerance shows that for any $k \ge 2$ there are infinitely many integers with $n+k$ not dividing $\binom{2n}{n}$, but the set of integers $n$ for which $n+k$ does divide $\binom{2n}{n}$ has density $1$. So for any fixed $r$, for a density $1$ set of values of $n$ one has that $(n+1)$, $\ldots$, $(n+r)$ all divide $\binom{2n}{n}$, which means that their lcm must divide $\binom{2n}{n}$. But one can check without too much difficulty that the lcm of $n+1$, $\ldots$, $n+r$ is a multiple of $\binom{n+r}{r}$, and so for fixed $r$ one deduces that $R(m,r)$ is an integer for a set of values $m$ with density $1$. (Actually, Pomerance mentions explicitly in (5) of his paper that $(n+1)(n+2)\cdots (n+r)$ divides $\binom{2n}{n}$ for a set of full density.) Finally let me show that $R(m,r)$ is not an integer infinitely often when $r \ge 2$ is fixed. Let $p$ be a prime with $r<p <2r$; such a prime exists by Bertrand's postulate. Now take $n=p^a - r$ for any natural number $a$. Then note that $p^a$ exactly divides $\binom{n+r}{r}$, and that the power of $p$ dividing $\binom{2n}{n}$ is
$$
\sum_{j=1}^{a} \Big( \Big\lfloor \frac{2(p^a-r)}{p^j} \Big\rfloor - 2\Big\lfloor \frac{p^a-r}{p^j}\Big \rfloor \Big) =a-1,
$$
since the term $j=1$ contributes $0$ and the other terms contribute $1$.
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|
304,557 |
Let $g(x) = e^x + e^{-x}$. For $x_1 < x_2 < \dots < x_n$ and $b_1 < b_2 < \dots < b_n$, I'd like to show that the determinant of the following matrix is positive, regardless of $n$: $\det \left (\begin{bmatrix}
\frac{1}{g(x_1-b_1)} & \frac{1}{g(x_1-b_2)} & \cdots & \frac{1}{g(x_1-b_n)}\\
\frac{1}{g(x_2-b_1)} & \frac{1}{g(x_2-b_2)} & \cdots & \frac{1}{g(x_2-b_n)}\\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{g(x_n-b_1)} & \frac{1}{g(x_n-b_2)} & \cdots & \frac{1}{g(x_n-b_n)}
\end{bmatrix} \right ) > 0$. Case $n = 2$ was proven by observing that $g(x)g(y) = g(x+y)+g(x-y)$,
and
$g(x_2 - b_1)g(x_1-b_2) = g(x_1+x_2 - b_1-b_2)+g(x_2-x_1+b_2-b_1) > g(x_1+x_2 - b_1-b_2)+g(x_2-x_1-b_2+b_1) = g(x_1-b_1)g(x_2-b_2)$ However, things get difficult for $n \geq 3$. Any ideas or tips? Thanks!
|
At first, we prove that the determinant is non-zero, in other words, the matrix is non-singular. Assume the contrary, then by the linear dependency of the columns there exist real numbers $\lambda_1,\dots,\lambda_n$, not all equal to 0, such that $F(x_i):=\sum_j \frac{\lambda_j}{g(x_i-b_j)}=0$ for all $i=1,2,\dots,n$. But the equation $F(x)=0$ is a polynomial equation with respect to $e^{2x}$ and the degree of a polynomial is less than $n$. So, it can not have $n$ distinct roots. Now we note that the matrix is close to an identity when $x_i=b_i$ and $b_i$'s are very much distant from each other, and the phase space of parameters $\{(x_1,\dots,x_n,b_1,\dots,b_n):x_1<\dots<x_n,b_1<b_2<\dots <b_n\}$ is connected. Thus the sign of the determinant is always plus.
|
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|
304,568 |
A professor of mine told me that this is true, but he doesn't remember what the proof was or where to find it, and I haven't been able to find a source for it yet. As such I am looking for one here. In the theorem as stated, $\mathbb{F}$ is any field and $T_n(\mathbb{F})$ denotes the algebra of upper triangular $n\times n$ matrices over $\mathbb{F}$. Theorem: Let $A,B\in T_n(\mathbb{F})$ be such that for all $X\in T_n(\mathbb{F})$, $$AX=XA\implies BX=XB$$ Then $B=p(A)$ for some $p\in \mathbb{F}[t]$. Does anyone know of a source for this result? I have searched Google, MSE, MO, and the like to no avail. If we replace $T_n(\mathbb{F})$ by $M_n(\mathbb{F})$, the question is answered in this paper . Unfortunately, the argument doesn't seem to translate directly, as I can't find a way to force the $M_i$ maps to be upper-triangular. Also, I have already asked this question here on MSE. As the question is for an undergraduate research project, it felt appropriate to ask it here as well. Thanks for any help! Edit on 9 July, 2018: It's probably worth mentioning that the following theorem is false , so an appeal to Jordan form won't work (at least, not as easily as we'd hope it would). Fake Theorem: If $A\in T_n(\mathbb{F})$, then there exists an invertible $T\in T_n(\mathbb{F})$ and a permutation matrix $P$ such that $P^{-1}T^{-1} ATP$ is in Jordan form. An explicit counterexample is
$$A=\left[\begin{array}{cccc} 0&1&0&0\\ &0&0&1\\ & &0&1\\ & & &0\end{array}\right]$$
and a more detailed demolishing of this theorem is given here , where the authors prove that if $n\geq 12$ and $\mathbb{F}$ is infinite, then there are infinite sets of nilpotent matrices in $T_n(\mathbb{F})$, none of which are conjugate (in $T_n(\mathbb{F})$) to any of the others. I mention this because I thought it was true for longer than I'd like to admit, and a few other people I've talked to thought it was true as well until told otherwise.
|
This is false! Let
$$A = \begin{bmatrix}
0&0&0&1 \\
&0&1&0 \\
&&0&0 \\
&&&0 \\
\end{bmatrix}.$$
Imposing that $XA=AX$ for upper triangular $X$ gives linear equations on the $10$ entries of $X$. Solving them, I get that this occurs precisely for $X$ of the form
$$X=\begin{bmatrix}
a&0&\ast&\ast \\
&b&\ast&\ast \\
&&b&0 \\
&&& a \\
\end{bmatrix}.$$
In turn, an upper triangular matrix commutes with all such $X$ if and only if it is of the form
$$B=\begin{bmatrix}
c&0&0&d \\
&c&e&0 \\
&&c&0 \\
&&&c \\
\end{bmatrix}.$$
But such a $B$ is only a polynomial in $A$ if $d=e$.
|
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|
304,571 |
When understanding about gerbe over a manifold $X$ from Hitchin - Lectures on special Lagrangian submanifolds it is said that We are basically in gerbe territory (for smooth manifolds) if any one of the following is being considered a cohomology class in $H^3(X,\mathbb{Z})$ ** ** In similar manner, When reading about gerbes over stacks what kind of cohomology do we come across? Can some one give me some outline of how and what cohomology comes in when studying about gerbes over stacks?
|
This is false! Let
$$A = \begin{bmatrix}
0&0&0&1 \\
&0&1&0 \\
&&0&0 \\
&&&0 \\
\end{bmatrix}.$$
Imposing that $XA=AX$ for upper triangular $X$ gives linear equations on the $10$ entries of $X$. Solving them, I get that this occurs precisely for $X$ of the form
$$X=\begin{bmatrix}
a&0&\ast&\ast \\
&b&\ast&\ast \\
&&b&0 \\
&&& a \\
\end{bmatrix}.$$
In turn, an upper triangular matrix commutes with all such $X$ if and only if it is of the form
$$B=\begin{bmatrix}
c&0&0&d \\
&c&e&0 \\
&&c&0 \\
&&&c \\
\end{bmatrix}.$$
But such a $B$ is only a polynomial in $A$ if $d=e$.
|
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|
304,908 |
I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$ , and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$ , $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$ , of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So: Is $\Bbb S^2 \times \Bbb S^4$ symplectic? Edit: after some time I came back to those calculations. I had missed the obvious case $(n_1,n_2) = (2,2)$ . The proof given in the answers can be adapted to show that products of the form $\Bbb S^2 \times \Bbb S^{n_2}$ for even $n_2>2$ are not symplectic. The conclusion of what happened here is the Theorem: Let $1 \leq n_1 \leq n_2$ be natural numbers. Then $\Bbb S^{n_1}\times \Bbb S^{n_2}$ is symplectic if and only if $n_1=n_2=1$ or $n_1=n_2=2$ .
|
No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ is a symplectic form on $S^2\times S^4$. Then $[\omega]=c[\pi^*\alpha]$ for some $c\in\mathbb R^\times$. Then $[\omega^3]=c^3[\pi^*\alpha^3]=0$, contradicting the requirement that $\omega^3$ is everywhere nondegenerate.
|
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|
304,917 |
I am looking for a proof, reference, comment of an inequality as follows: If $f(x)$ and $g(x)$ be two continuous derivative funcions in interval $[a, b]$. Such that: $f(a)=g(a)$ and $f(b)=g(b)$ $($$f''(x) > 0$ or $f''(x) < 0$$)$ and $($$g''(x) > 0$ or $g''(x) < 0$$)$ $$|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)|\ge|g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a)|$$ Then $$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$
|
No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ is a symplectic form on $S^2\times S^4$. Then $[\omega]=c[\pi^*\alpha]$ for some $c\in\mathbb R^\times$. Then $[\omega^3]=c^3[\pi^*\alpha^3]=0$, contradicting the requirement that $\omega^3$ is everywhere nondegenerate.
|
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|
306,134 |
Who first chose the names Alice and Bob for the players (or observers) A and B?
|
Quoted from Wikipedia : The Alice and Bob characters were invented by Ron Rivest, Adi Shamir, and Leonard Adleman in their 1978 paper "A method for obtaining digital signatures and public-key cryptosystems". Rivest, R. L.; Shamir, A.; Adleman, L. , A method for obtaining digital signatures and public-key cryptosystems , Commun. ACM 21, 120-126 (1978). ZBL0368.94005 .
|
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|
306,851 |
The title is a quote from a Jim Holt article entitled, "The Riemann zeta conjecture and the laughter of the primes" (p. 47). 1 His example of a "long-standing conjecture" is the Riemann hypothesis,
and he is cautioning "those who blithely assume the truth of the
Riemann conjecture." Q .
What are examples of long-standing conjectures in analysis that turned out to be false? Is Holt's adverb "often" justified? 1 Jim Holt. When Einstein Walked with Gödel: Excursions to the Edge of Thought .
Farrar, Straus and Giroux, 2018. pp.36-50.
( NYTimes Review .)
|
I don't know about analysis in general, but I think it's definitely fair to say "often" in functional analysis. My feeling is that we have a solid, thorough, elegant body of theory which usually leads to positive solutions rather quickly, when they exist. (The Kadison-Singer problem is a recent exception which required radically new tools for a positive solution.) Problems that stick around for a long time tend to do so not because there's a complicated positive solution but because there's a complicated counterexample. That's a gross overgeneralization but I think there's some truth to it. The first examples I can think of are: every separable Banach space has the approximation property and has a Schauder basis (counterexample by Enflo) every bounded linear operator on a Banach space has a nontrivial closed invariant subspace (counterexamples by Enflo and Read) every infinite dimensional Banach space has an infinite dimensional subspace which admits an unconditional Schauder basis (counterexample by Gowers and Maurey) every infinite dimensional Banach space $X$ is isomorphic to $X \oplus \mathbb{R}$; if $X$ and $Y$ are Banach spaces, each linearly homeomorphic to a subspace of the other, then they are linearly homeomorphic (counterexample by Gowers) I can't resist also mentioning some examples that I was involved with. Dixmier's problem: every prime C*-algebra is primitive (counterexample by me) Naimark's problem: if a C*-algebra has only one irreducible representation up to unitary equivalence, then it is isomorphic to $K(H)$ for some Hilbert space $H$ (counterexample by Akemann and me) every pure state on $B(l^2)$ is pure on some masa (counterexample by Akemann and me) every automorphism of the Calkin algebra is inner (counterexample by Phillips and me) The last three require extra set-theoretic axioms, so the correct statement is that if ordinary set theory is consistent, then it is consistent that these counterexamples exist. Presumably all three are independent of the usual axioms of set theory, but this is only known of the last one, where the consistency of a positive solution was proved by Farah.
|
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|
307,421 |
As the question title asks for, how do others "visualize" Dirichlet’s unit theorem? I just think of it as a result in algebraic number theory and not one in algebraic geometry. Bonus points for pictures.
|
What I think of as the standard proof also gives a pretty clear picture. Let $K$ be a number field with $\rho_1$ , $\rho_2$ , ..., $\rho_r: K \to \mathbb{R}$ the real places and $\sigma_1$ , $\sigma_2$ , ..., $\sigma_s: K \to \mathbb{C}$ representatives with the complex places up to conjugacy. Define $\lambda: \mathcal{O}_K \to \mathbb{R}^{r+s}$ by $$\lambda(x) = (\log |\rho_1(x)|, \ldots, \log |\rho_r(x)|, 2 \log |\sigma_1(x)|, \ldots, 2 \log |\sigma_s(x)|).$$ Let $H \subset \mathbb{R}^{r+s}$ be the hyperplane where the sum of the coordinates equals $0$ . Then $\lambda^{-1}(H)$ is the unit group of $\mathcal{O}_K$ . The kernel of $\lambda$ is the roots of unity, and Dirichlet's unit theorem states that $\lambda(\mathcal{O}_K) \cap H$ is a lattice of rank $r+s-1 = \dim H$ in the hyperplane $H$ . Here is an image of $\lambda(\mathbb{Z}[\sqrt{2}])$ , with the hyperplane $H$ in blue. Notice that translation by a vector in $\lambda(\mathcal{O}_K) \cap H$ is a symmetry of the image; this corresponds to multiplication by a unit.
|
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|
307,461 |
Let $A$ be a matrix and $x$ a fixed vector. How can we determine whether or not there exists a permutation matrix $P$ such that $APx=0$? Does this problem reduce to anything well-understood?
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Let's see if I can convince everyone that this problem is NP-complete. First: it is in NP because a permutation $P$ can be guessed and checked in polynomial time. I'll restate the problem: Given a vector $x$ and a vector space $V$ (the null-space of $A$), is there a permutation of $x$ that lies in $V$? I'll take the number of components of $x$ to be $2n$. I believe that any vector space is the null space of some matrix, so I'll take the vector space $V$ consisting of all vectors whose sum of the first $n$ components equals the sum of the other $n$ components. It has dimension $2n-1$. Now take $x$ to be a vector of integers. There is a permutation of $x$ that lies in $V$ iff the entries of $x$ can be divided into two halves of the same sum. This is the PARTITION problem that we all teach our students to be NP-complete.
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|
307,931 |
In mathematics we introduce many different kinds of notation, and sometimes even a single object or construction can be represented by many different notations. To take two very different examples, the derivative of a function $y = f(x)$ can be written $f'(x)$, $D_x f$, or $\frac{dy}{dx}$; while composition of morphisms in a monoidal category can be represented in traditional linear style, linearly but in diagrammatic order, using pasting diagrams, using string diagrams, or using linear logic / type theory. Each notation has advantages and disadvantages, including clarity, conciseness, ease of use for calculation, and so on; but even more basic than these, a notation ought to be correct , in that every valid instance of it actually denotes something, and that the syntactic manipulations permitted on the notation similarly correspond to equalities or operations on the objects denoted. Mathematicians who introduce and use a notation do not usually study the notation formally or prove that it is correct. But although this task is trivial to the point of vacuity for simple notations, for more complicated notations it becomes a substantial undertaking, and in many cases has never actually been completed. For instance, in Joyal-Street The geometry of tensor calculus it took some substantial work to prove the correctness of string diagrams for monoidal categories, while the analogous string diagrams used for many other variants of monoidal categories have, in many cases, never been proven correct in the same way. Similarly, the correctness of the "Calculus of Constructions" dependent type theory as a notation for a kind of "contextual category" took a lot of work for Streicher to prove in his book Semantics of type theory , and most other dependent type theories have not been analogously shown to be correct as notations for category theory. My question is, among all these notations which have never been formally proven correct, has any of them actually turned out to be wrong and led to mathematical mistakes? This may be an ambiguous question, so let me try to clarify a bit what I'm looking for and what I'm not looking for (and of course I reserve the right to clarify further in response to comments). Firstly, I'm only interested in cases where the underlying mathematics was precisely defined and correct, from a modern perspective, with the mistake only lying in an incorrect notation or an incorrect use of that notation. So, for instance, mistakes made by early pioneers in calculus due to an imprecise notion of "infinitesimal" obeying (what we would now regard as) ill-defined rules don't count; there the issue was with the mathematics, not (just) the notation. Secondly, I'm only interested in cases where the mistake was made and at least temporarily believed publically by professional (or serious amateur) mathematician(s). Blog posts and arxiv preprints count, but not private conversations on a blackboard, and not mistakes made by students. An example of the sort of thing I'm looking for, but which (probably) doesn't satisfy this last criterion, is the following derivation of an incorrect "chain rule for the second derivative" using differentials. First here is a correct derivation of the correct chain rule for the first derivative, based on the derivative notation $\frac{dy}{dx} = f'(x)$: $$\begin{align}
z &= g(y)\\
y &= f(x)\\
dy &= f'(x) dx\\
dz &= g'(y) dy\\
&= g'(f(x)) f'(x) dx
\end{align}$$ And here is the incorrect one, based on the second derivative notation $\frac{d^2y}{dx^2} = f''(x)$: $$\begin{align}
d^2y &= f''(x) dx^2\\
dy^2 &= (f'(x) dx)^2 = (f'(x))^2 dx^2\\
d^2z &= g''(y) dy^2\\
&= g''(f(x)) (f'(x))^2 dx^2
\end{align}$$ (The correct second derivative of $g\circ f$ is $g''(f(x)) (f'(x))^2 + g'(f(x)) f''(x)$.) The problem is that the second derivative notation $\frac{d^2y}{dx^2}$ cannot be taken seriously as a "fraction" in the same way that $\frac{dy}{dx}$ can, so the manipulations that it justifies are incorrect. However, I'm not aware of this mistake ever being made and believed in public by a serious mathematician who understood the precise meaning of derivatives, in a modern sense, but was only led astray by the notation. Edit 10 Aug 2018: This question has attracted some interesting answers, but none of them is quite what I'm looking for (though Joel's comes the closest), so let me clarify further. By "a notation" I mean a systematic collection of valid syntax and rules for manipulating that syntax. It doesn't have to be completely formalized, but it should apply to many different examples in the same way, and be understood by multiple mathematicians -- e.g. one person writing $e$ to mean two different numbers in the same paper doesn't count. String diagrams and categorical type theory are the real sort of examples I have in mind; my non-example of differentials is borderline, but could in theory be elaborated into a system of syntaxes for "differential objects" that can be quotiented, differentiated, multiplied, etc. And by saying that a notation is incorrect , I mean that the "understood" way to interpret the syntax as mathematical objects is not actually well-defined in general, or that the rules for manipulating the syntax don't correspond to the way those objects actually behave. For instance, if it turned out that string diagrams for some kind of monoidal category were not actually invariant under deformations, that would be an example of an incorrect notation. It might help if I explain a bit more about why I'm asking. I'm looking for arguments for or against the claim that it's important to formalize notations like this and prove that they are correct. If notations sometimes turn out to be wrong, then that's a good argument that we should make sure they're right! But oppositely, if in practice mathematicians have good enough intuitions when choosing notations that they never turn out to be wrong, then that's some kind of argument that it's not as important to formalize them.
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This might not quite count, but if you start with a principal $G$ -bundle $f:P\rightarrow B$ , there are two natural ways to put a $G\times G$ structure on the bundle $P\times G\rightarrow B$ given by $(p,b)\mapsto f(p)$ . Because it is standard notational practice to denote such a bundle by simply writing down the map $P\times G\rightarrow B$ , there is nothing in the notation to distinguish between these structures, and therefore the notation leads you to believe they're the same. By following this lead, Ethan Akin "proves" in the 1978 JPAA paper $K$ -theory doesn't exist ( https://doi.org/10.1016/0022-4049(78)90032-4 ) that the $K$ -theory of $B$ is trivial, for any base space $B$ . He reports that it took three Princeton graduate students (including himself) some non-trivial effort before they found the error. This might meet the letter of your criterion by virtue of having made it into print, but probably violates the spirit because the author had already discovered the error, and indeed the whole point of the paper was to call attention to it.
|
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307,947 |
I'll state my questions upfront and attempt to motivate/explain them afterwards. Q1: Is there a direct way of expressing the relation " $y$ is a function of $x$ " inside set theory?
More precisely: Can you provide a formula of first order logic + $\in$, containing only two free variables $y$ and $x$, which directly captures the idea that "$y$ is a function of $x$"? In case the answer to Q1 is negative, here's Q2: Do any other foundations of mathematics (like univalent) allow one to directly formalize the relation "$y$ is a function of $x$"?
Or have there been any attempts to formalize (parts of) mathematics with a language where the relation is taken as primitive/undefined? From discussing Q1 with colleagues I've learned that it's hard to convey what my problem is, causing frustration on both sides. I suspect this is to a certain extent because we all only learned the modern definition of function (which is not the answer to Q1) and because neither the people I talk to nor myself are experts in logic/type theory/category theory. So please bear with me in (or forgive me for) this lengthy attempt at an Explanation/Motivation: The relation " $y$ is a function of $x$ " between two things $y$ and $x$, was the original (and apparently only ) way of using the word function in mathematics until roughly 1925. The things $y$ and $x$ were traditionally called variable quantities, and the same relation was sometimes worded differently as "$y$ depends on $x$" , " $y$ is determined by $x$ " or " $y$ changes with $x$ ". This was used as a genuine mathematical proposition: something that could be proved, refuted or assumed. People would say "let $y$ be a function of $x$" just like today we might say "let $U$ be subgroup of $G$". I could cite more than 40 well known mathematicians from Bernoulli to Courant who gave definitions of this relation, but I'll limit myself to quote eight at the end of my question. As far as I can tell, these definitions cannot be directly translated into first order logic + $\in$. Although the word function assumed a different meaning with the rise of set theory an formal logic, the original relation is still used a lot among physicist, engineers or even mathematicians. Think of statements like "The pressure is a function of the volume", "The area of the circle is a function of its radius", "The number of computations is a function of the size of the matrix", "The fiber depends on the base point" etc. This even crops up in scientific communities where I wouldn't expect it. One finds it for instance in Pierce's Types and Programming Languages or Harper's Practical foundations of programming languages . So it seems that something being a function of something else (or something depending on something else) is a very natural notion for many people. In fact, I have the impression that for physicists, engineers and most scientist who apply mathematics, this relation is closer to the ideas they want to express, than the modern notion of function. Yet, I don’t see a direct way of formalizing the idea inside set theory. (The modern notion of a map $f\colon X \to Y$ is not what I'm looking for, since by itself it's not a predicate on two variables.) I know how to capture the idea at the meta-level, by saying that a formula of first order logic is a function of $x$ , iff its set of free variables contains at most $x$. But this is not a definition inside FOL. When a physicist says “The kinetic energy is a function of the velocity” he’s certainly making a physical claim and not a claim about the linguistic objects he uses to talk about physics. So this syntactic interpretation of “$y$ is a function of $x$” is not what I’m looking for. I also know a way to encode the idea inside set theory. But I’m not completely happy with it. First, here’s a naive and obviously wrong approach: Let $x\in X$ and $y\in Y$, call $y$ a function of $x$ , iff there exists a map $f:X\to Y$, such that $y=f(x)$. Since every such $y\in Y$ is a function of every $x$ (use a constant map $f=(u\mapsto y))$, this is not the right definition. Here’s a better approach: Let $x$ and $y$ be maps with equal domain, say $x:A\to X$ and $y: A \to Y$. Before giving the definition let me switch terminology: Instead of calling $x$ a "map from $A$ to $X$" I'll call it a " variable quantity of type $X$ in context $A$ ". (This change of terminology is borrowed from categorical logic/type theory. In categorical logic people say “$x$ is a generalized element of $X$ with stage of definition $A$”. But don’t assume from this, that I have a thorough understanding of categorical logic or type theory.) Definition: Let $x$ and $y$ be variable quantities of type $X$ and $Y$ in the same context $A$. We call $y$ a function of $x$ , iff there exists a map $f: X\to Y$ such that $y=f\circ x$. (It would be suggestive to switch notation from $f{\circ} x$ to $f(x)$, so we could write $y=f(x)$ when $y$ is a function of $x$. I'll refrain from doing this, since $f(x)$ has an established meaning in set theory.) Since my post is getting long, I won’t explain why this definition captures the original idea quite well. Let me only say why I don't consider it a direct way of capturing it: It seems backwards from a historical perspective. Mathematicians first had the notion of something being a function of something else, and only from there did they arrive at maps and sets. With this approach we first need to make sense of maps and sets, in order to arrive at the original idea. This might not be a strong counter argument, but if I wanted to use the original idea when teaching calculus, I would need a lot of preparation and overhead with this approach. What I’d like to have instead, is a formalization of mathematics where the relation can be used "out of the box". The other thing I don’t like about this (maybe due to my lack of knowledge of categorical logic) has to do with the context $A$ and what Anders Kock calls an “important abuse of notation”. To illustrate: suppose I have two variables quantities $x,y$ of type $\mathbb{R}$ in some context $A$. If I now assume something additional about these variables, like the equation $y=x^2$, this assumption should change the context from $A$ to a new context $B$. This $B$ is the domain of the equalizer $e:B\to A$ of the two maps $x^2,y\colon A\to \mathbb{R}$. The abuse of notation consist in denoting the "new" variable quantities $x\circ e, y\circ e$ in context $B$, with the same letters $x,y$. I suspect this abuse is considered important, since in everyday mathematics it's natural to keep the names of mathematical objects, even when additional assumptions are added to the context. In fact, if I’m not mistaken, in a type theory with identity types there is no abuse of notation involved when changing the context from $A\vdash (x,y) \colon \mathbb{R}^2$ to $A, e\colon y=x^2 \vdash (x,y) \colon \mathbb{R}^2$. So maybe type theorist also already know a language where one can talk of "functions of something" in a way that's closer to how way mathematicians did until the 1920's? Some historical definitions of "$y$ is a function of $x$": Johann Bernoulli 1718,
in Remarques sur ce qu’on a donné jusqu’ici de solutions des Problêmes sur les isoprimetres ,
p. 241: Definition. We call a function of a variable quantity, a quantity composed in any way whatsoever of the variable quantity and constants. (I'd call this the first definition. Leibniz, who initiated the use of the word function in mathematics around 1673, gave a definition even earlier. But his is not directly compatible with Bernoulli's, even though he later approved of Bernoulli's definition.) Euler , 1755 in Institutiones calculi differentialis , Preface p.VI: Thus when some quantities so depend on other quantities, that if the latter are changed the former undergo change, then the former quantities are called functions of the latter ; this definition applies rather widely, and all ways, in which one quantity could be determined by others, are contained in it. If therefore $x$ denotes a variable quantity, then all quantities, which depend upon $x$ in any way, or are determined by it, are called functions of it. Cauchy , 1821 in Cours d'analyse , p. 19: When variable quantities are related to each other such that the
values of some of them being given one can find all of the others, we
consider these various quantities to be expressed by means of several
among them, which therefore take the name independent variables . The
other quantities expressed by means of the independent variables are
called functions of those same variables. Bolzano , ca. 1830 in Erste Begriffe der allgemeinen Grössenlehre , The variable quantity $W$ is a function of one or more variable quantities $X,Y,Z$, if there exist certain propositions of the form: "the quantity $W$ has the properties $w,w_{1},w_{2}$,", which can be deduced from certain propositions of the form "the quantity $X$ has the properties $\xi,\xi',\xi''$, -- the quantity $Y$ has the properties $\eta,\eta',\eta''$; the quantity $Z$ has the properties $\zeta,\zeta',\zeta''$, etc. Dirichlet , 1837 in Über die Darstellung ganz willkürlicher Functionen durch Sinus- und Cosinusreihen : Imagine $a$ and $b$ two fixed values and $x$ a variable quantity, which continuously assumes all values between $a$ and $b$. If now a unique finite $y$ corresponds to each $x$, in such a way that when $x$ ranges continuously over the interval from $a$ to $b$, ${y=f(x)}$ also varies continuously, then $y$ is called a continuous function of $x$ for this interval. (Many historians call this the first modern definition of function. I disagree, since Dirichlet calls $y$ the function, not $f$.) Riemann , 1851 in Grundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse If one thinks of $z$ as a variable quantity, which may gradually assume all possible real values, then, if to any of its values corresponds a unique value of the indeterminate quantity $w$, we call $w$ a function of $z$. Peano , 1884 in Calcolo differenziale e principii di calcolo integrale p.3: Among the variables there are those to which we can assign arbitrarily and successively different values, called independent variables , and others whose values depend on the values given to the first ones. These are called dependent variables or functions of the first ones . Courant , 1934 in Differential and Integral Calculus Vol. 1 , p.14: In order to give a general definition of the mathematical concept of function, we fix upon a definite interval of our number scale, say the interval between the numbers $a$ and $b$, and consider the totality of numbers $x$ which belong to this interval, that is, which, satisfy the relation $$ a\leq x \leq b. $$ If we consider the symbol $x$ as denoting at will any of the numbers in this interval, we call it a (continuous) variable in the interval . If now to each value of $x$ in this interval there corresponds a single definite value $y$, where $x$ and $y$ are connected by any law whatsoever, we say that $y$ is a function of $x$ (It's funny how at after Cauchy many mathematicians talk of values of variables, which is not something we're allowed to do in set theory. (What's the value of a set or of the element of a set?). Yet, if one looks at modern type theory literature, it's full of talk of "values" again.)
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First of all, it seems to me as though the real question here is "what is a variable quantity?" Most of the definitions you quote from pre-20th century mathematicians assume that the notion of "variable quantity" is already understood. But this is already not a standard part of modern formalizations of mathematics; so it's unsurprising that definitions of a subsidiary notion, like when one variable quantity is a function of another one, are hard to make sense of. So what is a variable quantity? If we want to define the notion of variable quantity "analytically" inside some standard foundational system, then I think we cannot do better than your second suggestion: given a fixed "state space" $A$, an $R$-valued quantity varying over $A$ is a map $A \to R$. Far from worrying that this is historically backward, I think we should be proud that modern mathematics supplies a precise way to make sense of a previously vague concept, and we should not be surprised that in stumbling towards precision people took historically a more roundabout route than the eventual geodesic route that we now know. I think if you pressed any modern mathematician using the phrase "is a function of" to say what they mean by it, this is what they would say (for some suitable $A$, e.g. in "the area of a circle is the function of its radius" the space $A$ is the space of circles, while in "the number of computations is a function of the size of the matrix" the space $A$ is the space of matrices). However, you seem to be looking for something somewhat different, such as a formalism which the notion of "variable quantity" is basic rather than defined in terms of other things — a "synthetic theory of variable quantities" if you will. Here I think topos theory as well as type theory does indeed help. Given a fixed state space $A$, consider the category ${\rm Sh}(A)$ of sheaves on $A$; this is a topos and hence has an internal logic that is a type theory in which we can do arbitrary (constructive) mathematics. If inside this "universe of $A$-varying mathematics" we construct the (Dedekind) real numbers $R_A$, what we obtain externally is the sheaf of continuous real-valued functions on $A$. Thus, internally "a real number", i.e. a section of this sheaf, is externally a continuous map $A\to \mathbb{R}$, i.e. a real-valued quantity varying over $A$ in the analytic sense. So here we have a formalism in which all quantities are variable. (This point of view, that objects of an arbitrary topos should be regarded as "variable sets" has been promulgated particularly by Lawvere.) This isn't sufficient to define "function of", however, because as you point out, internally in this type theory, for any "variable quantities" $x,y:R$ there exists a map $f:R\to R$ such that $f(x)=y$, namely the constant map at $y$. If we rephrase this externally, it says that given $x:A\to \mathbb R$ and $y:A\to \mathbb R$, there always exists $f:A\times \mathbb R\to \mathbb R$ such that $f(a,x(a)) = y(a)$ for all $a$, namely $f(a,r) \equiv y(a)$. So the problem is that although $x$ and $y$ are variable quantities, we don't want the function $f$ to be a "variable function"! Thus we need a formalism in which not only are "variable quantities" basic, there is also a contrasting basic notion of "constant quantity". Categorically, a natural way to talk about this is to think about not just the category ${\rm Sh}(A)$, but the geometric morphism $\Gamma:{\rm Sh}(A)\leftrightarrows \rm Set: \Delta$, which compares the "variable world" ${\rm Sh}(A)$ with the "constant world" $\rm Set$. Just as a single topos has an internal logic that is a type theory, a geometric morphism has an internal logic that is a modal type theory, in which there are two "modes" of types (here the "variable" and "constant" ones) and operators that shift back and forth between them (here the "global sections" $\Gamma$ and the "constant/discrete" $\Delta$). Now inside this modal type theory, we can construct the object $R^v$ of "variable real numbers" and also the object $R^c$ of "constant real numbers", by copying the usual Dedekind construction in the variable and constant word, and there is a map $\Delta R^c \to R^v$ saying that every constant real number can be regarded as a "trivially" variable one. This gives us a way to say in modal type theory that $y:R^v$ is a function of $x:R^v$, namely that there exists a non-variable function $f:R^c\to R^c$ such that $\Delta f : \Delta R^c \to \Delta R^c$ extends to a function $\bar{f}:R^v\to R^v$ such that $\bar{f}(x)=y$. Or, equivalently, that there is a function $g:R^v\to R^v$ such that $g(x)=y$ and $g$ "preserves constant real numbers", i.e. restricts to a map $\Delta R^c \to \Delta R^c$. I'm not sure exactly what you hope to achieve with the issue involving assumptions like $y=x^2$ (maybe you can elaborate), but it seems to me that this setup also handles that problem just fine, in roughly the way you sketch: if variable quantities are just elements of $R^v$, then assuming some property of them, like $y= x^2$, doesn't change those elements themselves, internally.
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307,973 |
i.e. does there exist an integer $C > 0$ such that $11, 11 + C, ..., 11 + 10C$ are all prime?
|
Such an integer $C$ exists. The smallest $C$ with this property is $C=1536160080$. I found this $C$ by computing the analogous number $C$ for a $3$-term prime arithmetic progression beginning with $3$, a $5$-term prime arithmetic progression beginning with $5$ and a $7$-term prime arithmetic progression beginning with $7$. This gave me the numbers $2,6,150$. When I plugged these into OEIS I found that the next term in this sequence is $1536160080$. You can see the relevant OEIS page here .
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308,194 |
While my question topic is that of mathematical writing of papers, which is a broad subject, the particular question is specific. I am writing a paper, in which we have a section called "Outline of Proof". (It's Section 2.)
The outline is fairly informal, and we omit some technical details, making approximations.
Among these approximations, should we state (and label) important definitions and results (lemmas, equations, etc), with the intent of, later in the paper, referencing thes? This raises the point of redundencies: some people don't like things being stated twice precisely (including in the outline), so wouldn't want anything explained/stated (even in the outline) re-explained/stated. This seems ill-advised to me. When I read a paper, I rarely carefully read the outline:
I just read it, and try to get an overview (or 'outline') of the proof;
if there are parts that I don't really understand, I don't get hung up on them, trusting that with the more rigorous explanation later I'll be able to make sense of what the authors are saying. So my question is this: (a) is it standard to read an outline of a proof carefully? (b) is it standard (or at least not discouraged) to state in the outline precisely important, even key, results/definitions that will be referred back to in the main body of the paper when giving proofs?
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I am aware that my personal opinion is somewhat of an outlier, given the style of "minimalism" for many decades now in mathematical writing. But, by this time, I am unabashedly annoyed by internal references that require me to flip back to have _any_idea_ of what's being referred-to. The most hilarious case of this is Bourbaki's textbooks, which strive mightily to avoid naming anything, nor referring to anything by traditional names, nor even telling what the actual content is, but will give without telling you that it's ... oh, say, the intermediate value theorem. Sure, Bourbaki's texts are an extreme case, but they've made me sensitive to the issue, since (long ago) they were the best source for several things. Many of my colleagues have said that it's an irrelevant comparison, but I do prefer to think of mathematical papers as things to be read through, like novels. So overt necessities (or commands) to flip back seem perverse to me. Rather, I like "recall, from section (5.2), that blah=blah. Also, theorem (4.5) says that blah." It costs relatively little in terms of space, and (to my mind) adds hugely to the readability. I do think that it may be worthwhile to separate fetishism about minimalism from other considerations... Oop: to answer the original questions, ... I do take seriously any offered outlines, although, yes, also, I do try to skim through as rapidly as possible to see what's going on. Probably in part because I've seen quite a few things, it is not hard for me to sort the "usual" from the "novel", and this doesn't slow me down. When I was younger/less-experienced, I would have been very happy to have outlines and such, and to have repeated/reminder definitions/statements throughout any longish essay. Apart from a possibly outlier claim that we should require human-readable documents (as opposed to computer-verifiable), there is a still further possible-outlier claim that we should require documents readable by not-only world-class experts. The caricature of this requirement is that an attempt at communication is ... poor... if the message can only be understood by a recipient who already knows the message. In this guise, it sounds silly, but, in fact, many mathematics papers come very close to this. Makes things top-heavy and un-sustainable.
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308,354 |
The are some parts of math in which you encounter easily new structures,
obtained by modifying or generalizing existing ones. Recent examples
can be tropical geometry , or the theory around the field with one element . If one works in those areas, one cannot avoid the problem of naming new objects. When working with such a "new" notion, more general than an existing
one, you have different options to name it. Either the red herring option, like group without inverses , a brand new name, like monoid , a derived name, like semigroup (which is actually a group without inverses and
without an identity element), or no name at all, so a very long name,
i.e. the category $M$ of sets with an associative binary operation.
Or even you can also decide to use the old name with a new meaning.
(The examples I wrote don't pretend to have a historical justification). Although the red herring construction is used everywhere in math, I
feel that it is not a good practice. To use the old name with a
different meaning can be the origin of a lot of errors. And the
option of not giving any name at all is like you elude your
responsibility, so if someone needs to use it they will have to put a
name to it (maybe your name?). So my preferred options are to choose a derived name or a new name.
Derived names are quite common: e.g. quasicoherent, semiring,
pseudoprime, prescheme (which is an old term), and they
contain some information which is useful, but sometimes they are ugly,
and it could seem you don't really want to take a decision: you just
write quasi/semi/pseudo/pre in front of the name. But new names can be difficult to invent, to sell and to justify: if you decide to give the name jungle to a proposed prototype of tropical variety , because it sounds to you that in the tropics are plenty of jungles, it is a loose justification and probably will have no future (unless you are Grothendieck). My question is: Which do you think is the best option? In fact, the situation can be worse in some cases: what happens if
some name has already been used but you don't agree with the
choice? Is it adequate to modify it, or can it be seen as some sort of
offense? I could put some very concrete examples, even papers where they
introduce red herrings, new meanings for old names, new names and
no-names for some objects, all in the same paper. But my point is
not to criticize what others did but to decide what to do.
|
Let me mention as a counterpoint that there is less need for
new terminology than one might expect. Mathematical exposition
is often more successful and clearer without new terminology, and
one should consider whether one needs any new terminology at all. It seems to be a typical beginner's mistake to name everything in
sight, introducing all kinds of fancy names and cluttering one's
writing with unnecessary terminology and jargon. To be sure, this naming process is easy, as well as fun; one feels like Adam or Eve in the garden. I've succumbed to the attraction of it myself. But now I view this more negatively, for it imposes a
kind of tax on the reader. One opens the article and finds a
theorem stated there: Theorem. Every big-topped parade is heartily divisible. The jargon prevents it from having an immediate meaning, even for an expert in the subject, and one must
hunt down the definitions of the various terms. I am sure that many mathematicians have had this experience. Articles are almost
never read front to back, and so the definitions of new terms are often missed. The question of whether the article
will be read at all is often settled by browsing through it and
seeing if the theorems are interesting. The jargon tax is a tax
many readers are not willing to pay — when one can't find
meaningful mathematics easily enough, then one simply looks elsewhere, and one article (perhaps your article!) is dropped in favor of another. For this reason, I find it desirable, when possible , for one to state one's
theorems in a manner that can be readily understood with ordinary
terminology, even if some new-fangled jargon would make it slightly shorter or would perfectly express some extremely abstract connection. In time, of course, new objects and ideas find an established usage, and there will be a need for new names. My comment is merely a caution against over-exuberant naming.
|
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|
308,370 |
I have questions about the definition of representation variety. In François Labourie's book "Lectures on representations of surface groups", Section 3.5 , the author gives four models of the representation variety. I am confused about the model using the language of vector bundles. Definition 1 (representation variety): A representation variety of $S$ is gauge equivalences of pairs ($G$-vector bundles $L$ over the surface $S$, flat $G$-connection on $L$). Definition 2 (Gauge equivalence): Two connections on the same vector bundle are said to be gauge equivalent if the can be connected using the pullback of some lift of the identity map. What confused me is definition 2. In order to use gauge equivalence in the definition of representation variety, why do we need to restrict the definition to the same vector bundle? In other words, can we say that every flat $\bf R$-vector bundle over the surface is trivial (where $\bf R$ is the real number field)?
|
Let me mention as a counterpoint that there is less need for
new terminology than one might expect. Mathematical exposition
is often more successful and clearer without new terminology, and
one should consider whether one needs any new terminology at all. It seems to be a typical beginner's mistake to name everything in
sight, introducing all kinds of fancy names and cluttering one's
writing with unnecessary terminology and jargon. To be sure, this naming process is easy, as well as fun; one feels like Adam or Eve in the garden. I've succumbed to the attraction of it myself. But now I view this more negatively, for it imposes a
kind of tax on the reader. One opens the article and finds a
theorem stated there: Theorem. Every big-topped parade is heartily divisible. The jargon prevents it from having an immediate meaning, even for an expert in the subject, and one must
hunt down the definitions of the various terms. I am sure that many mathematicians have had this experience. Articles are almost
never read front to back, and so the definitions of new terms are often missed. The question of whether the article
will be read at all is often settled by browsing through it and
seeing if the theorems are interesting. The jargon tax is a tax
many readers are not willing to pay — when one can't find
meaningful mathematics easily enough, then one simply looks elsewhere, and one article (perhaps your article!) is dropped in favor of another. For this reason, I find it desirable, when possible , for one to state one's
theorems in a manner that can be readily understood with ordinary
terminology, even if some new-fangled jargon would make it slightly shorter or would perfectly express some extremely abstract connection. In time, of course, new objects and ideas find an established usage, and there will be a need for new names. My comment is merely a caution against over-exuberant naming.
|
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|
308,558 |
Is it possible to isometrically immerse the hyperbolic plane into a compact Riemannian manifold as a totally geodesic submanifold? Any nice examples? Edit: Although I did not originally say so, I was looking for injective immersions or at least for immersions that do not factor through a covering onto a compact surface. Thank you for your answers and comments, they've been very helpful.
|
Yes, it immerses isometrically into certain solvmanifolds. Take an Anosov map of $T^2$, such as $\left[\begin{array}{cc}2 & 1 \\1 & 1\end{array}\right]$. The mapping torus admits a locally homogeneous metric modeled on the 3-dimensional unimodular solvable Lie group . The matrix has two eigenspaces with eigenvalues $\frac{3\pm\sqrt{5}}{2}$, and the suspensions of lines on the torus parallel to these eigenspaces give immersed manifolds modeled on $\mathbb{H}^2$. If the eigenspace line contains a periodic point of the Anosov map, the mapping torus of it will be an annulus. But otherwise it will be an immersed injective totally geodesic hyperbolic plane, which I think is what you're asking for.
|
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|
308,797 |
More and more I am becoming convinced that one should know at least one programming language very well as a mathematician of this century. Is my conviction justified, or not applicable? If I am right, then please what languages should someone aspiring to be a mathematician learn? The number out there is so bewildering for a complete novice to judge, and no one else can judge suitable ones than a body of working mathematicians, hence my posting this question here specifically. In particular, this language should be very useful for mathematics applications, should be close in syntax and structure to mathematics, and be mathematically related in other relevant ways. Indeed you may suggest a language that you have found useful/important in other ways not mentioned, but please explain why you make these suggestions clearly. Thank you. PS. I couldn't find sufficiently relevant tags. Please improve as appropriate.
|
Python, so they can use Sage .
From their website: SageMath is a free open-source mathematics software
system licensed under the GPL. It builds on top of many existing
open-source packages: NumPy , SciPy , matplotlib , Sympy , Maxima , GAP , FLINT , R and many more. Access their
combined power through a common, Python-based language or directly via
interfaces or wrappers.
Mission: Creating a viable free
open source alternative to Magma, Maple, Mathematica and Matlab .
|
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|
309,013 |
Several good references dedicated to hyperbolic groups have been written until 1990, including: Hyperbolic groups , written by M. Gromov. Géométrie et théorie des groupes : les groupes hyperboliques de Gromov , written by M. Coornaert, A. Papadopoulos and T. Delzant. Sur les groupes hyperboliques de M. Gromov , edited by E. Ghys and P. de la Harpe. Since then, fundamental tools have been introduced to study hyperbolic groups. For instance, I have in mind JSJ decompositions. Are there textbooks or surveys on the subject? What are good references dedicated to modern developments on hyperbolic groups? By "modern", I mean subjects not contained in the references mentioned above. I am particularly interested in progresses made in the study of outer automorphism groups of hyperbolic groups. Edit: I am aware that the ideas involved in the study of hyperbolic groups have lots of applications, so that covering them in just a few references is hopeless. Consequently, it would be better to focus on hyperbolic themselves, excluding generalisations (such as relatively / acylindrically hyperbolic groups) or specific examples (such as right-angled Coxeter groups, (automorphisms of) free groups or (automorphisms of) surface groups).
|
I think this is a great question, as there is still a need for an authoritative reference about (word-)hyperbolic groups. Since the textbook doesn't exist, I'd like to take the question in a slightly different direction by listing some of the material I think it should cover. (This is inevitably a personal and biased account.) I'll try to include the best references I can think of. The seminal article of Gromov's mentioned in the question contained many assertions about hyperbolic groups, often extensions of Thurston's famous theorems about hyperbolic manifolds. Probably the simplest such statement (which is fundamental in the study of outer automorphism groups of free groups) is now known as Paulin's theorem (cf. 5.4.A of Hyperbolic groups ). Paulin's theorem: If $\Gamma$ is a torsion-free hyperbolic group that doesn't split over a cyclic subgroup then $\mathrm{Out}(\Gamma)$ is finite. Another such statement is 5.3.C' from Gromov's article. Subgroup Rigidity Theorem: Let $\Gamma$ be a hyperbolic group and $H$ a one-ended finitely presented group. Then there are only finitely many conjugacy classes of subgroups of $\Gamma$ isomorphic to $H$ . In Hyperbolic groups , Gromov suggests that statements like these can be proved using a generalisation of Thurston's arguments using the geodesic flow. Even the problem of constructing a candidate geodesic flow over a hyperbolic group is notoriously difficult. The problem was eventually solved by Mineyev but, to the best of my knowledge, no one so far been able to use Mineyev's geodesic flow to give the Thurstonian proofs of these result that Gromov suggested. Instead, the key tool in proving these results turned out to be the Rips machine : Rips' classification of certain actions of groups on real trees. For actions of finitely presented groups, the Rips machine was developed by Bestvina—Feighn. An account was also given by Misha Kapovich in his book, so the interested reader could look at either of the following. Bestvina & Feighn, Stable actions of groups on real trees. Invent. Math. 121 (1995), no. 2, 287–321 M. Kapovich, Hyperbolic manifolds and discrete groups. Progress in Mathematics, 183. Birkhäuser Boston, Inc., Boston, MA, 2001. xxvi+467 pp For some applications, one needs the Rips machine for finitely generated groups. This was developed initially by Sela and corrected and refined by Guirardel, so the relevant references here are: Sela, Acylindrical accessibility for groups. Invent. Math. 129 (1997), no. 3, 527–565. Guirardel, Actions of finitely generated groups on $\mathbb{R}$ -trees. Ann. Inst. Fourier (Grenoble) 58 (2008), no. 1, 159–211. The link with hyperbolic groups is made via what Sela called the Bestvina—Paulin method . Given infinitely many actions of a group $G$ on a suitably nice $\delta$ -hyperbolic space $X$ , one can pass to a limiting action on a space which is either $X$ itself or a real tree $T$ , and in the latter case Rips’ machine applies. As well as Paulin’s original paper, another account of the proof, also modulo the Rips machine, was given by Bridson—Swarup (who corrected a small mistake in Paulin’s proof), and Bestvina gave a very useful account in his survey article on $\mathbb{R}$ -trees. So one could look at: Paulin, Outer automorphisms of hyperbolic groups and small actions on $\mathbb{R}$ -trees., Arboreal group theory (Berkeley, CA, 1988) , 331–343, Math. Sci. Res. Inst. Publ. , 19, Springer, 1991. Bridson & Swarup, On Hausdorff-Gromov convergence and a theorem of Paulin, Enseign. Math. (2) 40 (1994), no. 3-4, 267–289. Bestvina, $\mathbb{R}$ -trees in topology, geometry, and group theory. Handbook of geometric topology , 55–91, North-Holland, Amsterdam, 2002. For Paulin’s theorem, the only thing one needs from the Rips machine is that it promotes a (nice) action on a real tree to a (nice) action on a simplicial tree. Deeper applications, such as the Subgroup Rigidity Theorem, tend to require a notorious trick called the shortening argument . The idea is that if the actions of $G$ on $X$ were all chosen to be `shortest’ in their conjugacy classes then either $G$ is a free product or the limiting action of $G$ on $T$ isn’t faithful. This trick is notoriously, er, tricky. The first reference is Rips—Sela’s original paper in which they prove the Subgroup Rigidity Theorem in the torsion-free case. (The case with torsion was later handled by Delzant.) Rips & Sela, Structure and rigidity in hyperbolic groups. I. GAFA , 1994. The shortening argument is Theorem 4.3, and the Subgroup Rigidity Theorem is Theorem 7.1. I gave an account of the shortening argument in Theorem 5.1 of Wilton, Solutions to Bestvina and Feighn's exercises on limit groups, Geometric and cohomological methods in group theory , pp. 30–62, LMS Lect. Note Ser. 358, CUP, 2009 Another account of the shortening argument (and many of the theorems mentioned here) adapted to the setting of toral relatively hyperbolic groups was given by Groves in: Groves, Limit groups for relatively hyperbolic groups. II. Makanin-Razborov diagrams. Geom. Topol. 9 (2005), 2319–2358. This toolkit has some further spectacular consequences for the structure of hyperbolic groups. The two biggest are probably the Hopf property and the isomorphism problem. Recall that a group $G$ is said to be non-Hopfian if there is a non-injective epimorphism $G\to G$ . In Sela, Endomorphisms of hyperbolic groups. I. The Hopf property. Topology 38 (1999), no. 2, 301–321. Sela proved the Hopf property for all torsion-free hyperbolic groups. The case with torsion is treated in a preprint of Reinfeldt—Weidmann. Sela solved the isomorphism problem for torsion-free rigid hyperbolic groups (such as the hyperbolic 3-manifold groups) in Sela, The isomorphism problem for hyperbolic groups. I. Ann. of Math. (2) 141 (1995), no. 2, 217–283. To extend this to the non-rigid case, one needs JSJ theory for groups, which was introduced by Rips—Sela in Rips & Sela, Cyclic splittings of finitely presented groups and the canonical JSJ decomposition. Ann. of Math. (2) , 146 (1997), no. 1, 53–109. As is already apparent from Ian’s answer, this theory has been developed a great deal, and the more recent approaches are frankly a lot simpler than the original Rips—Sela version. In the end, the torsion-free non-rigid case (as well as the toral relatively hyperbolic case) was dealt with in Dahmani & Groves, The isomorphism problem for toral relatively hyperbolic groups. Publ. Math. Inst. Hautes Études Sci. No. 107 (2008), 211–290. while the case with torsion was dealt with in Dahmani & Guirardel, The isomorphism problem for all hyperbolic groups. Geom. Funct. Anal. 21 (2011), no. 2, 223–300. Both of these papers followed Sela’s outline, although major technicalities needed to be overcome. NWMT has alreay said in comments that all this is too difficult for a textbook. In total this is true, but I can imagine an advanced reference book that describes the Rips machine, the Bestvina—Paulin method and the shortening argument, and gives Paulin’s theorem and the Subgroup Rigidity theorem as applications. These are still some of the most amazing and beautiful arguments in the theory of hyperbolic groups, and I’m a little concerned that a lot of current research seems to be moving away from them, rather than attempting to simplify or extend them.
|
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|
309,019 |
In my research I came across the following question : Is it true that for every real function $f:\mathbb{R}\to\mathbb{R}$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?
|
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $\varepsilon>0$ such that all points $x\in(a,b)$ except $c$ have $|f(x)-f(c)|>\varepsilon$. (Otherwise, by taking smaller and smaller intervals and $\varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $\varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $\varepsilon$. Fix such $a$, $b$, and $\varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $\varepsilon$ --- a contradiction because there isn't that much room in $\mathbb R$.
|
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|
309,044 |
Some of the simplest and most interesting unproved conjectures in mathematics are Goldbach's conjecture, the Riemann hypothesis, and the Collatz conjecture. Goldbach's conjecture asserts that every even number greater than or equal to 4 can be written as the sum of two prime numbers. It's pretty straightforward how to create a computer program which halts if and only if there exists a counterexample to Goldbach's conjecture: simply loop over all integers, test if each one is a counterexample, and halt if a counterexample is found. For the Riemann hypothesis, there's also a "known" computer program which halts if and only if there exists a counterexample. (Given the usual statement of the Riemann hypothesis, this is not so clear, but Jeffrey C. Lagarias' paper " An Elementary Problem Equivalent to the Riemann Hypothesis " shows that the Riemann hypothesis is equivalent to the statement that a certain sequence of integers $L$ is a lower bound for a certain sequence of real numbers $R$ . The sequence $L$ is computable, and $R$ is computable to arbitrary precision, so our computer program only needs to compute all elements of $R$ in parallel, and halt if any element is ever discovered to be smaller than its corresponding element in $L$ .) But how about the Collatz conjecture? The Collatz conjecture states that for all positive integers $n$ , the "hailstone sequence" $H_n$ eventually reaches $1$ . We could try to do the same thing we did with Goldbach's conjecture: loop over all positive integers $n$ and halt if a counterexample is ever found. But there's a problem here: with the Collatz conjecture, given a positive integer $n$ , it's not obvious that it's even decidable whether or not $n$ is a counterexample. We can't simply "check whether or not $n$ is a counterexample" like we can with Goldbach's conjecture. So is there a known Turing machine which halts if and only if the Collatz conjecture is false? Of course, a "known Turing machine" doesn't have to be a Turing machine that someone has actually explicitly constructed; if it's straightforward how to write a computer program that would do this, then that counts as a "known Turing machine". On the other hand, saying "it's either the machine which trivially halts, or it's the machine which trivially does not halt" doesn't count as a "known Turing machine"; I'm asking for an answer which mentions one single Turing machine $M$ (with no input or output), such that we know that $M$ halts if and only if the Collatz conjecture is false.
|
Let's note that this is not a question of whether Collatz is undecidable.
The statement $\neg\mathrm{Con}(PA)$ is undecidable (by $PA$ , assuming $PA$ is consistent) but nevertheless $\neg\mathrm{Con}(PA)$ is provably equivalent to a certain Turing machine halting (the one that searches for a proof of a contradiction in PA). Rather, the question is whether there is a $\Pi^0_1$ statement $\varphi$ such that the Collatz problem, which on its face is $\Pi^0_2$ , is already known to be equivalent to $\varphi$ . Here already known means in particular that we are not allowed to assume that Collatz is or is not provable or disprovable in any particular system, unless we already know that. The best evidence that there is no such $\varphi$ seems to be in the paper mentioned by @Burak: Kurtz, Stuart A.; Simon, Janos , The undecidability of the generalized Collatz problem , Cai, Jin-Yi (ed.) et al., Theory and applications of models of computation. 4th international conference, TAMC 2007, Shanghai, China, May 22--25, 2007. Proceedings. Berlin: Springer (ISBN 978-3-540-72503-9/pbk). Lecture Notes in Computer Science 4484, 542-553 (2007). ZBL1198.03043 , MR2374341 . Namely, they give a parametrized family of similar problems such that the collection of parameters for which Collatz-for-those-parameters is true, is $\Pi^0_2$ -complete and hence not $\Pi^0_1$ . They can do this without thereby solving the Collatz problem, just like Matiyasevich et al. could show that solvability of diophantine equation was $\Sigma^0_1$ -complete, without thereby solving any particular equation themselves. If Collatz could somehow be simplified to a $\Pi^0_1$ form then quite plausibly the generalized version could too by the same argument (whatever that hypothetical argument would be) but that Kurtz and Simon show will not happen.
|
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|
309,260 |
In the monograph Equivariant Stable Homotopy Theory , Lewis, May, and Steinberger cite a monograph "The homotopical foundations of algebraic topology" by Peter May, as "in preparation." It's their [107]. In his paper "When is the Natural Map $X \rightarrow \Omega\Sigma X$ a Cofibration?" Lewis also cited this monograph, and additionally wrote "Monograph London Math. Soc., Academic Press, New York (in preparation)." Did this monograph ever appear, perhaps under a different name? I could not find anything with the given title. The reason I ask is that LMS wrote in many places things like "details may be found in [107]" and I was looking into something where nitpicky detail might matter.
|
An anonymous source told me this question is here. Dylan gave the quick answer and Tyler referred to it. I'll use the question as an excuse to give a pontificating longer answer. When I first planned on writing that, maybe 45 or 50 years ago, I had not yet been converted to model category theory, let alone anything more modern, and what I had in mind would have been very plodding. I've thought hard about the pedagogy, perhaps not to good effect, and Concise and More (or less) Concise give what I came up with. The latter is not as concise as I would like in large part because so much relevant detail seemed missing from both the algebraic and topological foundations of localization and completion, especially in the proper generality of nilpotent rather than simple spaces. Both are available on my web page. http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf http://www.math.uchicago.edu/~may/TEAK/KateBookFinal.pdf As an historical note, in retrospect I came to the conclusion that the standard foundations for classical algebraic topology, as I understood it a half century ago, were given by the mixed model structure on spaces discovered by Mike Cole. The weak equivalences are the same as in the usual Quillen model category, but the fibrations are the Hurewicz fibrations rather than the Serre fibrations. Then the cofibrant spaces are the spaces of the homotopy types of CW complexes, which for sure was everybody's favorite category of spaces in which to work back then.
|
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|
309,515 |
Earlier today, I stumbled upon this article written by V. Voevodsky about the "philosophy" behind the Univalent Foundations program. I had read it before around the time of his passing, and one passage that I remember vividly is this, for which I have little in the way of rigorous justification: The greatest roadblock for me was the idea that categories are “sets in the next dimension.” I clearly recall the feeling of a breakthrough that I experienced when I understood that this idea is wrong. Categories are not “sets in the next dimension.” They are “partially ordered sets in the next dimension” and “sets in the next dimension” are groupoids. — Voevodsky, The Origins and Motivations of Univalent Foundations If I had to guess, I'd say that the noninvertibility of morphisms in categories corresponds to some form of partial ordering, whereas a groupoid carries no such information since all morphisms are in fact isomorphisms, but surely there's something deeper at play that caused someone like Voevodsky to consider this realisation a "breakthrough" that accompanied a significant "roadblock".
|
First, there is indeed nothing mathematically very deep in this observation, and I agree that the word "breakthrough" might be exaggerated. But on the other hand lots of very deep ideas look trivial once spelled out explicitly. Moreover being younger than Voevosky I have never been really exposed to the idea that categories were sets of higher dimension (but this indeed appears in the early work on higher categories, typically in Baez & Dolan's work), so I can't comment on how important it was to understand that this is not a good point of view. But I can give some context on what Voevodsky probably meant here. One thing to understand is that, like many mathematicians and most category theorists, Voevodsky is very attached to the " principle of equivalence " saying that when talking about categories you should only use concepts that are invariant under equivalence of categories. For example, in his work on contextual categories, he renamed them "C-systems" because he didn't want to call them categories as their definition is not invariant under equivalence of categories. Now, if you follow this principle of equivalence very strictly, talking about "the set of objects of a category" is not meaningful (i.e. break the principle of equivalence):
equivalent categories can have non-isomorphic sets of objects.
So saying that a category is "a set of objects together with a set of arrows having a certain structure" is incorrect from this perspective. It is true that if you have a set of objects and a set of arrows with the appropriate structure you get a category, and it is also true that any category can be obtained this way, but you cannot recover the set of objects and the set of arrows from the category without breaking the principle of equivalence. To some extent the set of objects and of arrows with the appropriate structure is a "presentation" of your category. What is meaningful though (i.e. respects the principle of equivalence) is that a category has a "groupoid of objects" $X$, with a bifunctor $Hom : X \times X \rightarrow Set$ From the point of view of the principle of equivalence, a category is really a groupoid with structure. Moreover this structure is a very natural categorification of the notion of poset: A poset is a set X with a function $X \times X \rightarrow Prop$ satisfying reflexivity, anti-symmetry and transitivity. A category is a groupoid $X$ with a functor $X \times X \rightarrow Sets$, satisfying some conditions. Reflexivity corresponds to the existence of an identity, transitivity corresponds to the composition operation, and anti-symmetry corresponds to the fact that in the end one wants $X$ to be the core groupoid of the category. But if you really take the principle of equivalence seriously, you cannot define what a "groupoid" is, you have to take it as a primitive notion that you axiomatize. But this is not really different from the fact that you cannot define what a "Set" is, you can only axiomatize what you can do with sets. It is in this sense that groupoids are "higher dimensional sets" and categories are groupoids with a structure.
This has been made even more clear with the theory of $(\infty,1)$-categories, where it is completely clear that $\infty$-groupoids play the role that sets played for ordinary categories. (The $(\infty,1)$-categorical Yoneda lemma is in terms of functors to the category of $\infty$-groupoids etc...) Another way to say this is that in the $n$-categorical hierarchy "0-groupoid"s are just sets while $0$-categories are posets (if you consider them as $(n,1)$-categories for varying $n$...)
|
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309,577 |
I have a preprint X that is sitting in the ArXiv for which I am not sure if it is still worth publishing. It turns out the paper I wrote has considerable overlap with another preprint Y after one of its authors informed me about it through email. Consider the following: Paper Y was posted in the ArXiv just a month before I posted mine, and I was not aware of its existence previously. Prior to posting paper X, I looked for as many papers with related results to include in my discussion, but due to the differences in the terminologies used, Google did not show paper Y in the results. In order to find paper Y on Google or ArXiv, one would have to use a different set of keywords. It appears that we have been working on the same problem, they just finished first and had a month of lead. I am aware that alternate proofs are sometimes of interest and are therefore worth publishing, and I am trying to determine if this is the case with my paper. Since I wrote paper X independently and without knowledge of the existence of paper Y, my proof was essentially different and indeed the two papers have very different motivations for the constructions used. However, while the methods used as different, they have the same "flavor" and the key results in both papers use different versions of the same pre-existing theorem. If there is anything drastically different from the results of both papers, it would be the length. My proofs are shorter than theirs, use less lemmas and mathematical machinery. Paper Y derives a few more corrolaries which are not in paper X, but even if one compares the length of both papers using only the parallel or similar portions, my paper is still significantly shorter than theirs. I can also make the argument, understandably a subjective one, that my methods are simpler. However, results of Paper Y are, to a certain degree, more general than mine and therefore, technically my results follow from theirs. My questions are as follows: Should I still submit my paper to a journal? If yes, how should I deal with the existence of paper Y in my own paper? I feel obliged to cite it, but I'm not really sure how to discuss the similarity and differences between the results. Should I tell the editor about the situation? Should I make other assumptions on the reason behind the author of Paper Y for sending me an email about their paper, other than to inform me that our papers have the same results? If you were in the shoes of the author of paper Y, what would you prefer that I do?
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I once wrote a paper with an undergraduate that I thought was very nice. After it was accepted for publication, we found a paper not only proving our results, but going a step further. We hadn't found it previously because, similar to your situation, they used different terminology. In our case, our proofs didn't add anything new, and our results were weaker, so we withdrew the paper before it was published. That same undergraduate went on to write more papers with me, and is currently a successful graduate student. I mention this to emphasize that not publishing is not the end of the world. If you decide to publish, you should definitely cite the work you found, and mention to the reader that their proofs are different and came first. Something like the following should be fine: "After a preprint version of this paper was made public I was made aware of [Y], which appeared first and obtained Theorems ###. As the methods this paper are completely different, we believe that the proofs still merit publication." It is not necessarily your job to explain those differences, unless you think it would help the reader. However, it is your job to convince (yourself, the readers, and) the referee that your paper is worthwhile. So you need to deeply understand the differences, and perhaps explicitly point out what your paper adds to the literature. I imagine that the authors of Y mentioned their paper because they have priority, and would appreciate citations to their work. After you have cited their work, mentioning their priority, you owe no more obligations to them. (Although, if they contact you again, you can treat their communications with respect!) Don't try to mind-read their intentions. Focus on your own beliefs, whether you personally think your proofs would help readers understand the topic more deeply and give them interesting techniques. If you like, you can send Y an updated version of your paper, which clearly gives them priority, and kindly ask if they have any further comments.
|
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|
309,946 |
This is a crosspost from this MSE question from a year ago. Finite groups are cancellable from direct products, i.e. if $F$ is a finite group and $A\times F \cong B\times F$, then $A \cong B$. A proof can be found in this note by Hirshon. In the same note, it is shown that $\mathbb{Z}$ is not cancellable, but if we only allow $A$ and $B$ to be abelian, it is (see here ). I would like to know if there are any groups that can be cancelled from free products rather than direct products. That is: Is there a non-trivial group $C$ such that $A*C \cong B*C$ implies $A \cong B$? It is certainly not true that every group is cancellable in free products. For example, if $A$, $B$, $C$ are the free groups on one, two, and infinitely many generators respectively, then $A*C \cong C \cong B*C$ but $A\not\cong B$. Many non-examples can be constructed this way, but they are all infinitely generated. As is discussed in the original MSE question, it follows from Grushko's decomposition theorem that if $A$, $B$, and $C$ are finitely generated, then $A*C \cong B*C$ implies $A \cong B$.
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For $C=\mathbb{Z}/2\mathbb{Z}$, it follows from the Kurosh subgroup theorem that $A\ast C \cong B\ast C$ implies that $A\cong B$. Denote $C_1 \cong C_2\cong \mathbb{Z}/2\mathbb{Z}$, and let $\varphi: A\ast C_1 \to B\ast C_2$ be an isomorphism. Then $\varphi(C_1)\leq B\ast C_2$ is an injection. By Kurosh (and since $C$ is freely indecomposable), there is $C_3\leq B$ or $C_4\leq C_2$ and $g\in B\ast C_2$ such that $\varphi(C_1)=gC_3g^{-1}$ or $\varphi(C_1)=gC_4g^{-1}$. Replacing the isomorphism $\varphi$ with $g^{-1}\varphi g$, we may assume that $\varphi(C_1)=C_3$ or $C_4$.
In the second case, we have $C_4\cong C_2$, and hence $\varphi(C_1)=C_2$. Taking the quotient by the normal closure of $C_1$ and $C_2$ on the left and right respectively, we see that $A\cong B$. In the case $\varphi(C_1)=C_3< B$, $\varphi^{-1}(C_2) = hC_5h^{-1}$, $C_5\leq A, h\in A\ast C_1$ (since otherwise we would be in the other case again). Then we see that $A \cong A\ast C_1 / \langle\!\langle C_1\rangle\!\rangle \cong B \ast C_2 /\langle\!\langle C_3\rangle\!\rangle \cong (B/\langle\!\langle C_3\rangle\!\rangle)\ast C_2$ (we use $G/\langle\!\langle H\rangle\!\rangle$ to denote the quotient by the normal subgroup generated by $H$). Similarly, $B\cong (A/\langle\!\langle C_5\rangle\!\rangle)\ast C_1$. Let $A_1=B/\langle\!\langle C_3\rangle\!\rangle, B_1=A/\langle\!\langle C_5\rangle\!\rangle$. Then $A\cong A_1\ast C_2, B\cong B_1\ast C_1$, and $\varphi: A_1\ast C_2\ast C_1 \to B_1\ast C_1\ast C_2$, taking $C_i$ to a conjugate of $C_i$. Take the quotient by the normal closures of $C_1$ and $C_2$ on both sides, we see that $A_1\cong B_1$, so $A\cong B$.
|
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|
309,974 |
I was recently reading Bui, Conrey and Young's 2011 paper " More than 41% of the zeros of the zeta function are on the critical line ", in which they improve the lower bound on the proportion of zeros on the critical line of the Riemann $\zeta$-function. In order to achieve this, they define the ( monstrous ) mollifier: $$
\psi=\sum_{n\leq y_1}\frac{\mu(n)P_1[n]n^{\sigma_0-\frac{1}{2}}}{n^s}+\chi(s+\frac{1}{2}-\sigma_0)\sum_{hk\leq y_2}\frac{\mu_2(h)h^{\sigma_0-\frac{1}{2}}k^{\frac{1}{2}-\sigma_0}}{h^sk^{1-s}}P_2(hk).
$$ This is followed by a nearly 20-page tour-de-force of grueling analytic number theory, coming to a grinding halt by improving the already known bounds from $40.88\%$ to... $41.05\%$. Of course any improvement of this lower bound is a highly important achievement, and the methods and techniques used in the paper are deeply inspirational. However, given the highly complex nature of the proof and the small margin of improvement obtained over already existing results, I was inspired to ask the following question: What are other examples of notably long or difficult proofs that only improve upon existing results by a small amount?
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This 28-page paper by Le Gall lowers the best-known estimate for the asymptotic complexity of matrix multiplication from $O(n^{2.3728642})$ to $O(n^{2.3728639})$, a reduction by 0.00001%. However, it simplifies the previous method, so it does not really qualify as "notably long or difficult proof". Maybe this definition applies more to the previous improvement in a 73-page paper by Vassilevska-Williams , which brought down the exponent from $O(n^{2.374})$ to $O(n^{2.3728642})$
|
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|
310,384 |
I will be teaching a course on algebraic topology for MSc students and this semester, unlike previous ones where I used to begin with the fundamental group, I would like to start with ideas of singular homology as in Vick's book. I am quite new to the ideas of persistent homology and have not done a single computations in this field. But, I like to learn on the subject. More is that I like to lead the course that I will be teaching so that towards the end, I can give some taste of persistent homology to students. But, I am not sure if there is any well written set of lecture notes on the material, or should we dive into the literature and start with some papers!?! The course involves of $3/2\times 30$ hours of lectures. Do you think this is possible or should I use some simplicial approaches instead? or you think it is more suitable to give this as a task to students to start as a project and discover the ideas for themsevles?!?! I would be very grateful for any advise in terms of addressing to main references on the subject. I also would be grateful if you can give me some advise on the history of the subject; for instance when people decided to use homology to study biological problems and whether or not the main stream researchers in biology or data analysis really consider these kind of tools?!
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Since this area is developing rather quickly, there is a dearth of canonical references that would satisfy basic pedagogical requirements. If I were teaching a course on this material right now, I would probably use Oudot's nice book if the students had sufficient background, and the foundational paper of Zomorodian-Carlsson if they did not. I haven't read Jose's recent article mentioned in Joe's answer, but here is what I remember of the good old days (with apologies to all the important stuff that got missed). 1992 : Frosini introduces " size functions ", which we would today consider equivalent to 0-dimensional persistent homology. 1995 : Mischaikow + Mrozek publish a computer-assisted proof of chaos in the Lorenz equations; a key step involves computing Conley indices, which are relative homology classes. This produces considerable interest in machine computation of homology groups of spaces from finite approximations (eg large cell complexes). 1999 : Robbins publishes this paper emphasizing that functoriality helps approximate the homology of an underlying space from Cech complexes of finite samples; meanwhile Kaczynski, Mischaikow and Mrozek publish their book on efficient homology computation via simple homotopy type reductions of cell complexes. 2002 : Edelsbrunner, Letscher and Zomorodian introduce persistence from a computational geometry viewpoint; as written, their algorithm works only for subcomplexes of spheres and only with mod-2 coefficients. 2005 : Zomorodian and Carlsson reinterpret persistence of a filtration via the representation theory of graded modules over graded pid's, thus giving an algorithm for all finite cell complexes over arbitrary field coefficients; they also introduce the barcode, which is a perfect combinatorial invariant of certain tame persistence modules. 2007 : de Silva and Ghrist use persistence to give a slick solution to the coverage problem for sensor networks. Edelsbrunner, Cohen-Steiner and Harer show that the map $$\text{[functions X to R]} \to \text{[barcodes]}$$ obtained by looking at sublevel set homology of nice functions on triangulable spaces is 1-Lipschitz when the codomain is endowed with a certain metric called the bottleneck distance. This is the first avatar of the celebrated stability theorem . 2008 : Niyogi, Smale and Weinberger publish a paper solving the homology inference problem for compact Riemannian submanifolds of Euclidean space from finite uniform samples. Carlsson, with Singh and Sexton, starts Ayasdi , putting his money where his math is. 2009 : Carlsson and Zomorodian use quiver representation theory to point out that getting finite barcodes for multiparameter persistence modules is impossible , highlighting dimension 2 as the new frontier for theoretical work in the field. 2010 : Carlsson and de Silva, by now fully immersed in the quiver-rep zone, introduce zigzag persistence . The first software package for computing persistence (Plex, by Adams, de Silva, Vejdemo-Johansson,...) materializes. 2011 : Nicolau, Levine and Carlsson discover a new type of breast cancer using 0-dimensional persistence on an old, and purportedly well-mined, tumor dataset. 2012 : Chazal, de Silva, Glisse and Oudot unleash this beastly reworking of the stablity theorem . Gone are various assumptions about tameness and sub-levelsets. They show that bottleneck distance between barcodes arises from a certain "interleaving distance" on the persistence modules. This opens the door for more algebraic and categorical interpretations of persistence, eg Bubenik-Scott . 2013 : Mischaikow and I retool the simple homotopy-based reductions to work for filtered cell complexes, thus producing the first efficient preprocessor for the Zomorodian-Carlsson algorithm along with a fast (at the time!) software package Perseus. 2015 : Lesnick publishes a comprehensive study of the interleaving distance in the context of multiparameter persistence modules. 2018 : MacPherson and Patel concoct bisheaves to attack multi-parameter persistence geometrically for fibers of maps to triangulable manifolds. Good luck with your course!
|
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|
310,391 |
This preprint claims that, for finite kinetic energy initial solutions, uniqueness of weak solutions to the Navier-Stokes equations doesn't hold: https://arxiv.org/abs/1709.10033 What's the current consensus of the community? Is the proof considered to be correct? Does this imply that the Navier-Stokes equations are not a valid model of fluid flow, or do we need a similar result for the smooth solutions of NS before we have to abandon/modify them?
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In regards to the question of the "consensus" or "correctness", I will only point out that Tristan Buckmaster has had a proven record of studying nonuniqueness problems for low-regularity solutions in incompressible fluids, and contributed significantly to the settling of Onsager's Conjecture on the nonuniqueness problem for incompressible Euler. In regards to Navier-Stokes: weak solutions are called weak for a reason. To put it in simplest terms: the "solvability" of a PDE depends on what you accept as a valid solution. (As a digression, this is not a problem unique to PDEs. Even in arithmetic if you work over $\mathbb{Q}$ the equation $x^2 = 2$ is not solvable, and if you work over $\mathbb{R}$ the equation $x^2 = -1$ is not solvable. Mathematics has a long history of "completing" the "space of admissible solutions" to solve previously unsolvable problems.) There's an obvious trade off: if you enlarge the admissible solution space, you make it easier to solve an equation. But by making it easier to find a solution, you risk making it possible to find more than one solution. (As an example, consider $x^3 = 3$. It is not solvable in $\mathbb{Q}$, it has a unique solution in $\mathbb{R}$, and it has three solutions in $\mathbb{C}$.) In some sense you can think of existence and uniqueness as competing demands; a lot of PDE theory is built on figuring out how to restrict to a reasonable set of "admissible solutions" while guaranteeing both existence AND uniqueness. In the context of Navier-Stokes, Leray (and Hopf) figured out a way to guarantee existence. People however have long suspected that their method does not guarantee uniqueness (in other words, that they are too generous when admitting something as a solution). Buckmaster and Vicol's work tries to carve away at this problem, by proving that for an even more generous notion of solution non-uniqueness can arise. So no, we are absolutely nowhere near saying anything useful about physics or engineering; we are merely calibrating PDE theory. As an aside, local existence and uniqueness for smooth solutions of NS hold. So a "similar result for smooth solutions" is in fact, impossible . This brings me back to the point of calibration: We know for sufficiently regular initial values, local-in-time existence and uniqueness of solutions to Navier-Stokes hold. We know that if we sufficiently relax the notion of solutions, global -in-time existence of solutions to Navier-Stokes hold. We know further that if an initial data admits a global weak solution that is in fact sufficiently regular, then that is the unique weak solution (in the sense of Leray-Hopf). The main question on Navier-Stokes existence and uniqueness can be reformulated as: does there exist a sense of weak solution which guarantees global, unique solutions for all initial data, or is there a dichotomy where a sense of weak solutions that guarantees global solutions for all initial data is always too weak to guarantee uniqueness, and any sense of solutions guaranteeing uniqueness of solutions is always too strong to guarantee global solutions.
|
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|
310,576 |
let $M$ be a smooth $n$-manifold with boundary $\partial M$; I denote by $M^o$ the internal part of $M$, that is $M \smallsetminus \partial M$.
The question is the same as in the title: let $M$ and $N$ be two compact orientable topological manifolds such that $M^o$ is homeomorphic to $N^o$. Does this imply that $M$ is homeomorphic to $N$? Can we say something about the connected components of the boundary? I think that this question should have a really easy answer, but I cannot find it. One thing that I could prove is that the Euler characteristic of $\partial M$ must be the same as the one of $\partial N$, this being an easy consequence of the fact that $M^o$ is homotopically equivalent to $M$. The next step i tried is to prove that the sequence of Euler characteristics of the boundary components must agree:
$$ ((\partial M)_1, \ldots , (\partial M)_r) = ((\partial N)_1, \ldots , (\partial N)_r) ;$$
but I do not see a way to prove this. I am specially interested in the case of 3-manifolds, and from this one could conclude that $\partial M$ is homeomorphic to $\partial N$, because Euler characteristic identifies closed surfaces. Do you have any tip or any simple proof?
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In general, this is wrong. Take $M=L(7,1)\times S^{2n}\times[0,1]$ and $N=L(7,2)\times S^{2n}\times[0,1]$. Their boundaries $L(7,1)\times S^{2n}$ and $L(7,2)\times S^{2n}$ are not homeomorphic but $h$-cobordant, as proven by Milnor in J. Milnor,
Two complexes which are homeomorphic but combinatorially distinct.
Ann. of Math. (2) 74 (1961) 575–590. It follows that $\mathring{M}=L(7,1)\times S^{2n}\times \mathbb R=L(7,2)\times S^{2n}\times \mathbb R=\mathring{N}$. However, in dimension $3$ this should be true, following from work of Edwards: C. Edwards,
Concentricity in 3-manifolds.
Trans. Amer. Math. Soc. 113 (1964) 406–423. He showed that two compact $3$-manifolds are homeomorphic if and only if their interiors are homeomorphic. A main ingerdient in his proof is (as you said) that oriented $2$-manifolds are determined by their Euler characteristic.
|
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|
310,821 |
The starting point of this question is the following: If $G$ is a group such that all elements have order at most $2$, then $G$ is commutative. If $G$ is any group, let $G_{>2}$ denote the set of elements $g\in G$ such that $g^2 \neq 1_G$ where $1_G$ denotes the neutral element of the group. Question. If $G$ is infinite, directly indecomposable, and $|G_{>2}|<|G|$, is $G$ necessarily commutative?
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Let $G$ be an infinite group in which the set of elements of order $\neq 2$ has cardinal $<|G|$. Then $G$ is 2-elementary abelian. Indeed, by contradiction, let $g\in G$ be of order $3\le d\le \infty$. So the conjugacy class of $g$ has cardinal $<|G|$, and hence the centralizer $C_g$ of $g$ has order $|G|$, and in turn, the set $C_g^{(2)}$ of elements of order 2 in $C_g$ has cardinal $|G|$. Then for every $h\in C_g^{(2)}$, the element $gh$ does not have order 2 (as $(gh)^2=g^2h^2=g^2\neq 1$). Since $h\mapsto gh$ is injective, this produces $|G|$ elements of order $\ge 3$, a contradiction.
|
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|
311,071 |
The question. Which mathematical objects would you like to see formally defined in the Lean Theorem Prover ? Examples. In the current stable version of the Lean Theorem Prover, topological groups have been done, schemes have been done, Noetherian rings got done last month, Noetherian schemes have not yet been done (but are probably not going to be too difficult, if anyone is interested in trying), but complex manifolds have not yet been done. In fact I think we are nearer to perfectoid spaces than complex manifolds -- maybe because algebra is closer to the axioms than analysis. But actually we also have Lebesgue measure (it's differentiability we're not too strong at), and today we got modular forms. There is a sort of an indication of where we are. What else should we be doing? What should we work on next? Some background. The Lean theorem prover is a computer program which can check mathematical proofs which are written in a sufficiently formal mathematical language. You can read my personal thoughts on why I believe this sort of thing is timely and important for the pure mathematics community. Other formal proof verification software exists (Coq, Isabelle, Mizar...). I am very ignorant when it comes to other theorem provers and feel like I would like to see a comparison of where they all are. Over the last year I have become increasingly interested in Lean's mathematics library , because it contains a bunch of what I as a number theorist regard as "normal mathematics". No issues with constructivism, the axiom of choice, quotients by equivalence relations, the law of the excluded middle or anything. My impression that most mathematicians are not particularly knowledgeable about what can actually be done now with computer proof checkers, and perhaps many have no interest. These paragraphs are an attempt to give an update to the community. Let's start by getting one thing straight -- formalising deep mathematical proofs is extremely hard . For example, it would cost tens of millions of dollars at least , i.e. many many person-years, to formalize and maintain (a proof is a computer program , and computer programs needs maintaining!) a complete proof of Fermat's Last Theorem in a theorem prover. It would certainly be theoretically possible, but it is not currently clear to me whether any funding bodies are interested in that sort of project. But formalising deep mathematical objects is really possible nowadays. I formalised the definition of a scheme earlier this year. But here's the funny thing. 15 months ago I had never heard of the Lean Theorem Prover, and I had never used anything like a theorem prover in my life. Then in July 2017 I watched a live stream (thank you Newton Institute!) of Tom Hales' talk in Cambridge , and in particular I saw his answer to Tobias Nipkow's question 48 minutes in. And here we are now, just over a year later, with me half way through perfectoid spaces , integrating Lean into my first year undergraduate teaching, and two of my starting second year Imperial College undergraduate students, Chris Hughes and Kenny Lau , both much better than me at it. The links are to their first year undergraduate projects, one a complete formal proof of Sylow's theorems and the other an almost finished formalization of the local Langlands conjectures for abelian algebraic groups over a p-adic field. It's all open source, we are writing the new Bourbaki in our spare time and I cannot see it stopping. I know many people don't care about Bourbaki, and I know it's not a perfect analogy, but I do care about Bourbaki. I want to know which chapters should get written next, because writing them is something I find really good fun. But why write Bourbaki in a computer language? Well whether you care or not, I think it's going to happen. Because it's there. Whether it happens in Lean or one of the other systems -- time will tell. Tom Hales' formal abstracts project plans to formalise the statements of new theorems (in Lean) as they come out -- look at his blog to read more about his project. But to formalise the statements of hard theorems you have to formalise the definitions first. Mathematics is built on rigorous definitions . Computers are now capable of understanding many more mathematical definitions than they have ever been told, and I believe that this is mostly because the mathematical community, myself included, just didn't ever realise or care that it was happening. If you're a mathematician, I challenge you to formalise your best theorem in a theorem prover and send it to Tom Hales! If you need hints about how to do that in Lean, come and ask us at the Lean Zulip chat. And if if it turns out that you can't do it because you are missing some definitions, you can put them down here as answers to this big list question. We are a small but growing community at the Lean prover Zulip chat and I am asking for direction.
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I would like to see that Statement of the Classification of Finite Simple Groups formalized. And then I would like to see the proof formalized. Many people use the theorem as a black box. The first generation human proof is too big for people who are not working on the second generation and third generation proofs to completely understand and it would be a huge service to the mathematical community to formalize this, as was done for the Odd Order Theorem.
|
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|
311,073 |
Tischler theorem states that the existence of a nowhere vanishing closed $1$ -form in a compact manifold $M$ implies that the manifold fibers over $S^1$ . Do you know any other diffential topology results of this kind? By this kind I mean $$ \text{Existence of some diffential form/s} \implies \text{Topological consequences on $M$}$$
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I would like to see that Statement of the Classification of Finite Simple Groups formalized. And then I would like to see the proof formalized. Many people use the theorem as a black box. The first generation human proof is too big for people who are not working on the second generation and third generation proofs to completely understand and it would be a huge service to the mathematical community to formalize this, as was done for the Odd Order Theorem.
|
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311,576 |
I've been wanting to ask this question, because I have no insights into the way other mathematicians prepare papers (for eventual publication). How much are editing, revising, updating, adding to, etc., part of the "normal" process of process of drafting math papers? Specifically, papers of moderate (10-15 pages) length. I've noticed I tend to do this for several months, and the thought has occurred to me that perhaps I'm being too "fussy" and that I'm wasting time.
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My quick thoughts on the topic: Most of the papers are not written carefully. Here are my major complains that apply to many (if not to most) of the papers that I have seen: Proofs are too sketchy and difficult to follow. Very often they are not correct. Even, if the mistakes are minor and easy to correct, very often the proofs are not correct the way they are written. Results often have complicated and abstract statements and there are no examples that would illustrate the main result. Authors often write something like that: ` It follows from [5] ' and they refer to a 500 pages long book without specifying any particular results. Introductions do not provide right motivation by placing the results within the existing literature. I think it is extremely rude and unprofessional to write papers in the manner described above. This are some of the reasons why: Reading a paper that is not well written takes much more time both for the referee and the readers. The author perhaps saved some of his or her precious time by writing a paper not very carefully, but he or she put a burden of filling details on the shoulders of those who want to read it. Writing: ` It follows from [5] ' means that the author did not bother to check what particular result needs to be used and the reader, not familiar with [5], needs to spend hours finding the right result. Very often no result actually applies, because the result from [5] that the author had in mind, needs to be modified before it can be applied to the particular situation. By writing a paper the author should (in my opinion): Keep in mind that most of the readers are graduate students who have a very limited knowledge and maturity. Papers should help them learn the subject and so the papers should have all necessary details and relevant comments placing the result in a broader framework. For example the introductions should be like a short and well written survey on the subject. At last, but not least, the papers should have no mistakes. In my own practice I do my best to follow the rules above. Whether I am successful or not, others will judge. When I write a paper, I always try to write it in a way accessible to a (talented and motivated) graduate student. When I have a result with a complete proof and if I could write a (10 pages long) paper within a week, it usually takes me at least a month (of hard work from early morning to late night) to write it, because of the standards that I try to follow. At the stage of writing a paper, I usually write about 30 pages of scrap paper for every single page in the paper. This is, because I check carefully every proof several times, each time from scratch. If something can be easily explained by adding a couple of lines, I do it. For example it is well known that every separable metric space can be isometrically embedded into $\ell^\infty$ . One could quote this result from the literature, but since the proof is 1-2 lines long, why not to add such a proof?
Of course, one needs to keep a certain balance and not add too many details. A good advice (in my opinion) is: if a paper could be written in 10 pages, it should be written on say, 13 pages. When I quote a result from the literature I usually state the result as a lemma (instead of saying: by Theorem 3.12 in [5]). If the statement is not exactly as in [5] I explain why the modified statement is true.
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311,921 |
Hoping that my question is appropriate for MO, I would like to ask the following question: I have sent one of the editors of a very good math journal a paper of mine which contains a main result, call it theorem A.
The editor wrote me that he sent my paper to referee and will contact me when he will get the referee's report (this was 7 weeks ago).
A week ago I have noticed that my Theorem A can be generalized to Theorem B, and Theorem B has a nice application, which Theorem A does not have. What should I do: (1) Should I be patient and wait for the referee's report, and only after receiving it decide what to do with my Theorem B? (2) Should I inform the editor about my Theorem B and let him decide whether to send it to the referee or not? On the one hand, I do not want to disturb the editor (and referee), but on the other hand, perhaps the referee will be glad to see my Theorem B? (if he is not sure if my Theorem A is good enough to be published in their very good math journal, but Theorem B is). This question is slightly similar, but it asks about fixing an error, not about generalizing a result. Thank you very much! Edit: I have just found this question , which is quite similar to mine, though there are differences between the two.
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I myself would do (2). It adds information, enhances interest and acceptability, and is basic openness. The goal here is to enlighten and advance our understanding of the subject matter. Good luck! You should probably also prepare a careful revision for the referee's consideration.
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312,118 |
Disclaimer: I wasn't sure if this was an appropriate question for MathOverflow, and so I've also asked this on StackExchange. There appears to be a discrepancy in the literature regarding the definition of a localisation of a category. Let $\mathcal{C}$ be a category and let $S$ be a class of morphisms. The classical literature, for example Gabriel & Zisman, define a localisation of $\mathcal{C}$ by $S$ as follows: GZ1) A category $\mathcal{C}[S^{-1}]$ along with a functor $Q: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ which makes the elements of $S$ isomorphisms. GZ2) If a functor $F: \mathcal{C} \longrightarrow \mathcal{D}$ makes the elements of $S$ isomorphisms, then there is a functor $G: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$ such that $F$ factors through $Q$ in the sense that $F = G \circ Q$ . Moreover, $G$ is unique up to natural isomorphism. Gabriel & Zisman then state the following lemma: For each category $\mathcal{D}$ , the functor $$- \circ Q: \text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D}) \longrightarrow \text{Fun}(\mathcal{C}, \mathcal{D})$$ is an isomorphism of categories from $\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})$ to the full subcategory of $\text{Fun}(\mathcal{C}, \mathcal{D})$ consisting of functors which make elements of $S$ invertible. Gabriel & Zisman then claims that this lemma is just a restatement of the conditions GS1 and GS2 above in more precise terms. On the other hand, Kashiwara & Shapira define a localisation of the category $\mathcal{C}$ by $S$ as follows: KS1) A category $\mathcal{C}[S^{-1}]$ along with a functor $Q: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ which makes the elements of $S$ isomorphisms; KS2) If a functor $F: \mathcal{C} \longrightarrow \mathcal{D}$ makes the elements of $S$ isomorphisms, then there is a functor $G: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$ such that $F$ factors through $Q$ in the sense that $F = G \circ Q$ . KS3) If $G_{1}$ and $G_{2}$ are two objects of $\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})$ , then the natural map $$
- \circ Q: \text{Hom}_{\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})}(G_{1}, G_{2}) \longrightarrow \text{Hom}_{\text{Fun}(\mathcal{C}, \mathcal{D})}(G_{1} \circ Q, G_{2} \circ Q)
$$ is a bijection. However, Kashiwara & Shapira then make the claim that condition KS3 implies that the $G$ in KS2 is unique up to unique isomorphism. These seem to be contradictory. Gabriel & Zisman claims that their definition makes $G$ unique up to isomorphism. Kashiwara & Shaipira claims that their definition makes $G$ unique up to unique isomorphism. Ordinarily this wouldn't be a problem - obviously one is free to define your terms however you please. But on the face of it, it would appear that GZ1+GZ2 is equivalent to KS1+KS2+KS3, yet each text makes a different claim about the uniqueness of G. When I attempted to prove the uniqueness of G, I was able to show that it was unique up to isomorphism, but not that the isomorphism was unique as Kashiwara & Shapira claim. In fact, if $F$ has an automorphism besides the identity, then it would seem that there would necessarily be multiple distinct isomorphisms between $G_{1}$ and $G_{2}$ . This is something that I have seen in a number of other texts besides these two. And even worse, some texts seem to use one definition or the other in their proofs. Am I just missing something and these are equivalent? Any clarification here is appreciated. Thanks
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Actually, both of these definitions look weird to me. I would say there are two natural ways to define the localization $C[S^{-1}]$ by a universal property, as follows. For any category $D$ , let ${\rm Fun}_S(C,D)$ denote the full subcategory of the functor category ${\rm Fun}(C,D)$ spanned by those functors that send the morphisms in $S$ to isomorphisms. Let $Q:C\to C[S^{-1}]$ be a functor that sends the morphisms in $S$ to isomorphisms; then there is an induced functor $(-\circ Q): {\rm Fun}(C[S^{-1}],D) \to {\rm Fun}_S(C,D)$ . I would say that $Q$ is a strict localization if this functor $(-\circ Q)$ is an isomorphism of categories, and a weak localization if $(-\circ Q)$ is an equivalence of categories. If we unravel that, then being a strict localization means that For any functor $F:C\to D$ sending $S$ -morphisms to isomorphisms, there exists a literally unique functor $G:C[S^{-1}]\to D$ such that $F = G\circ Q$ , and condition KS3. while being a weak localization means that For any functor $F:C\to D$ sending $S$ -morphisms to isomorphisms, there exists some functor $G:C[S^{-1}]\to D$ such that $F \cong G\circ Q$ (isomorphic, not equal!), and condition KS3. The definition you quoted from KS lies somewhere in between these two: it asks only that $G$ exists rather than being unique, but it asks this $G$ to factor $F$ strictly. This corresponds to asking that the functor $(-\circ Q)$ be a surjective-on-objects equivalence , which is a rather odd condition. (The nlab gives the "weak localization" definition. Note that being a strict localization is the same as the lemma you quoted from GZ, but that it is actually strictly stronger than their definition as you quoted it: saying that $(-\circ Q)$ is an isomorphism requires $G$ to be literally unique given $F$ , not merely unique up to isomorphism. Moreover, the up-to-isomorphism GZ definition, as Simon says, is not "categorically correct", and should not be written as the definition of localization. (I believe that GZ were writing before questions such as the difference between isomorphism and equivalence of categories was widely appreciated, though, which may somewhat excuse this sloppiness.) However, I believe that in this special case, it happens to be true that if a strict localization exists (which, in $\rm Cat$ , it always does), then any GZ-localization is equivalent to this strict localization. For applying both universal properties yields functors back and forth comparing the two localizations, which commute strictly with the localization functors from $C$ . Then the second clauses of the two universal properties imply that both composites of these functors are isomorphic to identities, hence form an equivalence of categories. So if you "only care about determining categories up to equivalence", then the two definitions are in fact equivalent — but if you only care about determining categories up to equivalence, then you should really be using the definition of weak localization, since it is the only one that's invariant under replacing everything by something equivalent. Regarding your problem with showing uniqueness of the isomorphism, note that "unique up to unique isomorphism" doesn't mean that there is a unique isomorphism $G_1\cong G_2$ period, it means that there is a unique such isomorphism that commutes with all the other data. The latter condition removes any dependency on automorphisms of the input. For instance, a product object $A\times B$ is unique up to unique isomorphism, even though any automorphism of $A$ or $B$ induces an automorphism of $A\times B$ : such nontrivial automorphisms don't commute with the projection maps. In the case of localization, the weak localization condition should be enough to show that there is a unique isomorphism that commutes with the other data.
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312,439 |
This question isn't related to any specific research. I've been thinking a bit about how existence theorems are generally proven, and I've identified three broad categories: constructive proofs, proofs involving contradiction/contrapositive, and proofs involving the axiom of choice. I'm convinced that there must be some existence theorem that can be proven without any of these techniques (and I'm fairly confident that I've probably encountered some myself in the past haha), but I can't come up with any examples at the moment. Can anyone else come up with one? I'd also like to stipulate the following conditions: The proof can't piggyback on another existence theorem whose proof involves one of the above-mentioned devices. It has to be a theorem of ZF - no exotic and "high power" axioms allowed! Now for the interesting question: is there any existence theorem (again in ZF) such that every one of its proofs is of this type? Has anyone investigated something like this? If so, what results exist? Edit: This issue came up a few times in the comments: here I use "constructive" in its weaker sense (i.e. a constructive proof is merely one that constructs an object and shows that it satisfies the required properties). The stronger sense - that the proof may not use the law of excluded middle or involve any infinite objects - is not what I'm invoking.
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There are probabilistic proofs of existence. Do they fall into one of your three categories? For example, prove the existence of a real number that is normal in all bases: To do it, we show that "almost all" real numbers (according to Lebesgue measure) have that property. Therefore at least one real number has the property. And the point is: this "almost all" proof is easier than constructing an explicit example. See some nice examples due to Erdős in the cited Wikipedia page which use only finite probability spaces. If we show that a probability is ${} > 0$ , then the set is not empty.
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313,374 |
Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$ . $\bullet$ Is $X=\{0,1,2,\ldots\}$ ? $\bullet$ If not what is a minimal set $X$ with the given property?
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The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $\{ 0, 1, 2, \ldots \} \setminus S $ for an infinite $S$ . For any $\varepsilon>0$ , Erdös and Nathanson proved the existence of a set $X$ with $|X \cap [0,n]| = O(n^{\frac{3}{4} +\varepsilon})$ , so that already provides an upper bound for your second question. The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X \cap [0,n]| = O(n^{1/2}\log^{1/2} n)$ . As the lower bound $|X \cap [0,n]| =\Omega(n^{1/2})$ is obvious, this leaves a very small gap for improvement. Spencer has found a different proof of Wirsing's result. Other relevant references may be found in the second page of a paper of Vu . Note that most of these proofs are probabilistic.
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313,961 |
First off I apologize if this question does not belong here, I would be happy to hear about any better locations to post this on. I am a (first year) undergraduate mathematics student, and I recently discovered some interesting properties hidden in certain families of sequences. I ran these ideas past my math professors, and they said these are interesting ideas and the kind of thing I may want to consider publishing. I did pretty thorough research through my school's library and I am highly confident at this point that these ideas have not been written on before. To give some context into the situation, I am attending a top 15 university, and my professors are all Harvard, Cambridge, Princeton, etc. educated, so I know they know what they are talking about. With that said, the ideas I discovered -- while interesting -- are really pretty trivial as they can be completely understood with background in Calculus III and some linear algebra, and all my proofs fit onto about 3 pages of LaTeX. I would love to go ahead with trying to write up a paper on these ideas and get them published as I think that would be a great experience, however I do not want to run the risk of coming off as an over-confident "crank" type early on and be stained by that. Would you recommend I keep my cards close to my chest so to speak, or do you think I should put my ideas out there and see what happens? If I was to pursue publication, I would just submit to undergraduate journals, or maybe something else a little more specialized.
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After writing a manuscript (which it seems you may have already), go through it and revise it a few times until you feel that it is in a polished form. Then you could ask your professors to read it and provide some feedback and revise accordingly. This revision/feedback process will be a good experience for practicing and getting a feel of what the process writing research mathematics is like. Plus, you may come across potential generalizations or other cases you may not have previously considered. It's important to stress here that you should listen to your professors . They will generally know what journals / media that your paper would be suitable for. For instance, they could advise you on whether it would be worth doing any of the following: Presenting the material at an undergraduate mathematics conference (e.g. the MAA Meetings ). Post the article on arXiv Publishing the article in an undergraduate journal (e.g. Involve , the Rose-Hulman Undergraduate Mathematics Journal , the Stanford Undergraduate Research Journal ) Publishing the article in a professional research journal These could definitely be valuable experiences to get a feel for mathematics research! Even if you don't publish your current research, the experience is usually more useful in the following two ways than the actual mathematics: Understanding the research process Motivating future mathematical study Regarding (1), use this process to see if research mathematics is something that you want to pursue! If anything, you'll learn how to present logical arguments cohesively. Regarding (2), explore the connections of what you are studying to fields of mathematics that you haven't learned yet. Perhaps when looking at generalizations or applications of your result, you'll find that you'll need some deeper mathematical theory X . Use this as motivation to go learn X ! Maybe after studying X , you'll have a much better understanding of your previous results when going back to it. Who knows, maybe the ideas from your initial research can be useful for something else much further down the line, regardless of whether it was publishable in its initial form!
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313,966 |
[I asked and bountied this question on Math SE, where it got several upvotes and a comment suggesting it was research-level, but no answers. So I'm reposting here with slight edits, but please feel free to close it if it's inappropriate. Also, I'm a physicist rather than a mathematician, so fancy answers might go over my head.] In this blog post , Terry Tao discusses the $n$ -fold tensor product of a one-dimensional vector space $V^L$ ( $L$ is just a non-numeric label, not an exponent). He claims that With a bit of additional effort (and taking full advantage of the one-dimensionality of the vector spaces), one can also define spaces with fractional exponents; for instance, one can define $V^{L^{1/2}}$ as the space of formal signed square roots $\pm l^{1/2}$ of non-negative elements $l$ in $V^L$ , with a rather complicated but explicitly definable rule for addition and scalar multiplication. ... However, when working with vector-valued quantities in two and higher dimensions, there are representation-theoretic obstructions to taking arbitrary fractional powers of [vectors]. What is the "rather complicated but explicitly definable rule for addition and scalar multiplication"? Is it easy to see why this construction doesn't work in higher than one dimension? (Not necessarily a rigorous proof, just intuition for what goes wrong.) Could one extend the construction to include irrational exponents? (a) Tao claims earlier in the blog post that the vector space needs to be totally ordered. Considering that the vector space is 1D, is this equivalent to the requirement that the underlying field be totally ordered? (b) What properties does the vector space (or underlying field) need to satisfy in order for this construction to work? Presumably it doesn't work for arbitrary ordered fields, because you certainly can't define a square root function $\mathbb{Q} \to \mathbb{Q}$ . Does it only work for real vector spaces?
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After writing a manuscript (which it seems you may have already), go through it and revise it a few times until you feel that it is in a polished form. Then you could ask your professors to read it and provide some feedback and revise accordingly. This revision/feedback process will be a good experience for practicing and getting a feel of what the process writing research mathematics is like. Plus, you may come across potential generalizations or other cases you may not have previously considered. It's important to stress here that you should listen to your professors . They will generally know what journals / media that your paper would be suitable for. For instance, they could advise you on whether it would be worth doing any of the following: Presenting the material at an undergraduate mathematics conference (e.g. the MAA Meetings ). Post the article on arXiv Publishing the article in an undergraduate journal (e.g. Involve , the Rose-Hulman Undergraduate Mathematics Journal , the Stanford Undergraduate Research Journal ) Publishing the article in a professional research journal These could definitely be valuable experiences to get a feel for mathematics research! Even if you don't publish your current research, the experience is usually more useful in the following two ways than the actual mathematics: Understanding the research process Motivating future mathematical study Regarding (1), use this process to see if research mathematics is something that you want to pursue! If anything, you'll learn how to present logical arguments cohesively. Regarding (2), explore the connections of what you are studying to fields of mathematics that you haven't learned yet. Perhaps when looking at generalizations or applications of your result, you'll find that you'll need some deeper mathematical theory X . Use this as motivation to go learn X ! Maybe after studying X , you'll have a much better understanding of your previous results when going back to it. Who knows, maybe the ideas from your initial research can be useful for something else much further down the line, regardless of whether it was publishable in its initial form!
|
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314,613 |
Let $1 \leq k < n$ be natural numbers. Given orthonormal vectors $u_1,\dots,u_k$ in ${\bf R}^n$ , one can always find an additional unit vector $v \in {\bf R}^n$ that is orthogonal to the preceding $k$ . My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,\dots,u_k$ , as the tuple $(u_1,\dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.) When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,\dots,u_k$ that is consistent with a chosen orientation on ${\bf R}^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise. It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$ , though I don't know how to calculate the space of such sections.
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$\def\RR{\mathbb{R}}$ This problem was solved by Whitehead, G. W. , Note on cross-sections in Stiefel manifolds , Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702 . Such sections exist only in the cases $(k,n) = (1,2m)$ , $(n-1, n)$ , $(2,7)$ and $(3,8)$ . All sections can be given by antisymmetric multilinear maps (and thus, in particular, can be taken to be smooth). The $(2,7)$ product is the seven dimensional cross product , which is octonion multiplication restricted to the octonions of trace $0$ . The $(3,8)$ product was computed by Zvengrowski, P. , A 3-fold vector product in $R^8$ , Comment. Math. Helv. 40, 149-152 (1966). ZBL0134.38401 to be given by the formula $$X(a,b,c) = -a (\overline{b} c) + a (b \cdot c) - b (c \cdot a) + c (a \cdot b)$$ where $\cdot$ is dot product while multiplication with no symbol and $\overline{b }$ have their standard octonion meanings. Note that, if $(a,b,c)$ are orthogonal, the last $3$ terms are all $0$ , so the expression simplifies to $- a (\overline{b} c)$ ; writing in the formula in the given manner has the advantage that $X(a,b,c)$ is antisymmetric in its arguments and perpendicular to the span of $a$ , $b$ and $c$ for all $(a,b,c)$ .
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314,981 |
I conjecture the following inequality: For $x > 1$ , and $n$ a positive integer, $$\sum_{k=1}^{n}\{kx\}\le\dfrac{n}{2}x.$$ For $n=1$ , the inequality becomes $$\{x\}\le\dfrac{x}{2}\Longleftrightarrow \{x\}\le [x];$$ and, for $n = 2$ , it becomes $$\{x\}+\{2x\}\le x\Longleftrightarrow \{2x\}\le [x],$$ which is obvious since $[x]\ge 1>\{2x\}$ . If $x\ge 2$ , the inequality is obvious, since $$\dfrac{n}{2}x\ge n\ge\sum_{k=1}^{n}\{kx\}.$$ However, I cannot prove it for $1 < x < 2$ .
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As მამუკა ჯიბლაძე said, the proof is a picture though I had trouble reconstructing the picture of his and came up with a slightly different one. Let $0<y<1$ . Consider the rectangles with bases $[k-1,k]$ of heights $[ky]$ for $k=1,2,\dots,n$ . They cover the triangle with the vertices $(0,0),(n,0),(n,ny)$ up to several triangles of height $\le 1$ with sum of bases $\le n$ . This proves the inequality $$
\sum_{k=1}^n[ky]\ge \frac{n^2y}2-\frac n2\,.
$$ Now, for $x=1+y\in(1,2)$ , we have $$
\sum_{k=1}^n\{kx\}=\sum_{k=1}^n\{ky\}=\sum_{k=1}^n ky-\sum_{k=1}^n[ky]
\\
\le \frac{n(n+1)y}2-\frac{n^2y}2+\frac n2=\frac n2(1+y)=\frac n2x\,.
$$
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315,060 |
So i recently learnt that there is now a certain software called ''Coq'' by which one can check the validity of mathematical proofs. My questions are: Are there limitations on the kinds of proofs that Coq can verify? How long on average does Coq take to verify a proof? Do math journals use Coq? How do I go about it if I want to verify a proof using Coq?
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Coq is a proof assistant , and not the only one. Other popular ones are Agda , Isabelle and the related HOL light . They all use type theory as a mathematical foundation (as opposed to first-order logic and set theory), although there is a version of Isabelle that builds ZFC on top of its type theory. And I should mention the venerable Mizar system, which is based on set-theory but it is organized in Bourbaki-style structuralism (I hope I am not misrepresenting Mizar too badly, as I never used it). Many of these tools were developed by theoretical computer scientists, often for the purpose of verifying proofs of correctness of software. This explains why there is such an emphasis on type theory, but also quite independently of any applications, type theory seems to also be well-suited for organization of mathematics. Anyhow, the point is that in order to use Coq, you have to learn a new language which sits somewhere between a logic and a programming language. This is not such a small up-front investment. Once you know a bit about Coq, you will discover that there exists a sizable collection of formalized mathematics , which however is minuscule compared to the entire body of mathematical knowledge. Chances are, that your particular topic of interest has not been formalized yet. This presents a problem because a typical working mathematician relies on a large amount of pre-existing knowledge. In fact, a typical mathematician never digs all the way down to the foundations of mathematics, and so is at a loss when presented with the task of building up from scratch their own branch of mathematics. Organization of mathematics is a serious problem, akin to software engineering, and is essentially unsolved. Let me also answer your specific questions: "Are there limitations on the kinds of proofs that Coq can verify?" No, not anything you would notice, unless you are a logician or set theorist who makes proofs that are sensitive to the details of the underlying foundations. Coq by default is intuitionistic, but it's easy enough to just add excluded middle and the axiom of choice to it. And you get a lot of universes to play with, in case you want very large collections of objects. "How long on average does Coq take to verify a proof?" That depends a little bit on what you are doing, but is somewhere between instantaneous and several seconds. Anything longer than that gets people nervous and they start optimizing the proof script. In a larger development long verification times can become a problem. There are a number of techniques to speed up verification, but they rarely have anything to do with mathematics. They're about the internals of Coq and how it goes about checking proofs. [Note: I originally misunderstood the question as asking how long it takes humans to develop Coq proofs. Here is the original answer.] That depends on how complicated the proof is and how well-versed the user is. Coq is an assistant , which means that a human user directs the proof by breaking it up into chunks that Coq can do by itself. The better you are at directing Coq, the more helpful it will be. A novice will struggle with something like $x + (y + 0) = y + x$ because they will not know which library to use. On the other side of the spectrum is large-scale formalization by teams of experts. I recommend reading the report on the six-year effort to formalize the Odd order theorem , and on the formal proof of the Kepler conjecture , which was another large formalization effort. "Do math journals use Coq?" Not to my knowledge, except for Formalized mathmatics . However, some conferences in theoretical computer science, e.g. POPL accept formalized proofs as supplementary material. "How do I go about it if I want to verify a proof using Coq?" The honest answer? Take a course on Coq from a computer science department. Contact some people who know how to use Coq for formalization of mathematics. If unlucky, spend some years formalizing your branch of mathematics. Prove your theorem. It's sad but true that formalized mathematics is not ready for the mainstream mathematician, and the mainstream mathematician is not ready for formalized mathematics.
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315,816 |
Morel and Voevoedsky developed what is now called motivic homotopy theory, which aims to apply techniques of algebraic topology to algebraic varieties and, more generally, to schemes. A simple way of stating the idea is that we wish to find a model structure on some algebraic category related to that of varieties or that of schemes, so as to apply homotopy theory in an abstract sense. The uninitiated will find the name of the theory intriguing, and will perceive the simple idea presented above as a very fair approach, but upon reading the details of the theory, he might get somewhat puzzled, if not worried, about some of its aspects. We begin with the following aspects, which are added for completeness. The relevant category we start out with is not a category of schemes, but a much larger category of simplicial sheaves over the category of smooth schemes endowed with a suitable Grothendieck topology. A relevant MathOverflow topic is here . The choice of Grothendieck topology is the Nisnevich topology, the reasons being discussed right over here . The choice of smooth schemes rather than arbitrary schemes has been discussed here . Let us now suppose the uninitiated has accepted the technical reasons that are mentioned inside the linked pages, and that he will henceforth ignore whatever aesthetic shortcomings he may still perceive. He continues reading through the introduction, but soon finds himself facing two more aspects which worries him even more. The resulting homotopy theory does not behave as our rough intuition would like. In particular, what we might reasonable want to hold, such as the fact that a space ought to be homotopy equivalent to the product of itself with the affine line, is false. We solve the issue by simply forcing them to be homotopy equivalent, and hope that whatever theory rolls out is more satisfactory. There are two intuitive analogues of spheres in algebraic geometry, and the theory does not manage to identify them. We solve the issue by just leaving both of them into the game, accepting that all homology and cohomology theories will be bigraded, and we hope that this doesn't cause issues. The fact that the resulting theory is satisfactory, has proved itself over time. But hopefully the reader will not find it unreasonable that the uninitiated perceives the two issues mentioned above as a warning sign that the approach is on the wrong track, if not the 'wrong' one altogether; moreover, he will perceive the solutions presented as 'naive', as though it were but a symptomatic treatment of aforementioned warning signs. Question. How would you convince the uninitiated that Voevodsky's approach of motivic homotopy theory is 'the right one'?
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(Don't be afraid about the word " $\infty$ -category" here: they're just a convenient framework to do homotopy theory in). I'm going to try with a very naive answer, although I'm not sure I understand your question exactly. The (un)stable motivic ( $\infty$ -)category has a universal property. To be precise the following statements are true Theorem: Every functor $E:\mathrm{Sm}_S\to C$ to a(n $\infty$ -)category $C$ that is $\mathbb{A}^1$ -invariant (i.e. for which $E(X\times \mathbb{A}^1)\to E(X)$ is an equivalence) satisfies "Mayer-Vietoris for the Nisnevich topology" (i.e. sends elementary Nisnevich squares to pushout squares) factors uniquely through the unstable motivic ( $\infty$ -)category. (see Dugger Universal Homotopy Theories , section 8) Theorem: Every symmetric monoidal functor $E:\mathrm{Sm}_S\to C$ to a pointed presentable symmetric monoidal ( $\infty$ -)category that is $\mathbb{A}^1$ -invariant satisfies "Mayer-Vietoris for the Nisnevich topology" sends the "Tate motive" (i.e. the summand of $E(\mathbb{P}^1)$ obtained by splitting off the summand corresponding to $E(\mathrm{Spec}S)\to E(\mathbb{P}^1)$ ) to an invertible object factors uniquely through the stable motivic ( $\infty$ -)category. (see Robalo K-Theory and the bridge to noncommutative motives , corollary 2.39) These two theorems are saying that any two functors that "behave like a homology theory on smooth $S$ -schemes" will factor uniquely through the (un)stable motivic ( $\infty$ -)category. Examples are $l$ -adic étale cohomology, algebraic K-theory (if $S$ is regular Noetherian), motivic cohomology (as given by Bloch's higher Chow groups)... Conversely, the canonical functor from $\mathrm{Sm}_S$ to the (un)stable motivic ( $\infty$ -)category has these properties. So the (un)stable motivic homotopy theory is in this precise sense the home of the universal homology theory. In particular all the properties we can prove for $\mathcal{H}(S)$ or $SH(S)$ reflect on every homology theory satisfying the above properties (purity being the obvious example). Let me say a couple of words about the two aspects that "worry you" $\mathbb{A}^1$ -invariance needs to be imposed. That's not surprising: we do need to do that also for topological spaces, when we quotient the maps by homotopy equivalence (or, more precisely, we need to replace the set of maps by a space of maps, where paths are given by homotopies: this more complicated procedure is also responsible for the usage of simplicial presheaves rather than just ordinary presheaves) Having more kinds of spheres is actually quite common in homotopy theory. A good test case for this is $C_2$ -equivariant homotopy theory. See for example this answer of mine for a more detailed exploration of the analogy. Surprisingly, possibly the most problematic of the three defining properties of $SH(S)$ is the $\mathbb{A}^1$ -invariance. In fact there are several "homology theories" we'd like to study that do not satisfy it (e.g. crystalline cohomology). I know some people are trying to find a replacement for $SH(S)$ where these theories might live. As far as I know there are some definitions of such replacements but I don't think they have been shown to have properties comparable to the very interesting structure you can find on $SH(S)$ , so I don't know whether this will bear fruit or not in the future.
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315,843 |
What are examples of well received mathematical papers in which the author provides detail on how a surprising solution to a problem has been found. I am especially looking for papers that also document the dead ends of investigation, i.e. ideas that seemed promising but lead nowhere, and where the motivation and inspiration that lead to the right ideas came from. By "surprising solution" I mean solutions that feel right at first reading and it isn't clear why they haven't been found earlier.
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Richard P. Stanley's How the Upper Bound Conjecture was proved ends with two morals: The shortest path may not be the best. Even if you don’t arrive at your destination, the journey can still be
worthwhile.
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"https://mathoverflow.net/questions/315843",
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|
316,505 |
Let $\omega=\sum_{i=1}^n dx_i\wedge dy_i\in\bigwedge^2(\mathbb{R}^{2n})^*$ be a standard symplectic form. The following result is due to Lefschetz: For $k\leq n$ , the Lefschetz operator $L^{n-k}:\bigwedge^k(\mathbb{R}^{2n})^*\to
\bigwedge^{2n-k}(\mathbb{R}^{2n})^*$ defined by $$
L^{n-k}\alpha=\alpha\wedge\underbrace{\omega\wedge\ldots\wedge\omega}_{n-k}=\alpha\wedge\omega^{n-k}
$$ is an isomorphism. This follows from Proposition 1.2.30 [1]. The proof is slick and it uses representation theory. Question: What are other standard references for this result? I want to quote it properly and I would like to see a reference to a
proof by brute force. A proof by brute force is not difficult but ugly
and perhaps in some reference I can find an elegant presentation of
such a proof. This result seems so fundamental that it should be available in many textbooks, but I am not aware of any except [1]. [1] D. Huybrechts, Complex geometry. An introduction. . Universitext. Springer-Verlag, Berlin, 2005. (MathsciNet review) .
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There is an elementary proof in our 2003 book Exterior Differential Systems and Euler-Lagrange Partial Differential Equations (Bryant, et al, University of Chicago Press). It does not use any representation theory and is not 'brute force'; it only takes a couple of paragraphs using elementary facts about exterior algebra. See Proposition 1.1a, with the proof on page 13. I learned that proof from Eugenio Calabi more than 30 years ago, and he told me that he had found it sometime back in the 50s.
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317,100 |
This question is motivated by one of the problem set from this year's Putnam Examination . That is, Problem. Let $S_1, S_2, \dots, S_{2^n-1}$ be the nonempty subsets of $\{1,2,\dots,n\}$ in some order, and let $M$ be the $(2^n-1) \times (2^n-1)$ matrix whose $(i,j)$ entry is $$M_{ij} = \begin{cases} 0 & \mbox{if }S_i \cap S_j = \emptyset; \\
1 & \mbox{otherwise.}
\end{cases}$$ Calculate the determinant of $M$ . Answer: If $n=1$ then $\det M=1$ ; else $\det(M)=-1$ . I like to consider the following variant which got me puzzled. Question. Preserve the notation from above, let $A$ be the matrix whose $(i,j)$ entry is $$A_{ij} = \begin{cases} 1 & \# (S_i \cap S_j) =1; \\
0 & \mbox{otherwise.}
\end{cases}$$ If $n>1$ , is this true? $$\det(A)=-\prod_{k=1}^nk^{\binom{n}k}.$$ Remark. Amusingly, the same number counts "product of sizes of all the nonempty subsets of $[n]$ " according to OEIS . POSTSCRIPT. If $B$ is the matrix whose $(i,j)$ entry is $B_{ij} = q^{\#(S_i\cap S_j)}$ then does this hold? $$\det(B)=q^n(q-1)^{n(2^{n-1}-1)}.$$
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$\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\suml}{\sum\limits}
\newcommand{\sumS}{\suml_{S \in P}}
\newcommand{\prodl}{\prod\limits}
\newcommand{\prodS}{\prodl_{S \in P}}
\newcommand{\subs}{\subseteq}
\newcommand{\sups}{\supseteq}$ Here are proofs for both the original question (Theorem 1 below) and for the postscript (Theorem 5 below) that follow my argument at https://artofproblemsolving.com/community/u432h1747709p11384734 as closely as possible. Sorry for their length, much of it due to exposition of matrix folklore. (If you know this, scroll all the way down to section 2.) 1. Matrices indexed by finite sets We are grown-ups and don't need to index the rows and the columns of a matrix by numbers. We can pick any set $P$ , and define a $P \times P$ -matrix (say, with rational entries) to be a family $\tup{a_{p, q}}_{\tup{p, q} \in P \times P} \in \QQ^{P \times P}$ of rational numbers indexed by pairs $\tup{p, q} \in P \times P$ . When the set $P$ is finite, such a $P \times P$ -matrix always has a well-defined determinant, which can be defined by \begin{equation}
\det \tup{ \tup{a_{p, q}}_{\tup{p, q} \in P \times P} }
= \sum_{\sigma \in S_P} \tup{-1}^{\sigma} \prod_{p \in P} a_{p, \sigma\tup{p}} .
\end{equation} Here, $S_P$ denotes the set of permutations of $P$ , and the notation $\tup{-1}^{\sigma}$ stands for the sign of a permutation $\sigma$ . If you think that this is problematic because the sign of a permutation is usually defined using a total order on $P$ , then take a break and convince yourself that it does not actually depend on this total order. (This is proven in maximum detail in the solution of Exercise 5.12 of my Notes on the combinatorial fundamentals of algebra , 10th of January 2019 .) Determinants of $P \times P$ -matrices behave as nicely as determinants of "usual" $n \times n$ -matrices, and arguably even better, since it is easier to define minors and cofactors when the rows and columns are indexed by an arbitrary finite set. Do keep in mind that the rows and the columns must be indexed by the same set; we cannot define the determinant of a $P \times Q$ -matrix for different $P$ and $Q$ , even when $\abs{P} = \abs{Q}$ . Of course, an $n \times n$ -matrix (for a nonnegative integer $n$ ) is the same as an $N \times N$ -matrix for $N = \set{1,2,\ldots,n}$ . Thus, our notion of $P \times P$ -matrices generalizes the classical notion of $n \times n$ -matrices. This generalization "respects" standard concepts like determinants -- e.g., the determinant of an $n \times n$ -matrix does not change if we instead consider it as an $N \times N$ -matrix. Does this generalization exhibit genuinely new behavior or does it just add convenience? Arguably it's the latter, because conversely, you can turn any $P \times P$ -matrix (for any finite set $P$ ) into an $n \times n$ -matrix. Namely, if you have a finite set $P$ and a $P \times P$ -matrix $A = \tup{a_{p, q}}_{\tup{p, q} \in P \times P} \in \QQ^{P \times P}$ , then you can set $n = \abs{P}$ and fix a bijection $\phi : \set{1,2,\ldots,n} \to P$ , and form the $n \times n$ -matrix $A_\phi := \tup{ a_{\phi\tup{i}, \phi\tup{j}} }_{\tup{i, j} \in \set{1,2,\ldots,n} \times \set{1,2,\ldots,n}} \in \QQ^{n \times n}$ , and observe that $A_\phi$ is "basically the same as" $A$ except that the rows and the columns have been reindexed. We shall refer to this operation (going from $A$ to $A_\phi$ ) as numerical reindexing . So $P \times P$ -matrices differ from $n \times n$ -matrices only in how we index their rows and columns. Nevertheless they are useful, because often there is no canonical bijection $\phi : \set{1,2,\ldots,n} \to P$ (exhibit A: this Putnam problem), and we are algebraists and want to make as few non-canonical choices as possible. Whenever $P$ is a finite set, we can multiply two $P \times P$ -matrices: The product $AB$ of two $P \times P$ -matrices $A = \tup{a_{p, q}}_{\tup{p, q} \in P \times P} \in \QQ^{P \times P}$ and $B = \tup{b_{p, q}}_{\tup{p, q} \in P \times P} \in \QQ^{P \times P}$ is defined to be the $P \times P$ -matrix $\tup{ \sum_{r \in P} a_{p, r} b_{r, q} }_{\tup{p, q} \in P \times P} \in \QQ^{P \times P}$ . This multiplication behaves just as nicely as usual multiplication of $n \times n$ -matrices. In particular, $\det\tup{AB} = \det A \cdot \det B$ whenever $A$ and $B$ are two $P \times P$ -matrices. And of course, you can prove this by numerical reindexing, because numerical reindexing preserves both products and determinants (assuming, of course, that you pick one bijection $\phi : \set{1,2,\ldots,n} \to P$ and stick with it). If you remember the definition of triangular $n \times n$ -matrices, then its generalization to $P \times P$ -matrices is straightforward: When $P$ is a totally ordered finite set , we say that a $P \times P$ -matrix $A = \tup{a_{p, q}}_{\tup{p, q} \in P \times P}$ is lower-triangular if \begin{equation}
a_{p, q} = 0 \qquad \text{for all } \tup{p, q} \in P \times P \text{ satisfying } p < q ;
\end{equation} and we say that a $P \times P$ -matrix $A = \tup{a_{p, q}}_{\tup{p, q} \in P \times P}$ is upper-triangular if \begin{equation}
a_{p, q} = 0 \qquad \text{for all } \tup{p, q} \in P \times P \text{ satisfying } p > q .
\end{equation} Again, lower-triangular $P \times P$ -matrices become lower-triangular $n \times n$ -matrices upon numerical reindexing, as long as we make sure to pick our bijection $\phi : \set{1,2,\ldots,n} \to P$ to be an order isomorphism (which, by the way, is unique). So we can derive properties of lower-triangular $P \times P$ -matrices from analogous properties of lower-triangular $n \times n$ -matrices by numerical reindexing. In particular, we can thus show that the determinant of a lower-triangular $P \times P$ -matrix equals the product of its diagonal entries; i.e., if $A = \tup{a_{p, q}}_{\tup{p, q} \in P \times P}$ is a lower-triangular $P \times P$ -matrix, then $\det A = \prodl_{p \in P} a_{p, p}$ . The same holds for upper-triangular $P \times P$ -matrices. 2. Answering the original question Now, let's not forget about the question. Fix a positive integer $n$ . Let $N$ be the set $\set{1,2,\ldots,n}$ . Let $P$ be the set of all the $2^n - 1$ nonempty subsets of $N$ . (We shall later totally order $P$ , but for now $P$ is just a finite set.) We shall also use the Iverson bracket notation , which will save us some ampersands and braces. This allows us to rewrite the definition of $A_{ij}$ in your question as $A_{ij} = \ive{ \abs{ S_i \cap S_j } = 1 }$ . But as we said, we drop the numerical indexing, and instead define a matrix whose rows and columns are indexed by $P$ (that is, by all the nonempty subsets of $N$ ). Thus, the claim of the exercise rewrites as follows: Theorem 1. Let $A$ be the $P \times P$ -matrix $\tup{ \ive{ \abs{ S \cap T } = 1 } }_{\tup{S, T} \in P}$ . Then, \begin{equation}
\det A = \tup{-1}^{\ive{n \neq 1}} \prod_{k=1}^n k^{\dbinom{n}{k}} .
\end{equation} The key to proving this is the following simple fact: Lemma 2. Let $S \in P$ and $T \in P$ . Then, \begin{equation}
\ive{ \abs{ S \cap T } = 1 } = \suml_{U \in P} \tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T } .
\end{equation} This, in turn, shall be derived from the following binomial identity: Lemma 3. Let $k$ be a nonnegative integer. Then, \begin{equation}
\sum_{i=1}^k \tup{-1}^{i-1} i \cdot \dbinom{k}{i} = \ive{k = 1}.
\end{equation} Proof of Lemma 3 (sketched). If $k = 0$ , then this equality holds for obvious reasons (indeed, the left hand side is an empty sum and thus vanishes, while the right hand side vanishes since $k = 0 \neq 1$ ). Hence, for the rest of this proof, we WLOG assume that $k \neq 0$ . Hence, $k$ is a positive integer. Thus, $k-1$ is a nonnegative integer. It is well-known that $\suml_{i=0}^n \tup{-1}^i \dbinom{n}{i} = \ive{ n = 0 }$ for every nonnegative integer $n$ . Applying this to $n = k-1$ , we obtain $\suml_{i=0}^{k-1} \tup{-1}^i \dbinom{k-1}{i} = \ive{ k - 1 = 0 } = \ive{k = 1}$ . Multiplying both sides of this equality by $k$ , we obtain $k \suml_{i=0}^{k-1} \tup{-1}^i \dbinom{k-1}{i} = k \ive{k = 1} = \ive{k = 1}$ (where the last equality sign is easy to check directly). Hence, \begin{align}
\ive{k = 1}
&= k \suml_{i=0}^{k-1} \tup{-1}^i \dbinom{k-1}{i}
= \suml_{i=0}^{k-1} \tup{-1}^i \cdot \underbrace{k \dbinom{k-1}{i}}_{\substack{= \tup{i+1} \dbinom{k}{i+1} \\ \text{(by a simple computation)}}} \\
&= \suml_{i=0}^{k-1} \tup{-1}^i \cdot \tup{i+1} \dbinom{k}{i+1}
= \suml_{i=1}^{k} \tup{-1}^{i-1} \cdot i \dbinom{k}{i}
\end{align} (here, we have substituted $i-1$ for $i$ in the sum). This proves Lemma 3. $\blacksquare$ Proof of Lemma 2 (sketched). If a subset $U \in P$ does not satisfy $U \subs S$ , then the Iverson bracket $\ive{ U \subs S }$ is zero and thus renders the whole product $\tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T }$ zero. Similarly, this product also is rendered zero when $U \in P$ does not satisfy $U \subs T$ . Thus, the product $\tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T }$ is zero unless $U \in P$ satisfies both $U \subs S$ and $U \subs T$ . Hence, the sum $\suml_{U \in P} \tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T }$ does not change its value when we restrict it to those $U$ that satisfy both $U \subs S$ and $U \subs T$ (because all the addends that we lose are zero anyway). In other words, \begin{align*}
\suml_{U \in P} \tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T }
&= \suml_{\substack{U \in P; \\ U \subs S \text{ and } U \subs T}} \tup{-1}^{\abs{U} - 1} \abs{U} \cdot \underbrace{\ive{ U \subs S }}_{=1} \underbrace{\ive{ U \subs T }}_{=1} \\
&= \suml_{\substack{U \in P; \\ U \subs S \text{ and } U \subs T}}
\tup{-1}^{\abs{U} - 1} \abs{U}.
\end{align*} Now, let $K$ be the intersection $S \cap T$ , and let $k = \abs{K}$ . Then, the sets $U \in P$ that satisfy both $U \subs S$ and $U \subs T$ are precisely the nonempty subsets of the $k$ -element set $K$ . Thus, we know how many such sets $U$ exist of each size: there are exactly $\dbinom{k}{1}$ ones of size $1$ , exactly $\dbinom{k}{2}$ ones of size $2$ , and so on, all the way up to $\dbinom{k}{k}$ ones of size $k$ . Hence, \begin{align*}
\suml_{\substack{U \in P; \\ U \subs S \text{ and } U \subs T}}
\tup{-1}^{\abs{U} - 1} \abs{U}
&= \dbinom{k}{1} \tup{-1}^{1-1} \cdot 1 + \dbinom{k}{2} \tup{-1}^{2-1} \cdot 2 + \cdots + \dbinom{k}{k} \tup{-1}^{k-1} \cdot k \\
&= \sum_{i=1}^k \dbinom{k}{i} \tup{-1}^{i-1} \cdot i
= \sum_{i=1}^k \tup{-1}^{i-1} i \cdot \dbinom{k}{i} \\
&= \ive{k = 1} \qquad \tup{ \text{by Lemma 3} } \\
&= \ive{ \abs{ S \cap T } = 1 }
\end{align*} (since $k = \abs{K} = \abs{ S \cap T }$ ).
Combining what we have, we obtain \begin{equation}
\suml_{U \in P} \tup{-1}^{\abs{U} - 1} \abs{U} \cdot \ive{ U \subs S } \ive{ U \subs T }
= \suml_{\substack{U \in P; \\ U \subs S \text{ and } U \subs T}}
\tup{-1}^{\abs{U} - 1} \abs{U}
= \ive{ \abs{ S \cap T } = 1 } .
\end{equation} This proves Lemma 2. $\blacksquare$ Lemma 4. (a) We have $\sumS \abs{S} = n 2^{n-1}$ . (b) We have $\sumS \tup{ \abs{S} - 1 } = n 2^{n-1} - 2^n + 1$ . (c) We have $\prodS \tup{-1}^{\abs{S} - 1} = \tup{-1}^{\ive{n \neq 1}}$ . (d) We have $\prodS \abs{S} = \prodl_{k=1}^n k^{\dbinom{n}{k}}$ . Proof of Lemma 4. (a) Fix $i \in \set{1,2,\ldots,n}$ . Then, exactly $2^{n-1}$ subsets of $\set{1,2,\ldots,n}$ contain $i$ (since we can freely choose which of the other $n-1$ elements such a subset should contain). Of course, all of these $2^{n-1}$ subsets are nonempty, and thus belong to $P$ . Hence, exactly $2^{n-1}$ subsets $S \in P$ contain $i$ . Hence, the sum $\sumS \ive{ i \in S }$ has exactly $2^{n-1}$ addends equal to $1$ , while all its other addends are $0$ . Therefore, this sum equals $2^{n-1}$ . In other words, we have \begin{equation}
\sumS \ive{ i \in S } = 2^{n-1} .
\label{darij1.pf.l4.1}
\tag{1}
\end{equation} Now, forget that we fixed $i$ . We thus have proven \eqref{darij1.pf.l4.1} for each $i \in \set{1,2,\ldots,n}$ . But we have $\abs{S} = \suml_{i=1}^n \ive{ i \in S }$ for each subset $S$ of $\set{1,2,\ldots,n}$ (because the sum $\suml_{i=1}^n \ive{ i \in S }$ contains exactly $\abs{S}$ many addends equal to $1$ , whereas all its other addends are $0$ ). Summing up this equation over all $S \in P$ , we obtain \begin{align}
\sumS \abs{S}
&= \sumS \suml_{i=1}^n \ive{ i \in S }
= \suml_{i=1}^n \underbrace{\sumS \ive{ i \in S }}_{\substack{= 2^{n-1} \\ \text{(by \eqref{darij1.pf.l4.1})}}}
= \suml_{i=1}^n 2^{n-1} = n 2^{n-1} .
\end{align} This proves Lemma 4 (a) . (b) We have $\abs{P} = 2^n - 1$ (since the set $P$ consists of all the $2^n$ subsets of $\set{1,2,\ldots,n}$ apart from the empty set). Now, \begin{align}
\sumS \tup{ \abs{S} - 1 }
= \underbrace{\sumS \abs{S}}_{\substack{= n 2^{n-1} \\ \text{(by Lemma 4 (a))}}}
- \underbrace{\sumS 1}_{= \abs{P} = 2^n - 1}
= n 2^{n-1} - \tup{2^n - 1} = n 2^{n-1} - 2^n + 1 .
\end{align} This proves Lemma 4 (b) . (c) If $n \neq 1$ , then $n \geq 2$ (since $n$ is a positive integer), and therefore both $2^{n-1}$ and $2^n$ are even integers. Hence, if $n \neq 1$ , then $n 2^{n-1} - 2^n + 1 \equiv 1 \mod 2$ . On the other hand, if $n = 1$ , then $n 2^{n-1} - 2^n + 1 = 1 \cdot 2^{1-1} - 2^1 + 1 = 0 \equiv 0 \mod 2$ . Combining the previous two sentences, we conclude that $n 2^{n-1} - 2^n + 1 \equiv \ive{n \neq 1} \mod 2$ . Hence, $\tup{-1}^{n 2^{n-1} - 2^n + 1} = \tup{-1}^{\ive{n \neq 1}}$ . Now, \begin{align}
\prodS \tup{-1}^{\abs{S} - 1}
&= \tup{-1}^{\sumS \tup{ \abs{S} - 1 }}
= \tup{-1}^{n 2^{n-1} - 2^n + 1}
\qquad \tup{ \text{by Lemma 4 (b)} } \\
&= \tup{-1}^{\ive{n \neq 1}} .
\end{align} This proves Lemma 4 (c) . (d) Recall that $P$ is the set of all nonempty subsets of $\set{1,2,\ldots,n}$ . Hence, among the elements of $P$ are exactly $\dbinom{n}{1}$ subsets of size $1$ , exactly $\dbinom{n}{2}$ subsets of size $2$ , and so on, and finally exactly $\dbinom{n}{n}$ subsets of size $n$ (and no other subsets). Therefore, the product $\prodS \abs{S}$ has exactly $\dbinom{n}{1}$ factors equal to $1$ , exactly $\dbinom{n}{2}$ factors equal to $2$ , and so on, and finally exactly $\dbinom{n}{n}$ factors equal to $n$ (and no other factors). Therefore, this product equals $\prodl_{k=1}^n k^{\dbinom{n}{k}}$ . This proves Lemma 4 (d) . $\blacksquare$ Proof of Theorem 1 (sketched). Define two $P \times P$ -matrices \begin{align*}
B &= \tup{ \tup{-1}^{\abs{T} - 1} \abs{T} \cdot \ive{ T \subs S } }_{\tup{S, T} \in P \times P} \qquad \text{and} \\
C &= \tup{ \ive{ S \subs T } }_{\tup{S, T} \in P \times P} .
\end{align*} Lemma 2 says that $A = BC$ . (More precisely, Lemma 2 says that the $\tup{S, T}$ -th entry of $A$ equals the corresponding entry of $BC$ for all $S \in P$ and $T \in P$ ; but this yields $A = BC$ .) Thus, $\det A = \det\tup{BC} = \det B \cdot \det C$ . Now, it remains to find $\det B$ and $\det C$ . Here we shall finally endow our set $P$ with a total ordering. Namely, we pick any total order on $P$ with the property that every pair of two subsets $S \in P$ and $T \in P$ satisfying $S \subs T$ must satisfy $S \leq T$ . There are many total orders that do the trick here; for example, we can order them by increasing size (resolving ties arbitrarily). Now, it is easy to see that the $P \times P$ -matrix $B$ is lower-triangular. Since the determinant of a lower-triangular $P \times P$ -matrix equals the product of its diagonal entries, we thus conclude that \begin{align}
\det B
&= \prodS \tup{ \tup{-1}^{\abs{S} - 1} \abs{S} \underbrace{\ive{ S \subs S }}_{=1} }
= \prodS \tup{ \tup{-1}^{\abs{S} - 1} \abs{S} } \\
&= \underbrace{\tup{ \prodS \tup{-1}^{\abs{S} - 1} }}_{\substack{= \tup{-1}^{\ive{n \neq 1}} \\ \text{(by Lemma 4 (c))}}}
\tup{ \prodS \abs{S} }
= \tup{-1}^{\ive{n \neq 1}} \tup{ \prodS \abs{S} } .
\end{align} Likewise, the $P \times P$ -matrix $C$ is upper-triangular. Since the determinant of an upper-triangular $P \times P$ -matrix equals the product of its diagonal entries, we thus conclude that \begin{equation}
\det C = \prodS \underbrace{\ive{ S \subs S }}_{=1}
= 1 .
\end{equation} Now, \begin{align}
\det A
&= \det B \cdot \underbrace{\det C}_{=1} = \det B = \tup{-1}^{\ive{n \neq 1}} \underbrace{\tup{ \prodS \abs{S} }}_{\substack{= \prodl_{k=1}^n k^{\dbinom{n}{k}} \\ \text{(by Lemma 4 (d))}}} \\
&= \tup{-1}^{\ive{n \neq 1}} \prod_{k=1}^n k^{\dbinom{n}{k}} .
\end{align} This proves Theorem 1. $\blacksquare$ In hindsight, the above proof could have been restated without introducing $P \times P$ -matrices, since picking a total order on a finite set $P$ is equivalent to picking a bijection $\phi : \set{1,2,\ldots,n} \to P$ . But I prefer to give the exercise an "invariant" formulation, and $P \times P$ -matrices are worth exposing anyway. A few more telegraphic remarks: The decomposition $A = BC$ constructed in the above proof of Theorem 1 is an LU-decomposition . The solution to the original Putnam question I found first was the same recursive argument appearing a few times in the AoPS thread linked above. But the structure of the recursion foreshadowed the existence of a slicker proof -- the block-row-reduction operations felt like multiplying by a lower-triangular (in a properly chosen order) matrix. So I dug deeper and found the above. This reminds me of Lemma 7.1 of Chris Godsil, An Introduction to the Moebius Function , arXiv:1803.06664v1 . 3. Answering the postscriptum Anyway, I wanted to prove another theorem, from the postscriptum to the original question: Theorem 5. Let $q \in \QQ$ . Let $D$ be the $P \times P$ -matrix $\tup{ q^{\abs{ S \cap T }} }_{\tup{S, T} \in P}$ . Then, \begin{equation}
\det D = q^n \tup{q-1}^{n\tup{2^{n-1}-1}} .
\end{equation} Just as our proof of Theorem 1 relied on Lemma 2, our proof of Theorem 5 will rely on the following lemma: Lemma 6. Let $S \in P$ and $T \in P$ . Let $q \in \QQ$ be nonzero. Then, \begin{equation}
q^{\abs{ S \cap T }} = \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } .
\end{equation} Note that the assumptions " $S \in P$ and $T \in P$ " in Lemma 6 can be replaced by the weaker assumption " $S$ and $T$ are subsets of $N$ ", but we will only use Lemma 6 in the case when $S$ and $T$ belong to $P$ . Proof of Lemma 6 (sketched). If a subset $U \in P$ does not satisfy $U \sups S$ , then the Iverson bracket $\ive{ U \sups S }$ is zero and thus renders the whole product $\tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T }$ zero. Similarly, this product also is rendered zero when $U \in P$ does not satisfy $U \sups T$ . Thus, the product $\tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T }$ is zero unless $U \in P$ satisfies both $U \sups S$ and $U \sups T$ . Hence, the sum $\suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T }$ does not change its value when we restrict it to those $U$ that satisfy both $U \sups S$ and $U \sups T$ (because all the addends that we lose are zero anyway). In other words, \begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= \suml_{\substack{U \in P; \\ U \sups S \text{ and } U \sups T}} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \underbrace{\ive{ U \sups S }}_{=1} q^{\abs{T}} \underbrace{\ive{ U \sups T }}_{=1} \\
&= \suml_{\substack{U \in P; \\ U \sups S \text{ and } U \sups T}}
\tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} q^{\abs{T}} \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \in P; \\ U \sups S \text{ and } U \sups T}}
\tup{q-1}^{n-\abs{U}} .
\label{darij1.pf.l6.1}
\tag{2}
\end{align} Now, let $G = S \cup T$ . This is a nonempty subset of $N$ (since $S$ and $T$ are nonempty subsets of $N$ ). Let $H = N \setminus G$ be the complement of $G$ in $N$ . Let $h = \abs{H}$ . Every subset $U$ of $N$ that satisfies $U \sups S$ and $U \sups T$ is automatically nonempty (since $U$ contains the nonempty set $S$ as a subset), and thus belongs to $P$ . Hence, the $U \in P$ that satisfy $U \sups S$ and $U \sups T$ are precisely the subsets $U$ of $N$ that satisfy $U \sups S$ and $U \sups T$ . Thus, we have \begin{align*}
&\ \set{ U \in P \mid U \sups S \text{ and } U \sups T } \\
= &\ \set{ U \subs N \mid U \sups S \text{ and } U \sups T } \\
= &\ \set{ U \subs N \mid U \sups \underbrace{S \cup T}_{=G} } \\
= &\ \set{ U \subs N \mid U \sups G } \\
= &\ \set{ U \subs N \mid N \setminus U \subs N \setminus G }
\end{align*} (because the condition " $U \sups G$ " on a subset $U$ of $N$ is equivalent to the condition " $N \setminus U \subs N \setminus G$ ").
Hence, the summation sign " $\suml_{\substack{U \in P; \\ U \sups S \text{ and } U \sups T}}$ " in \eqref{darij1.pf.l6.1} can be rewritten as " $\suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}$ ". Thus, \eqref{darij1.pf.l6.1} rewrites as \begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}
\tup{q-1}^{n-\abs{U}} .
\label{darij1.pf.l6.2}
\tag{3}
\end{align} Every subset $U$ of $N$ satisfies $n - \abs{U} = \abs{N \setminus U}$ (since $U \subs N$ yields $\abs{N \setminus U} = \underbrace{\abs{N}}_{=n} - \abs{U} = n - \abs{U}$ ). Thus, \eqref{darij1.pf.l6.2} becomes \begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \underbrace{\suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}}_{\substack{= \suml_{\substack{U \subs N; \\ N \setminus U \subs H}} \\ \tup{ \text{since } N \setminus G = H } }}
\underbrace{\tup{q-1}^{n-\abs{U}}}_{\substack{= \tup{q-1}^{\abs{N \setminus U}} \\ \tup{ \text{since } n - \abs{U} = \abs{N \setminus U} } }} \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ N \setminus U \subs H}}
\tup{q-1}^{\abs{N \setminus U}} \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ U \subs H}}
\tup{q-1}^{\abs{U}}
\label{darij1.pf.l6.3}
\tag{4}
\end{align} (here, we have substituted $U$ for $N \setminus U$ in the sum, since the map \begin{align}
\set{U \subs N} \to \set{U \subs N}, \qquad
U \mapsto N \setminus U
\end{align} is a bijection). The summation sign " $\suml_{\substack{U \subs N; \\ U \subs H}}$ " on the right hand side of \eqref{darij1.pf.l6.3} can be replaced by " $\suml_{U \subs H}$ ", since $H$ is a subset of $N$ . Thus, \eqref{darij1.pf.l6.3} rewrites as \begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{U \subs H} \tup{q-1}^{\abs{U}} .
\label{darij1.pf.l6.4}
\tag{5}
\end{align} The sum on the right hand side of this equality is easy to simplify. The set $H$ has size $\abs{H} = h$ . Thus, it has exactly $\dbinom{h}{0}$ subsets of size $0$ , exactly $\dbinom{h}{1}$ subsets of size $1$ , and so on, all the way up to $\dbinom{h}{h}$ subsets of size $h$ . Hence, the sum $\suml_{U \subs H} \tup{q-1}^{\abs{U}}$ has exactly $\dbinom{h}{0}$ many terms equal to $\tup{q-1}^0$ , exactly $\dbinom{h}{1}$ many terms equal to $\tup{q-1}^1$ , and so on, all the way up to $\dbinom{h}{h}$ many terms equal to $\tup{q-1}^h$ . Hence, this sum can be rewritten as follows: \begin{align}
& \suml_{U \subs H} \tup{q-1}^{\abs{U}} \\
& = \dbinom{h}{0} \tup{q-1}^0 + \dbinom{h}{1} \tup{q-1}^1 + \cdots + \dbinom{h}{h} \tup{q-1}^h \\
& = \sum_{k=0}^h \dbinom{h}{k} \tup{q-1}^k = \tup{\tup{q-1}+1}^h \\
& \qquad \tup{ \text{since the binomial formula yields } \tup{\tup{q-1}+1}^h = \sum_{k=0}^h \dbinom{h}{k} \tup{q-1}^k } \\
& = q^h .
\end{align} Thus, \eqref{darij1.pf.l6.4} becomes \begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \underbrace{\suml_{U \subs H} \tup{q-1}^{\abs{U}}}_{= q^h} \\
&= q^{\abs{S}-n} q^{\abs{T}} q^h = q^{\tup{\abs{S}-n} + \abs{T} + h} .
\label{darij1.pf.l6.5}
\tag{6}
\end{align} But $H$ is the complement of $G$ in $N$ . Hence, $\abs{H} = \underbrace{\abs{N}}_{=n} - \abs{G} = n - \abs{G}$ . Therefore, $\abs{G} = n - \underbrace{\abs{H}}_{= h} = n - h$ . Recall the basic fact that $\abs{S} + \abs{T} = \abs{S \cup T} + \abs{S \cap T} = \abs{G} + \abs{S \cap T}$ (since $S \cup T = G$ ). Solving this equation for $\abs{S \cap T}$ , we obtain \begin{align}
\abs{S \cap T} = \abs{S} + \abs{T} - \underbrace{\abs{G}}_{= n - h} = \abs{S} + \abs{T} - \tup{n - h} = \tup{\abs{S}-n} + \abs{T} + h .
\end{align} Therefore, $q^{ \abs{S \cap T} } = q^{\tup{\abs{S}-n} + \abs{T} + h}$ . Comparing this with \eqref{darij1.pf.l6.5}, we obtain \begin{equation}
q^{\abs{ S \cap T }} = \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } .
\end{equation} This proves Lemma 6. $\blacksquare$ Proof of Theorem 5 (sketched). Both $\det D$ and $q^n \tup{q-1}^{n\tup{2^{n-1}-1}}$ are polynomials in $q$ with integer coefficients (since $2^{n-1}-1$ is a nonnegative integer). Thus, the equality we need to prove is an equality between two polynomials in $q$ . Hence, in order to prove it for all $q \in \QQ$ , it suffices to prove it for infinitely many values of $q$ (because an equality between two polynomials in $q$ that holds for infinitely many values of $q$ must automatically hold for all $q \in \QQ$ ). Thus, in particular, it suffices to prove it for all nonzero $q \in \QQ$ .
Thus, let us WLOG assume that $q$ is nonzero.
Define two $P \times P$ -matrices \begin{align*}
E &= \tup{ \tup{q-1}^{n-\abs{T}} q^{\abs{S}-n} \ive{ T \sups S } }_{\tup{S, T} \in P \times P} \qquad \text{and} \\
F &= \tup{ q^{\abs{T}} \ive{ S \sups T } }_{\tup{S, T} \in P \times P} .
\end{align*} Lemma 6 says that $D = EF$ . (More precisely, Lemma 6 says that the $\tup{S, T}$ -th entry of $D$ equals the corresponding entry of $EF$ for all $S \in P$ and $T \in P$ ; but this yields $D = EF$ .) Thus, $\det D = \det\tup{EF} = \det E \cdot \det F$ . Now, it remains to find $\det E$ and $\det F$ . Recall that $\abs{P} = 2^n - 1$ (as we proved during our proof of Lemma 4 (b) ). Also, $\sumS \abs{S} = n 2^{n-1}$ (by Lemma 4 (a) ). Thus, \begin{align}
\sumS \tup{n - \abs{S}}
&= \underbrace{\sumS n}_{= \abs{P} \cdot n} - \underbrace{\sumS \abs{S}}_{= n 2^{n-1}}
\\
&= \underbrace{\abs{P}}_{= 2^n - 1} \cdot n - n 2^{n-1} = \tup{2^n - 1} \cdot n - n 2^{n-1} \\
&= n \tup{2^n - 1 - 2^{n-1}} = n \tup{2^{n-1} - 1}
\end{align} (since $\underbrace{2^n}_{= 2 \cdot 2^{n-1}} - 1 - 2^{n-1} = 2 \cdot 2^{n-1} - 1 - 2^{n-1} = 2^{n-1} - 1$ ). Endow our set $P$ with the same total ordering that we used in the proof of Theorem 1. Now, it is easy to see that the $P \times P$ -matrix $E$ is upper-triangular. Since the determinant of an upper-triangular $P \times P$ -matrix equals the product of its diagonal entries, we thus conclude that \begin{align}
\det E
&= \prodS \tup{ \underbrace{\tup{q-1}^{n-\abs{S}} q^{\abs{S}-n}}_{= \tup{\dfrac{q-1}{q}}^{n-\abs{S}} } \underbrace{\ive{ S \sups S }}_{= 1} }
\\
&= \prodS \tup{\dfrac{q-1}{q}}^{n-\abs{S}}
= \tup{\dfrac{q-1}{q}}^{\sumS \tup{n - \abs{S}}}
= \tup{\dfrac{q-1}{q}}^{n 2^{n-1} - n}
\end{align} (since $\sumS \tup{n - \abs{S}} = n \tup{2^{n-1} - 1} = n 2^{n-1} - n$ ). Likewise, the $P \times P$ -matrix $F$ is lower-triangular. Since the determinant of a lower-triangular $P \times P$ -matrix equals the product of its diagonal entries, we thus conclude that \begin{align}
\det F
&= \prodS \tup{ q^{\abs{S}}
\underbrace{\ive{ S \sups S }}_{= 1} }
= \prodS q^{\abs{S}}
= q^{\sumS \abs{S}}
= q^{n 2^{n-1}}
\end{align} (since $\sumS \abs{S} = n 2^{n-1}$ ). Now, \begin{align}
\det D
&= \underbrace{\det E}_{= \tup{\dfrac{q-1}{q}}^{n 2^{n-1} - n}}
\cdot
\underbrace{\det F}_{= q^{n 2^{n-1}}}
= \tup{\dfrac{q-1}{q}}^{n 2^{n-1} - n} q^{n 2^{n-1}} \\
&= \dfrac{\tup{q-1}^{n 2^{n-1} - n}}{q^{n 2^{n-1} - n}} q^{n 2^{n-1}}
= \tup{q-1}^{n 2^{n-1} - n} q^{n 2^{n-1} - \tup{n 2^{n-1} - n}} \\
&= q^{n 2^{n-1} - \tup{n 2^{n-1} - n}} \tup{q-1}^{n 2^{n-1} - n}
= q^n \tup{q-1}^{n \tup{2^{n-1} - 1}}
\end{align} (since $n 2^{n-1} - \tup{n 2^{n-1} - n} = n$ and $n 2^{n-1} - n = n \tup{2^{n-1} - 1}$ ).
This proves Theorem 5. $\blacksquare$
|
{
"source": [
"https://mathoverflow.net/questions/317100",
"https://mathoverflow.net",
"https://mathoverflow.net/users/66131/"
]
}
|
317,222 |
Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?
|
This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point": Example. Let $k$ be an infinite field, let $R = k[x]$ , and for each $\alpha \in k$ let $R_\alpha = R[y_\alpha]/((x-\alpha)y_\alpha,y_\alpha^2)$ . Then $R_\alpha$ is an affine line with an embedded prime $\mathfrak p_\alpha = (x-\alpha,y_\alpha)$ at $x = \alpha$ , sticking out in the $y_\alpha$ -direction. Finally, let $$R_\infty = \bigotimes_{\alpha \in k} R_\alpha = \operatorname*{colim}_{\substack{\longrightarrow\\I \subseteq k\\\text{finite}}} \bigotimes_{\alpha \in I} R_\alpha$$ be their tensor product over $R$ ( not over $k$ ); that is $$R_\infty = \frac{k[x]\left[\{y_\alpha\}_{\alpha \in k}\right]}{\sum_{\alpha \in k}((x-\alpha)y_\alpha, y_\alpha^2)}.$$ This is not a Noetherian ring, because the radical $\mathfrak r = (\{y_\alpha\}_{\alpha \in k})$ is not finitely generated. But $\operatorname{Spec} R_\infty$ agrees as a topological space with $\operatorname{Spec} R_\infty^{\operatorname{red}} = \mathbb A^1_k$ , hence $|\!\operatorname{Spec} R_\infty|$ is a Noetherian topological space. On the other hand, the map $R \to R_\alpha$ is an isomorphism away from $\alpha$ , and similarly $R_\alpha \to R_\infty$ induces isomorphisms on the stalks at $\alpha$ . Thus, the stalk $(R_\infty)_{\mathfrak q_\alpha} = (R_\alpha)_{\mathfrak p_\alpha}$ at $\mathfrak q_\alpha = \mathfrak p_\alpha R_\infty + \mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $\mathfrak r$ is just $R_{(0)} = k(x)$ . Thus, we conclude that all the stalks of $R_\infty$ are Noetherian. $\square$ Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 \subseteq I_1 \ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$ . Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$ . Then you can try to use the Noetherian rings $\mathcal O_{X,x}$ for the generic points $x$ of $V$ . However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3 ] is that the inclusion $I_{i-1} \subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$ . Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$ . Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $\mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.
|
{
"source": [
"https://mathoverflow.net/questions/317222",
"https://mathoverflow.net",
"https://mathoverflow.net/users/52982/"
]
}
|
317,256 |
What would you do/have you done in such a situation? Hand out the improvement for free in your report Wait until the result is published and then submit elsewhere Inform the editor about the situation and ask for advice The paper is not posted publicly so contacting the authors directly informing them and asking what they want to do is out of the question.
|
Option (1) is definitely the professional course of action in this case. As pointed out in the remarks, it is likely to lead to an offer of co-authorship from the original author, but that is purely within the author's discretion. If you feel that your improvement is really substantial and you are worried about credit you can try to increase the chances of co-authorship by asking the editor to put you in contact with the author (after explaining the situation to the editor). You may then discuss this with the author directly and suggest co-authorship, a situation in which the author is more likely to accept (but they still may insist to refuse, in which case you should give them the idea "for free"). If your improvement is sufficiently significant and novel that not getting credit for it seems an unacceptable injustice, then what you can do is wait for the paper to be published (or accepted and online) and then write to the author with your idea of improvement and suggest co-authorship for a second paper. Here of course if they refuse you can publish alone. In any case, you should not submit your own paper without giving the original author a chance of co-authorship. That would be rather unprofessional.
|
{
"source": [
"https://mathoverflow.net/questions/317256",
"https://mathoverflow.net",
"https://mathoverflow.net/users/130882/"
]
}
|
317,376 |
Let $G$ be a finite solvable group of order $n$ , and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $\sum a_i$ is relatively prime to $n$ . Consider $\mathbb{C}[G]$ , the group ring of $G$ with complex coefficients. Does the element $\sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)
|
This is false for the cyclic group of order $6$ . Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$ -th root of $1$ , and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$ . Generalizing this example, the statement is false for the cyclic group of order $pq$ , with $p$ and $q$ two different primes. Let $g$ generate this group and let $\chi: G \to \mathbb{C}^{\ast}$ be a character with $\chi(g)$ a primitive $pq$ -th root of unity. Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$ -th cyclotomic polynomial. So the element $\Phi_{pq}(g)= \sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\Phi_{pq}(\chi(g)) =0$ on the representation $\chi$ . Thus $\sum c_k g^k$ is not a unit. On the other hand, $\sum c_k = \Phi_{pq}(1)= 1$ . (To compute the last, note that $\Phi_{pq}(x) = \frac{(x^{pq}-1)(x-1)}{(x^q-1)(x^p-1)}$ and take the limit as $x \to 1$ .) I claim further that, if $G$ has any element of non-prime-power order, then $G$ fails to have this condition. Let $g$ be an element of order $pq$ and let $\chi: \langle g \rangle \to \mathbb{C}^{\ast}$ be an injective character. Let $V = \mathrm{Ind}_{\langle g \rangle}^G \chi$ . Then $V$ restricted to $\langle g \rangle$ has $\chi$ as a summand, and this summand is in the kernel of $\Phi_{pq}(g)$ acting on $V$ . So $\Phi_{pq}(g)$ acts non-injectively on a representation of $G$ , and thus is not a unit. So the only groups for which this might be right are groups where every element has prime power order. These were classified by Higman , so you can dig into his paper if you care enough. On the positive side, the statement is true whenever $G$ is a $p$ -group. Let $\alpha = \sum c_g g \in \mathbb{Z}[G]$ . I will show that the determinant of $\alpha$ acting on $\mathbb{Z}[G]$ is $\left( \sum c_g \right)^{|G|} \bmod p$ , and hence is not $0$ if $\sum c_g \not \equiv 0 \bmod p$ . Reducing $\mathbb{Z}[G]$ modulo $p$ , we get an action of $\alpha$ on $\mathbb{F}_p[G]$ . More generally, I claim that $\alpha$ acts on any $G$ -representation $V$ over $\mathbb{F}_p$ by $\left( \sum c_g \right)^{\dim V}$ . This is simple: $V$ has a filtration whose associated graded is a $\dim V$ -dimensional trivial representation. Passing to the associated graded doesn't change determinant, and $\alpha$ acts on the $1$ -dimensional trivial representation by $\sum c_g$ .
|
{
"source": [
"https://mathoverflow.net/questions/317376",
"https://mathoverflow.net",
"https://mathoverflow.net/users/55916/"
]
}
|
317,643 |
The problem comes from a problem I encountered when I wrote the article Find all positive integer $m$ such $$2^{m}+1\mid5^m-1$$ it seem there no solution. I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.
If $m$ is odd then $2^m+1$ is divisible by 3 but $5^m-1$ is not.
so $m$ be even, take $m=2n$ , then $$4^n+1\mid25^n-1.$$ if $n$ is odd,then $4^n+1$ is divisible by $5$ , but $25^n-1$ is not
so $n$ is even,take $n=2p$ , we have $$16^p+1\mid625^p-1.$$
|
Here is a proof. Theorem. $2^m+1$ never divides $5^m-1$ . Assume that there is some $m$ such that $2^m+1$ divides $5^m-1$ . We already know that $m$ must be divisible by $4$ .
Let $m = 2^n a$ with an odd integer $a$ and $n \ge 2$ .
The $n$ th Fermat number $$F_n = 2^{2^n} + 1$$ is congruent
to $2$ mod $5$ (this uses $n \ge 2$ ), so it has a prime
divisor $p$ such that $$p \equiv \pm 2 \pmod 5.$$ We know that $p-1 = 2^{n+1}k$ for some integer $k$ .
Since $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1$ ,
we have that $$5^{2^n k} = 5^{(p-1)/2} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well. In particular, $$5^m \not\equiv 1 \pmod p.$$ On the other hand, $$2^m = (2^{2^n})^a \equiv (-1)^a = -1 \pmod p.$$ Thus $p$ divides $2^m+1$ , but does not divide $5^m-1$ ,
a contradiction.
|
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"source": [
"https://mathoverflow.net/questions/317643",
"https://mathoverflow.net",
"https://mathoverflow.net/users/38620/"
]
}
|
318,839 |
There are some questions on mathoverflow such as What out-of-print books would you like to see re-printed? Old books still used with answers that tell us things such as: Mathematicians prefer to use older books because of some old books are full of amazing ideas and some of them are comprehensive (such as books of Spivak). Question: What older books (with low quality typesetting) would you like to see reprinted with high quality typesetting? My question is not just a question. We are a group of math students (most of them are geometry students) that want to re-write popular old books using $\mathrm{\LaTeX}$ . One can search for most cited books such as: Curvature and Betti numbers (K Yano, S Bochner) or Einstein manifolds (AL Besse). Update: We have some rules: After sending $\LaTeX$ and PDF file of rewritten books to main author or publisher, we delete it from our computer. We never publish it anywhere on internet (If publisher or author give an answer for re-typing). We don't want to earn money by selling these books (If publisher or author didn't accept to pay for our work we have no way but creating a donation page after author or publisher approval).
|
I have some experience resurrecting old math books and I want to make a few comments about copyright. First, it is definitely true that except for very old books, someone owns the copyright. Typically it's the publisher, although sometimes it's the author. (If it's a collection of articles by multiple authors then the copyright may be shared in some complicated way.) In some cases, it's not actually clear who owns the copyright, e.g., because the publisher was bought out by another publisher and some of the paperwork was misplaced. But in any case, usually you should start by presuming that the publisher owns the copyright. What are the implications of copyright? First, there's really nothing stopping you from creating a $\mathrm{\LaTeX}$ version of a book for your own personal use. It's only when you want to post it on the web or share it with someone else that copyright issues rear their head. So one approach you can take is to do all the work, and then approach the copyright holder and hope that they will agree to publish your new version. Note that if you do this, then the copyright holder is under no obligation to pay you for your work or give you royalties or anything like that. Another possibility is to approach the copyright holder before doing any work and reach some sort of agreement ahead of time. The advantage of doing this is that you know what you are getting yourself into before you put in a lot of work. Be aware that even if the book gets republished and it sells well, you're unlikely to see much if any of that money. Either way, be aware that the copyright holder is under no obligation to do you any favors. If they elect not to republish the book then legally there's not much you can do about that. If you've already created the $\mathrm{\LaTeX}$ , they could demand that you hand it over ( EDIT in response to comments: Such a demand will typically not be legally enforceable but they may issue it anyway as an intimidation tactic), and if you comply, they may then sit on it without publishing it or releasing the copyright to anyone else. Having said all this, I don't mean to say that you shouldn't go ahead with your plans. I have successfully managed to get a couple of old math books republished. It was more work than I initially expected (even though I didn't have to do any typesetting) and I didn't ask for or receive a dime, but I did get the satisfaction of seeing the books resurrected. Finally, as others have already mentioned, if you're going to all this trouble then you might want to consider not just re-typesetting but also correcting as many errors as possible.
|
{
"source": [
"https://mathoverflow.net/questions/318839",
"https://mathoverflow.net",
"https://mathoverflow.net/users/90655/"
]
}
|
318,847 |
The set of real $n \times n$ matrices forms a vector space over the reals. Given any set $S$ of $n \times n$ matrices, there is a basis $S' \subseteq S$ of size at most $n^2$ such that any $x \in S \setminus S'$ can be written as a linear combination of elements in $S'$ . I am looking for a property $\Pi$ of matrices, such that any set of $n \times n$ matrices with property $\Pi$ has a basis of size $O(n)$ instead of $n^2$ . In other words: I'm looking for subspaces of the $n \times n$ matrices that have dimension $O(n)$ . Taking $\Pi$ to be the the property of being a diagonal matrix works. Similarly, taking $\Pi$ to be the property of 'the nonzeros are restricted to at most one column' works; or more generally, considering any subset of $O(n)$ positions and having $\Pi$ require that all other positions are zeros, works. However, I am looking for less restrictive properties that yield a basis of size $O(n)$ for 'more interesting reasons'. Taking $\Pi$ to be the positive semi-definite matrices does not work; the space of PSD matrices has dimension $\binom{n+1}{2}$ . Are there any nontrivial properties beyond diagonality that ensure dimension $O(n)$ ?
|
I have some experience resurrecting old math books and I want to make a few comments about copyright. First, it is definitely true that except for very old books, someone owns the copyright. Typically it's the publisher, although sometimes it's the author. (If it's a collection of articles by multiple authors then the copyright may be shared in some complicated way.) In some cases, it's not actually clear who owns the copyright, e.g., because the publisher was bought out by another publisher and some of the paperwork was misplaced. But in any case, usually you should start by presuming that the publisher owns the copyright. What are the implications of copyright? First, there's really nothing stopping you from creating a $\mathrm{\LaTeX}$ version of a book for your own personal use. It's only when you want to post it on the web or share it with someone else that copyright issues rear their head. So one approach you can take is to do all the work, and then approach the copyright holder and hope that they will agree to publish your new version. Note that if you do this, then the copyright holder is under no obligation to pay you for your work or give you royalties or anything like that. Another possibility is to approach the copyright holder before doing any work and reach some sort of agreement ahead of time. The advantage of doing this is that you know what you are getting yourself into before you put in a lot of work. Be aware that even if the book gets republished and it sells well, you're unlikely to see much if any of that money. Either way, be aware that the copyright holder is under no obligation to do you any favors. If they elect not to republish the book then legally there's not much you can do about that. If you've already created the $\mathrm{\LaTeX}$ , they could demand that you hand it over ( EDIT in response to comments: Such a demand will typically not be legally enforceable but they may issue it anyway as an intimidation tactic), and if you comply, they may then sit on it without publishing it or releasing the copyright to anyone else. Having said all this, I don't mean to say that you shouldn't go ahead with your plans. I have successfully managed to get a couple of old math books republished. It was more work than I initially expected (even though I didn't have to do any typesetting) and I didn't ask for or receive a dime, but I did get the satisfaction of seeing the books resurrected. Finally, as others have already mentioned, if you're going to all this trouble then you might want to consider not just re-typesetting but also correcting as many errors as possible.
|
{
"source": [
"https://mathoverflow.net/questions/318847",
"https://mathoverflow.net",
"https://mathoverflow.net/users/90005/"
]
}
|
319,805 |
First, I have to admit that I have already asked the same question on MSE several days ago. If I am bending any rules, I apologize for that and moderator can delete or close this question without warning. The problem has received a number of upvotes on MSE but with no suggestions or hints provided from the community. I need to prove the following: $$\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$$ ...with $p$ being an odd prime number. I'm pretty much sure that the statement is correct (I have tested it by computer for many values of $p$ and did not find an exception). What puzzles me is the fact that the statement is trivial and obviously true for $\pmod p$ . The left-hand side is congruent to $-1 \pmod p$ by Fermat's little theorem, and the right-hand side is also congruent to $-1 \pmod p$ by Wilson's theorem. However, I am unable to "lift the exponent" and go from $\pmod p$ to $\pmod {p^2}$ . Maybe the sum on the left could somehow be transformed using the existence of a primitive root $\pmod {p^2}$ . Thanks and happy holidays!
|
The result can be easily proved without using Bernoulli numbers. If $a$ and $b$ are integers not divisible by an odd prime $p$ , then \begin{align}(ab)^{p-1}-1=&b^{p-1}(a^{p-1}-1)+(b^{p-1}-1)
\\\equiv& (a^{p-1}-1)+(b^{p-1}-1)\pmod {p^2}.\end{align} Thus \begin{align*}\sum_{n=1}^{p-1}(n^{p-1}-1)\equiv& \prod_{n=1}^{p-1}n^{p-1}-1=((p-1)!+1-1)^{p-1}-1
\\\equiv &(p-1)((p-1)!+1)(-1)^{p-2}\equiv(p-1)!+1\pmod{p^2}\end{align*} and hence the desired congriuence follows.
|
{
"source": [
"https://mathoverflow.net/questions/319805",
"https://mathoverflow.net",
"https://mathoverflow.net/users/134054/"
]
}
|
319,980 |
In ${\rm\LaTeX}$ typesetting, when we repeat a long and complex formula in long documents, it is appropriate to create a new command that just by calling this new command we get the desired output. For example, I have used the following math expression in my previous document frequently: $$\{a^1,a^2,\ldots,a^n\}$$ For doing this in usual way, we need to press 22 keys on keyboard (and think about $(\frac{\partial}{\partial x^1}, \cdots,\frac{\partial}{\partial x^n})$ and other terrible formulas). Of course we can do this by copy and paste from similar one in the text. it is much better to define the following new command on preamble \newcommand{\set}[1]{\setaux#1\relax}
\def\setaux#1#2#3\relax{%
\{ {#1}#2 1,
\ifnum\pdfstrcmp{#3}{3}=0
{#1}#2 2
\else
\ldots
\fi
, {#1}#2{#3} \}
} and just by typing \set{a^n} in our text we get the same output. Question: What are your favorite ${\rm\LaTeX}$ tricks that save your time in long document typesetting?
|
Related to the question. I showed the following web-pages to a table-neighbor at a conference and literally got the "You just saved 3 days of my life" reaction. Doi2Bib ISBN2Bib Arxiv2Bib These web-sites are exactly as described: given say a DOI, it produces the bibtex entry for that particular article. I use this all the time.
|
{
"source": [
"https://mathoverflow.net/questions/319980",
"https://mathoverflow.net",
"https://mathoverflow.net/users/90655/"
]
}
|
320,019 |
The question below was posted on Mathematics Stack Exchange . It received no answer, and I do not expect any direct answer to it here. However, the question seems to me a natural one. Thus I wonder whether it has been posed before and, if so, whether there is any history to it. The question is: Is there any function $f:\Bbb R^2\to\Bbb R$ which has no representation in the following form? $$f(x,y)=\sum_{n=1}^{\infty} g_n(x)h_n(y)\quad(x,y\in \Bbb R).$$ As a candidate for such a function, I suggest $f(x,y)=\min\{x,y\}$ .
|
In "Representation of functions of two variables as sums of rectangular functions, I" , Roy O. Davies shows that under the continuum hypothesis every function has a representation of this form. More precisely, he shows that we can get a representation with the additional property that for all $x,y$ , the sum has only finitely many non-zero terms.
Conversely, he proves that if there is a representation with this additional property for $(x,y) \mapsto e^{xy}$ , then the continuum hypothesis holds. In his survey Set Theoretic Real Analysis (p. 18), Krzysztof Ciesielski mentions this problem (and a lot of interesting similar results), it seems that it is open whether or not the continuum hypothesis is equivalent to the existence of weak representations . It is not open, cf. Santi Spadaro's comment below, or Péter Komjáth's answer.
|
{
"source": [
"https://mathoverflow.net/questions/320019",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7458/"
]
}
|
320,190 |
Reading through Bhatt-Morrow-Scholze's "Topological Hochschild Homology and Integral p-adic Hodge Theory" I encountered the following statement. We use only “formal” properties of THH throughout the paper, with the
one exception of Bökstedt’s computation of $THH(\mathbb{F}_p)$ ... As a reminder, Bökstedt (and Breen?) computed $\pi_{*}THH(\mathbb{F}_p)$ as $\mathbb{F_p}[\sigma]$ , where $\sigma$ is a polynomial generator in degree 2. Whereas ordinary (derived) Hochschild homology of $\mathbb{F_p}$ is easily seen to be a divided power algebra generated in the same degree, $\mathbb{F}_p\langle x\rangle.$ Note that in the latter algebra, writing $x^p = p! \frac{x^p}{p!}$ gives $x^p = 0,$ quite a different flavor than the topological theory. At some point I looked this up and the computation appeared to rest on some not particularly formal spectral sequence manipulations which I was unable to follow. Is this still the situation? Has anyone revisited these computations recently?
|
Alright, here is the promised answer. First, the Hopkins-Mahowald theorem states that $\mathbb{F}_p$ is the free $\mathbb{E}_2$ -ring with $p=0$ , i.e. it is the homotopy pushout in $\mathbb{E}_2$ -rings of $S^0 \leftarrow \mathrm{Free}_{\mathbb{E}_2}(x) \rightarrow S^0$ where we have $x \mapsto 0$ for the first arrow and $x \mapsto p$ for the second arrow.
Equivalently (after $p$ -localization), if we let $S^1 \to \mathrm{BGL}_1(S^0_{(p)})$ be adjoint to $1-p \in \pi_0(S^0)_{(p)}^{\times}$ , extend to a double loop map $\Omega^2S^3 \to \mathrm{BGL}_1(S^0_{(p)})$ and take the Thom spectrum, we get $\mathrm{H}\mathbb{F}_p$ . At the prime 2 there are many references, and the original proof was due to Mahowald- it appears, for example, in his papers "A new infinite family in $\pi_*S^0$ " and in "Ring spectra which are Thom complexes". At odd primes this is due to Hopkins-Mahowald, but it's hard to trace down the "original" reference- it appears, for example, in Mahowald-Ravenel-Shick "The Thomified Eilenberg-Moore Spectral Sequence" as Lemma 3.3. In both cases the result is stated in the Thom spectrum language. And, again, let me reiterate: this is a non-formal result and essentially equivalent to Bokstedt's original computation. One always needs to know Steinberger's result (proved independently in Bokstedt's manuscript) that $Q_1\tau_i = \tau_{i+1}$ mod decomposables in $\mathcal{A}_*$ , where $Q_1$ is the top $\mathbb{E}_2$ -operation. (I suppose one could get away with slightly less: one needs to know that $\tau_0$ generates $\mathcal{A}_*$ with $\mathbb{E}_2$ -Dyer-Lashof operations). Anyway, given this result, let's see how to compute THH. In Blumberg-Cohen-Schlichtkrull ( https://arxiv.org/pdf/0811.0553.pdf ) they explain how to compute THH of any $\mathbb{E}_1$ -Thom spectrum, with increasingly nice answers as we add multiplicative structure. For an $\mathbb{E}_2$ -Thom spectrum (i.e. one arising from a double loop map) we have that $\mathrm{THH}(X^{\xi}) \simeq X^{\xi} \wedge BX^{\eta B\xi}$ where $\eta B\xi$ is the composite $BX \to B(BGL_1(S^0)) \stackrel{\eta}{\to} BGL_1(S^0)$ . So, in our case we have that $THH(\mathbb{F}_p) \simeq \mathbb{F}_p \wedge (\Omega S^3)^{\eta B\xi}$ . Thom spectra are colimits, and smash products commute with those, so we can compute this smash product as the Thom spectrum of $BX \to B(BGL_1(S^0)) \to BGL_1(S^0) \to BGL_1(\mathbb{F}_p)$ . But $\eta \mapsto 0$ in $\pi_*\mathrm{H}\mathbb{F}_p$ so this is the trivial bundle, and we deduce the result we're after: $\mathrm{THH}(\mathbb{F}_p) \simeq \mathbb{F}_p \wedge \Omega S^3_+ \simeq \mathbb{F}_p[x_2]$ . It's worth giving the same argument again but in a more algebraic way, and, as a bonus, you can basically see why the Blumberg-Cohen-Schlichtkrull result holds. The point (which is very well explained in a more general setting in Theorem 5.7 of Klang's paper here: https://arxiv.org/pdf/1606.03805.pdf ) is that $\mathbb{F}_p$ , as a module over $\mathbb{F}_p \wedge \mathbb{F}_p^{op}$ , is obtained by extending scalars from $S^0$ as an $S^0[\Omega S^3_+]$ -module. Indeed, start with the pushout of $\mathbb{E}_2$ -algebras $S^0 \leftarrow \mathrm{Free}_{\mathbb{E}_2}(x) \to S^0$ and smash with $\mathbb{F}_p$ to get a pushout of $\mathbb{E}_2$ - $\mathbb{F}_p$ -algebras. But now both maps in the pushout are the augmentation, and we see that the map $\mathbb{F}_p \wedge \mathbb{F}_p \simeq \mathbb{F}_p \wedge \mathbb{F}_p^{op} \to \mathbb{F}_p$ is equivalent to the map $\mathrm{Free}_{\mathbb{E}_2-\mathbb{F}_p}(\Sigma x) \to \mathbb{F}_p$ which is just the augmentation. This is tensored up from the map $\mathrm{Free}_{\mathbb{E}_2}(\Sigma x) \to S^0$ , so we get that $\mathrm{THH}(\mathbb{F}_p) \simeq \mathbb{F}_p \otimes_{\mathbb{F}_p \wedge \mathbb{F}_p} \mathbb{F}_p \simeq S^0 \otimes_{\mathrm{Free}_{\mathbb{E}_2}(\Sigma x)} \mathbb{F}_p \simeq S^0 \otimes_{\mathrm{Free}_{\mathbb{E}_2}(\Sigma x)} S^0 \otimes_{S^0} \mathbb{F}_p$ . The left hand factor is given by $\mathrm{Free}_{\mathbb{E}_1}(\Sigma^2x)$ , which you can see in various ways. For example, this bar construction is the suspension spectrum of the relative tensor product in spaces $* \otimes_{\Omega^2 S^3} *$ , which is the classifying space construction and yields $B(\Omega^2S^3) \simeq \Omega S^3$ .
|
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"https://mathoverflow.net/questions/320190",
"https://mathoverflow.net",
"https://mathoverflow.net/users/134242/"
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|
320,210 |
I am adding some context here. I am reading Introduction to Differentiable Stacks by Gregory Ginot. In page no $7$ , just before the remark $2.2$ he says the following. One shall be careful that fibre product of differentiable stacks is not necessarily a differentiable stack (though it is a stack over $\text{Diff}$ ). So, one question we can think of is, when does a fibre product of differentiable stacks a differentiable stack? This seems to be too difficult to answer. So, I look for a simpler question. When does $B\mathcal{G}\times_{B\mathcal{H}} B\mathcal{G}$ a differentiable stack where the fibre product is coming from a given morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ . Question in context of Lie groups is as follows. Suppose $G,H$ be Lie groups and $F:BG\rightarrow BH$ be a map of stacks. When can we say that the fibered product $BG\times_{BH}BG$ is a differentiable stack i.e., of the form $B\mathcal{K}$ for some Lie groupoid $\mathcal{K}$ ? Assume that this $F:BG\rightarrow BH$ is coming from a morphism of Lie groups $\theta:G\rightarrow H$ . Then, we can ask following question. When can we say that the fibered product $BG\times_{BH}BG$ is a differentiable stack coming from Lie group i.e., of the form $BK$ for some Lie group $K$ ? Any comments are welcome.
|
Alright, here is the promised answer. First, the Hopkins-Mahowald theorem states that $\mathbb{F}_p$ is the free $\mathbb{E}_2$ -ring with $p=0$ , i.e. it is the homotopy pushout in $\mathbb{E}_2$ -rings of $S^0 \leftarrow \mathrm{Free}_{\mathbb{E}_2}(x) \rightarrow S^0$ where we have $x \mapsto 0$ for the first arrow and $x \mapsto p$ for the second arrow.
Equivalently (after $p$ -localization), if we let $S^1 \to \mathrm{BGL}_1(S^0_{(p)})$ be adjoint to $1-p \in \pi_0(S^0)_{(p)}^{\times}$ , extend to a double loop map $\Omega^2S^3 \to \mathrm{BGL}_1(S^0_{(p)})$ and take the Thom spectrum, we get $\mathrm{H}\mathbb{F}_p$ . At the prime 2 there are many references, and the original proof was due to Mahowald- it appears, for example, in his papers "A new infinite family in $\pi_*S^0$ " and in "Ring spectra which are Thom complexes". At odd primes this is due to Hopkins-Mahowald, but it's hard to trace down the "original" reference- it appears, for example, in Mahowald-Ravenel-Shick "The Thomified Eilenberg-Moore Spectral Sequence" as Lemma 3.3. In both cases the result is stated in the Thom spectrum language. And, again, let me reiterate: this is a non-formal result and essentially equivalent to Bokstedt's original computation. One always needs to know Steinberger's result (proved independently in Bokstedt's manuscript) that $Q_1\tau_i = \tau_{i+1}$ mod decomposables in $\mathcal{A}_*$ , where $Q_1$ is the top $\mathbb{E}_2$ -operation. (I suppose one could get away with slightly less: one needs to know that $\tau_0$ generates $\mathcal{A}_*$ with $\mathbb{E}_2$ -Dyer-Lashof operations). Anyway, given this result, let's see how to compute THH. In Blumberg-Cohen-Schlichtkrull ( https://arxiv.org/pdf/0811.0553.pdf ) they explain how to compute THH of any $\mathbb{E}_1$ -Thom spectrum, with increasingly nice answers as we add multiplicative structure. For an $\mathbb{E}_2$ -Thom spectrum (i.e. one arising from a double loop map) we have that $\mathrm{THH}(X^{\xi}) \simeq X^{\xi} \wedge BX^{\eta B\xi}$ where $\eta B\xi$ is the composite $BX \to B(BGL_1(S^0)) \stackrel{\eta}{\to} BGL_1(S^0)$ . So, in our case we have that $THH(\mathbb{F}_p) \simeq \mathbb{F}_p \wedge (\Omega S^3)^{\eta B\xi}$ . Thom spectra are colimits, and smash products commute with those, so we can compute this smash product as the Thom spectrum of $BX \to B(BGL_1(S^0)) \to BGL_1(S^0) \to BGL_1(\mathbb{F}_p)$ . But $\eta \mapsto 0$ in $\pi_*\mathrm{H}\mathbb{F}_p$ so this is the trivial bundle, and we deduce the result we're after: $\mathrm{THH}(\mathbb{F}_p) \simeq \mathbb{F}_p \wedge \Omega S^3_+ \simeq \mathbb{F}_p[x_2]$ . It's worth giving the same argument again but in a more algebraic way, and, as a bonus, you can basically see why the Blumberg-Cohen-Schlichtkrull result holds. The point (which is very well explained in a more general setting in Theorem 5.7 of Klang's paper here: https://arxiv.org/pdf/1606.03805.pdf ) is that $\mathbb{F}_p$ , as a module over $\mathbb{F}_p \wedge \mathbb{F}_p^{op}$ , is obtained by extending scalars from $S^0$ as an $S^0[\Omega S^3_+]$ -module. Indeed, start with the pushout of $\mathbb{E}_2$ -algebras $S^0 \leftarrow \mathrm{Free}_{\mathbb{E}_2}(x) \to S^0$ and smash with $\mathbb{F}_p$ to get a pushout of $\mathbb{E}_2$ - $\mathbb{F}_p$ -algebras. But now both maps in the pushout are the augmentation, and we see that the map $\mathbb{F}_p \wedge \mathbb{F}_p \simeq \mathbb{F}_p \wedge \mathbb{F}_p^{op} \to \mathbb{F}_p$ is equivalent to the map $\mathrm{Free}_{\mathbb{E}_2-\mathbb{F}_p}(\Sigma x) \to \mathbb{F}_p$ which is just the augmentation. This is tensored up from the map $\mathrm{Free}_{\mathbb{E}_2}(\Sigma x) \to S^0$ , so we get that $\mathrm{THH}(\mathbb{F}_p) \simeq \mathbb{F}_p \otimes_{\mathbb{F}_p \wedge \mathbb{F}_p} \mathbb{F}_p \simeq S^0 \otimes_{\mathrm{Free}_{\mathbb{E}_2}(\Sigma x)} \mathbb{F}_p \simeq S^0 \otimes_{\mathrm{Free}_{\mathbb{E}_2}(\Sigma x)} S^0 \otimes_{S^0} \mathbb{F}_p$ . The left hand factor is given by $\mathrm{Free}_{\mathbb{E}_1}(\Sigma^2x)$ , which you can see in various ways. For example, this bar construction is the suspension spectrum of the relative tensor product in spaces $* \otimes_{\Omega^2 S^3} *$ , which is the classifying space construction and yields $B(\Omega^2S^3) \simeq \Omega S^3$ .
|
{
"source": [
"https://mathoverflow.net/questions/320210",
"https://mathoverflow.net",
"https://mathoverflow.net/users/118688/"
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|
321,348 |
Making soufflé tonight, I wondered if the six yolks took on the
optimal circle packing configuration.
They do not. It is only with seven congruent circles that the optimal
packing places one in the center. Q .
Why don't the yolks in a bowl follow the optimal packing of congruent
circles in a circle? Six yolks in a bowl. Image from Wikipedia .
Optimal packings for $5,6,7$ circles.
|
The system doesn't try to minimise the radius of the enclosing circle, but its potential energy. We can idealise this as non-overlapping disks in a convex rotationally symmetric potential $V$ with $V(0) = 0$ . The configuration that was physically realised then has potential energy $5 V(d)$ (with $d$ the diameter of the yolks) while the configuration from Wikipedia would have potential energy $6 V(d)$ .
|
{
"source": [
"https://mathoverflow.net/questions/321348",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6094/"
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}
|
321,547 |
There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic). Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware). Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P). On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group: $$G_{(1,2)} = \langle a, b | a^{a^b}= a^2 \rangle$$ in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly. So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?
|
An earlier reference for groups with this property is J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78. There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 \subset E_1 \subset E_2 \subset \cdots$ , where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials. The above paper describes constructions of finitely presented groups $G_n$ for $n \ge 3$ , in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$ ,
|
{
"source": [
"https://mathoverflow.net/questions/321547",
"https://mathoverflow.net",
"https://mathoverflow.net/users/122275/"
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|
321,603 |
I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry . Looking at the part regarding the orientation of a topological $n$ -manifold $M^n$ , at page 341 we find the following well-known result, with its usual proof (Proposition 7.1): So far, so good. However, after five pages we find what follows: This makes me confused, for at least two reasons: Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$ -torsion part of the abelian group $G$ , so if $G$ is torsion-free (for instance, if $G=\mathbf{Z}$ ) then ${}_2G=0$ . This is clearly very different from the free-product $G \ast \mathbf{Z_2}$ (here $\ast$ seems to denote the free-product, see pages 158-159). Point 2. In Corollary 7.11, take $A=\{x\}$ and $G=\mathbf{Z}$ . Then, when $M$ is not orientable one finds $H_n(M, \, M-\{x\}, \, \mathbf{Z})=0$ , and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, \, M-\{x\}, \, \mathbf{Z})= \mathbf{Z}$ . Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?
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Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := \operatorname{Tor}_{\Bbb Z}(A, B)$ . Usually it is clear from the context whether free product or torsion product is meant.
|
{
"source": [
"https://mathoverflow.net/questions/321603",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7460/"
]
}
|
321,897 |
Let $\mathrm{Graph}$ be the category of simple, undirected graphs with graph homomorphisms . For any graphs $G, H$ we denote by $\text{Hom}(G, H)$ the set of graph homomorphisms $f:G\to H$ . (Note that $\text{Hom}(G, H)$ can be empty, for instance if $\chi(G) > \chi(H)$ .) We can make $\text{Hom}(G, H)$ into a graph by saying that $f, g: G\to H$ form an edge if and only if $\{f(v), g(v)\}\in E(H)$ for all $v\in V(G)$ . Is $\mathrm{Graph}$ cartesian-closed?
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There are many categories of graphs, so perhaps it's best to take a synoptic view (though far from exhaustive). The table below surveys several categories of directed multigraphs (DM), directed graphs (DG), undirected multigraphs (UM), and undirected graphs (UG). The asterisks indicate the ones which are not cartesian closed. There are many other interesting categories of graphs, some of which are mentioned in the comments. The category of simple graphs is (16) in the chart and list below. It is not cartesian closed. But a slight modification of this category -- (15) below -- with equivalent objects but where morphisms are allowed to collapse an edge to a point, is cartesian closed. However, the description of the internal hom given in the question is incorrect for any of the categories below (though it's a reasonable guess!). The internal hom $H^G$ has for vertices all functions $V(G) \to V(H)$ between vertex sets. An edge from $f$ to $g$ is a rule which assigns, to every edge $x \to y$ in $G$ , an edge $f(x) \to g(y)$ in $H$ (of course, in graphs which are not multigraphs, there is always at most one edge $f(x) \to g(y)$ to choose, so that this is just a condition). This should be compared to the internal hom in chain complexes, $G$ -representations, etc, where maps from the terminal / unit object are not quite the same as "points" or "elements" in the most obvious sense and the internal hom ends up bigger than one might think. The graph homomorphisms may be recovered from this by homming in from the terminal object. A subtle point is that e.g. if we dress up a simple graph to be a reflexive graph (moving from (16) to (15) or eqivalently (14) as suggested above), it's important to remember that we've done this when computing the internal hom via the above description, or when mapping in from the terminal object to recover the homomorphisms. \begin{array}{lcccc}
& DM & DG & UM & UG \\
\text{Loops allowed}
& 1 & 5 & 9* & 13\\
\text{Loops allowed and reflexive}
& 2 & 6 & 10* & 14\\
\text{strictly reflexive}
& 3* & 7 & 11* & 15 \\
\text{Loops not allowed}
& 4* & 8* & 12* & 16*
\end{array} The concepts of reflexive and strictly reflexive graphs are distinct for multigraphs, but the same for graphs, which I forgot when writing. Nevertheless, the following should be accurate, just with unnecessarily complicated discussion for (7) (equivalent to (6)) and (15) (equivalent to (14)). A particular salient point to keep in mind is: Even if you're interested in, say, simple graphs, the notion of graph homomorphism might not be the correct notion of morphism to use. Changing the morphisms may be equivalent to equipping your graph with a bit of "dummy data" and using the resulting standard notion of homomorphism sometimes results in a better-behaved notion of morphism. Warmup: We're going to systematically go through all these different categories of graphs. To motivate the approach, consider for example the category of (1) directed multigraphs where loops are allowed . Such a graph $G$ comprises two sets $V(G),E(G)$ of vertices and edges respectively, and two functions $E(G) \rightrightarrows V(G)$ , one of which sends an edge to its starting vertex, and the other to its ending vertex. This is data is nicely summarized by saying that we have a presheaf on the category $\mathcal I$ with two objects and two nonidentity morphisms which looks like this [1]: $$V \rightrightarrows E$$ Presheaves come with a natural notion of morphism, and it agrees with the usual notion of a graph homomorphism. So we see that the category of directed graphs where loops are allowed is equivalent to the category of presheaves on the category $\mathcal I$ . Therefore, as Dmitri Pavlov observes in the comments, it is cartesian closed. Categories of Presheaves: In the following, we'll consider presheaves on $\mathcal I$ as well as three other categories which we'll call $\mathcal R, \mathcal U, \mathcal T$ , each of which has two objects $V,E$ and a finite number of morphisms. The categories of presheaves on these categories will be denoted $Psh(\mathcal I), Psh(\mathcal R),Psh(\mathcal U),Psh(\mathcal T)$ . The category $\mathcal I$ which we've already seen, can alternatively be described as (a skeleton of) the category of linearly ordered sets of cardinality 1 or 2, and all injective order-preserving maps. This category has 2 objects and 2 nonidentity morphisms. A presheaf on $\mathcal I$ is a directed graph. The category $\mathcal R$ , (a skeleton of) the category of linearly ordered sets of cardinality 1 or 2 and all order-preserving maps. This category has 2 objects and 5 nonidentity morphisms. A presheaf on $\mathcal R$ is a directed graph equipped with a distinguished loop at each vertex (and homomorphisms are required to carry distinguished loops to distinguished loops). Such a graph is called a (2) reflexive, directed multigraph where loops are allowed . The category $\mathcal U$ , (a skeleton of) the category of sets of cardinality 1 or 2 and all injective maps. This category has 2 objects and 3 nonidentity morphisms. A presheaf on $\mathcal U$ is a directed graph equipped with an involution on its edges which swaps the endpoints. We'll see that such structures are related to undirected multigraphs. The category $\mathcal T$ , (a skeleton of) the catgeory of sets of cardinality 1 or 2 and all maps. This category has 2 objects and 6 nonidentity morphisms. A presheaf on $\mathcal T$ is like a presheaf on $\mathcal U$ , in addition has a designated loop at each vertex which is fixed by the involution. We'll see that such structures are related to reflexive, undirected multigraphs. Generalities on Presheaf Categories: We will use the following facts: Every presheaf category is cartesian closed. Let $\mathcal D$ be a full reflective subcategory of a presheaf category $Psh(\mathcal C)$ containing the representable presheaves. Then $\mathcal D$ is cartesian closed if and only if $\mathcal D$ is an exponential ideal (i.e. $Y \in \mathcal D, X \in Psh(\mathcal C) \Rightarrow Y^X \in \mathcal D$ ). Let $S \subseteq Mor (Psh(\mathcal C))$ be a set of morphisms, and let $\mathcal D \subseteq Psh(\mathcal C)$ be the right orthogonal complement to $S$ . Suppose that $\mathcal D$ contains the representables. Then $\mathcal D$ is cartesian closed if and only if, for every representable $c$ , every $i \in I$ , and every $Y \in \mathcal D$ , $Y$ is right orthogonal to $i \times 1_c$ . Below, all the categories we consider are full subcategories of presheaf categories conatining the representables (there are always just two representables -- the free point $V$ and the free edge $E$ ) and defined by explicit orthogonality conditions. In each case, to check the last condition for cartesian closedness, it will suffice to consider the case where $c$ is the free edge $E$ . This can always be done by hand. Since it's hard to create the appropriate images and to post them here, I will just report what the verdict ends up being for cartesian closure. Directed Graphs as Presheaves: As we've seen, the category of directed multigraphs with loops allowed and homomorphisms is just $Psh(\mathcal I)$ . As a presheaf category, it is cartesian closed. This is the category theorist's default notion of "graph". As we've seen, the category of reflexive directed multigraphs with loops allowed is just $Psh(\mathcal R)$ . As a presheaf category, it is cartesian closed. This type of graph is also commonly used in category theory. The category of strictly reflexive directed multigraphs is the full subcategory of $Psh(\mathcal R)$ where there are no loops besides the designated ones (whenver I say "strictly reflexive" in this answer, that is what I mean -- reflexive, and the only loops are the designated ones). This is the right orthogonal complement to the map from the free loop to the terminal object. The product of this map with the free edge is not orthogonal to all strictly reflexive multigraphs, so strictly reflexive multigraphs are not cartesian closed. The category of loop-free directed multigraphs may be regarded as a full subcategory of the category of directed multigraphs where loops are allowed. It lacks a terminal object, so is not cartesian closed. Note that the objects of (3) and (4) are equivalent -- even if you don't like loops, you might pretend they're there for the sake of defining morphisms. The category of directed graphs where loops are allowed is the full subcategory of directed multigraphs which is right orthogonal to the map from the graph that looks like this $\bullet \rightrightarrows \bullet$ to the free edge $\bullet \to \bullet$ . This is an exponential ideal, so this category is cartesian closed. The category of reflexive directed graphs where loops are allowed is the full subcategory of reflexive directed multigraphs which is right orthogonal to a similar map as in (5) (if we take the convention that designated loops are suppressed in drawings). This is an exponential ideal, so this category is cartesian closed. The category of strictly reflexive directed graphs is the full subcategory of reflexive directed multigraphs which is right orthogonal to both the morphism of (6) and the morphism of (3). The morphism of (3), after taking the product with the free edge, was not orthogonal to every strictly reflexive directed multigraph, but it is orthogonal to every strictly reflexive directed graph. The other morphism also works out as in (6), so this category is an exponential ideal and hence cartesian closed. The category of loop-free directed graphs may be regarded as a full subcategory of the category of directed graphs. It lacks a terminal object, and so is not cartesian closed. Note again that the objects of (7) and (8) are equivalent. I think graph theorists are generally more likely to think about graphs as described in (8), but this time the trick of changing the morphisms to those of (7) buys us cartesian closure. Undirected Graphs as Presheaves: The category of undirected multigraphs where loops are allowed may be embedded into $Psh(\mathcal U)$ in at least two ways. The most obvious way is to do the obvious thing for vertices, and for each undirected edge, to put in two directed edges, exchanged by the involution. Unfortunately, this yields "spurious" morphisms which exchange the two copies we get of each loop. To fix this, we modify the construction by putting in only one directed edge for each loop, and have it fixed by the involution. This gives the right notion of morphism. Thus undirected multigraphs where loops are allowed form the full subcateory of $Psh(\mathcal U)$ right orthogonal to the map from the free loop (which is a point with two loops exchanged by the involution) to the terminal object (which is a point with a loop which is fixed by the involution). This is not an exponential ideal, and hence not cartesian closed. The category of reflexive undirected multigraphs where loops are allowed is the full subcategory of $Psh(\mathcal T)$ defined by a similar orthogonality condition to (9). This category is not an exponential ideal, and so not cartesian closed. The category of strictly reflexive undirected multigraphs is the full subcategory of $Psh(\mathcal T)$ satisfying the orthogonality condition of (10) as well as one similar to (3). It is not an exponential ideal, and so not cartesian closed. The category of loop-free undirected multigraphs is the full subcategory of $Psh(\mathcal T)$ satisfying the orthogonality conditions of (11), plus one similar to (4). This category lacks a terminal object, and so is not cartesian closed. Again the objects of (11) and (12) are equivalent. The category of undirected graphs where loops are allowed is the full subcategory of $Psh(\mathcal U)$ which is right orthogonal to the map of (9) plus a map analogous to (5). As in (7), the non-multi-ness comes to the rescue -- this is an exponential ideal, and hence cartesian closed. The category of reflexive undirected graphs where loops are allowed is the full subcategory of $Psh(\mathcal T)$ which is right orthogonal to maps analogous to (13). Similarly to (13), this is an exponential ideal, and hence cartesian closed. The category of strictly reflexive undirected graphs is the full subcategory of $Psh(\mathcal T)$ which is right orthogonal to the maps of (14), plus the map of (11). This is an exponential ideal, and hence cartesian closed. The category of loop-free undirected graphs may be regarded as a full subcategory of the category of undirected graphs where loops are allowed. This category lacks a terminal object, and hence is not cartesian closed. (16) is the category of the question. As Wojowu observed in the comments, it lacks a terminal object, and so is not cartesian closed for trivial reasons. But (15) has equivalent objects to (16), with the morphisms differing only in that we allow an edge to be contracted to a point. With this change, the category becomes cartesian closed. [1] Note the reversal in direction of the arrows because a presheaf is a functor $(V\rightrightarrows E)^{op} \to Set$ -- the reversal comes from the "op". The "op" is a matter of convention, but in this case is nice because it allows the definitions of the categories we use to be a bit slicker.
|
{
"source": [
"https://mathoverflow.net/questions/321897",
"https://mathoverflow.net",
"https://mathoverflow.net/users/8628/"
]
}
|
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