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https://mathoverflow.net/questions/9134 | 43 | Thinking of arbitrary tensor products of rings, $A=\otimes\_i A\_i$ ($i\in I$, an arbitrary index set), I have recently realized that $Spec(A)$ should be the product of the schemes $Spec(A\_i)$, a priori in the category of affine schemes, but actually in the category of schemes, thanks to the string of equalities (where $X$ is a not necessarily affine scheme)
$$ Hom\_{Schemes} (X, Spec(A))= Hom\_{Rings}(A,\Gamma(X,\mathcal O))=\prod\_ {i\in I}Hom\_{Rings}(A\_i,\Gamma(X,\mathcal O)) $$
$$ =\prod\_ {i\in I}Hom\_{Schemes}(X,Spec(A\_i))$$
Since this looks a little too easy, I was not quite convinced it was correct but a very reliable colleague of mine reassured me by explaining that the correct categorical interpretation of the more down to earth formula above is that the the category of affine schemes is a reflexive subcategory of the category of schemes.
(Naturally the incredibly category-savvy readers here know that perfectly well, but I didn't at all.)
And now I am stumped: I had always assumed that infinite products of schemes don't exist and I realize I have no idea why I thought so!
Since I am neither a psychologist nor a sociologist, arguments like "it would be mentioned in EGA if they always existed " don't particularly appeal to me and I would be very grateful if some reader could explain to me what is known about these infinite products.
| https://mathoverflow.net/users/450 | Arbitrary products of schemes don't exist, do they? | Let me rephrase the question (and Ilya's answer). Given an arbitrary collection $X\_i$ of schemes, is the functor (on affine schemes, say)
$Y \mapsto \prod\_i Hom(Y, X\_i)$
representable by a scheme? If the $X\_i$ are all affine, the answer is yes, as explained in the statement of the question. More generally, any filtered inverse system of schemes with essentially affine transition maps has an inverse limit in the category of schemes (this is in EGA IV.8). The topology in that case is the inverse limit topology, by the way.
It is easy to come up with examples of infinite products of non-separated schemes that are not representable by schemes. This is because any scheme has a locally closed diagonal. In other words, if $Y \rightrightarrows Z$ is a pair of maps of schemes then the locus in $Y$ where the two maps coincide is locally closed in $Y$.
Suppose $Z$ is the affine line with a doubled origin. Every distinguished open subset of an affine scheme $Y$ occurs as the locus where two maps $Y \rightrightarrows Z$ agree. Let $X = \prod\_{i = 1}^\infty Z$. Every countable intersection of distinguished open subsets of $Y$ occurs as the locus where two maps $Y \rightarrow X$ agree. Not every countable intersection of open subsets is locally closed, however, so $X$ cannot be a scheme.
Since the diagonal of an infinite product of separated schemes is closed, a more interesting question is whether an infinite product of separated schemes can be representable by a scheme. Ilya's example demonstrates that the answer is no.
Let $Z = \mathbf{A}^2 - 0$. This represents the functor that sends $Spec A$ to the set of pairs $(x,y) \in A^2$ generating the unit ideal. The infinite product $X = \prod\_{i = 1}^\infty Z$ represents the functor sending $A$ to the set of infinite collections of pairs $(x\_i, y\_i)$ generating the unit ideal. Let $B$ be the ring $\mathbf{Z}[x\_i, y\_i, a\_i, b\_i]\_{i = 1}^\infty / (a\_i x\_i + b\_i y\_i = 1)$. There is an obvious map $Spec B \rightarrow X$. Any (nonempty) open subfunctor $U$ of $X$ determines an open subfunctor of $Spec B$, and this must contain a distinguished open subset defined by the invertibility of some $f \in B$. Since $f$ can involve at most finitely many of the variables, the open subset determined by $f$ must contain the pre-image of some open subset $U'$ in $\prod\_{i \in I} Z$ for some finite set $I$. Let $I'$ be the complement of $I$. If we choose a closed point $t$ of $U'$ then $U$ contains the pre-image of $t$ as a closed subfunctor. Since the pre-image of $t$ is $\prod\_{i \in I'} Z \cong X$ this shows that any open subfunctor of $X$ contains $X$ as a closed subfunctor.
In particular, if $X$ is a scheme, any non-empty open affine contains a scheme isomorphic to $X$ as a closed subscheme. A closed subscheme of an affine scheme is affine, so if $X$ is a scheme it is affine.
Now we just have to show $X$ is not an affine scheme. It is a subfunctor of $W = \prod\_{i = 1}^\infty \mathbf{A}^2$, so if $X$ is an affine scheme, it is locally closed in $W$. Since $X$ is not contained in any closed subset of $W$ except $W$ itself, this means that $X$ is open in $W$. But then $X$ can be defined in $W$ using only finitely many of the variables, which is impossible.
**Edit:** Laurent Moret-Bailly pointed out in the comments below that my argument above for this last point doesn't make sense. Here is a revision: Suppose to the contrary that $X$ is an affine scheme. Then the morphism $p : X \rightarrow X$ that projects off a single factor is an affine morphism. If we restrict this map to a closed fiber then we recover the projection from $Z$ to a point, which is certainly not affine. Therefore $X$ could not have been affine in the first place.
| 37 | https://mathoverflow.net/users/32 | 9161 | 6,266 |
https://mathoverflow.net/questions/9037 | 76 | My apologies if this is too elementary, but it's been years since I heard of this paradox and I've never heard a satisfactory explanation. I've already tried it on my fair share of math Ph.D.'s, and some of them postulate that something deep is going on.
The Problem:
You are on a game show. The host has chosen two (integral and distinct) numbers and has hidden them behind doors A and B. He allows you to open one of the doors, thus revealing one of the numbers. Then, he asks you: is the number behind the other door bigger or smaller than the number you have revealed? Your task is to answer this question correctly with probability strictly greater than one half.
The Solution:
Before opening any doors, you choose a number $r$ at random using any continuous probability distribution of your choice. To simplify the analysis, you repeat until $r$ is non-integral. Then you open either door (choosing uniformly at random) to reveal a number $x$. If $r < x$, then you guess that the hidden number $y$ is also smaller than $x$; otherwise you guess that $y$ is greater than $x$.
Why is this a winning strategy? There are three cases:
1) $r$ is less than $x$ and $y$. In this case, you guess "smaller" and win the game if $x > y$. Because variables $x$ and $y$ were assigned to the hidden numbers uniformly at random, $P(x > y) = 1/2$. Thus, in this case you win with probability one half.
2) $r$ is greater than $x$ and $y$. By a symmetric argument to (1), you guess "larger" and win with probability one half.
3) $r$ is between $x$ and $y$. In this case, you guess "larger" if $x < y$ and "smaller" if $x > y$ -- that is, you always win the game.
Case 3 occurs with a finite non-zero probability $\epsilon$, equivalent to the integral of your probability distribution between $x$ and $y$. Averaging over all the cases, your chance of winning is $(1+\epsilon)/2$, which is strictly greater than half.
The Paradox:
Given that the original numbers were chosen "arbitrarily" (i.e., without using any given distribution), it seems impossible to know anything about the relation between one number and the other. Yet, the proof seems sound. I have some thoughts as to the culprit, but nothing completely satisfying.
Insightful members, could you please help me out with this one? Thanks!
| https://mathoverflow.net/users/2614 | How is it that you can guess if one of a pair of random numbers is larger with probability > 1/2? | After Bill's latest clarifications in the commentary on Critch's answer, I think the question is interesting again. My take:
One thing that always seemed to fall through the cracks when I learned about probability theory is that probability is intricately tied to information, and probabilities are only defined in the context of information. Probabilities aren't absolute; two people who have different information about an event may well disagree on its probability, even if both are perfectly rational. Similarly, if you get new information relevant to a certain event, then you should probably reevaluate what you think is the probability that it will occur. Your particular problem is interesting because the new information you get isn't enough for you to revise that probability by purely mathematical considerations, but I'll get to that in good time.
With the previous paragraph in mind, let's compare two games:
G1. You are given two closed doors, A and B, with two numbers behind them, and your goal is to choose the door with the higher number. You are given no information about the doors or numbers.
G2. You are given two closed doors, A and B, with two numbers behind them, and your goal is to choose the door with the higher number. **You are allowed to look behind one of the doors and *then* make your choice.**
For the first game, by symmetry, you clearly can't do better than choosing a door randomly, which gives you a success probability of exactly 1/2. However, the second game has a chance of being better. You are playing for the same goal with strictly more information, so you might expect to be able to do somewhat better. [I had originally said that it was obviously better, but now I'm not so sure that it's obvious.] The tricky thing is quantifying how much better, since it's not clear how to reason about the relationship between two numbers if you know one of the numbers and have no information about the other one. Indeed, it isn't even possible to quantify it mathematically.
"But how can that be?" you may ask. "This is a mathematical problem, so how can the solution not be mathematically definable?" There's the rub: part of the issue is that the problem *isn't* formulated in a mathematically rigorous way. That can be fixed in multiple ways, and any way we choose will make the paradox evaporate. The problem is that we're asked to reason about "the probability of answering the question correctly," but it's not clear what context that probability should be computed in. (Remember: probabilities aren't absolute.) In common probability theory problems and puzzles, this isn't an issue because there is usually an unambiguous "most general applicable context": we should obviously assume exactly what's given in the problem and nothing else. We can't do that here because the most general context, in which we assume nothing about how the numbers $x$ and $y$ are chosen, **does not define a probability space at all** and thus the "probability of answering the question correctly" is not a meaningful concept.
Here's a simpler ostensible probability question that exhibits the same fallacy: "what's the probability that a positive integer is greater than 1,000,000?" In order to answer that, we have to pick a probability distribution on the positive integers; the question is meaningless without specifying that.
As I said, there are multiple ways to fix this. Here are a couple:
I1. (Tyler's interpretation.) We really want the probability of answering the question correctly *given a particular $x$ and $y$* to be greater than 1/2. (The exact probability will of course depend on the two numbers.)
I2. (Critch's interpretation.) More generally, we want the probability of answering correctly *given a particular probability distribution for $(x,y)$* to be greater than 1/2. (The exact probability will of course depend on the distribution.)
(Those two are actually equivalent mathematically.) Clearly, if we knew what that distribution was, we could cook up strategies to get a success probability strictly above 1/2. That's pretty much obvious. It is not nearly as obvious that *a single* strategy (such as the one in the statement of the question) can work for *all* distributions of $(x,y)$, but it's true, as Bill's proof shows. It's an interesting fact, but hardly paradoxical now.
Let me summarize by giving proper mathematical interpretations of the informal statement "there is a strategy that answers the question correctly with probability strictly greater than 1/2," with quantifiers in place:
>
> (1a) $\exists \text{ strategy } S: \forall x, y: \exists \delta > 0$: $S$ answers correctly on $x$, $y$ with probability at least $1/2 + \delta$.
>
>
> (1b) $\exists \text{ strategy } S: \forall \text{ probability distributions } P \text{ on } \mathbb{N}^2: \exists \delta > 0$: $S$ answers correctly, when $x$, $y$ are chosen according to $P$, with probability at least $1/2 + \delta$.
>
>
>
I think with the proper quantifiers and the dependence on $x$ and $y$ explicit, it becomes a cool mathematical result rather than a paradox. Actually, based on my arguments at the beginning, it's not even that surprising: we *should* expect to do better than random guessing, since we are given information. However, simply knowing one number doesn't seem very useful in determining whether the other number is bigger, and that's reflected in the fact that we can't improve our probability by any fixed positive amount without more context.
**Edit:** It occurred to me that the last part of my discussion above has a nonstandard-analytical flavor. In fact, using the first version of the formula for simplicity (the two being equivalent), and the principle of idealisation, I think we immediately obtain:
>
> (2) $\exists \text{ strategy } S: \exists \delta > 0: \forall \text{ standard }x, y:$ $S$ answers correctly on $x$, $y$ with probability at least $1/2 + \delta$.
>
>
>
(Please correct me if I'm wrong.)
The number $\delta$ is not necessarily standard, and a basic argument shows that it must actually be smaller than all standard positive reals, i. e., infinitesimal. Thus, we can say that being able to look behind one door gives us an unquantifiably small, infinitesimal edge over random guessing. That actually meshes pretty well with my intuition! (It might still a nontrivial observation that the strategy $S$ can be taken standard; I'm not sure about that...)
| 54 | https://mathoverflow.net/users/302 | 9164 | 6,267 |
https://mathoverflow.net/questions/9166 | 7 | Given two vectors of size $n$
$$u = [u\_1, u\_2, u\_3, ..., u\_n ] $$
and
$$v = [v\_1, v\_2, v\_3, ..., v\_n ] $$
What is the name of the operation "$u ? v$" such that the result is a vector of size $n$ of the form
$$u ? v = [v\_1 \times u\_1, v\_2 \times u\_2, v\_3\times u\_3, ..., v\_n \times u\_n ]$$
For want of a better name, I have termed it "piecewise vector multiplication".
What is this operation normally known as in the literature?
| https://mathoverflow.net/users/2644 | Is there an existing name for "piecewise vector multiplication" | It's pointwise product. See Wikipedia articles [here](http://en.wikipedia.org/wiki/Matrix_multiplication#Hadamard_product%20%22here%22) and [here](http://en.wikipedia.org/wiki/Pointwise_product)
| 7 | https://mathoverflow.net/users/1888 | 9167 | 6,268 |
https://mathoverflow.net/questions/9129 | 12 | The following problem arose for my collaborators and me when studying the computational complexity of the Maximum-Cut problem.
Let $f : \mathbb{R} \to \mathbb{R}$ be an odd function. Let $\rho \in [0,1]$. Let $X$ and $Y$ be standard Gaussians with covariance $\rho$. Prove that $\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[f(X)^2 \mathrm{sgn}(X) \mathrm{sgn}(Y)]$.
The quantity on the left-hand side arises naturally in many contexts; e.g., it is the integral of $f(x)f(y)$ against the Mehler kernel (with parameter $\rho$).
I have some reason to believe this inequality is true. For one piece of evidence, suppose $f$'s range is $\pm 1$. Then the inequality reduces to
$\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[\mathrm{sgn}(X) \mathrm{sgn}(Y)]$;
i.e., it's saying that $\mathrm{sgn}$ is the $\pm 1$-valued odd function maximizing the LHS. This is indeed true; it follows from a result of Christer Borell ("Geometric bounds on the Ornstein-Uhlenbeck velocity process"), proved by Ehrhard symmetrization. It was also given a different proof by Beckner, deducing it from a rearrangment inequality on the sphere.
The second inequality generalizes to the case of functions $f : \mathbb{R}^n \to$ {$-1,1$}, but I believe the first inequality, which I would like to prove, is inherently $1$-dimensional.
Any ideas, or pointers to literature that might help?
Thanks!
| https://mathoverflow.net/users/658 | Inequality in Gaussian space -- possibly provable by rearrangement? | Using the antisymmetry of $f$ and $\mathrm{sgn}$ to bring the expectations to integral expressions
over $[0, \infty) \times [0, \infty)$, the first expectation takes the form:
$const\times \int f(x) f(y) \exp\left(-\frac{x^2+y^2}{2(1-\rho^2)}\right)\sinh\left(\frac{2\rho xy}{2(1-\rho^2)}\right) dx dy$
while for the second case $f(y)$ is replaced by $f(x)$.
The proof becomes an application of the Cauchy–Schwarz inequality.
| 5 | https://mathoverflow.net/users/1059 | 9175 | 6,273 |
https://mathoverflow.net/questions/9162 | 14 | I have posted this [elsewhere](http://www.mathlinks.ro/viewtopic.php?t=198806) and got only a partial reply. I don't know whether this qualifies the question for an open-problem tag; if it does, please anyone insert it.
>
> Let $L$ be a field, and $K$ a subfield of $L$. Let $n$ and $m$ be two nonnegative integers.
>
>
> For any $n\times n$ matrix $A \in \operatorname{M}\_n\left(K\right)$ and any $m\times m$-matrix $B \in \operatorname{M}\_m\left(K\right)$, and any field $S$ containing $K$, we define
> \begin{align}
> \rho\_{S}\left( A,B\right)
> := \max\left\{\operatorname{Rank} Q
> \mid Q\in\operatorname{M}\_{n,m}\left( S\right) ;
> \ AQ = QB\right\}.
> \end{align}
> We can call this number $\rho\_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$.
>
>
>
(Note that if $n = m$, then this conjugacy rank is directly connected with conjugacy -- i.e., similarity -- of matrices: Namely, in this case, we have $\rho\_{S}\left( A,B\right) = n$ if and only if the matrices $A$ and $B$ are conjugate to each other in the ring $\operatorname{M}\_{n}\left( S\right)$.)
>
> My question is: Do we have $\rho\_{K}\left( A,B\right) = \rho\_{L}\left( A,B\right)$ for any two matrices $A \in \operatorname{M}\_n\left(K\right)$ and $B \in \operatorname{M}\_m\left(K\right)$ ?
>
>
>
This can be shown in the case of $n = m \leq \left\vert K\right\vert$ by a "polynomials which vanish everywhere must be identically $0$" argument. Besides, in the case of $\rho\_{L}\left( A,B\right) = n = m$, it can be shown using the rational canonical form. I am interested in the most general case of the problem -- neither restricting $\left\vert K\right\vert$, nor $\rho\_{L}\left( A,B\right)$, nor requiring $n = m$.
What also might be of help: For any field $S$ containing $K$, the space
$$R\_{S}\left( A,B\right) = \left\{ Q\in\operatorname{M}\_{n,m}\left( S\right) \mid AQ = QB\right\}$$
is a subspace of the vector space $\operatorname{M}\_{n,m}\left(S\right)$.
Besides, every basis of the space $R\_{K}\left( A,B\right)$ is also a basis of the space $R\_{S}\left( A,B\right)$ for every field $S$ containing $K$. However, *this alone is not enough*; you can easily construct a subspace of $\operatorname{M}\_{n}\left(\mathbb{F}\_p\right)$ that consists of singular matrices only but loses this property when extended into $\operatorname{M}\_{n}\left(\mathbb{F}\_{p^2}\right)$.
| https://mathoverflow.net/users/2530 | "Conjugacy rank" of two matrices over field extension | I think this is true, and can be proved by brute force: write an explicit formula for conjugacy rank. I'll prefer to restate the problem in terms of modules.
To an $n\times n$ matrix $A$ over a field $K$, associate the $K[x]$-module $M$ that is $K^n$
as a vector space, while $x$ acts as $A$. Everywhere below, all $K[x]$-modules are finite-dimensional as $K$-vector spaces. Then your definition becomes as follows:
Let $M$ and $N$ be two $K[x]$-modules. Define their conjugacy rank $\rho(M,N)$ to be
the maximal dimension (over $K$) of a $K[x]$-module that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$. We aim to prove that $\rho(M,N)$ is stable under field extensions of $K$.
By structure theorem for modules over PID, we can write $M\simeq\bigoplus K[x]/f\_i$,
where invariant factors $f\_i=f\_i(M)\in K[x]$ satisfy $f\_{i+1}|f\_i$. (We set $f\_i = 1$ when $i$ is larger than the number of invariant factors.) It is easy to check the following claim:
Lemma: $M'$ is isomorphic to a quotient of $M$ if and only $f\_i(M')|f\_i(M)$. The same criterion holds for $M'$ being isomorphic to a submodule of $M$.
Corollary: There is unique up to isomorphism maximal-dimensional module $M'$ that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$; its invariant factors are given by $f\_i(M')=gcd(f\_i(M),f\_i(N))$.
Since the formula for $M'$ is stable under field extensions of $K$, the claim follows.
| 9 | https://mathoverflow.net/users/2653 | 9187 | 6,281 |
https://mathoverflow.net/questions/9185 | 14 | How do I feasibly generate a random sample from an $n$-dimensional $\ell\_p$ ball? Specifically, I'm interested in $p=1$ and large $n$. I'm looking for descriptions analogous to the statement for $p=2$: Take $n$ standard gaussian random variables and normalize.
| https://mathoverflow.net/users/825 | How to generate random points in $\ell_p$ balls? | For arbitrary p, [this paper](http://arxiv.org/abs/math/0503650) does exactly what you want. Specifically, pick $X\_1,\ldots,X\_n$ independently with density proportional to $\exp(-|x|^p)$, and $Y$ an independent exponential random variable with mean 1. Then the random vector
$$\frac{(X\_1,\ldots,X\_n)}{(Y+\sum |X\_i|^p)^{1/p}}$$
is uniformly distributed in the unit ball of $\ell\_p^n$.
The paper also shows how to generate certain other distributions on the $\ell\_p^n$ ball by modifying the distribution of $Y$.
| 30 | https://mathoverflow.net/users/1044 | 9192 | 6,285 |
https://mathoverflow.net/questions/9190 | 5 | There are lots of "Ext groups" in homological algebra which measure extensions of various things. I'm sure there must be a homological algebra machine for computing the following, and I'm hoping that someone out there knows about it.
I'm interested in the following situation. Let R and S be commutative rings and fix a ring homomorphism $f:R \to S$. Also fix a commutative S-algebra A. I'm interested in understanding/classifying those R-algebras B, together with a (surjective?) ring homomorphism $g: B \to A$ which intertwines the algebra structures in the following sense:
$ g(rb) = f(r) g(b)$
for all $r \in R$, and $ b \in B$. Is there a homological algebra way to do this?
A particular example that I am interested in is when we have the equality $B \otimes\_R S = A$, but I am also interested in other cases as well.
| https://mathoverflow.net/users/184 | Classifying Algebra Extensions over a fixed extension? | I think your problem is not constrained enough to have an interesting answer. Notice that your intertwining condition can be rephrased by saying that $g: B \to A$ is a homomorphism of $R$-algebras, where $A$ is given the structure of $R$-algebra given by $f$. In these terms, what you are looking for is the comma category $\mathcal{C} = (\mathbf{Alg}\_R \downarrow A)$, whose objects are precisely the pairs $(B, g)$ as above, and whose morphisms $\operatorname{Hom}\_{\mathcal{C}}((B, g), (B', g'))$ are the $R$-algebra homomorphisms $h: B \to B'$ such that $g = g' \circ h$. I am not sure it is possible to capture this beast with a cohomology group of any sort.
What you can do is restrict the class of objects that you are looking at. For example, you can classify square-zero extensions: fixing an $A$-module $I$, you can look at $R$-algebras $B$ such that you have a short exact sequence $0 \to I \to B \to A \to 0$; the name square-zero comes from the fact that $I$ is an ideal of $B$ with $I^2 = 0$. You can read about them in the first chapter of Sernesi's *Deformations of Algebraic Schemes*.
| 3 | https://mathoverflow.net/users/1797 | 9194 | 6,287 |
https://mathoverflow.net/questions/9209 | 6 | I thought about asking this question a while ago, but decided against it. But now I see a question about Eichler's "modular forms" quote, so while I guess it's probably still, um, questionable, what the hey.
So when Serre won the Fields Medal in 1954, Hermann Weyl (I guess) presented the award and described Serre's work. The Wikipedia article on Serre describes it thus: "...Weyl praised Serre in seemingly extravagant terms, and also made the point that the award was for the first time awarded to an algebraist."
If you're still not sure what I'm talking about, this is the speech where Weyl says something like "Never before have I seen such a rapid or bright ascension of a star in the mathematical sky as yours," if you've heard that quote.
Anyway, I've been trying to find a full version of Weyl's remarks (just out of curiosity), but to no avail. I would guess it would probably be in the congress proceedings somewhere, but I don't really have a copy on me. Anyone know where else I could find it?
| https://mathoverflow.net/users/382 | Where can I find the text of Weyl's Fields Medal speech for Serre? | Google Books snippet view shows you different little snippets of the speech. The beginning of the speechs has "by study and information we became convinced that Serre and Kodaira had not only made highly original and important..." If you do a search for the phrase "convinced that Serre" (with quotes) in Google Books, you'll find several books that have the speech: "International Mathematical Congresses: An Illustrated History", the ICM Proceedings, a French "Compte Rendu" of the Proceedings, and the "Gesammelte Abhandlungen" of Hermann Weyl. The search comes up empty elsewhere, which strongly suggests that you need to get one of these books from the library for the whole speech.
The Intelligencer has the last paragraph, which has been reprinted many times and is surely fair use:
>
> Here ends my report. If I omitted essential parts or misrepresented others, I ask for your pardon, Dr. Serre and Dr. Kodaira; it is not easy for an older man to follow your striding paces. Dear Kodaira: Your work has more than one connection with what I tried to do in my younger years; but you reached heights of which I never dreamt. Since you came to Princeton in 1949 it has been one of the greatest joys of my life to watch your mathematical development. I have no such close personal relation to you, Dr. Serre, and your research; but let me say this: that never before have I witnessed such a brilliant ascension of a star in the mathematical sky as yours. The mathematical community is proud of the work you both have done. It shows that the old gnarled tree of mathematics is still full of sap and life. Carry on as you began!
>
>
>
| 19 | https://mathoverflow.net/users/1450 | 9215 | 6,303 |
https://mathoverflow.net/questions/9221 | 17 | The standard way to show that a problem **is** NP-complete is to show that another problem known to be NP-complete reduces to it. That much is clear. Given a problem in NP, what's known about how to show that it is **not** NP-complete?
(My real question is likely to be inappropriate for this site for one or more reasons; I'm curious about what a proof that factoring isn't NP-complete would look like.)
| https://mathoverflow.net/users/290 | What techniques exist to show that a problem is not NP-complete? | There are much stronger versions of the P vs. NP conjecture that complexity theorists often take as axioms, and that imply that many problems are not NP-complete. The most standard class of assumptions is the conjecture that the NP hierarchy does not collapse. You can define NP as the analysis of polynomially bounded solitaire games with complete information. For instance, generalized Sudoku is such a game and it is known to be NP-complete. The nth level of the polynomial hierachy $\Sigma^n P$ and $\Pi^n P$ can be similarly defined by games with $n$ half moves. For instance, suppose that in generalized Sudoku, there is an initial board, and I can first try to make you lose by filling in some of the squares, with restrictions. (Like say only the "red" squares, and only with certain numbers.) After that you can move. Then whether I can win is a natural problem in $\Sigma^2 P$.
In particular, the assertion that the polynomial hierarchy PH does not collapse implies that NP does not equal co-NP. If a problem is both in NP and co-NP, it cannot be NP-complete without collapse. (Proof: If solitaire were equivalent to co-solitaire, than a game with two half moves would also be equivalent to solitaire.) A good near example is the graph isomorphism problem, which is in NP and co-AM. AM is like NP but with randomness; it is the model in which Arthur gets an adaptive proof in response to a randomized question and becomes statistically convinced. AM is not quite NP, but it is conjectured to be the same. So if you put two standard conjectures together, graph isomorphism is not NP-complete either. **Edit:** Ryan and Harrison both point out that [Boppana, Håstad, and Zachos](http://portal.acm.org/citation.cfm?id=31202) proved that if NP is contained in co-AM, then PH is contained in $\Sigma^2 P$. I.e., the hierarchy would collapse at the second level whether or not AM = NP. In particular this applies to graph isomorphism.
Problems in BQP, such as factoring, are strongly suspected not to be NP-complete either, but it is an open problem to show that that would imply that the polynomial hierarchy collapses. However, decision-based problems from factoring, such as whether a number is square-free, are known to be in both NP and co-NP. This was known earlier, but it follows particularly quickly from the fact that primality is in P, since that certifies a factorization. **Addendum:** Certification of factorization is equivalent to [certification of primality](http://en.wikipedia.org/wiki/Primality_certificate), which as David pointed was first proved by Vaughan Pratt.
| 11 | https://mathoverflow.net/users/1450 | 9225 | 6,309 |
https://mathoverflow.net/questions/9220 | 45 | Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series
$$ A(x) = \frac{x}{1 - e^{-x}} = \sum\_{m=0}^\infty \left( -\sum\_{n=1}^\infty \frac{(-x)^n}{(n+1)!} \right)^m = 1 + \frac12 x + \frac1{12}x^2 + 0x^3 - \frac1{720}x^4 + \dots $$
where I might have made an arithmetic error in expanding it out.
1. Are all the coefficients *egyptian*, in the sense that they are given by $A^{(n)}(0)/n! = 1/N$ for $N$ an integer? The answer is no, unless I made an error, e.g. the third coefficient. But maybe every non-zero coefficient is egyptian?
2. If all the coefficients were positive eqyptian, then the sequence of denominators might count something — one hopes that the $n$th element of any sequence of nonnegative integers counts the number of ways of putting some type of structure on an $n$-element set.
Of course, generating functions really come in two types: ordinary and exponential. The difference is whether you think of the coefficients as $\sum a\_n x^n$ or as $\sum A^{(n)} x^n/n!$. If it makes more sense as an exponential generating function, that's cool too.
So my question really is: is there a way of computing the $n$th coefficient of $A(x)$, or equivalently of computing $A^{(n)}(0)/n!$, without expanding products of power series the long way?
### Where you might have seen this series
Let $\xi,\psi$ be non-commuting variables over a field of characteristic $0$, and let $B(\xi,\psi) = \log(\exp \xi \exp \psi)$ be the Baker-Campbell-Hausdorff series. Fixing $\xi$ and thinking of this as a power series in $\psi$, it is given by
$$B(\xi,\psi) = \xi + A(\text{ad }\xi)(\psi) + O(\psi^2)$$
where $A$ is the series above, and $\text{ad }\xi$ is the linear operator given by the commutator: $(\text{ad }\xi)(\psi) = [\xi,\psi] = \xi\psi - \psi\xi$.
More generally, $B$ can be written entirely in terms of the commutator, and so makes sense as a $\mathfrak g$-valued power series on $\mathfrak g$ for any Lie algebra $\mathfrak g$. It converges in a neighborhood of $0$ when $\mathfrak g$ is finite-dimensional over $\mathbb R$, in which case $\mathfrak g$ is a (generally noncommutative) "partial group".
(More generally, you can consider the "formal group" of $\mathfrak g$. Namely, take the commutative ring $\mathcal P(\mathfrak g)$ of formal power series on $\mathfrak g$; then $B$ defines a non-cocommutative comultiplication, making $\mathcal P = \mathcal P(\mathfrak g)$ into a Hopf algebra. Or rather, $B(\mathcal P)$ does not land in the algebraic tensor product $\mathcal P \otimes \mathcal P$. Instead, $\mathcal P$ is *cofiltered*, in the sense that it is a limit $\dots \to \mathcal P\_2 \to \mathcal P\_1 \to \mathcal P\_0 = 0$, where (over characteristic 0, anyway) $\mathcal P\_n = \text{Poly}(\mathfrak g)/(\mathfrak g \text{Poly}(\mathfrak g))^n$, where $\text{Poly}(\mathfrak g)$ is the ring of polynomial functions on $\mathfrak g$, and $\mathfrak g \text{Poly}(\mathfrak g)$ is the ideal of functions vanishing at $0$. Then $B$ lands in the *cofiltered tensor product*, which is just what it sounds like. (In arbitrary characteristic, $\mathcal P$ is the cofiltered dual of the filtered Hopf algebra $\mathcal S \mathfrak g$, the symmetric algebra of $\mathfrak g$, filtered by degree.))
### Why I care
When $\mathfrak g$ is finite-dimensional over $\mathbb R$, and $U$ is the open neighborhood of $0$ in which $B$ converges, then $\mathfrak g$ acts as left-invariant derivations on $U$, where by *left-invariant* I mean under the multiplication $B$. Hence there is a canonical identification of the universal enveloping algebra $\mathcal U\mathfrak g$ with the algebra of left-invariant differential operators on $U$. Since $\mathfrak g$ is in particular a vector space, the "symbol" map gives a canonical identification between the algebra of differential operators on $U$ and the algebra of functions on the cotangent bundle $T^{\ast} U$ that are polynomial (of uniformly bounded degree) in the cotangent directions. Left-invariance then means that the operators are uniquely determined by their restrictions to the fiber $T^{\ast}\_0\mathfrak g = \mathfrak g^{\ast}$, and the space of polynomials on $\mathfrak g^{\ast}$ is canonically the symmetric algebra $\mathcal S \mathfrak g$. This gives a canonical PBW map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$, a fact I learned from J. Baez and J. Dolan.
(In the formal group language, the noncocommutative cofiltered Hopf algebra $\mathcal P(\mathfrak g)$ is precisely the cofiltered dual to the filtered algebra $\mathcal U\mathfrak g$, whereas with its cocommutative Hopf structure $\mathcal P(\mathfrak g)$ is dual to $\mathcal S \mathfrak g$. But as algebras these are *the same*, and unpacking the dualizations gives the PBW map $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$, and explains why it is actually an isomorphism of coalgebras.)
Anyway, in one direction, the isomorphism $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$ is easy. Namely, the map $\mathcal S \mathfrak g \to \mathcal U \mathfrak g$ is given on monomials by the "symmetrization map" $\xi\_1\cdots \xi\_n \mapsto \frac1{n!} \sum\_{\sigma \in S\_n} \prod\_{k=1}^n \xi\_{\sigma(k)}$, where $S\_n$ is the symmetric group on $n$ letters, and the product is ordered. (In this direction, the isomorphism of coalgebras is obvious. In fact, the corresponding symmetrization map into the full tensor algebra is a coalgebra homomorphism.)
In the reverse direction, I can explain the map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ as follows. On a monomial $\xi\_1\cdots \xi\_n$, it acts as follows. Draw $n$ dots on a line, and label them $\xi\_1,\dots,\xi\_n$. Draw arrows between the dots so that each arrow goes to the right (from a lower index to a higher index), and each dot has either 0 or 1 arrow out of it. At each dot, totally order the incoming arrows. Then for each such *diagram*, evaluate it as follows. What you want to do is collapse each arrow $\psi\to \phi$ into a dot labeled by $[\psi,\phi]$ at the spot that was $\phi$, but never collapse $\psi\to \phi$ unless $\psi$ has no incoming arrows, and if $\phi$ has multiple incoming arrows, collapse them following your chosen total ordering. So at the end of the day, you'll have some dots with no arrows left, each labeled by an element of $\mathfrak g$; multiply these elements together in $\mathcal S\mathfrak g$. Also, multiply each such element by a numerical coefficient as follows: for each dot in your original diagram, let $m$ be the number of incoming arrows, and multiply the final product by the $m$th coefficient of the power series $A(x)$. Sum over all diagrams.
Anyway, the previous paragraph is all well and cool, but it would be better if the numerical coefficient could be read more directly off the diagram somehow, without having to really think about the function $A(x)$.
| https://mathoverflow.net/users/78 | What does the generating function $x/(1 - e^{-x})$ count? | Two people have pointed it out already, but somehow I can't resist: your formal power series is precisely the defining power series of the Bernoulli numbers:
<http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function>
Accordingly, they are far from Egyptian: as came up recently in response to the question
[When does the zeta function take on integer values?](https://mathoverflow.net/questions/7586/when-does-the-zeta-function-take-on-integer-values)
the odd-numbered terms (except the first) are all zero, whereas the even-numbered terms alternate in sign and grow rapidly in absolute value, so only finitely many are reciprocals of integers.
I find it curious that you are looking at this sequence from such a sophisticated perspective and didn't know its classical roots. I feel like there should be a lesson here, but I don't know exactly what it is. Here's a possibility: every young mathematician should learn some elementary number theory regardless of their primary interests. Comments?
| 31 | https://mathoverflow.net/users/1149 | 9230 | 6,314 |
https://mathoverflow.net/questions/9224 | 1 | I know at least one method of constructing a convex valuation ring of rank $n$ (but it is rather complicated). What are the easiest methods of doing this? Given a natural number $n$ I want to have a valuation ring (preferably convex, defined below) whose rank is $n$. I have heard you can do this with polynomials and power series, but I am not really sure how this is done.
Let $A$ be a valuation ring with quotient field $K$. Suppose furthermore that $K$ is totally ordered by $<$. The ring $A$ is said to be convex if for any $x,y$ in $A$ and $z$ in $K$ with $x<z<y$ one has $z$ is in $A$.
| https://mathoverflow.net/users/1245 | Constructing a convex valuation ring/ordered group of rank $n$ | Here is a way to build a convex valuation ring with valuation group any ordered abelian group. This is inspired by Gerald Edgar's notes on transseries, which I learned about [here](https://mathoverflow.net/questions/3057/is-there-a-topology-on-growth-rates-of-functions/3105#3105). If we take $(G, \prec)$ to be $\mathbb{Z}^n$ ordered by lexicographic order, we will get a valuation of rank $n$.
Let $G$ be any ordered abelian group. Define $\mathbb{R}((G))$ to be the field of formal power series of the form
$$\sum\_{0 \leq k\_1, \ldots, k\_r \leq \infty} a\_{k\_1 \ldots k\_r} \ t^{h+k\_1 g\_1 +\cdots k\_r g\_r}.$$
Here $r$ can be any positive integer, $a\_{k\_1 \ldots k\_r}$ is any sequence of real numbers, $h$ is any element of $G$ and $g\_1$, ... $g\_r$ are any positive elements of $G$. The symbol $t$ is a formal parameter.
For example, $\mathbb{R}((\mathbb{Z}))$ is the field of formal Laurent series.
We add, multiply and divide such formal series by the obvious formal manipulations. We leave it to the reader to check that the sum or product of such power series is another such power series, as is the reciprocal of such a series, and that we never need to add more than finitely many coefficients together, so there are no issues of convergence.
There is a valuation $v:\mathbb{R}((G)) \to G$, sending a power series to the least $g$ for which $t^G$ appears with nonzero coefficient. The valuation ring of this valuation is clearly $\mathbb{R}[[G]]$; the ring of power series where $h$ (above) is nonnegative.
Finally, I have to tell you the ordering on $\mathbb{R}((G))$. A power series $f$ is positive if the coefficient of $t^{v(f)}$ is positive.
| 7 | https://mathoverflow.net/users/297 | 9231 | 6,315 |
https://mathoverflow.net/questions/9233 | 6 | Consider the basic axioms of planar incidence geometry, which allow us to speak of in-betweeness, collinearity and concurrency. These axioms per se are not complete, since for example, Desargues theorem may not always hold. in fact, Desargues theorem holds if and only if the model of incidence geometry can be coordinatized by a field, i.e. KP^2 serves as a model for some field K.
My question, then, is whether the theory of planar incidence geometry together with Desargues theorem is complete? (Call this theory IG + D)
If it is not, then what time is true in RP^2 (resp CP^2) that is indepedent of the theory IG + D?
| https://mathoverflow.net/users/nan | Is the theory of incidence geometry complete? | As Greg explains, the theory of projective planes obeying Desargues is basically equivalent to the theory of division rings, while the theory of projective planes obeying Desargues and Pappus as equivalent to the theory of fields. I haven't seen an axiomitization of projective planes with betweenness, but I assume that this would turn into the theory of ordered fields.
To finish the answer, one should say that none of these theories are complete. For example, the [Fano plane](http://en.wikipedia.org/wiki/Fano_plane) is realizable in $KP^2$ if $K$ has characteristic $2$, but not otherwise. There is an example, which I am too lazy to draw, of an arrangement of points and lines which is true in $KP^2$ if and only if $K$ contains a square root of $5$. Thus, $\mathbb{R}P^2$ can be distinguished from $\mathbb{Q}P^2$, even though both obey Desargues and Pappus, and presumably whatever axioms of betweenness you want to impose.
You should be able to adapt the proof of [Mnev's universality theorem](http://www.pdmi.ras.ru/~mnev/bhu.html) ([see also](http://arxiv.org/abs/math.AG/0411469)) in order to show that, if $K$ and $L$ are fields which can be distinguished by some first order property, then the projective planes $KP^2$ and $LP^2$ can be similarly distinguished.
| 9 | https://mathoverflow.net/users/297 | 9239 | 6,319 |
https://mathoverflow.net/questions/9234 | 1 | Reading "Monte Carlo Statistical Methods" by Robert and Casella, they mention that if
$f(x) = h(x) \exp(\langle \theta, x \rangle - A(\theta))$
defines a family of distributions for $X$, parametrized by $\theta$, then $A$ is the cumulant generating function of $h(X)$. It seems like this should be easy to prove if it's true, but I don't see how to proceed. Any ideas/references?
| https://mathoverflow.net/users/2586 | for a natural exponential family, A is the cumulant function of h? | Integrate with respect to x. The LHS is one. The RHS consists of the product two terms, one being the inverse of the exponential of the cumulant generating function of h, the other being $e^{-A(\theta)}$.
(Also plausible that I am asleep, in which case I apologize)
| 3 | https://mathoverflow.net/users/2282 | 9242 | 6,322 |
https://mathoverflow.net/questions/8933 | 7 | For quantum $\operatorname{SU}(2)$, Woronowicz gave a well differential calculus. If we denote the generators of quantum $\operatorname{SU}(2)$ by $a$, $b$, $c$, $d$, then the ideal of $\ker(\epsilon)$ corresponding to this calculus is
$$
\langle a+ q^2d - (1+q^2),b^2,c^2,bc,(a-1)b,(d-1)c\rangle.
$$
This calculus can be shown to generalise the classical calculus on $\operatorname{SU}(2)$ when $q=1$. Does anyone know of a (good) calculus (and its ideal) for quantum $\operatorname{SU}(3)$?
| https://mathoverflow.net/users/1867 | Is there a good differential calculus for quantum SU(3)? | I do not know whether it fits all of your requirements, but at least going by the abstract, some version of Woronowicz' result was generalized to all of the quantum groups of classical type in [Differential calculus on quantized simple Lie groups](https://doi.org/10.1007/BF00403543), by Branislav Jurčo.
| 3 | https://mathoverflow.net/users/1450 | 9243 | 6,323 |
https://mathoverflow.net/questions/9255 | 21 | How do I test whether a given undirected graph is the [1-skeleton of a polytope](https://en.wikipedia.org/wiki/Geometric_graph_theory)?
How can I tell the dimension of a given 1-skeleton?
| https://mathoverflow.net/users/2672 | Can you determine whether a graph is the 1-skeleton of a polytope? | A few comments:
In general, you can't tell the dimension of a polytope from its graph. For any $n \geq 6$, the complete graph $K\_n$ is the edge graph of both a $4$-dimensional and a $5$-dimensional polytope. (Thanks to dan petersen for correcting my typo.) The term for such polytopes is "neighborly".
On the other hand, you can say that the dimension is bounded above by the lowest vertex degree occurring anywhere in the graph.
A beautiful paper of [Gil Kalai](https://www.ams.org/mathscinet-getitem?mr=964396) shows that, given a $d$-regular graph, there is at most one way to realize it as the graph of a $d$-dimensional polytope, and gives an explicit algorithm for reconstructing that polytope. You could try running his algorithm on your graph. (Or a more efficient version recently found by [Friedman](https://www.ams.org/mathscinet-getitem?mr=2471873).) This algorithm will output some face lattice; that is to say, it will tell you which collections of vertices should be $2$-faces, which should be $3$-faces and so forth.
Unfortunately, going from the face lattice to the polytope is very hard. According to the MathSciNet review, [Richter-Gebert](https://www.ams.org/mathscinet-getitem?mr=1482230) has shown that it is NP-hard to, given a lattice of subsets of a finite set, decide whether it is the face lattice of a polytope. Note that this is a lower bound for the difficulty of your problem.
---
Let me be more explicit about the last statement. Richter-Gebert shows that, given a collection $L$ of subsets of $[n]$, it is NP-hard to determine whether there is a polytope with vertices labeled by $[n]$ whose edges, $2$-faces and $3$-faces are the given sets. (Here $[n] = \{ 1,2, \ldots, n \}$.)
Suppose we had an algorithm to decide whether a graph could be the edge graph of a polytope. Take our collection $L$ and look at the two-element sets within it. These form a graph with vertex set $[n]$. Run the algorithm on it. If the output is NO, then the answer to Richter-Gebert's problem is also no. If the answer is YES, then we have the problem that our algorithm might have found a polytope whose $2$-faces and $3$-faces differ from those prescribed by $L$. If our graph is $4$-regular, this problem doesn't come up by Kalai's result. But, not having read Richter-Gebert myself, I don't know whether the problem is still NP-hard when we restrict to $4$-regular graphs.
However, even if Richter-Gebert's result doesn't apply directly, I find it difficult to imagine that there could be an efficient algorithm to solve the graph realization problem, since there isn't one to solve the face lattice problem.
| 24 | https://mathoverflow.net/users/297 | 9268 | 6,342 |
https://mathoverflow.net/questions/9272 | 2 | Let K be an algebraically closed field of char. 0, let A\_n(K) be the Weyl algebra. Let I in A\_n(K) be a left ideal generated by p elements. Set M := A\_n / I.
Does the following then hold?
dim Ch(M) \geq 2n - p
Here Ch(M) is the characteristic variety of M. (I know that the answer is yes if n = 1, and also if I is generated by homogeneous elements.)
Thank you.
| https://mathoverflow.net/users/2682 | Lower bound for characteristic variety | The isn't true: there's a theorem of Stafford which says that any left ideal in the Weyl algebra is generated by two elements so if your claim was true, then the singular support would have to have dimension at least 2n-2 always, which of course isn't the case if n>2.
| 8 | https://mathoverflow.net/users/1878 | 9275 | 6,346 |
https://mathoverflow.net/questions/9267 | 0 | In principle a sequence in a non-Hausdorff space can converge to two points simultaneously.
Can anyone give me an explicit example of the above?
Or tell me any method of generating such kinds of examples?
| https://mathoverflow.net/users/2678 | What is an explicit example of a sequence converging to two different points? | Let $X = \mathbb{R} \setminus \{0 \} \cup \{ a,b\}$. Hence $X$ is the real line sans the origin with two points $a\neq b$, both not in $\mathbb{R}$, thrown in. The topology is generated by the open intervals in $\mathbb{R} \setminus \{0\}$ along with sets of the form $(u,0)\cup \{a\} \cup (0,v)$ and $(u,0)\cup \{b\} \cup (0,v)$, where $u < 0 < v$. $X$ is not Hausdorff because $a$ and $b$ cannot be separated by disjoint open sets. Every sequence that converges to $a$ also converges to $b$. Eg. $1/n \to a$ and $1/n \to b$.
| 12 | https://mathoverflow.net/users/2683 | 9282 | 6,351 |
https://mathoverflow.net/questions/9235 | 17 | Let $K\subseteq L$ be number fields over the field of rationals $\Bbb Q$.
with rings of integers $\mathcal{O}\_K\subseteq \mathcal{O}\_L$.
Let $P$ be a prime ideal of $\mathcal{O}\_L$, let $p$ be a prime ideal of $\mathcal{O}\_K$, such that $P$ is over $p$.
The residue class degree $f$ is defined to be $f=[\mathcal{O}\_L/P:\mathcal{O}\_K/p]$.
The norm of $P$ is the ideal $N(P)=p^f$.
This is the usual definition of norm of an ideal. (See Serre's *Local fields* and Serge Lang's *Algebraic tumber theory*.)
Swinnerton-Dyer's *A brief guide to algebraic number theory* has a different definition of norm of an ideal (page 25).
Namely if $A$ is an ideal of $\mathcal{O}\_L$, it is defined as $N(A)$ = ideal in $\mathcal{O}\_K$ generated by elements $N(a)$ where $a\in A$.
I don't know why these two definitions are the same. Swinnerton-Dyer claims
so in his book. Can anyone here give a hint, an explanation or anything
else?
| https://mathoverflow.net/users/2666 | A problem in algebraic number theory, norm of ideals | **[Edit: this answer is incomplete/incorrect, see rather [this one](https://mathoverflow.net/a/97720/14094)]**
Ok, here's the argument:
First recall that the usual norm for non-zero elements of a field is transitive in towers; thus the same is true for your second definition of the norm of an ideal. In particular, $N\_{K|Q}\circ N\_{L|K} = N\_{L|Q}$. The fact that the norm $N\_{L|Q}(\mathfrak{P}) = [\mathcal{O}\_L:\mathfrak{P}] \cdot \mathbb{Z}$ is easy to see for a prime $\mathfrak{P}$ in $\mathcal{O}\_L$; edit: and thus the same is true for any integral ideal $\mathfrak{a}$. Now let $\mathfrak{p} = \mathcal{O}\_K \cap \mathfrak{P}$ and $(p) = \mathbb{Z} \cap \mathfrak{P}$.
We have $N\_{L|Q}(\mathfrak{P}) = p^{f(\mathfrak{P}|p)} = N\_{K|Q}N\_{L|K}\mathfrak{P}$. In particular, we deduce that $N\_{L|K}\mathfrak{P} = \mathfrak{p}^d$ for some $d$. Moreover, we know that
$N\_{K|Q}\mathfrak{p}^d = p^{d \cdot f(\mathfrak{p}|p)} = p^{f(\mathfrak{P}|p)}.$
But then $d = f(\mathfrak{P}|p) / f(\mathfrak{p}|p) = f(\mathfrak{P}|\mathfrak{p})$ as required.
| 4 | https://mathoverflow.net/users/nan | 9286 | 6,354 |
https://mathoverflow.net/questions/9284 | 8 | Let H be a (finite-dimensional) Hermitian matrix with algebraic numbers for its entries, all of which lie in some minimal field extension of the rational numbers; call this field ℚ(H) for short. Let's assume that the eigenvalues of H are distinct, and let D be the diagonal matrix of eigenvalues of H in non-increasing order, say. Since H is Hermitian with a non-degenerate spectrum, there is a unique unitary matrix U that diagonalizes H. The normalized eigenvectors of H comprise the columns of U. Finally, let ℚ(U,D) be the field extension of the rationals containing the matrix elements of U and D.
Is there any relationship between ℚ(H) and ℚ(U,D)? In particular, I would like to know a bound on the degree of ℚ(U,D)/ℚ(H). Also if possible, is there a way to take H (without diagonalizing it!) and calculate ℚ(U,D)? I'd also be happy with something that contains ℚ(U,D) and isn't that much bigger. (I mean, its only bigger by a factor that is constant or depends on the dimension, not on the particular H.)
| https://mathoverflow.net/users/1171 | Field extension containing the eigenvectors of a Hermitian matrix | Do you expect something smaller than $n!\cdot 2^{n-1}$, where $n$ is the size of our matrix $H$ ? We can get $n!\cdot 2^n$ the following way: In order to get the entries of $U$ and $D$ into our field, we extend our field step by step: First, we adjoin all the eigenvalues of the matrix (they are roots of a polynomial of degree $n$, namely the characteristic polynomial of $H$, so we need an extension of degree $n!$ in the worst case), then find the corresponding eigenvectors using Gauss' elimination algorithm (this requires no field extension), and then normalize these eigenvectors (requiring a field extension of degree $2$ for each eigenvector, as we have to divide by a square root). There is a subtle point here - we need to know that during each step of our stepwise field extension, the field that we get is symmetric with respect to the real axis (i. e., for every complex number $z$ lying in our field, its conjugate $z^{\ast}$ lies in the field as well). This is clear for our initial field $\mathbb{Q}\left(H\right)$ (in fact, the matrix $H$ is Hermitian, so for each entry it contains, it also contains its conjugate), and remains correct during all the extension steps that we perform (in fact, each of these steps consists of adjoining \_all\_roots of some given *real* polynomial (in particular, the characteristic polynomial of $H$ is real, and adjoining the square root of the length of a vector means adjoining all roots of a real quadratic), and it is clear that such a step won't destroy the symmetry with respect to the $x$-axis). Adding the observation that only the first $n-1$ eigenvectors need a quadratic extension of the base field to be normalized (because the $n$-th vector can then be taken as the (generalized to $n$ dimensions) cross product of the first $n-1$ ones), we can lower the $n!\cdot 2^n$ bound to $n!\cdot 2^{n-1}$. At least for $n=2$, this is also optimal. Any ideas on higher $n$?
| 4 | https://mathoverflow.net/users/2530 | 9290 | 6,357 |
https://mathoverflow.net/questions/9297 | 2 | Let $S\overset{\pi}{\to} E$ be a ruled surface over an elliptic curve over complex field. Clearly, there are rational curves and elliptic curves on $S$. Is there any higher genus curves on $S$. Are all the elliptic curves isomorphic. The reason for the second question is that, all the smooth sections are isomorphic elliptic curves, except this sections is there any other elliptic curves?If so, how do I find them?
| https://mathoverflow.net/users/2348 | Curves on elliptic ruled surfaces? | Any surface has lots of curves of high genus. Just take a generic hypersurface section of high degree.
Any other elliptic curve will be isogenous to $E$ with $\pi$ inducing the isogeny. I think you can embed any isogenous curve in $S$.
| 6 | https://mathoverflow.net/users/2290 | 9304 | 6,365 |
https://mathoverflow.net/questions/9293 | -4 | The largest complete graph that embeds in 2 dimensions is $K\_4$, while the largest complete graph that embeds in 3 dimensions is $K\_{\infty}$, right? However, I don't know any constructive proof of it.
**Informal Explanation**:
What is the max number of points in $\mathbb{R}^3$, interconnected by lines of any curvature, such that no line intersects any other line? Each point is connected with all other points.
For $\mathbb{R}^2$ it is only 4 points (smth. like Mercedes symbol) - why 4 and not 3 or 5? How many points are possible to connect in such way in $\mathbb{R}^3$? (I suggest, infinite number, but it is interesting to look at a proof). What are some special properties of the Euclidean $\mathbb{R}^3$ such that the number of interconnected points jumps from 4 in $\mathbb{R}^2$ to infinity in $\mathbb{R}^3$?
**PS**: I don't understand why my question has got already 4 downvotes? No comments, no critics, why? English is not my native language, that's why?
| https://mathoverflow.net/users/2266 | What is the max number of points in R^3, interconnected by generic curves? | Take straight lines connecting the points $(t, t^2, t^3), t \in \mathbb{N}$. As far as I can tell you can also boost this to $t \in \mathbb{R}$. The point here is that two distinct lines between points on this curve intersect if and only if the four points involved lie on a plane (or there are only three points involved, but in that case you already know what the intersection is), but any plane $ax + by + cz = d$ intersects the curve in at most three points because $at + bt^2 + ct^3 = d$ has at most three roots.
| 14 | https://mathoverflow.net/users/290 | 9305 | 6,366 |
https://mathoverflow.net/questions/9158 | 2 | Let $\pi:X\to\mathcal X$ be a presentation of an Artin stack $\mathcal X$ of finite type over a field $k,$ and let $f:Y\to X$ be a finite \'etale covering. Does there exist a finite \'etale covering $Y'\to X$ factoring through $Y,$ such that $Y'$ can be given descent structure, i.e. there exists an isomorphism $pr\_1^\*Y'\cong pr\_2^\*Y'$ over $X\times\_{\mathcal X}X$ satisfying cocycle condition, so that $Y'\to X$ descend to a finite \'etale covering $\mathcal Y\to\mathcal X$?
| https://mathoverflow.net/users/370 | Descend finite etale algebras | I don't think so (finite etale covers cannot be localized in smooth topology in the sense that you describe). Say, $\mathcal{X}$ is a point, and $X$ is a smooth variety with non-trivial fundamental group, say, an elliptic curve (or $\mathbb{A}^1-\{0\}$). Then $\pi$ is a presentation. Let $f:Y\to X$ be a non-trivial finite etale cover, say, the cover of the elliptic curve by an isogeneous elliptic curve. Then your question becomes: `is there a trivial (i.e., lifted from $\mathcal{X}$) cover $Y'$ of $X$ with a map to $Y$? This is of course not true.
| 4 | https://mathoverflow.net/users/2653 | 9313 | 6,369 |
https://mathoverflow.net/questions/7772 | 10 | What are some good graduate level books on applied mathematics which explain in-depth the general modern problem-solving methods of the real-world typical hard problems?
There is a lot of books on numerical methods, engineering math, but I do not know any good modern book, which emphasizes algorithmic complexity of the discussed problems.
| https://mathoverflow.net/users/2266 | Applied mathematics Books (graduate level) | Since the question was tagged with "algorithms", I will give an algorithms recommendation. (You don't say specifically what type of problems you want to solve, but you do mention "algorithmic complexity.") For a book that was written to motivate the theory of algorithms from real-world problems, I would recommend Algorithm Design by Kleinberg and Tardos. It discusses many problem-solving methods. From the website for the book:
>
> Algorithm Design introduces algorithms by looking at the real-world problems that motivate them. In a clear, straight-forward style, Kleinberg and Tardos teaches students to analyze and define problems for themselves and from this, to recognize which design principles are appropriate for a given situation. The text encourages a greater understanding of the algorithm design process and an appreciation of the role of algorithms in the broader field of computer science.
>
>
>
Amazon link: [http://www.amazon.com/Algorithm-Design-Jon-Kleinberg/dp/0321295358](http://rads.stackoverflow.com/amzn/click/0321295358)
| 5 | https://mathoverflow.net/users/2618 | 9314 | 6,370 |
https://mathoverflow.net/questions/9309 | 24 | Recall the two following fundamental theorems of mathematical logic:
Completeness Theorem: A theory T is syntactically consistent -- i.e., for no statement P can the statement "P and (not P)" be formally deduced from T -- if and only if it is semantically consistent: i.e., there exists a model of T.
Compactness Theorem: A theory T is semantically consistent iff every finite subset of T is semantically consistent.
It is well-known that the Compactness Theorem is an almost immediate consequence of the
Completeness Theorem: assuming completeness, if T is inconsistent, then one can deduce
"P and (not P)" in a finite number of steps, hence using only finitely many sentences of T.
The traditional proof of the completeness theorem is rather long and tedious: for instance, the book *Models and Ultraproducts* by Bell and Slomson takes two chapters to establish it, and Marker's *Model Theory: An Introduction* omits the proof entirely. There is a quicker proof due to Henkin (it appears e.g. on Terry Tao's blog), but it is still relatively involved.
On the other hand, there is a short and elegant proof of the compactness theorem using ultraproducts (again given in Bell and Slomson).
So I wonder: can one deduce completeness from compactness by some argument which is easier than Henkin's proof of completeness?
As a remark, I believe that these two theorems are equivalent in a formal sense: i.e., they are each equivalent in ZF to the Boolean Prime Ideal Theorem. I am asking about a more informal notion of equivalence.
---
UPDATE: I accepted Joel David Hamkins' answer because it was interesting and informative. Nevertheless, I remain open to the possibility that (some reasonable particular version of) the completeness theorem can be easily deduced from compactness.
| https://mathoverflow.net/users/1149 | In model theory, does compactness easily imply completeness? | There are indeed many proofs of the Compactness theorem. Leo Harrington once told me that he used a different method of proof every time he taught the introductory graduate logic course at UC Berkeley. There is, of course, the proof via the Completeness Theorem, as well as proofs using ultrapowers, reduced products, Boolean-valued models and so on. (In my day, he used Boolean valued models, but that was some time ago, and I'm not sure if he was able to keep this up since then!)
Most model theorists today appear to regard the Compactness theorem as the significant theorem, since the focus is on the models---on what is true---rather than on what is provable in some syntactic system. (Proof-theorists, in contast, may focus on the Completeness theorem.) So it is not because Completness is too hard that Marker omits it, but rather just that Compactness is the important fact. Surely it is the Compactness theorem that has deep applications (or at least pervasive applications) in model theory. I don't think formal deductions appear in Marker's book at all.
But let's get to your question. Since the exact statement of the Completeness theorem depends on which syntactic proof system you set up---and there are a huge variety of such systems---any proof of the Completeness theorem will have to depend on those details. For example, you must specify which logical axioms are formally allowed, which deduction rules, and so on. The truth of the Completness Theorem depends very much on the details of how you set up your proof system, since if you omit an important rule or axiom, then your formal system will not be complete. But the Compactness theorem has nothing to do with these formal details. Thus, there cannot be hands-off proof of Completeness using Compactness, that does not engage in the details of the formal syntactic proof system. Any proof must establish some formal properties of the formal system, and once you are doing this, then the Henkin proof is not difficult (surely it fits on one or two pages). When I prove Completeness in my logic courses, I often remark to my students that the fact of the theorem is a foregone conclusion, because at any step of the proof, if we need our formal system to be able to make a certain kind of deduction or have a certain axiom, then we will simply add it if it isn't there already, in order to make the proof go through.
Nevertheless, Compactness can be viewed as an abstract Completness theorem. Namely, Compactness is precisely the assertion that if a theory is not satisfiable, then it is because of a finite obstacle in the theory that is not satisfiable. If we were to regard these finite obstacles as abstract formal "proofs of contradiction", then it would be true that if a theory has no proofs of contradiction, then it is satisfiable.
The difference between this abstract understanding and the actual Completness theorem, is that all the usual deduction systems are highly effective in the sense of being computable. That is, we can computably enumerate all the finite inconsistent theories by searching for formal syntactic proofs of contradiction. This is the new part of Completness that the abstract version from Compactness does not provide. But it is important, for example, in the subject of Computable Model Theory, where they prove computable analogues of the Completeness Theorem. For example, any consistent decidable theory (in a computable language) has a decidable model, since the usual Henkin proof of Completeness is effective when the theory is decidable.
---
Edit: I found in [Arnold Miller's lecture
notes](https://people.math.wisc.edu/%7Eawmille1/old/m771-98/logintro.pdf)
an entertaining account of an easy proof of (a fake version of) Completeness from Compactness (see page 58). His system amounts to the abstract formal system I describe above. Namely, he introduces the MM proof system (for Mickey Mouse),
where the axioms are *all* logical validities, and the only
rule of inference is Modus Ponens. In this system, one can
prove Completeness from Compactness easily as follows: We
want to show that $T$ proves φ if and only if every model
of $T$ is a model of φ. The forward direction is
Soundness, which is easy. Conversely, suppose that every
model of $T$ is a model of $\phi$. Thus, $T+\lnot\phi$ has no
models. By Compactness, there are finitely many axioms
$\phi\_0,\ldots,\phi\_n$ in $T$ such that
there is no model of them plus $\lnot\phi$. Thus,
$\phi\_0\land\ldots\land\phi\_n\implies\phi$ is a logical validity. And from this, one can easily
make a proof of $\phi$ from $T$ in MM. QED!
But of course, it is a joke proof system, since the
collection of validities is not computable, and Miller uses this example to illustrate the point as follows:
>
>
> >
> > The poor MM system went to the Wizard of OZ and said, “I
> > want to be more like all the other proof systems.” And the
> > Wizard replied, “You’ve got just about everything any other
> > proof system has and more. The completeness theorem is easy
> > to prove in your system. You have very few logical rules
> > and logical axioms. You lack only one thing. It is too hard
> > for mere mortals to gaze at a proof in your system and tell
> > whether it really is a proof. The difficulty comes from
> > taking all logical validities as your logical axioms.” The
> > Wizard went on to give MM a subset Val of logical
> > validities that is recursive and has the property that
> > every logical validity can be proved using only Modus
> > Ponens from Val.
> >
> >
> >
>
>
>
And he then goes on to describe how one might construct
Val, and give what amounts to a traditional proof of
Completeness.
| 37 | https://mathoverflow.net/users/1946 | 9317 | 6,371 |
https://mathoverflow.net/questions/9308 | 11 | I am asking because the literature seems to contain some inconsistencies as to the definition of a braided monoidal category, and I'd like to get it straight. According Chari and Pressley's book ``A guide to quantum groups," a braided monoidal category is a monoidal category $\mathcal{C}$ along with a natural system of isomorphisms $\sigma\_{U,V}: U \otimes V \rightarrow V \otimes U$ for all pairs of objects $U$ and $V$, such that
(i) The ``Hexagon" axioms (two commutative diagrams) hold.
(ii) The ``identity object" axioms: $\rho\_V= \lambda\_V \circ \sigma\_{{\bf 1},V}: {\bf 1} \otimes V \rightarrow V$
and
$\lambda\_V= \rho\_V \circ \sigma\_{V, {\bf 1}}: {V} \otimes {\bf 1} \rightarrow V$,
where $\lambda\_V$ and $\rho\_V$ are the isomorphisms of $V \otimes {\bf 1}$ and ${\bf 1} \otimes V$ with $V$ that are part of the definition of monoidal category. See Chari-Pressley Definitions 5.2.1 and 5.2.4. They use the term "quasitensor category," but note on p153 that the term "braided monoidal category" is equivalent.
However, in some references (ii) seems to have been dropped. I am thinking in particular of Definition 3.1 is this
[expository paper](http://arxiv.org/abs/0804.4688),
and the
[wikipedia article](http://en.wikipedia.org/w/index.php?title=Braided_monoidal_category&oldid=309075821).
The wikipedia article goes further, and suggests that (ii) somehow follows from (i) and the axioms of a monoidal category. So, my questions are.
1) Is (ii) needed? That is if we do not impose (ii), does it follow from (i) and the axioms of a monoidal category?
2) If (ii) is needed, can someone provide an example demonstrating why? That is, provide an example of a monoidal category $\mathcal{C}$ along with maps $\sigma\_{U,V}$ such that (i) holds but (ii) fails. Alternatively, if (ii) is not needed, I'd like a proof (or reference to a proof) that it follows from other axioms.
| https://mathoverflow.net/users/1799 | Are the “identity object axioms” in the definition of a braided monoidal category needed? (Answered: No) | This is Proposition 1 in the seminal paper "Braided Monoidal Categories" by Joyal and Street. Relation (ii) is implied by the others.
| 7 | https://mathoverflow.net/users/184 | 9329 | 6,380 |
https://mathoverflow.net/questions/9322 | 7 | Is there a definable (in Zermelo Fraenkel set theory with choice) collection of non measurable sets of reals of size continuum? More verbosely: Is there a class A = {x: \phi(x)} such that ZFC proves "A is a collection, of size continuum, consisting of non Lebesgue measurable subsets of reals"?
| https://mathoverflow.net/users/2689 | Definable collections of non measurable sets of reals | (Edit.) With a closer reading of your question, I see that you asked for a very specific notion of definability.
If you allow the family to have size larger than continuum, there is a trivial **Yes** answer. Namely, let $\phi(x)$ be the assertion "$x$ is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are at least continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that $\{x\mid\phi(x)\}$ is a family of non-measurable sets of size *at least* continuum.
But if you insist that the family have size exactly the continuum, as your question clearly states, then this trivial answer doesn't work. Indeed, one can't even take the class of all Vitali sets in this case, since there are $2^{\mathfrak c}$ many sets of reals that contain exactly one point from each equivalence class for rational translations.
Qiachu Yuan's suggestion about translations of a single Vitali set does have size continuum, but there is little reason to expect the Vitali set to be definable in the way that you have requested, and so it does not provide the desired definable family.
In my earlier posted answer, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.
I pointed out that it is surely consistent with ZFC that there is the desired definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals and no sets of reals. So you can take any family of non-measurable sets that you like and go to a forcing extension where this family is definable.
Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula $\phi(x,z)$, such that $A\_z=\{x\mid\phi(x,z)\}$ is always non-measurable for any $z$ and all $A\_z$ are different. Such a formula would be a strong example of the phenomenon you seek.
The first answer to this way of asking the question is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.
The second answer to this version of the question, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.
In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced.)
| 7 | https://mathoverflow.net/users/1946 | 9330 | 6,381 |
https://mathoverflow.net/questions/8938 | 16 | The Torelli map $\tau\colon M\_g \to A\_g$ sends a curve C to its Jacobian (along with the canonical principal polarization associated to C); see [this](https://mathoverflow.net/questions/7505/are-jacobians-principally-polarized-over-non-algebraically-closed-fields/7513#7513) question for a description which works for families.
>
> Theorem (Torelli): If $\tau(C) \cong \tau(C')$, then $C \cong C'$.
>
>
>
If one prefers to work with coarse spaces (instead of stacks) it is okay to just say that $\tau$ is injective.
>
> Question: Is $\tau$ an immersion?
>
>
>
(One remark: $\tau$ isn't a closed immersion -- the closure of its image consists of products of Jacobians!)
| https://mathoverflow.net/users/2 | Is the Torelli map an immersion? | Respectfully, I disagree with Tony's answer. The infinitesimal Torelli problem fails for $g>2$ at the points of $M\_g$ corresponding to the hyperelliptic curves. And in general the situation is trickier than one would expect.
The tangent space to the deformation space of a curve $C$ is $H^1(T\_C)$, and the tangent space to the deformation space of its Jacobian is $Sym^2(H^1(\mathcal O\_C))$. The infinitesimal Torelli map is an immersion iff the map of these tangent spaces
$$ H^1(T\_C) \to Sym^2( H^1(\mathcal O\_C) )$$
is an injection. Dually, the following map should be a surjection:
$$ Sym^2 ( H^0(K\_C) ) \to H^0( 2K\_C ), $$
where $K\_C$ denotes the canonical class of the curve $C$. This is a surjection iff $g=1,2$ or $g=3$ and $C$ is not hyperelliptic; by a result of Max Noether.
Therefore, for $g\ge 3$ the Torelli map OF STACKS $\tau:M\_g\to A\_g$ is not an immersion. It is an immersion outside of the hyperelliptic locus $H\_g$. Also, the restriction $\tau\_{H\_g}:H\_g\to A\_g$ is an immersion.
On the other hand, the Torelli map between the coarse moduli spaces IS an immersion in char 0. This is a result of Oort and Steenbrink "The local Torelli problem for algebraic curves" (1979).
F. Catanese gave a nice overview of the various flavors of Torelli maps (infinitesimal, local, global, generic) in "Infinitesimal Torelli problems and counterexamples to Torelli problems" (chapter 8 in "Topics in transcendental algebraic geometry" edited by Griffiths).
P.S. "Stacks" can be replaced everywhere by the "moduli spaces with level structure of level $l\ge3$ (which are fine moduli spaces).
P.P.S. The space of the first-order deformations of an abelian variety $A$ is $H^1(T\_A)$. Since $T\_A$ is a trivial vector bundle of rank $g$, and the cotangent space at the origin is $H^0(\Omega^1\_A)$, this space equals $H^1(\mathcal O\_A) \otimes H^0(\Omega^1\_A)^{\vee}$ and has dimension $g^2$.
A polarization is a homomorphism $\lambda:A\to A^t$ from $A$ to the dual abelian variety $A^t$. It induces an isomorphism (in char 0, or for a principal polarization) from the tangent space at the origin $T\_{A,0}=H^0(\Omega\_A^1)^{\vee}$ to the tangent space at the origin $T\_{A^t,0}=H^1(\mathcal O\_A)$. This gives an isomorphism
$$ H^1(\mathcal O\_A) \otimes H^0(\Omega^1\_A)^{\vee} \to
H^1(\mathcal O\_A) \otimes H^1(\mathcal O\_A). $$
The subspace of first-order deformations which preserve the polarization $\lambda$ can be identified with the tensors mapping to zero in $\wedge^2 H^1(\mathcal O\_A)$, and so is isomorphic to $Sym^2 H^1(\mathcal O\_A)$, outside of characteristic 2.
| 23 | https://mathoverflow.net/users/1784 | 9338 | 6,386 |
https://mathoverflow.net/questions/9335 | 16 | That is, for any symplectomorphism $\psi: D^2 \to D^2$, there should be a time-dependent Hamiltonian *Ht* on *D2* such that the corresponding flow at time 1 is equal to $\psi$.
I found this in claim a paper, and I think it should be easy, but nothing comes to mind. I'd be happy with a reference to a page in McDuff-Salomon, but I couldn't find this there immediately.
Thanks!
| https://mathoverflow.net/users/2467 | Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? | It is a theorem of Smale that the group of orientation-preserving diffeomorphisms of $D^2$, rel boundary, is contractible. If the diffeomorphisms can move the boundary, you can establish a homotopy equivalence between that and the circle. The diffeomorphisms do not have to preserve area. Then, a [theorem of Moser](https://mathoverflow.net/questions/7817/normal-coordinates-for-a-manifold-with-volume-form) establishes a deformation retract from diffeomorphisms to volume-preserving diffeomorphisms. Moser's result is easier to see if you have a closed manifold, but it extends to manifolds with boundary with the doubling trick. Together, this indirectly gives you a curve of symplectomorphisms connecting the identity to $\psi$, since in two dimensions the symplectic structure is just a volume structure. Finally if you have a smooth curve of area-preserving diffeomorphisms of a disk, I think there is a time-dependent Hamiltonian obtained by integrating the corresponding vector field.
---
I shared the same concern that Ilya expresses in the comments, but after considering it, here is why I think that it works. To have a clean view of the boundary conditions, let's double the disk to the sphere and let everything be equivariant with respect to reflection across the equator.
Moser's theorem truly is a deformation retraction. Let $M$ be a Riemannian manifold, let $\mu$ be a volume form on $M$ (not necessarily Riemannian volume), and let $\phi\_\alpha:M \to M$ be a family of diffeomorphisms of $M$ that may or may not preserve $\mu$. Then $\mu\_\alpha = (\phi\_\alpha)\_\*(\mu)$ is "wrong". Let $\mu\_{\alpha,t}$ be a family of volume forms defined as the weighted geometric mean of $\mu\_\alpha$ and $\mu$:
$$\mu\_{\alpha,t} = \mu^t \mu\_\alpha^{1-t}.$$
Then there is a corresponding Moser flow $\phi\_{\alpha,t}$ such that $\phi\_{\alpha,0} = \phi\_\alpha$ and $\phi\_{\alpha,1}$ is volume-preserving. Moreover, $\phi\_{\alpha,t} = \phi\_\alpha$ for all $t$ if $\phi\_\alpha$ is already volume-preserving for some fixed $\alpha$.
In particular, if $\phi\_t$ is a curve of diffeomorphisms as produced by Smale's theorem with $\phi\_0$ the identity, then Moser gives you an improvement $\phi\_{t,s}$ such that $\phi\_{t,1}$ is then what you want. What worried us is whether $\phi\_{1,1} = \phi\_1$; if $\phi\_1$ is area-preserving, then it is true.
| 15 | https://mathoverflow.net/users/1450 | 9339 | 6,387 |
https://mathoverflow.net/questions/9336 | 3 | I'm aware that h/w problems are frowned upon (understandably) here. However - this really is just inspired by some h/w related confusion, so hopefully that's ok.
Anyway, can one have a smooth projective plane curve be hyperelliptic (i.e admitting a double cover of the projective line and of genus greater than 1)? It is easy enough to get affine curves that double cover the affine line but upon taking the closure in the projective plane I always encounter singularities. It is my understanding that there is a way to resolve such singularities but I imagine that this needn't result in something isomorphic to a plane curve (merely something birationally equivalent to one.)
Thanks in advance for any help given.
| https://mathoverflow.net/users/2691 | plane hyperelliptic curves | I don't know how to answer this question at homework level. If you have a plane curve of degree $d$, it has lots of maps to $P^1$ of degree $d-1$ by projecting from points. If the curve is also hyperelliptic, it has a map of degree two to $P^1$. For at least one of the maps of degree $d-1$, the conditions of the Castelnuovo genus bound (you'll have to look that up) is satisfied and we get that the genus satisfies $g \le d-2$. Now, if your plane curve is smooth (which I had not previously assumed) then $g = (d-1)(d-2)/2$, which combined with the previous bound gives, not surprisingly, $d \le 3$.
| 3 | https://mathoverflow.net/users/2290 | 9342 | 6,390 |
https://mathoverflow.net/questions/9347 | 1 | I understand how weights are defined for a Lie algebra representation.
How are weight spaces defined for a Lie group action (with respect to a fixed torus)?
I know this is a very embarrassing basic question, but i've looked through Harris+Fulton with no satisfactory explanation, and the only thing I can think of is using the exponential map somehow to reduce it to a Lie algebra, which seems unefficient computationally. Surely there must be a better way.
| https://mathoverflow.net/users/2623 | weight space for a Lie group representation | In the case of a finite dimensional representation of a compact Lie group, one picks a basis in which the action of a maximal torus T is diagonal. The weight associated to a vector in this basis is the homomorphism
lambda: T-->T^1 : t\_1^lambda\_1 . t\_2^lambda\_2 ... . t\_n^lambda\_n
by which the maximal torus acts on the vector. The weight space of the group representation is the set of weights of the representation in the charcter lattice of the maximal torus.
A clear exposition of this material can be found in Pressley and Segal : Loop groups chapter 2.
| 3 | https://mathoverflow.net/users/1059 | 9353 | 6,396 |
https://mathoverflow.net/questions/9351 | 6 | Given a partition $\lambda$ of $n$, consider the orbit closure $\overline{ \mathcal{O}\_{\lambda}}$ of the nilpotent orbit corresponding to that partition. My question, is how to explicitly construct the affine coordinate ring of the (singular) variety that is the closure of this orbit?
My second question, is the same but for the orbit closure of an orbit in the enhanced nilpotent cone (see, for instance, Achar-Henderson's ``Orbit closures in the enhanced nilpotent cone"). How would you explicitly find equations generating the ideal killing that variety in this case?
At least the first question is probably well-known and an exercise in combinatorics/linear algebra, but I am having trouble finding a reference - the closest I could find is a paper describing the Springer correspondence for types other than A via the definition of the Weyl group acting on the weight $0$ space, but I couldn't find the answer there.
| https://mathoverflow.net/users/2623 | the affine coordinate ring of orbit closures in the ordinary nilpotent cone | For your first question, I take it that you are interested in orbit closures of nilpotent $n \times n$ matrices. I don't know anything about nilpotent orbits for other Lie algebras, but some stuff is in the references below.
As to your first question, it depends on what you want. Do you want an ideal of polynomials vanishing on these orbits? Or do you want the radical of this ideal? The first case can be done without too much work, but the second question is difficult (as far as I know).
We need a way to index these orbits. They are of course indexed by partitions like you say, given by the block sizes of the Jordan blocks. One way is to say that the orbit corresponding to the partition $\lambda$ is given by the conditions $\lbrace A \mid \ker A^i = \lambda\_1 + \cdots + \lambda\_i\rbrace$, in which case an ideal which set-theoretically defines the orbit closure is given by the equations
$\dim \ker A^i \ge \lambda\_1 + \cdots + \lambda\_i$
To get explicit equations, let A be a generic matrix of variables $x\_{i,j}$. The inequality above is the same as saying that
$\operatorname{rank} A^i \le n - (\lambda\_1 + \cdots + \lambda\_i)$,
which we can write as polynomial equations by requiring that the $(N\_i+1) \times (N\_i+1)$ minors of $A^i$ are all 0, where $N\_i = n-(\lambda\_1 + \cdots + \lambda\_i)$.
As for finding the radical of this ideal, one possible reference for this stuff is Chapter 8 of Jerzy Weyman's book *Cohomology of Vector Bundles and Syzygies*. Some of this is based on the material in a paper in which he calculates the radical of the ideal for certain nilpotent orbits (and proves some results about them in general): <http://arxiv.org/abs/math/0006232> . What you're interested in should be in the paper, though it's heavy on sheaf cohomology (hopefully you like that). For algebraic properties of these coordinate rings like normality, Gorensteinness, rational singularities, see the book.
EDIT: I should have linked to the paper that David mentions since it contains more complete results than the one I included. However, the calculating of generators for the radical ideal (but not minimal ones) is done in Section 8.2 of the book, so I still recommend looking at it. In particular, all of the relevant techniques are treated from scratch in the earlier chapters. In particular, Chapter 5 is crucial.
| 7 | https://mathoverflow.net/users/321 | 9354 | 6,397 |
https://mathoverflow.net/questions/9352 | 7 | * Let $\mathcal{D}$ be the $n$th Weyl algebra $ \mathcal{D} :=k[x\_1,...,x\_n,\partial\_1,...,\partial\_n] $, where $\partial\_ix\_i-x\_i\partial\_i=1$.
* Let $\widetilde{\mathcal{D}}$ be its Rees algebra, which is $ \mathcal{D} :=k[t, x\_1,...,x\_n,\partial\_1,...,\partial\_n] $, where $\partial\_ix\_i-x\_i\partial\_i=t$, and $t$ is central.
* Let $\mathcal{O}\_X$ denote the polynomial algebra $k[x\_1,...,x\_n]$, which is a left $\widetilde{\mathcal{D}}$-module, where $t$ and all the $\partial\_i$ act by zero. NOTE: this is different than the homogenization of the standard $\mathcal{D}$-module structure on $\mathcal{O}\_X$.
The question I am interested in is, **how many generators does a left ideal $M$ in $\widetilde{\mathcal{D}}$ need before $Hom\_{\widetilde{\mathcal{D}}}(\mathcal{O}\_X,\widetilde{\mathcal{D}}/M)$ can be non-zero?** My conjecture is that $M$ needs at least $n+1$ generators. NOTE: Savvy Weyl algebra veterans will know every left ideal in $\mathcal{D}$ can be generated by two elements; however, this is *not* true in $\widetilde{\mathcal{D}}$. There can be ideals generated by $n+1$ elements and no fewer.
The functor $Hom\_{\widetilde{\mathcal{D}}}(\mathcal{O}\_X,-)$ acts as a relative analog of the more familiar functor $Hom\_R(k,-)$ (where $R=k[t,y\_1,...y\_n]$). Therefore, the above question is analogous to asking "how many generators must an ideal $I\subseteq R$ have before $R/I$ can have depth zero?" The answer here is $n+1$, which follows from Thm 13.4, pg 98 of Matsumura (essentially a souped up version of the Hauptidealsatz).
In the noncommutative case, if you try to make this work with the $\widetilde{\mathcal{D}}$-module $k$ (where $t$, $x\_i$ and $\partial\_i$ all act by zero), it doesn't work. The natural conjecture would be that $Hom\_{\widetilde{D}}(k,\widetilde{\mathcal{D}}/M)\neq 0$ implies $M$ had at least $2n+1$ generators, except this fails even for the first Weyl algebra and $M=\widetilde{\mathcal{D}}x\_1+\widetilde{\mathcal{D}}\partial\_1$ (since $\widetilde{\mathcal{D}}/M=k$).
However, it seems that things might work right for the relative module $O\_X$, based on a fair amount of experimentation. It is easily true in the first Weyl algebra. Oh, and equivalent condition is to ask when $Hom\_{\overline{\mathcal{D}}}(\mathcal{O}\_X,\overline{M})\neq 0 $, where $\overline{\mathcal{D}}=\widetilde{\mathcal{D}}/t$, and $\overline{M}$ is $M/Mt$.
| https://mathoverflow.net/users/750 | Depth Zero Ideals in the Homogenized Weyl Algebra | I am not sure I understand the analogue correctly, but in the commutative case, one can get to depth zero with 3 generators. That is because any second syzygy of a module of depth at least $1$ is isomorphic to a second syzygy of a 3-generated ideal by a [result](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=42575&vfpref=html&r=81&mx-pid=399074) of Bruns. It is even implemented here (be warned that the statement misses the at least depth $1$ part):
<http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Bruns/html/>
You can take the N to be second syzygy of $m=(t,y\_1,...,y\_n)$, so $N$ has depth 3. Produce a three generated ideal $I$ such that $syz^2(I)\cong N$. So $depth I =3-2=1$, and $depth R/I=0$.
I think Theorem 13.4 shows that $dim R/I=0$ implies $I$ is at least $n+1$-generated.
| 5 | https://mathoverflow.net/users/2083 | 9357 | 6,398 |
https://mathoverflow.net/questions/9356 | 4 | Let R be a real closed field, and let U be a semialgebraic subset of $R^n$. Let $S^0(U)$ be the ring of continuous R-valued semialgebraic functions. Also let $\tilde{U}$ be the subset of Spec$\_r (R[X\_1, \ldots, X\_n])$ corresponding to U.
What does the real spectrum of $S^0(U)$ look like? Is it related to $\tilde{U}$ in some way?
| https://mathoverflow.net/users/1709 | Real spectrum of ring of continuous semialgebraic functions | I don't agree with the preceding answer.
When $U$ is a locally compact semialgebraic set, then $\widetilde{U}$ equipped with its sheaf of semi-algebraic continuous functions is isomorphic to the affine scheme $\mathrm{Spec}(S^0(U))$. This is proposition 6 in Carral, Coste : Normal spectral spaces and their dimensions, J. Pure Appl. Algebra 30 (1983) 227-235. In particular $\widetilde{U}$ is homeomorphic to the prime spectrum of $S^0(U)$, which is homeomorphic to its real spectrum. In case $U$ is not locally compact, the situation is different; there are more points in $\mathrm{Spec}(S^0(U))$.
| 5 | https://mathoverflow.net/users/2630 | 9363 | 6,401 |
https://mathoverflow.net/questions/9332 | 8 | Are there such things as recurrence equations with random variable coefficients. For example, $$W\_n=W\_{n-1}+F\cdot W\_{n-1}$$ where $F$ is a random variable. I tried to see if I could make sense of it using the simplest possible case of $F$ being a uniform discrete random variable on 2 points but I didn't get far because even if the initial data is not random the succeeding terms in the sequence are and each term seems to live on a different space. I couldn't figure out what space $W\_n$ and $W\_{n-1}$ should live on. A google search turned up nothing for the obvious keywords "random recurrence equation".
Edit in response to Alekk's answer: More specifically suppose I wanted to find the probability $P(W\_{200}>3000)$. Is there a way to compute the distribution of $W\_{200}$ explicitly given the distribution of $F$ and some non-random initial data $W\_0$?
Edit: $F$ does not depend on $n$ and to make things even more explicit lets say $F$ has the distribution $P(F=2f)=\dfrac{1}{2}, P(F=-f)=\dfrac{1}{2}$.
| https://mathoverflow.net/users/nan | linear recurrence relations with random coefficients | So this is the product of IID random variables $1+F\_n$, so you could take logarithms and do the more conventional sums of IID random variables $\log(1+F\_n)$. Perhaps the logarithms are complex numbers.
| 3 | https://mathoverflow.net/users/454 | 9366 | 6,404 |
https://mathoverflow.net/questions/9274 | 8 | Definition: A polytope has **property X** iff there is a function f:N+ → R+ such that for each pair of vertices vi, vj the following holds:
disteuclidean(vi, vj) = f(distcombinatorial(vi, vj))
with distcombinatorial(vi, vj) = shortest path of edges between vi and vj.
That means: for each vi1, vj1, vi2, vj2:
disteuclidean(vi1, vj1) = disteuclidean(vi2, vj2)
iff
distcombinatorial(vi1, vj1) = distcombinatorial(vi2, vj2)
Question 1: Is property X already named? What's its common name?
Question 2: Which polytopes have property X? The regular polytopes seem to have it, but are there more?
| https://mathoverflow.net/users/2672 | Combinatorial distance ≡ Euclidean distance | Another way of describing your property X is to say that concentric spheres in the shortest path metric in the graph of the polytope are mapped into concentric Euclidean spehres. I have never heard of these polytopes before, but it is a very natural question. My suspicion is that there are not very many "unsymmetric" ones.
Here is a plan for enumerating them in dimension 3. As mentioned by Martin M. W., in dimension two the only such polytopes are the regular polygons: all edges must be equal, and all angles must also be equal, because of vertices at distance 2 in the graph must have the same Euclidean distance. So in dimension 3, each facet of such a polytope must be a regular polygon. Because a regular polygon fixes the second-smallest distance, all non-triangular facets must be congruent. The 3-polytopes with each face a regular polygon are known: they are the 5 [Platonic solids](http://en.wikipedia.org/wiki/Platonic_solid), the 13 [Archimedean solids](http://en.wikipedia.org/wiki/Archimedean_solid), the infinite family of [prisms](http://en.wikipedia.org/wiki/Prism_%28geometry%29), the infinite family of [antiprisms](http://en.wikipedia.org/wiki/Antiprism), and the 92 [Johnson solids](http://en.wikipedia.org/wiki/Johnson_solid).
Only two prisms satisfy this property, the triangular one and the square one (aka the cube). Most likely, the only antiprism will be the triangular one (aka the octahedron), but this will need some calculations. This leaves finitely many, each of which can be checked (more calculations).
In principle, the same programme could be carried through in dimension 4. Then each facet will be one in the finite list (not) enumerated above, and because of the various distances realized by each possible 3-polytope, not many of them could co-exist. So it could be that the possibilities are even more restricted in dimension 4. Or there could be a combinatorial explosion. It's not clear to me that there would be only finitely many such polytopes (up to similarity) in a fixed dimension $\geq 4$.
Anyone for a computational project? (Note: [Zalgaller's enumeration](http://www.ams.org/mathscinet-getitem?mr=240719) of the Johnson solids takes up almost 100 pages.)
| 5 | https://mathoverflow.net/users/932 | 9368 | 6,406 |
https://mathoverflow.net/questions/9369 | 8 | What is the explanation of the apparent randomness of high-level phenomena in nature?
For example the distribution of females vs. males in a population (I am referring to randomness in terms of the unpredictability and not in the sense of it necessarily having to be evenly distributed).
1. Is it accepted that these phenomena are not really random, meaning that given enough information one could predict it? If so isn't that the case for all random phenomena?
2. If there is true randomness and the outcome cannot be predicted - what is the origin of that randomness? (is it a result of the randomness in the micro world - quantum phenomena etc...)
where can i find resources about the subject?
| https://mathoverflow.net/users/2705 | randomness in nature | This is, of course, a very important problem. One (extreme) point of view is that any form of classical (=commutative) randomness reflects "only" human uncertainty and does not have an "objective" physical meaning.
(Further answers to this question and more discussion are welcome on [the posting entitled "Randomness in nature" on my blog "Combinatorics and More"](http://gilkalai.wordpress.com/2009/12/27/randomness-in-nature/). Here is a link to a [subsequent post with further discussion](http://gilkalai.wordpress.com/2010/01/20/randomness-in-nature-ii/).) Some related material can be found in the site of the conference "[The Probable and the Improbable: The Meaning and Role of Probability in Physics](http://www.vanleer.org.il/probability/1-home.htm)".
| 7 | https://mathoverflow.net/users/1532 | 9371 | 6,408 |
https://mathoverflow.net/questions/9393 | -2 | Let {X,T} be a topology, T the set of open subsets of X.
---
Definition: Three points x, y, z of X are in *relation N* (Nxyz, read "x is nearer to y than to z") iff
1. there is a basis **B** of T and **b** in **B** such that x and y are in **b** but z is not and
2. there is no basis **C** of T and **c** in **C** such that x and z are in **c** but not y.
---
For some topologies there are no points x, y, z in relation N, for example if T = {Ø,X} or T = P(X), but for others there are (e.g. for ones induced by a metric [my claim]).
---
Definition: A topology has *property M1* iff
(x)(y) ((z) (z ≠ x & z ≠ y) → Nxyz) → x = y
(This is an analogue of d(xy) = 0 → x = y, the best one I can imagine).
---
Definition: A topology has *property M2* iff
(x)(y)(z) Nxyz & Nyzx → Nzyx
(This is a kind of an analogue of d(xy) = d(yx), the best one I can imagine)
---
First (bunch of) question(s):
1. Properties *M1* and *M2* do not capture the whole of the corresponding conditions of a metric. Can anyone figure out "better" definitions (e.g. an analogon of x = y → d(xy) = 0)?
2. Can anyone figure out a *property M3* that is an analogue of the triangle equality?
If it can be shown that no such property M3 is definable, the following becomes obsolete.
If such a definition can be made, we define:
---
Definition: A topology has *property M* (read "induces a metric") iff it has properties M1, M2, M3.
---
Second question:
Which topologies have property M, i.e. induce a metric? Are these "accidentally" exactly those that are induced by a metric?
| https://mathoverflow.net/users/2672 | Can topologies induce a metric? | Your condition 1 is satisfied for all triples $x,y,z\in X$ such that $z\not\in\{x,y\}$ if the space is [$T\_1$](http://en.wikipedia.org/wiki/T1_space).
Maybe reading a bit about uniform spaces and the corresponding metrizability results will be of help.
| 13 | https://mathoverflow.net/users/1409 | 9395 | 6,426 |
https://mathoverflow.net/questions/9396 | 1 | I'm trying to find the correct term for a specific kind of totally ordered space:
Let $S$ be a totally ordered space with strict total order $<$.
Property: For any two $s\_{1}$ and $s\_{2}$ in $S$ where $s\_1 < s\_2$, there must exist some $s\_{3}$ such that $s\_{1} < s\_{3}$ and $s\_{3} < s\_{2}$.
What is the name of this property? Thank you!
| https://mathoverflow.net/users/1998 | The proper name for a kind of ordered space | [Dense order](http://en.wikipedia.org/wiki/Dense_order) is one name that concept goes by.
| 6 | https://mathoverflow.net/users/1409 | 9397 | 6,427 |
https://mathoverflow.net/questions/9378 | 6 | Hi! I would like to know if there is an explicit classification of the algebraic (i.e., Zariski closed) subgroups of the symplectic group Sp(4,R) and/or more generally Sp(2n,R) somewhere in the literature.
| https://mathoverflow.net/users/1568 | What's the classification of the algebraic subgroups of Sp(4,R)? | On the one hand, I could not find a published answer with a cursory search. On the other hand, as Ben says, you could work out the answer "by hand". Instead of writing down a sheer list, which might be complicated (and I haven't done the work), I'll write down the main ingredients.
A Zariski-closed subgroup $H$ of any connected semisimple Lie group $G$ has three pieces: (1) finite, (2) connected semisimple, and (3) connected solvable. The Zariski topology forces $H$ to have only finitely many components; if $H\_0$ is the connected subgroup, then $H/H\_0$ is the finite piece. Then the Lie algebra of $H\_0$ has a Levi decomposition, so that you get the other two pieces. The way to analyze the question is to chase down the possibilities for all three pieces.
I think that the finite part always lifts to a slightly larger finite subgroup of $H$. This is not true for groups in general, but I think that it is true in context. Then this finite group is contained in a maximal compact group of $G$. Happily, the compact core of $\text{Sp}(4,\mathbb{R})$ is $\text{SU}(2)$, and the finite subgroups are classified by simply laced Dynkin diagrams.
A semisimple, connected subgroup of $G$ corresponds to a semisimple Lie subalgebra, and that complexifies. The Lie algebra $\text{sp}(4,\mathbb{C})$ does not have very many inequivalent semisimple subalgebras. From looking a rank, they are isomorphic to $\text{sl}(2,\mathbb{C})$ or $\text{sl}(2,\mathbb{C}) \oplus \text{sl}(2,\mathbb{C})$. I am confusing myself a little with the possible positions of the former, although I know there are only a few. The latter embeds in only one way. Then you would work backwards to get the real forms of these complex subalgebras; again there wouldn't be very many.
Finally the solvable part also complexifies and I think that it is contained in a Borel subalgebra at the Lie algebra level.
As for the more general question, for $\text{Sp}(2n,\mathbb{R})$, there is a tidy converse answer that also shows you that you can't expect a tidy answer for all fixed $n$. Namely, if $G$ is any algebraic group, you can classify its anti-self-dual (or symplectically self-dual) representations. Every algebraic group will have some, because every algebraic group has representations in $\text{GL}(n,\mathbb{R})$. A more interesting case is when $G$ has an irreducible symplectically self-dual representation. For that purpose, you check that the irreducible representation is real, and then check the Frobenius–Schur indicator.
| 9 | https://mathoverflow.net/users/1450 | 9399 | 6,429 |
https://mathoverflow.net/questions/9401 | 10 | This is a question for all you number theorists out there...based on my skimming of number theory textbooks and survey articles, it seems like most of the applications of geometry and complex variables to number theory are restricted to surfaces and the theory of a single complex variable. My questions are
1) Is this impression indeed accurate?
and, if so
2) Why is this? Is it because the theories of surfaces and of a single complex variable are "easier"(in the sense that they're simple enough to have a single unified theory,) or is there some deeper reason? And if the former is the case, are there some deep conjectures in number theory that could be solved using higher-dimensional geometry/complex variables?
| https://mathoverflow.net/users/2497 | Number Theory and Geometry/Several Complex Variables | I have heard algebraic number theory called "algebraic geometry in one dimension". (Or maybe you could call it arithmetic geometry in one dimension.) There is a natural emphasis in algebraic number theory on elliptic curves, function fields, etc. The reason is that algebraic geometry in one dimension is relatively well controlled, so that you can instead focus on arithmetic issues. I might describe the philosophy this way, even though it is not my area and not my philosophy: If number theory is like playing music, and algebraic geometry is like juggling, then general arithmetic geometry is like playing music while juggling the instruments. I.e., it's a great thing to do, but it combines two problems that already hard enough separately.
To the extent that you adopt this emphasis, it makes sense that you would learn more from complex curves than from higher-dimensional complex geometry. But without the one-dimensional emphasis, it is not true. For instance, the Weil conjectures are a deep result in higher-dimensional number theory, if you can call finite fields number theory, and they are both motivated by and informed by higher-dimensional complex geometry. (If number theory is music, finite fields could be music mainly in one note. If I had to juggle instruments, I might rather juggle tambourines or triangles.)
I should add that restricting to one dimension isn't a completely consistent vision of geometry. To give two examples, if $C$ is a complex curve of genus $g > 1$, it comes from a higher-dimensional moduli space, and it has a higher-dimensional Jacobian variety. These issues also arise in positive characteristic. The philosophy that algebraic number theory is one-dimensional is there, but it is not a completely serious philosophy, and not just because higher-dimensional generalizations exist. The Langlands program eventually leads you to the opposite philosophy.
Finally there are also connections between number theory and manifolds with three *real* dimensions, because of the fact that the boundary of hyperbolic space $\mathbb{H}^3$ is a Riemann sphere and $\text{Isom}(\mathbb{H}^3) = \text{PSL}(2,\mathbb{C})$.
| 15 | https://mathoverflow.net/users/1450 | 9405 | 6,433 |
https://mathoverflow.net/questions/9406 | 9 | The wikipedia page [Covering groups of the alternating and symmetric groups](http://en.wikipedia.org/wiki/Covering_groups_of_the_alternating_and_symmetric_groups) gives explicit presentations for the double covers of the symmetric group Sn (n ≥ 4). Can someone provide a similar presentation, or better yet an explicit combinatorial description, of the double cover of the alternating group An? (I really only care about the case of large n, in case it matters.)
| https://mathoverflow.net/users/126667 | Presentation for the double cover of A_n | Yeah, Schur did this a long time ago. Let $\tilde \Sigma\_n \to \Sigma\_n$ be a double cover (there are two) -- lets denote them $\tilde \Sigma\_n = \Sigma\_n^\epsilon$ where $\epsilon \in \{+1, -1\}$.
Schur uses the notation $[a\_1 a\_2 \cdots a\_k]$ for a specific lift of the cycle $(a\_1 a\_2 \cdots a\_k) \in \Sigma\_n$ to $\Sigma\_n^\epsilon$ -- might as well call these $k$-cycles. Then his presentation goes like this:
$$[a\_1 a\_2 \cdots a\_k] = [a\_1 a\_2 \cdots a\_i][a\_i a\_{i+1} \cdots a\_k] \ \ \forall 1 < i< k$$
and all $k$-cycles, $k>1$.
$$[a\_1 a\_2 \cdots a\_k]^{[b\_1 b\_2 \cdots b\_j]} = (-1)^{j-1}[\phi(a\_1) \phi(a\_2) \cdots \phi(a\_k)]$$
where $\phi$ is the cycle $(b\_1 b\_2 \cdots b\_j)$
$$[a\_1 a\_2 \cdots a\_k]^k = \epsilon$$
for all $k$-cycles -- ie this is always $+1$ or $-1$ depending on which extension of $\Sigma\_n$ you're interested in. And:
$$[a\_1 a\_2 \cdots a\_k][b\_1 b\_2 \cdots b\_j] = (-1)^{(k-1)(j-1)}[b\_1b\_2 \cdots b\_j][a\_1 a\_2 \cdots a\_k]$$
provided the cycles $(a\_1 a\_2 \cdots a\_k)$ and $(b\_1 b\_2 \cdots b\_j)$ are disjoint.
The map $\tilde \Sigma\_n \to \Sigma\_n$ sends $[a\_1 \cdots a\_k]$ to $(a\_1 \cdots a\_k)$. So this gives you a corresponding presentation of the double of $A\_n$ -- take your favourite presentation of $A\_n$, lift the relators and see what happens using the above relations.
A small extra tidbit -- think of $\Sigma\_n$ as being the group of orientation-preserving isometries of $\mathbb R^{n}$ that preserves a regular (n-1)-simplex. Then if you lift this group to $Spin(n)$, the extension you want is the one where $[a\_1 a\_2 \cdots a\_k]^k = -1$.
| 12 | https://mathoverflow.net/users/1465 | 9407 | 6,434 |
https://mathoverflow.net/questions/8388 | 14 | Let A, B and C be finitely supported probability distributions with at most d nonzero probabilities each. Now consider the following simultaneous equations using p-norms, for each value of p≥1, given by
||A||p + ||B||p = ||C||p
where A, B and C are still non-negative, but we relax normalization on A and B. Imagine that C is fixed and, without loss of generality, normalized. We want to solve for A and B.
First, note that one obvious family of solutions is
A = (1-x) C , B = x C , 0≤x≤1 .
Question: Ignoring the obvious permutation symmetries, are these the *only* solutions?
Edit: By p-norm, I mean the vector p-norm: ||A||p = (∑j |aj|p )1/p. Although we don't really need the absolute values, since the aj are all non-negative.
| https://mathoverflow.net/users/1171 | Are two probability distributions uniquely constrained by the sum of their p-norms? | Here is a proof that Steve's rescaling gives you all solutions, together with the trivial operation of permuting the components of $A$, $B$, and $C$ if you view them as vectors with positive coeifficients. (If you view them this way, then Steve's notation $||A||\_p$ is just the usual $p$-norm.)
I first tried what Alekk tried: You can take the limit as $p \to \infty$ and eventually obtain certain power series expansions in $1/p$. Or you can take the limit $p \to 0$ and obtain certain power series expansions in $p$. The problem with both approaches is that the information in the terms of these expansions is complicated. To help understand the second limit, I observed that the two sides of Steve's equation are analytic in $p$, but it only helped so much.
Then I realized that when you have a complex analytic function of one variable, you can get a lot of information from looking at singularities. So let's look at that. Let
$\alpha\_k = \ln a\_k$, so that
$$||A||\_p = \exp\left( \frac{\ln \bigl[\exp(\alpha\_1 p) + \exp(\alpha\_2 p) + \cdots + \exp(\alpha\_d p) \bigr]}{p} \right).$$
The expression inside the logarithm has been called an exponential polynomial in the literature, which I'll call $a(p)$. As indicated, $||A||\_p$ has a logarithmic singularity when $a(p) = 0$. $||A||\_p$ has another kind of singularity when $p = 0$, but won't matter for anything. Also $a(p)$ is an entire function, which means in particular that it is univalent and has isolated zeroes. Also, none of the zeroes of $a(p)$ are on the real axis. Let $b(p)$ and $c(p)$ be the corresponding exponential polynomials for $B$ and $C$.
Suppose that you follow a loop that starts on the positive real axis, encircles an $m$-fold zero of $a(p)$ at $p\_0$, and then retraces to its starting point. Then the value of $||A||\_p$, which is non-zero for $p > 0$, gains a factor of $\exp(2m\pi i/p\_0)$. Thus Steve's equation is not consistent unless all three of $a(p)$, $b(p)$ and, $c(p)$ have the same zeroes with the same multiplicity. (Since $\exp(2m\pi i/p\_0)$ cannot have norm 1, geometric sequences with this ratio but with different values of $m$ are linearly independent.)
At this point, the problem is solved by a very interesting paper of Ritt, [On the zeros of exponential polynomials](http://www.jstor.org/pss/1989556). Ritt reviews certain results of Tamarkin, Polya, and Schwengler, which imply in particular that if an exponential polynomial $f(z)$ does not have any zeroes, then it is a monomial $f\_\alpha \exp(\alpha z)$. Ritt's own theorem is that if $f(z)$ and $g(z)$ are exponential polynomials, and if the roots of $f(z)$ are all roots of $g(z)$ (with multiplicity), then their ratio is another exponential polynomial. Thus in our situation $a(p)$, $b(p)$, and $c(p)$ are all proportional up to a constant and an exponential factor. Thus, $A$, $B$, and $C$ must be the same vectors up to permutation, repetition, and rescaling of the coordinates. Repetition is an operation that hasn't yet been analyzed. If $A^{\oplus n}$ denotes the $n$-fold repetition of $A$, then $||A^{\oplus n}||\_p = n^{1/p}||A||\_p$. Again, since geometric sequences with distinct ratios are linearly independent, Steve's equation is not consistent if $A$, $B$, and $C$ are repetitions of the same vector by different amounts.
The same argument works for the generalized equation
$$x\_1||A\_1||\_p + x\_2||A\_2||\_p + \cdots + x\_n||A\_n||\_p = 0.$$
The result is that any such linear dependence trivializes, after rescaling the vectors and permuting their coordinates.
**Update** (by *J.O'Rourke*): Greg's paper based on this solution was just published:
>
> "Norms as a function of $p$ are linearly
> independent in finite dimensions," *Amer. Math. Monthly*, Vol. 119, No. 7, Aug-Sep 2012, pp. 601-3
> ([JSTOR link](http://www.jstor.org/stable/10.4169/amer.math.monthly.119.07.601)).
>
>
>
| 12 | https://mathoverflow.net/users/1450 | 9408 | 6,435 |
https://mathoverflow.net/questions/9269 | 32 | In
>
> Lawvere, F. W., 1966, “The Category of
> Categories as a Foundation for
> Mathematics”, Proceedings of the
> Conference on Categorical Algebra, La
> Jolla, New York: Springer-Verlag,
> 1–21.
>
>
>
Lawvere proposed an elementary theory of the category of categories which can serve as a foundation for mathematics.
So far I have heard from several sources that there are some flaws with this theory so that it does not completely work as proposed.
So my question is whether there is currently any (accepted) elementary theory of the category of categories that is rich enough so that one can formulate, say, the following things in the theory:
* The category of sets.
* Basic notions of category theory (functor categories, adjoints, Kan extensions, etc.).
* Other important categories (like the category of rings or the category of schemes).
The elementary theory I am looking for should allow me to identify what should be called a category of commutative rings (at best I would like to see this category defined by a universal 2-categorical property) or how to work with this category. I am not interested in defining groups, rings, etc. as special categories as this seems to be better done in an elementary theory of sets.
P.S.: The same question has an analogue one level higher. Assume that we have constructed an object in the category of categories (=: CAT) which can serve as a, say, category C of spaces. Classically, we can associate to each space X in C the sheaf topos over it. In the picture I have in mind, one should ask whether there is a similar elementary theory of the category of 2-categories (=: 2-CAT). Then one should be able to lift the object C from CAT to 2-CAT (as one is able to form the discrete category from a set), define an object T in 2-CAT that serves as the 2-category of toposes, and a functor C -> T in 2-CAT.
| https://mathoverflow.net/users/1841 | Category of categories as a foundation of mathematics | My personal opinion is that one should consider the *2-category* of categories, rather than the 1-category of categories. I think the axioms one wants for such an "ET2CC" will be something like:
* Firstly, some exactness axioms amounting to its being a "2-pretopos" in the sense I described here: <http://ncatlab.org/michaelshulman/show/2-categorical+logic> . This gives you an "internal logic" like that of an ordinary (pre)topos.
* Secondly, the existence of certain exponentials (this is optional).
* Thirdly, the existence of a "classifying discrete opfibration" $el\to set$ in the sense introduced by Mark Weber ("Yoneda structures from 2-toposes") which serves as "the category of sets," and internally satisfies some suitable axioms.
* Finally, a "well-pointedness" axiom saying that the terminal object is a generator, as is the case one level down with in ETCS. This is what says you have a 2-category of categories, rather than (for instance) a 2-category of stacks.
Once you have all this, you can use finite 2-categorical limits and the "internal logic" to construct all the usual concrete categories out of the object "set". For instance, "set" has finite products internally, which means that the morphisms $set \to 1$ and $set \to set \times set$ have right adjoints in our 2-category Cat (i.e. "set" is a "cartesian object" in Cat). The composite $set \to set\times set \to set$ of the diagonal with the "binary products" morphism is the "functor" which, intuitively, takes a set $A$ to the set $A\times A$. Now the 2-categorical limit called an "inserter" applied to this composite and the identity of "set" can be considered "the category of sets $A$ equipped with a function $A\times A\to A$," i.e. the category of magmas.
Now we have a forgetful functor $magma \to set$, and also a functor $magma \to set$ which takes a magma to the triple product $A\times A \times A$, and there are two 2-cells relating these constructed from two different composites of the inserter 2-cell defining the category of magmas. The "equifier" (another 2-categorical limit) of these 2-cells it makes sense to call "the category of semigroups" (sets with an associative binary operation). Proceeding in this way we can construct the categories of monoids, groups, abelian groups, and eventually rings.
A more direct way to describe the category of rings with a universal property is as follows. Since $set$ is a cartesian object, each hom-category $Cat(X,set)$ has finite products, so we can define the category $ring(Cat(X,set))$ of rings internal to it. Then the category $ring$ is equipped with a forgetful functor $ring \to set$ which has the structure of a ring in $Cat(ring,set)$, and which is universal in the sense that we have a natural equivalence $ring(Cat(X,set)) \simeq Cat(X,ring)$. The above construction then just shows that such a representing object exists whenever Cat has suitable finitary structure.
One can hope for a similar elementary theory of the 3-category of 2-categories, and so on up the ladder, but it's not as clear to me yet what the appropriate exactness properties will be.
| 27 | https://mathoverflow.net/users/49 | 9412 | 6,439 |
https://mathoverflow.net/questions/9418 | 38 | Why does a space with finite homotopy groups [for every n] have finite homology groups? How can I proof this [not only for connected spaces with trivial fundamental group]? The converse is false. $\mathbb{R}P^2$ is a counterexample.
Do finitely generated homotopy groups imply finitely generated homology groups? I can proof this only for connected spaces with trivial fundamental group. The converse is false. $S^1\vee S^2$ is a counterexample.
| https://mathoverflow.net/users/2625 | Why do finite homotopy groups imply finite homology groups? | (This answer has been edited to give more details.)
Finitely generated homotopy groups do not imply finitely generated homology groups. Stallings gave an example of a finitely presented group $G$ such that $H\_3(G;Z)$ is not finitely generated. A $K(G,1)$ space then has finitely generated homotopy groups but not finitely generated homology groups. Stallings' paper appeared in Amer. J. Math 83 (1963), 541-543. Footnote in small print: Stallings' example can also be found in my algebraic topology book, pp.423-426, as part of a more general family of examples due to Bestvina and Brady.
As Stallings noted, it follows that any finite complex $K$ with $\pi\_1(K)=G$ has $\pi\_2(K)$ nonfinitely generated, even as a module over $\pi\_1(K)$. This is in contrast to the example of $S^1 \vee S^2$.
Finite homotopy groups do imply finite homology groups, however. In the simply-connected case this is a consequence of Serre's mod C theory, but for the nonsimply-connected case I don't know a reference in the literature. I asked about this on Don Davis' algebraic topology listserv in 2001 and got answers from Bill Browder and Tom Goodwillie. Here's the link to their answers:
<http://www.lehigh.edu/~dmd1/tg39.txt>
The argument goes as follows. First consider the special case that the given space $X$ is $BG$ for a finite group $G$. The standard model for $BG$ has finite skeleta when $G$ is finite so the homology is finitely generated. A standard transfer argument using the contractible universal cover shows that the homology is annihilated by $|G|$, so it must then be finite.
For a general $X$ with finite homotopy groups one uses the fibration $E \to X \to BG$ where $G=\pi\_1(X)$ and $E$ is the universal cover of $X$. The Serre spectral sequence for this fibration has $E^2\_{pq}=H\_p(BG;H\_q(E))$ where the coefficients may be twisted, so a little care is needed. From the simply-connected case we know that $H\_q(E)$ is finite for $q>0$. Since $BG$ has finite skeleta this implies $E\_{pq}^2$ is finite for $q>0$, even with twisted coefficients. To see this one could for example go back to the $E^1$ page where $E\_{pq}^1=C\_p(BG;H\_q(E))$, the cellular chain group, a finite abelian group when $q>0$, which implies finiteness of $E\_{pq}^2$ for $q>0$. When $q=0$ we have $E\_{p0}^2=H\_p(BG;Z)$ with untwisted coefficients, so this is finite for $p>0$ by the earlier special case. Now we have $E\_{pq}^2$ finite for $p+q>0$, so the same must be true for $E^\infty$ and hence $H\_n(X)$ is finite for $n>0$.
Sorry for the length of this answer and for the multiple edits, but it seemed worthwhile to get this argument on record.
| 85 | https://mathoverflow.net/users/23571 | 9422 | 6,445 |
https://mathoverflow.net/questions/9415 | 7 | In "Random Matrices and Random Permutations" by Okounkov it says, "It is classically known that every problem about the combinatorics of a covering has a translation into a problem about permutations which arise as the monodromies around the ramification points." Apparently, this is called the "Hurwitz encoding" but I never got a clear explanation of how it works.
I'd like to understand what it means for a covering of the sphere to be branched and how to identify its ramification points. Perhaps I should learn about how monodromy permutations are computed. Also, how to engineer branched covers which encode problems about the permutation group?
| https://mathoverflow.net/users/1358 | Hurwitz Encoding | **What it means for a covering of a sphere to be branched:** Let $f:X \to Y$ be a map of Riemann surfaces. We are particularly interested in the case that $Y$ is $\mathbb{CP}^1$; in this case, $Y$ has the topology of a sphere. At most points $y$ in $Y$, there will be a neighborhood $V$ of $y$ so that $f^{-1}(V)$ is just a union of $n$ disjoint copies of $U$, each mapping isomorphically to $V$. These are the points where there is no branching.
At a few points of $y$, something different will happen. Let $x$ be a preimage of $y$, let $V$ be a small neighborhood of $y$ and let $U$ be the connected component of $f^{-1}(V)$ containing $x$. At these points, the map $f$ looks like $t \mapsto t^e$, as a map from the unit disc in $\mathbb{C}$ to itself. This is called branching.
The generic situation is that, at finitely many points of $Y$, one of the preimages is branched with $e=2$ and the other $n-2$ preimages are unbranched.
Algebraically, if $t$ is a local coordinate on $X$, then we have branching where $\partial f/\partial t$ vanishes. Even if we can't explicitly write $f$ as a function of $t$, if we can find a polynomial relation $P(f,t)=0$, then we have $(\partial P/\partial f)(\partial f/\partial t) = \partial P/\partial t$, so branching will occur when $\partial P/\partial t=0$.
**The relation between branched covers and the symmetric group**: Let $f: X \to \mathbb{CP}^1$ be a branched cover. Let $R \subset \mathbb{CP}^1$ be the points over which branching occurs. Then $f^{-1}(\mathbb{CP}^1 \setminus R) \to \mathbb{CP}^1 \setminus R$ is a cover, in the sense of algebraic topology.
As you probably know, connected covers of a space $U$ are classified by subgroups of $\pi\_1(U)$. In particular, degree $n$ covers are classified by index $n$ subgroups. I like to recast this and say that degree $n$ connected covers of $U$ are classified by transitive actions of $\pi\_1(U)$ on an $n$-element set. This has the advantage that, more generally, we can say that degree $n$ covers of $U$ are classified by actions of $\pi\_1(U)$ on an $n$-element set.
Now, in this case, $\pi\_1(\mathbb{CP}^1 \setminus R)$ is isomorphic to the group generated by $R$, modulo the relation $\prod{r \in R} [r]=1$. You should be warned that this isomorphism depends on some choices. First of all, we need to choose a base point $y$ in $\mathbb{CP}^1 \setminus R$! Even once we've done that, we need to choose loops based at $y$, circling each of the elements of $R$, and disjoint from each other away from $y$. These will then be the classes $[r]$. The order that the product above is taken is related to the circular order at which these loops come in to $y$.
So, to covers of $\mathbb{CP}^1 \setminus R$ correspond to maps from this group to $S\_n$. To give such a map, we choose an element $b(r)$ of $S\_n$ for each $r$ in $R$; these must obey $\prod\_{r \in R} b(r)=1$. (I am being very sloppy about when two such maps give isomorphic covers, and, indeed, what it means to say two covers are isomorphic.)
The relation between the geometry of the cover, and the map $b(r)$ is the following: If the permutation $b(r)$ has cycles of lengths $e\_1$, $e\_2$, ..., $e\_k$, then $f^{-1}(r)$ contains $k$ points, which are branched with degrees $e\_1$, $e\_2$, ..., $e\_k$.
**Useful facts to know:** The cover is connected if and only if the action of the $b(r)$ on $[n]$ is transitive.
The genus of $X$ is given by, the Riemmann-Hurwitz formula:
$$2g-2 = -2n+\sum\_{r \in R} (n-\#\mbox{cycles of $r$}).$$
It is very difficult to obtain an explicit equation for $X$ from the data of $R$ and $b:R \to S\_n$.
| 10 | https://mathoverflow.net/users/297 | 9424 | 6,447 |
https://mathoverflow.net/questions/8731 | 49 | Can there be a foundations of mathematics using only category theory, i.e. no set theory? More precisely, the definition of a category is a class/set of objects and a class/set of arrows, satisfying some axioms that make commuting diagrams possible. So although in question 7627, where psihodelia asked for alternative foundations for mathematics without set theory, there Steven Gubkin said that Lawvere and McCarthy did some work in reformulation the set theoretic axioms ZFC as the axioms of elementary topoi, this manner of foundations is still not complete since a category is still ultimately a set!
J Williams in his answer below noted that via metacategories, we can have a first order axiomatization of categories. However, this does not provide a foundations of mathematics using only category theory, since set theory permeates the formulation of first order logic. In first order logic, structures are sets together with constants, functions and relations. Here constants, functions and relations are also sets. So even if we say that categories are first order axiomatizable, at the very end, categories are still defined in terms of sets.
I admit in wanting foundations totally in terms of categories, then there will be some kind of recursiveness. However, this recursiveness should not be seen as a problem since as described above, first order axiomatization of sets like ZFC, are written in the language of first order logic which (at least in a meta-level) are sets themselves. In fact, this recursiveness is very much a feature of symbolic logic and is partially responsible for the successful that a single primitve concept of set/set-membership can describe so much (or all?) of mathematics.
I'm aware also in certain proofs of equivalence of categories in mainstream math, like GAGA theorems by Serre, there is a need to use categories where the objects are of classes of different levels, like the NBG set theory. In the end, the reasons provided for why the argument of using classes can be pushed down to essentially small category, this in the end invokes NBG set theory.
| https://mathoverflow.net/users/nan | Categorical foundations without set theory | On the subject of categorical versus set-theoretic foundations there
is too much complicated discussion about structure that misses the
essential point about whether "collections" are necessary.
It doesn't matter exactly what your personal list of mathematical
requirements may be -- rings, the category of them, fibrations,
2-categories or whatever -- developing the appropriate foundational
system for it is just a matter of "programming", once you understand
the general setting.
The crucial issue is whether you are taken in by the Great Set-Theoretic
Swindle that mathematics depends on collections (completed infinities).
(I am sorry that it is necessary to use strong language here in order to
flag the fact that I reject a widely held but mistaken opinion.)
Set theory as a purported foundation for mathematics does not and cannot
turn collections into objects. It just axiomatises some of the intuitions
about how we would like to handle collections, based on the relationship
called "inhabits" (eg "Paul inhabits London", "3 inhabits N"). This
binary relation, written $\epsilon$, is formalised using first order
predicate calculus, usually with just one sort, the universe of sets.
The familiar axioms of (whichever) set theory are formulae in first order
predicate calculus together with $\epsilon$.
(There are better and more modern ways of capturing the intuitions about
collections, based on the whole of the 20th century's experience of algebra
and other subjects, for example using pretoposes and arithmetic universes,
but they would be a technical distraction from the main foundational issue.)
Lawvere's "Elementary Theory of the Category of Sets" axiomatises some
of the intuitions about the category of sets, using the same methodology.
Now there are two sorts (the members of one are called "objects" or "sets"
and of the other "morphisms" or "functions"). The axioms of a category
or of an elementary topos are formulae in first order predicate calculus
together with domain, codomain, identity and composition.
Set theorists claim that this use of category theory for foundations
depends on prior use of set theory, on the grounds that you need to start
with "the collection of objects" and "the collection of morphisms".
Curiously, they think that their own approach is immune to the same
criticism.
I would like to make it clear that I do NOT share this view of Lawvere's.
Prior to 1870 completed infinities were considered to be nonsense.
When you learned arithmetic at primary school, you learned some rules that
said that, when you had certain symbols on the page in front of you,
such as "5+7", you could add certain other symbols, in this case "=12".
If you followed the rules correctly, the teacher gave you a gold star,
but if you broke them you were told off.
Maybe you learned another set of rules about how you could add lines and
circles to a geometrical figure ("Euclidean geometry"). Or another one
involving "integration by parts". And so on. NEVER was there a "completed
infinity".
Whilst the mainstream of pure mathematics allowed itself to be seduced
by completed infinities in set theory, symbolic logic continued and
continues to formulate systems of rules that permit certain additions
to be made to arrays of characters written on a page. There are many
different systems -- the point of my opening paragraph is that you can
design your own system to meet your own mathematical requirements --
but a certain degree of uniformity has been achieved in the way that they
are presented.
* We need an inexhaustible supply of VARIABLES for which we may substitute.
* There are FUNCTION SYMBOLS that form terms from variables and other terms.
* There are BASE TYPES such as 0 and N, and CONSTRUCTORS for forming new
types, such as $\times$, $+$, $/$, $\to$, ....
* There are TRUTH VALUES ($\bot$ and $\top$), RELATION SYMBOLS ($=$)
and CONNECTIVES and QUANTIFIERS for forming new predicates.
* Each variable has a type, formation of terms and predicates must respect
certain typing rules, and each formation, equality or assertion of a
predicate is made in the CONTEXT of certain type-assignments and
assumptions.
* There are RULES for asserting equations, predicates, etc.
We can, for example, formulate ZERMELO TYPE THEORY in this style. It has
type-constructors called powerset and {x:X|p(x)} and a relation-symbol
called $\epsilon$. Obviously I am not going to write out all of the details
here, but it is not difficult to make this agree with what ordinary
mathematicians call "set theory" and is adequate for most of their
requirements
Alternatively, one can formulate the theory of an elementary topos is this
style, or any other categorical structure that you require. Then a "ring"
is a type together with some morphisms for which certain equations are
provable.
If you want to talk about "the category of sets" or "the category of rings"
WITHIN your tpe theory then this can be done by adding types known as
"universes", terms that give names to objects in the internal category
of sets and a dependent type that provides a way of externalising
the internal sets.
So, although the methodology is the one that is practised by type theorists,
it can equally well be used for category theory and the traditional purposes
of pure mathematics. (In fact, it is better to formalise a type theory
such as my "Zermelo type theory" and then use a uniform construction to
turn it into a category such as a topos. This is easier because the
associativity of composition is awkward to handle in a recursive setting.
However, this is a technical footnote.)
A lot of these ideas are covered in my book "Practical Foundations of
Mathematics" (CUP 1999), <http://www.PaulTaylor.EU/Practical-Foundations>
Since writing the book I have written things in a more type-theoretic
than categorical style, but they are equivalent. My programme called
"Abstract Stone Duality", <http://www.PaulTaylor.EU/ASD> is an example of the
methodology above, but far more radical than the context of this question
in its rejection of set theory, ie I see toposes as being just as bad.
| 46 | https://mathoverflow.net/users/2733 | 9428 | 6,451 |
https://mathoverflow.net/questions/9420 | 4 | Let |V| be a (incomplete) linear series on a nonsingular projective surface. Hironaka says that there is a resolution of the singularities of |V| along smooth centers. If the base locus of |V| is just a collection of points, does it mean I can acheive this resolution by a series of blow ups at points?
| https://mathoverflow.net/users/nan | resolution of singularities on surfaces | I may be misunderstanding something but this question does not seem to have anything to do with Hironaka's desingularization. You are asking if you can resolve the indeterminacy of a rational map, right? If this is the question, then you can do it with finitely many blow-ups at points.
Suppose you have a rational map $f$ from a smooth projective surface $S$ to $\mathbb{P}^{n}$ given by some linear system $V \subset |D|$. Without loss of generality you may assume that the image of $f$ is non-degenerate, i.e. is not contained in any hyperplane. In that case we can subtract the fixed component of $V$ to make the base locus of $V$ at most finite. Now take a base point of $V$ and blow it up $g : \widehat{S} \to S$. The pullback $g^{\\*}V$ will consist of divisors in $|g^{\\*}D|$ which contain a positive multiple $nE$ of the exceptional divisor $E$. So $g^{\\*}V$ has a fixed component and therefore is of the form $\widehat{V} + nE$, where $\widehat{V} \subset |g^{\\*}D - nE|$. The linear system $\widehat{V}$ gives a rational map $\hat{f}$ from $\widehat{S}$ to $\mathbb{P}^{n}$ which has no fixed component and coincides with the original $f$ on $\widehat{S} - E$. Now set $\widehat{D} = g^{\\*}D - nE$. Note that the fact that $\widehat{D}$ has no fixed component implies that $\widehat{D}^{2} = D^{2} - n^{2} < D^{2}$. If $\hat{f}$ has no base point, then we are done. If it does have base points, then we can repeat the process. Since the degree of the linear system decreases on each step the process will terminate.
| 9 | https://mathoverflow.net/users/439 | 9430 | 6,452 |
https://mathoverflow.net/questions/9449 | 8 | Not every orientable 3-manifold is a double cover of $S^3$ branched over a link. For example, the 3-torus isn't. However, in 1975 Montesinos conjectured (Surjery on links and double branched covers of $S^3$, in: "Knots, groups and 3-manifolds", papers dedicated to the memory of R. Fox) that every orientable 3-manifold is a double branched cover of a sphere with handles i.e. the connected sum of a certain number of copies of $S^1\times S^2$ (this number can be zero, in which case we get $S^3$). Notice that this time $T^3$ does not provide a counter-example since if we take the quotient of $T^3$ by the involution $(x,y,z)\mapsto (x^{-1}, y^{-1},z)$ we get $S^2\times S^1$.
I was wondering what the status of this conjecture is.
| https://mathoverflow.net/users/2349 | A conjecture of Montesinos | It is false. For example, there are closed, orientable, aspherical 3-manifolds that admit no nontrivial action of a finite group whatsoever. The first examples were due to F. Raymond and J. Tollefson in the 1970s, I believe.
| 11 | https://mathoverflow.net/users/1822 | 9452 | 6,467 |
https://mathoverflow.net/questions/9466 | 6 | Wikipedia tells me that:
Gaussian curvature is the limiting difference between the circumference of a geodesic circle and a circle in the plane:
$K = \lim\_{r \rightarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}$
Gaussian curvature is the limiting difference between the area of a geodesic circle and a circle in the plane:
$K = \lim\_{r \rightarrow 0} (\pi r^2 - \mbox{A}(r)) \cdot \frac{12}{\pi r^4}$
Can anyone explain to me why we are dividing by the factors $\frac{3}{\pi r^3}$ and $\frac{12}{\pi r^4}$ respectively? I don't understand why we are dividing by these particular factors?
| https://mathoverflow.net/users/2011 | Why these particular numerical factors in the definition of Gaussian curvature? | First, I guess it should say "geodesic disc" rather than "circle". At least to me, a geodesic circle is a closed geodesic loop in your surface, whereas a geodesic disc of radius r is all the points distance r from a fixed point (at least for r smaller than the injectivity radius). Note the boundary of a geodesic disc is *not* a geodesic.
As for the factors in those formulae, well, there's no absolute scale for Gaussian curvature. People have just agreed on the convention that the curvature of the unit sphere should be 1. (**EDIT:** As Greg Kuperberg points out in his answer, there are some good reasons for this convention. E.g., Gauss-Bonnet.) That then forces those factors to be what they are. It amounts to the statement that, for a small geodesic disc on the unit sphere of radius r ,
$$C(r) \sim 2\pi\left( r - \frac{1}{6} r^3\right),$$
and a similar formula for the area. There really is no deeper reason than that.
So, to see if the factors are right (and you should never trust what you read on the internet!) I would suggest doing exactly those calculations for the unit sphere. I've checked the first formula involving the circumference and it looks good to me. If you have problems with the calculation, leave a comment and I'll write my version down for you, but I have a feeling it's best to do these things yourself.
| 9 | https://mathoverflow.net/users/380 | 9467 | 6,477 |
https://mathoverflow.net/questions/9468 | 15 | I come from a background of having done undergraduate and graduate courses in General Relativity and elementary course in riemannian geometry.
Jurgen Jost's book does give somewhat of an argument for the the statements below but I would like to know if there is a reference where the following two things are proven explicitly,
1. That the sectional curvature of a 2-dimensional subspace of a tangent space at a point on the Riemannian manifold is independent of the choice of basis. That is the definition of the sectional curvature depends only on the choice of the 2-dim subspace.
2. That the sectional curvature determines the Riemannian curvature fully.
Secondly can one give me a reference where I can see how in practice is sectional curvature computed. To a first timer to this subject it is not obvious how one does a calculation on "all" 2-dimensional subspaces of a high-dimensional space. Especially when people talk of manifolds with "constant sectional curvature". How are they realized?
I would like to see some explicit examples to understand this point.
Further some studies about homogeneous spaces (needed to understand some issues in Quantum Field Theory) got me to the following 4 very non-trivial ideas in Riemannian manifolds which I am stating in my own way here ,
1. That the isometry group of a Riemannian manifold is always a lie group.
2. The isotropy subgroup of any point on a Riemannian manifold under the
smooth transitive action of its own isometry group on itself is a compact
subgroup. (The context being what is called a "Riemannian Homogeneous Space")
3. *{This point was earlier framed in a way which made the bi-implication false as pointed out by some people}* The formulation should be as follows.
A *Riemannian Homogeneous Space* is a riemannian manifold on which the isometry group acts transitively. Now the theorem is that such a space is compact IFF its isometry group is compact.
Thats the statement whose intuition I am looking for.
Apologies for the confusion caused.
4. *{ This question too was not framed properly. Basically I could not figure out how to write the nabla for the connection! It should be as Jose has pointed out.}*
A riemannian manifold is locally symmetric if and only if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection.
Can one give me the intuition behind these or give me specific references where these are proven in explicit details?
| https://mathoverflow.net/users/2678 | Riemannian Geometry | *To get a better feel of the Riemann curvature tensor and sectional curvature:*
1. Work through one of the definitions of the Riemann curvature tensor and sectional curvature with a $2$-dimensional sphere of radius $r$.
2. Define the hyperbolic plane as the space-like "unit sphere" of $3$-dimensional Minkowski space, defined using an inner product with signature $(-,+,+)$. Work out the sectional and Riemann curvature of that
3. Repeat #1 and #2 for the $n$-dimensional sphere and hyperbolic space, as well as flat space
*Sectional curvature determines Riemann curvature:*
That the sectional curvature uniquely determines the Riemann curvature is a consequence of the following:
1. The Riemann curvature tensor is a quadratic form on the vector space of $\Lambda^2T\_xM$
2. The sectional curvature function corresponds to evaluating the Riemann curvature tensor (as a quadratic form) on decomposable elements of $\Lambda^2T\_xM$
3. There is a basis of $\Lambda^2T\_xM$ consisting only of decomposable elements
*Added in response to Anirbit's comment*
Perhaps you shouldn't try to compute the curvature too soon. First, make sure you understand the Riemannian metric of the unit sphere and hyperbolic space inside out. There are many ways to do this. But the most concrete way I know is to use stereographic projection of the sphere onto a hyperplane orthogonal to the last co-ordinate axis. Either the hyperplane through the origin or the one through the south pole works fine. This gives you a very nice set of co-ordinates on the whole sphere minus one point. Work out the Riemannian metric and the Christoffel symbols. Also, work out formulas for an orthonormal frame of vector fields and the corresponding dual frame of 1-forms. Figure out the covariant derivatives of these vector fields and the corresponding dual connection 1-forms.
After you do this, do everything again with hyperbolic space, which is the hypersurface
$-x\_0^2 + x\_1^2 + \cdots + x\_n^2 = -1$ with $x\_0 > 0$
in Minkowski space with the Riemannian metric induced by the flat Minkowski metric. You can do stereographic projection just like for the sphere but onto the unit $n$-disk given by
$x\_1^2 + \cdots + x\_n^2 = 1$ and $x\_0 = 0$,
where the formula for the hyperbolic metric looks just like the spherical metric in stereographic co-ordinates but with a sign change in appropriate places. This is the standard conformal model of hyperbolic space.
After you understand this inside out, you can use these pictures to figure out why the $n$-sphere and its metric is given by $O(n+1)/O(n)$ and hyperbolic space by $O(n,1)/O(n)$ and why the metrics you've computed above correspond to the natural invariant metric on these homogeneous spaces. You can then check that the formulas for invariant metrics on homogeneous spaces give you the same answers as above.
Use references only for the general formulas for the metric, connection (including Christoffel symbols), and curvature. I recommend that you try to work out these examples by hand yourself instead of trying to follow someone else's calculations. If possible, however, do it with another student who is also trying to learn this at the same time.
If, however, you want to peek at a reference for hints, I recommend the book by Gallot, Hulin, and Lafontaine. I suspect that the book by Thurston is good too (I studied his notes when I was a student). For invariant Riemannian metrics on a homogeneous space, I recommend the book by Cheeger and Ebin (available cheap from AMS! When I was a student, I had to pay a hundred dollars for this little book but it was well worth it).
But mostly when I was learning this stuff, I did and redid the same calculations many times on my own. I was never able to learn much more than a bare outline of the ideas from either books or lectures. Just try to get a rough idea of what's going on from the books, but do the details yourself.
| 12 | https://mathoverflow.net/users/613 | 9473 | 6,481 |
https://mathoverflow.net/questions/9484 | 11 | Let $F(k,n)$ be the number of permutations of an n-element set that fix exactly $k$ elements.
We know:
1. $F(n,n) = 1$
2. $F(n-1,n) = 0$
3. $F(n-2,n) = \binom {n} {2}$
...
4. $F(0,n) = n! \cdot \sum\_{k=0}^n \frac {(-1)^k}{k!}$ (the subfactorial)
The summation formula is obviously
$\displaystyle\sum\_{k=0}^n F(k,n) = n!$
A recursive definition of $F(k,n)$ is (my claim):
$$F(k,n) = \binom {n} {k} \cdot \Big( k! - \displaystyle\sum\_{i=0}^{k-1} F(i,k) \Big)$$
Question 1: Is there a common name for the "generalized factorial" $F(k,n)$?
Question 2: Does anyone know a closed form for $F(k,n)$ or have an idea how to get it from the recursive definition? (generating function?)
| https://mathoverflow.net/users/2672 | Number of permutations with a specified number of fixed points | The "semi-exponential" generating function for these is
$\sum\_{n=0}^\infty \sum\_{k=0}^n {F(k,n) z^n u^k \over n!} = {\exp((u-1)z) \over 1-z}$
which follows from the exponential formula.
These numbers are apparently called the [rencontres numbers](https://oeis.org/A008290) although I'm not sure how standard that name is.
Now, how do we get a formula for these numbers out of this? First note that
$$exp((u-1)z) = 1 + (u-1)z + {(u-1)^2 \over 2!} z^2 + {(u-1)^3 \over 3!} z^3 + \cdots $$
and therefore the "coefficient" (actually a polynomial in $u$) of $z^n$ in $exp((u-1)z)/(1-z)$ is
$$ P\_n(u) = 1 + (u-1) + {(u-1)^2 \over 2!} + \cdots + {(u-1)^n \over n!} = \sum\_{j=0}^n {{(u-1)^j } \over j!} $$
since division of a generating function by $1-z$ has the effect of taking partial sums of the coefficients.
The coefficient of $u^k$ in $P\_n(u)$ (which I'll denote $[u^k] P\_n(u)$, where $[u^k]$ denotes taking the $u^k$-coefficient) is then
$$ [u^k] P\_n(u) = \sum\_{j=0}^n [u^k] {(u-1)^j \over j!} $$
But we only need to do the sum for $j = k, \ldots, n$; the lower terms are zero, since they are the $u^k$-coefficient of a polynomial of degree less than $k$. So
$$ [u^k] P\_n(u) = \sum\_{j=k}^n [u^k] {(u-1)^j \over j!} $$
and by the binomial theorem,
$$ [u^k] P\_n(u) = \sum\_{j=k}^n {(-1)^{j-k} \over k! (j-k)!} $$
Finally, $F(k,n) = n! [u^k] P\_n(u)$, and so we have
$$ F(k,n) = n! \sum\_{j=k}^n {(-1)^{j-k} \over k!(j-k)!} $$
| 13 | https://mathoverflow.net/users/143 | 9486 | 6,489 |
https://mathoverflow.net/questions/9465 | 28 | Gabor Toth's [Glimpses of Algebra and Geometry](http://rads.stackoverflow.com/amzn/click/0387982132) contains the following beautiful proof (perhaps I should say "interpretation") of the formula $\displaystyle \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$, which I don't think I've ever seen before. Given a non-negative integer $r$, let $N(r)$ be the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 \le r^2$, i.e. the number of lattice points in the ball of radius $r$. Then if $r\_2(n)$ is the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 = n$, it follows that $N(r^2) = 1 + r\_2(1) + ... + r\_2(r^2)$.
On the other hand, once one has characterized the primes which are a sums of squares, it's not hard to show that $r\_2(n) = 4(d\_1(n) - d\_3(n))$ where $d\_i(n)$ is the number of divisors of $n$ congruent to $i \bmod 4$. So we want to count the number of divisors of numbers less than or equal to $r^2$ congruent to $i \bmod 4$ for $i = 1, 3$ and take the difference. This gives
$\displaystyle \frac{N(r^2) - 1}{4} = \left\lfloor r^2 \right\rfloor - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$
and now the desired result follows by dividing by $r^2$ and taking the limit.
**Question:** Does a similar proof exist of the formula $\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ...$?
By "similar" I mean one first establishes a finitary result with a clear number-theoretic or combinatorial meaning and then takes a limit.
| https://mathoverflow.net/users/290 | Is there a "finitary" solution to the Basel problem? | I think that the 14th and last proof in [Robin Chapman's collection](http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf) is just that. It relies on the formula for the number of representations of an integer as a sum of four squares, which is kind of overkill, but anyway.
| 26 | https://mathoverflow.net/users/25 | 9497 | 6,498 |
https://mathoverflow.net/questions/9490 | 19 | If C is a small category, we can consider the category of simplicial presheaves on C. This is a model category in two natural ways which are compatible with the usual model structure on simplicial sets. These are called the *injective* and *projective* model structures, and in both the weak equivalences are the *levelwise weak equivalences*, i.e. those natural transformations $X \to Y$ such that $X(c) \to Y(c)$ is a weak equivalence for all $c \in C$.
The cofibrations in the injective model structure are the levelwise cofibrations. The injective fibrations are then defined as those maps which have the right-lifting property with respect to all cofibrations which are also weak equivalences. The injective *fibrant objects* are those simplicial presheaves in which the map to the terminal presheaf is an injective fibration.
**My Question**: Is there a better characterization of the fibrant objects? If I have an object are there any shortcuts which I can use to test if it is fibrant? An obvious necessary condition is that each simplicial set $X(c)$ must be fibrant (i.e. a Kan complex), but I want sufficient conditions. I'm willing to put restrictions on C.
Dually, we can look at the projective model structure where the fibrations are the levelwise fibrations, and cofibrations have the extension property with respect to these. I'd also be interested in knowing about the cofibrant objects in this model structure, although for now I'm more interested in the injective model structure.
| https://mathoverflow.net/users/184 | What are the fibrant objects in the injective model structure? | In the introduction to his paper "Flasque Model Structures for Presheaves" (in fact simplicial presheaves) Isaksen states on the top of page 2 that his model structure has a nice characterisation of fibrant objects and that "This is entirely unlike the injective model structures, where there is no explicit description of the fibrant objects". This would answer your question. It might be my ignorance, but I think there is no justification for Isaksen quoted statement except that no characterisation is known as yet.
| 9 | https://mathoverflow.net/users/2146 | 9503 | 6,504 |
https://mathoverflow.net/questions/9361 | 2 | I understand if a partition $\lambda$ has all parts even and all multiplicities even, then the nilpotent orbit corresponding to $\lambda$ splits up into two orbits. By the nilpotent orbit corresponding to $\lambda$, what I mean is the set of all orthogonal nilpotent matrices with Jordan type $\lambda$; by 'splitting' I mean that this set is not a single orbit under the action of $SO\_{n}(\mathbb{C})$, but is the union of two orbits (this is a theorem from Collingwood & McGovern - "Nilpotent orbits in Semisimple Lie Algebras").
How can we describe explicitly this ''splitting'' - i.e. since we know the affine coordinate ring of the closure of $\mathcal{O}\_{\lambda}$ (this is the set of all orthogonal nilpotent matrices for the type $\lambda$), by having the equation for a matrix to be orthogonal, together with the equations for it to be nilpotent of type $\lambda$, how can we describe algebraically each of the two orbits that it splits up into?
| https://mathoverflow.net/users/2623 | How can we describe the splitting of nilpotent orbit for "very even" partitions in the special orthogonal group? | OK, I have an answer to the question of how to distinguish different $SO\_{2n}$ orbits that have the same Jordan form. I also have a proof that it is correct, but that is much longer, using the ideas in Ben's [excellent answer](https://mathoverflow.net/questions/9361/how-can-we-describe-the-splitting-of-nilpotent-orbit-for-very-even-partitions-i/9501#9501). So, the short answer for now, and the long answer if it is needed.
Let $N$ be a nilpotent matrix acting on a vector space $V$. Define the vector space $W$ to be the subspace of $V$ given by
$$N \ker N^2 + N^2 \ker N^4 + N^3 \ker N^6 + \cdots.$$
**Exercise:** If $N$ is skew-self-adjoint with respect to an inner product $\langle \ , \ \rangle$, then $W$ is isotropic for $\langle \ , \rangle$.
In terms of Jordan decomposition, $W$ is the "latter half" of each Jordan block.
In particular, suppose all of the Jordan blocks of $N$ have unit size. Then $W$ has dimension $(1/2) \dim V$. It is well known that the Lagrangian grassmannian has two connected components. The orbit of $N$ is determined by which component $W$ is in.
So, how do we determine which component of the Lagrangian grassmannian $W$ is in? First, change bases so that $\langle e\_i, e\_j \rangle = \delta\_{i (j+n)}$, where the indices are modulo $2n$. Consider the $2^n$ subsets $I$ of $\{ 1,2, \ldots, 2n \}$ which contain exactly one of $\{ i, n+i \}$ for each $i$. For half of these $I$, the corresponding Plucker coordinate $p\_I(W)$ is zero. Which half? That tells you which component we are in.
| 2 | https://mathoverflow.net/users/297 | 9516 | 6,513 |
https://mathoverflow.net/questions/9525 | 15 | Where can I find a concrete description of mapping class group of surfaces? I know the mapping class group of the torus is $SL(2, \mathbb{Z})$. Perhaps, there is a simple description for the sphere with punctures or the torus with punctures. Also, I would appreciate any literature reference for an arbitrary surface of genus g with n punctures.
Mapping class groups come up in my reading about billiards and the geodesic flow on flat surfaces. I wonder: the moduli space of complex structures on the torus is $\mathbb{H}/SL(2, \mathbb{Z})$, is it a coincidence the mapping class group appears here?
| https://mathoverflow.net/users/1358 | Mapping Class Groups of Punctured Surfaces (and maybe Billiards) | 1) Let me start by dealing with punctures and higher genus mapping class groups.
Aside from a few low-genus cases, there is no easy description of the mapping class group. As you said, the mapping class group of a torus is $SL\_2(\mathbb{Z})$, and adding one puncture to a torus does not change its mapping class group (adding a boundary component, however, turns it into the 3-strand braid group).
In general (ie for $(g,n)$ not equal to the degenerate cases of $(1,1)$ or $(0,k)$ with $k$ at most $3$), you can relate the mapping class group $\Gamma(g,n)$ of a genus $g$ surface with $n \geq 1$ punctures to the mapping class group of a surface with fewer punctures via the Birman exact sequence. Two forms of it are:
$$1 \longrightarrow \pi\_1(S\_{g,n-1}) \longrightarrow \Gamma(g,n) \longrightarrow \Gamma(g,n-1) \longrightarrow 1$$
and
$$1 \longrightarrow B(g,n) \longrightarrow \Gamma(g,n) \longrightarrow \Gamma(g,0) \longrightarrow 1$$
Here $B(g,n)$ is the $n$-strand braid group on a genus $g$ surface. The map to the cokernel comes from "forgetting" punctures, and the kernel comes from "dragging" punctures around the surface.
A good reference for this material is the book "A Primer on mapping class groups" by Farb and Margalit, which is available [here](http://www.math.utah.edu/~margalit/primer/).
2) As far as moduli space goes, the moduli space of complex structures on a genus $g$ surface with $n$ punctures (as long as $g$ and $n$ are not too small) is isomorphic to the quotient of Teichmuller space by the mapping class group. It is thus no accident that the mapping class group appears in the description of moduli space. Again, Farb and Margalit's book is a nice source.
| 25 | https://mathoverflow.net/users/317 | 9527 | 6,517 |
https://mathoverflow.net/questions/9512 | 21 | **(If you know basics in theoretical computer science, you may skip immediately to the dark box below. I thought I would try to explain my question very carefully, to maximize the number of people that understand it.)**
We say that a *Boolean formula* is a propositional formula over some 0-1 variables $x\_1,\ldots,x\_n$ involving AND and OR connectives, where the atoms are *literals* which are instances of either $x\_i$ or $\neg x\_i$ for some $i$. That is, a Boolean formula is a "monotone" formula over the $2n$ atoms $x\_1,\ldots, x\_n, \neg x\_1,\ldots, \neg x\_n$. For example, $$((\neg x\_1 ~OR~ x\_2) ~AND~ (\neg x\_2 ~OR~ x\_1)) ~OR~ x\_3$$ is a Boolean formula expressing that either $x\_1$ and $x\_2$ take on the same value, or $x\_3$ must be $1$. We say that a 0-1 assignment to the variables of a formula is *satisfying* if the formula evaluates to $1$ on the assignment.
The Cook-Levin theorem says that the general *satisfiability of Boolean formulas* problem is $NP$-complete: given an arbitrary formula, it is $NP$-hard to find a satisfying assignment for it. In fact, even satisfying Boolean formulas where each variable $x\_i$ appears at most three times in the formula is $NP$-complete. (Here is a reduction from the general case to this case: Suppose a variable $x$ appears $k > 3$ times. Replace each of its occurrences with fresh new variables $x^1, x^2, \ldots, x^k$, and constrain these $k$ variables to all take on the same value, by ANDing the formula with $$(\neg x^1 ~OR~ x^2) ~AND~ (\neg x^2 ~OR~ x^3) ~AND~ \cdots ~AND~ (\neg x^{k-1} ~OR~ x^k) ~AND~ (\neg x^k ~OR~ x^1).$$ The total number of occurrences of each variable $x^j$ is now three.) On the other hand, if each variable appears only once in the formula, then the satisfiability algorithm is very easy: since we have a monotone formula in $x\_1,\ldots, x\_n, \neg x\_1,\ldots, \neg x\_n$, we set $x\_i$ to $0$ if $\neg x\_i$ appears, otherwise we set $x\_i$ to $1$. If this assignment does not get the formula to output $1$, then no assignment will.
>
> My question is, suppose every variable appears at most *twice* in a general Boolean formula. Is the satisfiability problem for this class of formulas $NP$-complete, or solvable in polynomial time?
>
>
>
**EDIT: To clarify further, here is an example instance of the problem:
$$((x\_1 ~AND~ x\_3) ~OR~ (x\_2 ~AND~ x\_4 ~AND~ x\_5)) ~AND~ (\neg x\_1 ~OR~ \neg x\_4) ~AND~ (\neg x\_2 ~OR~ (\neg x\_3 ~AND~ \neg x\_5))$$**
Note that when we restrict the class of formulas further to conjunctive normal form (i.e. a depth-2 circuit, an AND of ORs of literals) then this "at most twice" problem is known to be solvable in polynomial time. In fact, applying the "resolution rule" repeatedly will work. But it is not clear (at least, not to me) how to extend resolution for the class of general formulas to get a polytime algorithm. Note when we reduce a formula to conjunctive normal form in the usual way, this reduction introduces variables with *three* occurrences. So it seems plausible that perhaps one might be able to encode an $NP$-complete problem in the additional structure provided by a formula, even one with only two occurrences per variable.
My guess is that the problem is polynomial time solvable. I'm very surprised that I could not find a reference to this problem in the literature. Perhaps I'm just not looking in the right places.
**UPDATE: Please think about the problem before looking below. The answer is surprisingly simple.**
| https://mathoverflow.net/users/2618 | Satisfiability of general Boolean formulas with at most two occurrences per variable | A theorem in a paper of Peter Heusch, "The Complexity of the Falsifiability Problem for Pure Implicational Formulas" (MFCS'95), seems to suggest the problem is NP-hard. I repeat the first part of its proof here:
By reduction from the restricted version of 3SAT where every variable occurs at most 3 times.
Given such a CNF, WOLOG every variable with 3 occurrences occurs once positively and twice negatively. If $x$ is such a variable, let $C\_1$, $C\_2$ be the clauses such that $C\_1 = \neg x \vee C\_1'$ and $C\_2 = \neg x \vee C\_2'$. Introduce new variables $x'$ and $x''$ and replace $C\_1$ and $C\_2$ by $\neg x \vee (x' \wedge x'')$, $\neg x' \vee C\_1'$, $\neg x'' \vee C\_2'$. Repeat.
| 20 | https://mathoverflow.net/users/658 | 9529 | 6,519 |
https://mathoverflow.net/questions/9504 | 13 | I’m studying some category theory by reading Mac Lane linearly and solving exercises.
In question 5.9.4 of the second edition, the reader is asked to construct left adjoints for each of the inclusion functors $\mathbf{Top}\_{n+1} $ in $\mathbf{Top}\_n$, for $n=0, 1, 2, 3$, where $\mathbf{Top}\_n$ is the full subcategory of all $T\_n$-spaces in Top, with $T\_4$=Normal, $T\_3$=Regular, etc.
For $n=0, 1, 2$, it seems to me that I can use the AFT, with the solution set constructed similarly to the one constructed for proving that Haus (=$\mathbf{Top}\_2$) is a reflective subcategory of Top (Proposition 5.9.2, p. 135 of Mac Lane).
But I can’t figure out what should I do with the case of $n=3$, that is, with the inclusion functor $\mathbf{Top}\_4$ in $\mathbf{Top}\_3$: $\mathbf{Top}\_4$ doesn’t even have products, so it seems that I cannot use the AFT.
Is there some direct construction of this left adjoint (by universal arrows, perhaps)? Answers including a reference would be especially helpful.
| https://mathoverflow.net/users/2734 | Is Top_4 (normal spaces) a reflective subcategory of Top_3 (regular spaces)? | I think that MacLane made a mistake. I think that he just forgot that the category of $T\_4$ spaces lacks closure properties.
**Claim:** If $\mathcal{A} \subseteq \mathcal{C}$ is a (full) reflective subcategory and $\mathcal{C}$ has finite products, then $\mathcal{A}$ is closed under $\mathcal{C}$'s finite products, up to isomorphism.
If $\mathcal{A}$ is reflective, it means that an object $X \in \mathcal{C}$ has an "$\mathcal{A}$-ification" $X'$, e.g., an abelianization in the case where $\mathcal{A}$ is abelian groups and $\mathcal{C}$ is groups. If $A, B \in \mathcal{A}$ are objects, then they have a product $A \times B$ in $\mathcal{C}$, and then that has an $\mathcal{A}$-ification $(A \times B)'$. There is an $\mathcal{A}$-ification morphism $A \times B \to (A \times B)'$, and there are also projection morphisms $A \times B \to A, B$. Since $A, B \in \mathcal{A}$, the projection morphisms factor through $(A \times B)'$, and then the universal property gives you a morphism $(A \times B)' \to A \times B$. So you get canonical morphisms in both directions between $A \times B$ and $(A \times B)'$, and I think that some routine bookkeeping shows that they are inverses.
So the Sorgenfrey plane (which is the non-normal Cartesian square of the Sorgenfrey line) does not have a "$T\_4$-ification".
I got onto this track after I found the paper, [Reflective subcategories and generalized covering spaces](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-12/issue-3/Full-reflective-subcategories-and-generalized-covering-spaces/10.1215/ijm/1256054104.full), by Kennison. Kennison lists closure under products as a necessary condition for reflectiveness.
| 15 | https://mathoverflow.net/users/1450 | 9530 | 6,520 |
https://mathoverflow.net/questions/9523 | 6 | Is there such a thing as "recursively dependent types"? Specifically, I would like a dependent type theory containing a type $A(x)$ which depends on a variable $x: A(z)$, where $z$ is a particular constant of type $A(z)$.
This may be more "impredicative" than some type-theorists would like, but from the perspective of semantics in locally cartesian closed categories, I can't see any reason it would be a problem: the type $A$ comes with a display map to $A\_z$, while $A\_z$ itself is the pullback of this display map along a particular morphism $z \colon 1\to A\_z$. But I want to know whether a corresponding syntax exists.
| https://mathoverflow.net/users/49 | Recursively dependent types? | If $z$ is a constant, it's completely unproblematic, but it's troublesome if $z$ is a variable. Here's a simple example: suppose $A$ is a type operator of kind $\mathbb{N} \to \star$, defined as follows:
$\matrix{
A(z) & = & \mathbb{N} \\\
A(n + 1) & = & \mathbb{N} \times A(n) }$
Then it's obviously the case that $z : A(z)$.
OTOH, if $z$ is a variable, on the other hand, then you'll run into the difficulty that the standard well-formedness rule for contexts says that for $\Gamma, x:A$ to be a well-formed context, then $\Gamma \vdash A$ -- that is, $A$ should be well-formed with respect to the variables in $\Gamma$.
Now, there's nothing semantically wrong about adding fixed point operators at every kind. That is, have type operators with kinds like $\mu : (\star \to \star) \to \star$, or $\mu' : ((\star \to \star) \to (\star \to \star)) \to \star \to \star$ or $\mu'' : ((\mathbb{N} \to \star) \to (\mathbb{N} \to \star)) \to \mathbb{N} \to \star$. This kinds of dependency, where the dependent index can vary at every level of a data structure, are very useful when programming in type theory. For example, we might define the type of lists of type $A$, indexed by length in the following way:
$\array{nil & : & list(z) \\\
cons & : & \forall n:\mathbb{N}.\; A \to list(n) \to list(n+1)}
$
So here, $list$ is the least fixed point of a type operator of kind $(\mathbb{N} \to \star) \to (\mathbb{N} \to \star)$. However, most type theories avoid adding the generic operator (like $\mu''$ above) in favor of only permitting inductive types as primitive definitions.
This is partly for philosophical reasons, and partly for pragmatic reasons involving not wanting to require supplying a well-ordering at each elimination of a recursive type. This is necessary to avoid being able to use a recursive type like $\mu \alpha.\; \alpha \to \alpha$ to introduce general recursion and inconsistency into the type theory. However, this complicates typechecking quite a bit -- if you're not very careful, you can lose decidability of typechecking. (In particular, in an inconsistent context, you can cook up with a bogus well-order using the local contradiction, and then use that to tip a conversion rule based on blind $\beta$-reduction into going into an infinite loop. This is not a problem for consistency, but it can annoy users.)
If you're okay with impredicativity, I don't think there are any semantic issues related to consistency, though.
| 4 | https://mathoverflow.net/users/1610 | 9535 | 6,523 |
https://mathoverflow.net/questions/9541 | 14 | I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space and let $S \subseteq \mathrm{End}\_k(V)$ be a subset of pairwise commuting (i.e. $\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds:
1. If all $f \in S$ are diagonalizable, then there exist maps $\chi\_i:S \rightarrow k$, $i=1,\ldots,r$, such that $V = \bigoplus\_{i=1}^r E\_{\chi\_i}(S)$, where $E\_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$.
2. The maps $\chi\_i$ in 1 are unique.
3. 1 is equivalent to the existence of a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M\_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is diagonal. (I believe that this might not be true)
4. If all $f \in S$ are trigonalizable, then there exists a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M\_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is upper triangular and for each diagonalizable $f \in S$ the matrix $M\_{\mathcal{B}}(f)$ is diagonal.
I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the $\chi$-maps when I try to prove this by induction on $\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think.
Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :)
One additional question: Suppose that $k$ is algebraically closed and that $G$ is an affine commutative algebraic group over $k$ which coincides with its semisimple part, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi\_i:G \rightarrow \mathbb{G}\_{m}$ morphisms of algebraic groups?
| https://mathoverflow.net/users/717 | Simultaneous diagonalization | All of these are true. First note that the space of endomorphisms of $V$ is finite-dimensional, so even an infinite $S$ can just be replaced by finitely many matrices that have the same span (it's really more elegant to think about the span of $S$ as a Lie algebra, rather than $S$ itself). You actually may want to look at some discussion of abelian Lie algebras, since really your question is about the natural structure theorem for semi-simple representations of abelian Lie algebras (if you think in this language question 4) is obvious from the first 3, since any representation has a flag whose successive quotients are semi-simple).
The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E\_\chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.
For 3) => 1), you have associated to each basis vector a $\chi$, given by looking at how the elements of $S$ act on it. $E\_\chi$ is just the span of all vectors associated to the particular map $\chi$.
For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-\lambda I)^i$ for some scalar $\lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.
| 13 | https://mathoverflow.net/users/66 | 9543 | 6,528 |
https://mathoverflow.net/questions/9321 | 25 | Quillens higher K-groups of rings can be realized as πnK(C) - the Waldhausen K-Theory of a suitable Waldhausen category C. Is this also true for Milnor K-Theory of Rings? Is there a functor F from rings to waldhausen categories s.t. $K^M\_n(R)\cong \pi\_n(K(F(R))$?
| https://mathoverflow.net/users/2146 | Does Milnor K-Theory arise from Waldhausen K-Theory | I don't know if there any evidence for this to be true. Note that Quillen K-groups *are defined* as homotopy groups of some space (+-construction, Q-construction, Waldhausen construction etc), whereas Milnor K-groups were defined in terms of generators and relations,
which generalize generators and relations for classical K\_2.
More invariantly Milnor K-groups can be constructed using homology of GL\_n (paper of Suslin and Nesterenko) or as certain motivic cohomology groups of a field (Suslin-Voevodsky).
However, these constructions are unrelated to any homotopy groups.
Also, I'm not sure how you define Milnor K-theory for a general ring R?
(I was interpreting your question with "ring R" replaced by "field F".)
| 6 | https://mathoverflow.net/users/2260 | 9545 | 6,529 |
https://mathoverflow.net/questions/9264 | 2 | Suppose $Q$ is an atomless countable boolean algebra, and $B$ is an arbitrary atomless boolean algebra. $Q$ is unique modulo isomorphisms. There is a subalgebra in $B$ that is isomorphic to $Q$. There is probably a mapping from $B$ to $Q$ that preserves all boolean operations, but I need something different. Let $f$ be an epimorphism (cover) from $B$ to $Q$ with the following properties:
1. If $b\_1\le b\_2$ then $f(b\_1)\le f(b\_2)$. (previously there was requirement that $\wedge$ should be preserved, but actually I don't need it)
2. $f(b\_1\vee b\_2)=f(b\_1)\vee f(b\_2)$
3. $f(b)=0$ iff $b=0$
4. $f(1)=1$
The negation operation does not need to be preserved.
Does $f$ exist for any $B$?
Example:
$B = \{\textrm{periodic sequences of non-negative reals with integer period}\}\subseteq 2^{\{r\ge 0\}}$
$Q = \{\textrm{periodic sequences of non-negative integers}\}\subseteq 2^{\{k\ge 0\}}$
$f(\alpha)=\{i : [i,i+1)\cap\alpha\ne\varnothing\}$
| https://mathoverflow.net/users/200 | Countable atomless boolean algebra covered by a larger boolean algebra | The answer to the revised version of the question is **Yes**. In fact, there is no need to assume that B is atomless, but rather, only that it is infinite.
Suppose that B is any infinite Boolean algebra. It follows that there is a countable maximal antichain A subset B. The idea of the proof is to map A arbitrarily into your countable atomless Boolean algebra Q, and then extend to B in a way I will describe. Enumerate the maximal antichain A = { an | n in ω } and the nonzero elements of Q as { qn | n in ω}. We will associate an with qn. In order to define f, suppose that b is any element of B. Let Ab = { qn | b ∧ an not = 0 } be the associated set in Q. Define the function f:B to Q by f(b) = ∨ Ab, if Ab is finite, and otherwise f(b)=1.
We now make several observations about this function f. First, the function is clearly onto, since f(an)=qn. Also, f(1)=1, since 1 meets every element of A, and f(0)=0 since 0 meets no elements of A. Moreover, f(b)=0 iff b=0, since no nonzero element of B has zero meet with every element of A, as A was a *maximal* antichain.
Because (b ∨ c) ∧ a = (b ∧ a) ∨ (c ∧ a), it follows that Ab ∨ c = Ab union Ac. From this, it follows that f(b ∨ c) = f(b) ∨ f(c), since if either set is infinite, then the answer is 1, and if they are finite, we are taking the join of two finite joins. Thus, f is join-preserving.
It follows that f is an order-homomorphism, since b <= c implies b ∨ c = c implies f(b) ∨ f(c) = f(b ∨ c) = f(c) implies f(b) <= f(c).
So f has all the desired properties.
Note that f definitely does not respect negation, since f(neg an) = 1 for every n. And f definitely does not respect meet, since any two elements of A have meet 0, but the corresponding qn must sometimes be nonzero.
This construction has some affinity with your example. Namely, if you take the various half-open unit characteristic functions as the elements of the maximal antichain (and use the corresponding qn's), then your f and my construction are the same.
| 1 | https://mathoverflow.net/users/1946 | 9559 | 6,537 |
https://mathoverflow.net/questions/8537 | 19 | There are models of differential geometry in which the intermediate value theorem is not true but every function is smooth. In fact I have a book sitting on my desk called "Models for Smooth Infinitesimal Analysis" by Ieke Moerdijk and Gonzalo E. Reyes in which the actual construction of such models is carried out. I'm quite new to this entire subject and I only stumbled upon it because I was trying to find something like non-standard analysis for differential geometry.
Already I'm liking the more natural formulations for differentials and tangent vectors in the new setting although I can see that true mastery of all the intricacies will require more background in category theory like Grothendieck topologies. So my questions are a bit philosophical. Suppose some big conjecture is refuted in one of these models but proven to be true in the classical setting then what exactly would that mean for classical differential geometry? Is such a state of affairs possible or am I missing something that rules out such a possibility like a metatheorem that says anything that can be proven in the new models can be proven in the usual classical model? More specifically what is the exact relationship between the new models and the classical one? Could one even make any non-trivial comparisons? References for such discussions are welcome. I'm asking the question here because I suspect there might be some experts familiar with synthetic differential geometry that will be able to illuminate the connection to the classical theory.
**Edit**: Found a very lively and interesting discussion by John Baez, Andrew Stacy, Urs Schreiber, Tom Leinster and many others on n-category cafe called [Comparative Smootheology](http://golem.ph.utexas.edu/category/2008/04/comparative_smootheology_ii.html) although I couldn't make out the exact relation to SDG.
| https://mathoverflow.net/users/nan | synthetic differential geometry and other alternative theories | Perhaps I can make the implications of what Harry said a bit more explicit. A well-adapted model of SDG embeds smooths manifolds fully and faithfully. This in particualar means that the SDG model and the smooth manifolds "believe" in the same smooths maps between smooth manifolds (but SDG model contains generalized spaces which do no correspond to any manifold), and moreover precisely the same equations hold in the SDG model and in smooth manifolds. In this sense SDG is conservative: the model will never validate an invalid equation involving smooth maps between smooth manifolds.
The situation is really quite similar to other situations where we have to distinguish between truth and meaning inside a model and truth and meaning outside the model. For example, there are models of set theory which violate the axiom of choice, but these models are built in a setting where the axiom of choice holds. This is no mystery or magic, as long as we remember that a statement can have a different meanings inside the model and outside. The same applies to SDG: when the internal meaning of statements in the model is appropriately interpreted on the outside, nothing can go wrong (that's what a model is, after all).
| 15 | https://mathoverflow.net/users/1176 | 9569 | 6,545 |
https://mathoverflow.net/questions/9566 | 1 | Perhaps this will be a trivial question. For this post, everything is over your favorite field of characteristic $0$.
### Definitions and notation
Recall that a *Lie algebra* is a vector space $\mathfrak g$ along with a map $\beta: \mathfrak g^{\wedge 2} \to \mathfrak g$ satisfying the Jacobi identity. One way to write the Jacobi identity is as follows: extend $\beta$ to $\mathfrak g^{\otimes 2} \to \mathfrak g$ via the usual projection $\mathfrak g^{\otimes 2} \to \mathfrak g^{\wedge 2}$, consider the map $\beta \circ (1 \otimes \beta): \mathfrak g^{\otimes 3} \to \mathfrak g$; then the restriction of this map to $\mathfrak g^{\wedge 3} \subseteq \mathfrak g^{\otimes 3}$ vanishes. (Because $\beta$ vanishes on the symmetric product $\mathfrak g^{\vee 2}$, the Jacobi identity is equivalent to $\beta \circ (1 \otimes \beta)$ vanishing on $\mathfrak g^{\vee 3}$.)
A *Lie coalgebra* is a vector space $\mathfrak g$ with a map $\delta: \mathfrak g \to \mathfrak g^{\wedge 2}$, satisfying the coJacobi identity, which asserts that the map $(\delta \otimes 1) \circ \delta: \mathfrak g \to \mathfrak g^{\wedge 3}$ vanishes. A vector space $\mathfrak g$ that is both a Lie algebra (under $\beta$) and a Lie coalgebra (under $\delta$), is a *Lie bialgebra* if $\beta$ and $\delta$ satisfy an additional relationship. Namely, let $\sigma: \mathfrak g^{\otimes 2} \to \mathfrak g^{\otimes 2}$ be the usual "flip" map; then the bialgebra identity is that $\delta \circ \beta$ and $(1 \otimes \beta)\circ (\delta \otimes 1) + (\beta \otimes 1) \circ (1 \otimes \delta) + (\beta \otimes 1) \circ (1\otimes \sigma) \circ (\delta \otimes 1) + (1 \otimes \beta) \circ (\sigma \otimes 1) \circ (1 \otimes \delta)$ are equal as maps $\mathfrak g^{\otimes 2} \to \mathfrak g^{\otimes 2}$.
### My question
In a calculation I'm doing, I'm led to consider the map $\mathfrak g^{\otimes 2} \to \mathfrak g^{\vee 3}$ given by $(1 \otimes \beta \otimes 1) \circ (\delta \otimes \delta)$. (I mean, $(1 \otimes \beta \otimes 1) \circ (\delta \otimes \delta)$ lands in $\mathfrak g^{\otimes 3}$, but I want the composition with the natural projection $\mathfrak g^{\otimes 3} \to \mathfrak g^{\vee 3}$.) In particular, for the calculation to come out right, I'd like for this map to vanish. Does it?
| https://mathoverflow.net/users/78 | Is this an identity in Lie bialgebras? | (Hopefully this time I did not mess up the indices in the QYBE :/ )
Let $\mathfrak{sl}\\_2$ be spanned by $e$, $f$ and $h$ with $[h,e]=2e$, $[h,f]=-2f$ and $[e,f]=h$, as usual. Let $r=e\wedge f\in\Lambda^2\mathfrak{sl}\\_2$ and let $\delta=[\mathord-,r]:\mathfrak g\to\Lambda^2\mathfrak g$ be the inner derivation corresponding to $r$. Of couse $\delta$ is a $1$-cocycle, and one checks by hand that it is a cobracket; explicitely, $\delta(h)=0$, $\delta(e)=e\wedge h$ and $\delta(f)=f\wedge h$. Let $\gamma:\mathfrak g\otimes\mathfrak g\to S^3\mathfrak g$ be the map in question. Then $\gamma(e\otimes f)\neq0$.
| 4 | https://mathoverflow.net/users/1409 | 9570 | 6,546 |
https://mathoverflow.net/questions/9572 | 0 | I have a web application that prompts users to answer a question when the computer they are using is not recognized. A user complained today saying she is always prompted for the same question. I explained to her that the **pool of questions was only 3**, so the likelihood of her being prompted for the same question was high, can some of you guru's tell me what the exact probability is?
We have the questions the user has set up (3 questions) stored in a database. When choosing which question to display, the developer who initially wrote the code invokes the TSQL [NewID() method](http://msdn.microsoft.com/en-us/library/ms190348.aspx) which gives him three unique identifiers, he sorts these random guids in ascending order and returns the first question associated with the top most guid.
Is it possible for me to provide the probability of to this person of her question being repeated constantly? She said she has been prompted for the same challenge question 12 times in a row. I just want to prove it mathematically.
| https://mathoverflow.net/users/2756 | Random values and their probability of reoccuring | Call your three questions A, B, C.
The probability that A gets chosen twelve times in a row is 1/(3^12), or 1 in 531441; similarly for B and C.
The probability that some question gets chosen twelve times in a row is thus 3/(3^12), or 1/(3^11), or 1 in 177147.
Personally, I think this seems like low enough a probability that if you knew for sure it actually happened, I'd be suspicious that your code isn't doing what you think it does. But it's also possible that your user really didn't get this question 12 times in a row and is just misremembering.
| 0 | https://mathoverflow.net/users/143 | 9573 | 6,547 |
https://mathoverflow.net/questions/9571 | 21 | Every category admits a Grothendieck topology, called canonical, which is the finest topology which makes representable functor into sheaves.
Is there a concrete description of the canonical topology on the category of schemes? By Grothendieck's results on descent this is at least as fine as the fpqc topology, but I don't even know if the two actually coincide. If not, what is known about it?
| https://mathoverflow.net/users/828 | Canonical topology on the category of schemes? | Proposition 3.4 in Orlov's paper
Quasicoherent sheaves in commutative and noncommutative geometry. (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 67 (2003), no. 3, 119--138; translation in Izv. Math. 67 (2003), no. 3, 535--554.
describes the canonical topology and the universally strict epimorphisms on the category of affine schemes. In the same paper Orlov characterizes the quasicoherent sheaves on the small Zariski site in terms of the canonical topology. He also studies an important subcanonical topology - the effective descent topology - which seems to be finer than the flat topology.
| 11 | https://mathoverflow.net/users/439 | 9578 | 6,550 |
https://mathoverflow.net/questions/9581 | 10 | One can define the $G$-equivariant cohomology of a space $X$ as being the ordinary singular cohomology of $X \times\_G EG$ --- I think this is due to Borel? (See e.g. section 2 of [these notes](http://arxiv.org/abs/0709.3615))
Alternatively if $X$ is a manifold, we also have $G$-equivariant de Rham cohomology, defined in terms of $G$-equivariant differential forms --- I think this is due to Cartan? (See e.g. section 3 of loc. cit.)
I suspect this is extremely standard or obvious, but if it is, I don't know where it's written down: Is it possible to define equivariant cohomology of a topological space in terms of some notion of "equivariant singular cochains", that is, *without* using the Borel construction?
| https://mathoverflow.net/users/83 | Equivariant singular cohomology | Here's an answer which I learned from Goresky-Kottwitz-MacPherson's paper on equivariant cohomology and Koszul duality: they use some notion of geometric chain which is probably something like subanalytic chains, but anyway, the idea is as follows.
Suppose $G$ is a compact Lie group of dimension d. An abstract equivariant $k$-chain $c$ is a $(k+d)$-dimensional chain in some $\mathbb R^n$ (or perhaps it's better to say $\mathbb R^\infty$) equipped with a free action of $G$. Then if $X$ is a $G$-space, an equivariant chain in $X$ is a $K$-equivariant map from an abstract chain to $X$. You can obviously form a chain complex out of these things, and the result gives you the equivariant cohomology of $X$ (in the Borel construction say).
| 11 | https://mathoverflow.net/users/1878 | 9583 | 6,553 |
https://mathoverflow.net/questions/9592 | 19 | We have the usual analogy between infinitesimal calculus (integrals and derivatives) and finite calculus (sums and forward differences), and also the generalization of infinitesimal calculus to fractional calculus (which allows for real and even complex powers of the differential operator). Have people worked on a "fractional finite" calculus, where instead of a differintegral we had some sort of "differsum"? I don't know much about it, but I was thinking maybe the answer might come from umbral calculus?
To give a motivating example/special case for this question: the [Wikipedia article on fractional calculus](http://en.wikipedia.org/wiki/Fractional_calculus#Half_derivative_of_a_simple_function) uses the example of the $\frac{1}{2}$th derivative, which when applied twice gives the standard derivative. What is the operator $D$ on sequences such that, when applied twice, it gives the forward difference of the original sequence?
Also, I have perhaps a related question: The solution to $\frac{d}{dx}f=f$ is $f=e^x$, while the solution to $\Delta f = f$ is $f=2^x$. Is the fact that $e$ is close to 2 a coincidence, or is there something connecting these results? Is there more generally some sort of spectrum of calculi lying between "finite" and "infinitesimal" each with its own "$e$"?
EDIT: After looking around some more I found [time scales](http://en.wikipedia.org/wiki/Time_scale_calculus), which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!
| https://mathoverflow.net/users/1916 | Generalizations of "standard" calculus | I don't know if you have seen this but there are papers devoted to "discrete fractional calculus". Like this one for example
<http://arxiv.org/abs/0911.3370> or <http://www.math.u-szeged.hu/ejqtde/sped1/103.pdf>
. Like in fractional calculus, of course the discrete fractional integral is easier to define than the discrete fractional derivative.
| 9 | https://mathoverflow.net/users/2384 | 9595 | 6,560 |
https://mathoverflow.net/questions/9584 | 1 |
>
> **Possible Duplicate:**
>
> [What is the max number of points in R^3, interconnected by generic curves?](https://mathoverflow.net/questions/9293/what-is-the-max-number-of-points-in-r3-interconnected-by-generic-curves)
>
>
>
Given a set of points connected by edges lying on an euclidean plane,
I'd like to find which is the smaller dimension of the euclidean space where the graph can lie without an overlapping of the edges. Is it a standard problem? Which mathematical tools I have to know to manage with this kind of problems?
I can obviously guess that I could always take $d=v-1$ where $v$ is the number of verticies
but I can't understand which is the smaller dimension.
Thanks
| https://mathoverflow.net/users/2758 | How many dimensions I need to embed a graph? | As Charles points out, you can always embed a graph in three dimensions. The interesting question is how complicated a **surface** one needs to embed a graph into. The number of handles one has to attach to a spehere in order for a graph to become embeddable is called the **genus** of the graph, see [graph embedding](http://en.wikipedia.org/wiki/Graph_embedding) on Wikipedia, which offers other useful information.
| 6 | https://mathoverflow.net/users/1176 | 9600 | 6,562 |
https://mathoverflow.net/questions/9576 | 63 | Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?
**EDIT** [Bjorn gave additional information in a comment below, which I am recopying here. -- Pete L. Clark]
Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$ ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial.
| https://mathoverflow.net/users/2757 | Smooth proper scheme over Z | Hey Bjorn. Let me try for a counterexample. Consider a hypersurface in projective $N$-space, defined by one degree 2 equation with integral coefficients. When is such a gadget smooth? Well the partial derivatives are all linear and we have $N+1$ of them, so we want some $(N+1)$ times $(N+1)$ matrix to have non-zero determinant mod $p$ for all $p$, so we want the determinant to be +-1. The determinant we're taking is that of a symmetric matrix with even entries down the diagonal (because the derivative of $X^2$ is $2X$) and conversely every symmetric integer matrix with even entries down the diagonal comes from a projective quadric hypersurface. So aren't we now looking for a positive-definite (to stop there being any Q-points or R-points) even unimodular lattice?
So in conclusion I think that the hypersurface cut out by the quadratic form associated in this way to e.g. the $E\_8$ lattice or the Leech lattice gives a counterexample!
| 68 | https://mathoverflow.net/users/1384 | 9605 | 6,565 |
https://mathoverflow.net/questions/9601 | 4 | I have 2 questions - the first is what the title refers to, and the second is something I want a reference on (I thought I'd include them in one post since they are very strongly related). Sorry this post is a bit long, I tried to put as much as detail as I could ..
$1$-st question: I'm interested only in the group $GL\_n(F\_q)$. In Carter's book "Finite Groups of Lie Type: Conjugacy Classes and Complex Characters", in Chapter 7 "The generalized characters of Deligne-Lusztig", the construction of the virtual representations $R\_{T, \theta}$ as alternating sums of $l$-adic cohomology of Deligne-Lusztig varieties is given in some details, and a series of formulae are proved in the chapter about these ($T$ a torus, and $\theta$ a character of $T^{F}$). It says that if $\theta \in \widehat{T^{F}}$ is in general position, then $\pm R\_{T, \theta}$ is irreducible. The following formula is given (also in <http://en.wikipedia.org/wiki/Deligne%E2%80%93Lusztig_theory>), where $g=su=us$, $s,u$ being the semisimple and unipotent parts, and $Q\_{T}(u) = R\_{T, 1}(u)$, $C^{0}(s)$ being the identity connected component of the centralizer of $s$, and $F$ the Frobenius endomorphism.
$ R\_{T, \theta}(g) = \frac{1}{ | C^{0}(s)^{F} |} \sum\_{ x \in G^{F}, x^{-1}sx \in T^{F} } \theta ( x^{-1} s x) Q\_{x T x^{-1}}^{C^{0}(s)} (u) $
The book then says that $Q\_{T}(u)$ is a Green function, depends only on the torus (I understand it will not change if we conjugate the torus in $G^F$ either so essentially corresponds to an element of $S\_n$ for the group general linear group of size $n$, which is what I'm most curious about; unless I'm mistaken). The book does not give an explicit formulae for these $Q\_{T}(u)$, but it does give orthogonality relations and such - explicit formulae is what I"m looking for:
**Question**: What's an explicit formulae for these $Q\_{T}(u)$? How does this relate to the Green function that I've been studying from in Macdonald's book "Symmetric Functions and Hall polynomials", in the chapter "Characters of $GL\_n$ over a finite field" -i.e., how do I express the character $ \pm R\_{T, \theta}$ as a sum of the irreducible characters described by Green functions in Macdonald's book (or a single irreducible character in the case where $\theta$ is in general position)?
In that book, I've learnt that the polynomials correspond to symmetric functions $S\_{\lambda}$, via a correspondence that maps $A$, the sums of the representation ring of for all $n$, to $B$, an algebra generated by elementary symmetric functions in independent variables $X\_{i,f}$ ($f$ ranges over all irreducible polynomials in $\mathbb{F}\_{q}[t]$). I'm sorry I'm being a bit vague right here - it would take pages to define precisely all the notation that Macdonald uses in his book; feel free to work with any alternative explicit definitions of these Green functions (but please include a reference so I know where to look it up).
$2$-nd question: I have looked through Carter's book and Digne&Michel's book on the same topic, but I have been unable to find a reference which gives the representing matrices for these virtual representations $\pm R\_{T, \theta}$ of these finite Lie type groups (the fact that they are defined with alternating sum complicates matters somewhat). I'm not so interested in the entries of the representing matrices as such, just a construction for the module which enables you to find the representing matrices. Can anyone suggest a good reference for this? The closest I can find is Lusztig's original book "Characters of reductive groups over finite fields", where it mentions that $l$-adic intersection homology can be used as a substitute (this was from what I can see in googlebooks preview); but I hear this book is horrible to learn from, and I'm not entirely certain if what's given there is what I'm looking for (I don't have a copy of the book at present).
| https://mathoverflow.net/users/2623 | Relating Deligne-Lusztig virtual representation characters to Green functions | The Green function $Q\_T(u)$ is the value of the Deligne-Lusztig character $R\_T^\theta$ at $u$ (a unipotent element), which turns out not to depend on $\theta$, hence the notation. Conjugacy classes of rational tori in $GL\_n$ are parametrized by conjugacy classes in the symmetric group, so this means you have one Green's function for each partition of $n$. For $GL\_n$ they are (I assume) what Macdonald calls Green functions (I don't have a copy of the book to hand), but regardless they can be made combinatorially explicit in various ways.
To relate to Macdonald's book/Green's paper, you can use the orthogonality formulas for the $R\_T^1$s to show that taking appropriate linear combinations of them according to character values of the symmetric group, you get an orthonormal set of class functions. You can then check that these are irreducible characters, and in fact they are just the representations of $GL\_n(\mathbb F\_q)$ which you get as constituents of $\text{Ind}\_B^G(1)$, where $B$ is the set of upper triangular matrices, the so-called unipotent representations. Thus the change of basis matrix for class functions on the unipotents between Green functions and irreducible unipotent characters is just given by the character table of $S\_n$. I think this is worked out in the book by Digne and Michel in one of the later chapters on examples, if you need a reference.
| 4 | https://mathoverflow.net/users/1878 | 9610 | 6,566 |
https://mathoverflow.net/questions/9611 | 4 | This question is an addition to my [question](https://mathoverflow.net/questions/9541/simultaneous-diagonalization) on simultaneous diagonalization from yesterday and it is probably also obvious but I just don't know this: Let $G$ be a commutative affine algebraic group over an algebraically closed field $k$. Let $G\_s$ be the semisimple part of $G$. Let $\rho:G \rightarrow GL\_n(V)$ be an embedding. Then $\rho(G\_S)$ is a set of commuting diagonalizable endomorphisms and I know from yesterday that I have unique morphisms of algebraic groups $\chi\_i: \rho(G\_s) \rightarrow \mathbb{G}\_m$, $1 \leq i \leq r$, and a decomposition $V = \bigoplus \_{i=1}^r E \_{\chi\_i}$, where $E\_{\chi\_i} = \lbrace v \in V \mid fv = \chi\_i(f)v \ \forall f \in \rho(G\_s) \rbrace$. Now, my question is: are the morphisms $\chi\_i$ independent of $\rho$ so that I get well-defined morphisms $\chi\_i:G\_s \rightarrow \mathbb{G}\_m$?
If somebody knows what I'm talking about, then please change the title appropriately! :)
| https://mathoverflow.net/users/717 | "Eigenvalue characters" | Unless I drastically misunderstand your question, of course the characters $\chi\_i$ depend on the representation $\rho$. Try looking at the simplest nontrivial case: $G = \mathbb{G}\_m$ acting on a one-dimensional vector space. In this case, there is exactly one $\chi\_i$ and it is simply a character of $\mathbb{G}\_m$, i.e., is of the form $x \mapsto x^n$ for a unique
integer $n$. This integer $n$ is determined by (and determines) $\rho$.
| 5 | https://mathoverflow.net/users/1149 | 9612 | 6,567 |
https://mathoverflow.net/questions/9623 | 1 | If I have a singular matrix $X$ with components $X\_{\mu\nu}$:
$t^{\nu}X\_{\mu\nu}=0$
By considering now $X\_{\mu\nu}$'s as components of a 2-form can I say that:
$X\wedge X=0$ ?
If yes, how?
| https://mathoverflow.net/users/2597 | Singular matrix and wedge product | Your condition on $X$ is that it has a kernel, and that by itself does not mean that
$X \wedge X$ doesn't have to vanish. For instance in five dimensions, you could have
$$X = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\\\ -1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & -1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$
Then letting $t = (0,0,0,0,1)$, your condition is satisfied, but $X \wedge X$ is not zero.
However, it is true in $2n$ dimensions that a non-zero $t$ exists if and only if $X^{\wedge n} = 0$. That's because $X^{\wedge n}$ is proportional to the [Pfaffian](http://en.wikipedia.org/wiki/Pfaffian) of $X$, which is a certain square root of the determinant of $X$. (In odd dimensions, the determinant of an antisymmetric matrix is zero by calculation, while the Pfaffian is set to zero by definition. So $t$ always exists in this case.)
| 5 | https://mathoverflow.net/users/1450 | 9624 | 6,574 |
https://mathoverflow.net/questions/9628 | 20 | Here are two questions about finitely generated and finitely presented groups (FP):
1. Is there an example of an FP group that does not admit a homomorphism to $\operatorname{GL}(n,C)$ with trivial kernel for any $n$?
The second question is modified according to the [suggestion](https://mathoverflow.net/a/9635) of Greg below.
2. For which $n$ given two subgroups of $\operatorname{GL}(n,C)$ generated by explicit lists of matrices, together with finite lists of relations and the promise that they are sufficient, is there an algorithm to determine if they are isomorphic as groups?"
In both cases we don't impose any conidtion on the group (apart from being FP), in particular it need not be discrete in $\operatorname{GL}(n,C)$.
| https://mathoverflow.net/users/943 | Finitely presented sub-groups of $\operatorname{GL}(n,C)$ | Here is a more complete picture to go with David's and Richard's answers.
It is a theorem of Malcev that a finitely presented group $G$ is residually linear if and only if it is residually finite. The proof is very intuitive: The equations for a matrix representation of $G$ are algebraic, so there is an algebraic solution if there is any solution. Then you can reduce the field of the solution to a finite field, as long as you avoid all primes that occur in the denominators of the matrices.
The same proof shows that $G$ has no non-trivial linear representations if and only if it has no subgroups of finite index. So Higman's group has this property.
A refined question is to find a finitely presented group which is residually finite, but nonetheless isn't "linear" in the sense of having a single faithful finite-dimensional representation. It seems that the automorphism group of a finitely generated free group, $\text{Aut}(F\_n)$, is an example. [Nielsen](https://doi.org/10.1112/jlms/s2-10.3.265 "McCool, J. On Nielsen’s presentation of the automorphism group of a free group. J. Lond. Math. Soc., II. Ser. 10, 265-270 (1975). zbMATH review at https://zbmath.org/0338.20029") found a finite presentation for this group, it is also [known to be residually finite](https://web.archive.org/web/20100629064551/http://www.math.lsu.edu/%7Ekasten/Papers/abs_lerf.html "Dasbach, Oliver T.; Mangum, Brian S. The automorphism group of a free group is not subgroup separable. Gilman, Jane (ed.) et al., Knots, braids, and mapping class groups. AMS/IP Stud. Adv. Math. 24, 23-27 (2001). zbMATH review at https://zbmath.org/1006.20027"), yet [Formanek and Procesi](https://doi.org/10.1016/0021-8693(92)90029-L "Formanek, Edward; Procesi, Claudio. The automorphism group of a free group is not linear. J. Algebra 149, No. 2, 494-499 (1992). zbMATH review at https://zbmath.org/0780.20023") showed that it is not linear when $n \ge 3$. More recently, [Druţu and Sapir](https://arxiv.org/abs/math/0405470 "Druţu, Cornelia; Sapir, Mark. Non-linear residually finite groups. J. Algebra 284, No. 1, 174-178 (2005), doi:10.1016/j.jalgebra.2004.06.025. zbMATH review at https://zbmath.org/1076.20021") found an example with two generators and one relator.
| 29 | https://mathoverflow.net/users/1450 | 9635 | 6,582 |
https://mathoverflow.net/questions/9627 | 6 | [In what follows $0^{0}$= 1 by convention.]
Is there some closed infinite dimensional linear subspace $F$ of $L^{2}(0,1)$
such that $\left\lvert f\right\rvert^{\left\lvert f\right\rvert}$ belongs to $L^{2}(0,1)$
for all $f$ in $F$ ?
This problem is related to the Erdős–Shapiro–Shields paper [ESS].
From this paper it follows that the answer is negative if $\left\lvert f\right\rvert^{\left\lvert f\right\rvert}$ is replaced by $\left\lvert f\right\rvert^{\left\lvert f\right\rvert^{2}}$.
Some thoughts. Suppose that such an $F$ exists, and take some $p > 2$.
Let $f$ be in $F$.
Then clearly $g:=(p/2)\cdot f$ is in $F$, too, hence $h :=\left\lvert g\right\rvert^{\left\lvert g\right\rvert}$
belongs to $L^{2}(0,1)$.
Next, it is easy to see that $\left\lVert f\right\rVert \_{p}^{p}\leq1+\left\lVert h\right\rVert \_{2}^{2}<+\infty$.
Therefore, $F$ is contained in $L^{p}(0,1)$ as a linear subspace
(i.e., algebraically).
Now, applying the Closed Graph Theorem to the natural linear embedding
$j:(F, \lVert{.}\rVert\_{2})\rightarrow L^{p}(0,1)$, it follows that $j$ is
continuous. Consequently, the Hilbertian 2-norm and the $p$-norm
are equivalent on $F$. Moreover, it follows that $F$ is complete
w.r.t. the $p$-norm, and, in turn, it is a closed subspace of $L^{p}(0,1)$.
And this is true for all $p > 2$.
[ESS] [Erdős, Shapiro, and Shields - Large and small subspaces of Hilbert space](http://doi.org/10.1307/mmj/1028999306).
| https://mathoverflow.net/users/2508 | Subspaces of $L^{2}$ | The classical lacunary series example allows you to integrate anything that is $e^{O(|f|^2)}$, so it works for your question. It seems that for reasonable (say, positive, increasing, and convex) functions $\Phi$, the complete answer is the following: A closed infinite-dimensional subspace $F$ such that $\Phi(|f|)$ is integrable for all $f\in F$ exists if and only if $\log\Phi(x)=O(x^2)$. The reason is that you can always create a random function in the unit ball that is pointwise almost Gaussian by taking a random linear combination of large number $n$ of the elements in the orthonormal basis of $F$ with random coefficients of size $n^{-1/2}$.
| 4 | https://mathoverflow.net/users/1131 | 9638 | 6,584 |
https://mathoverflow.net/questions/8854 | 6 | This can be considered as a relative of [Splitting a space into positive and negative parts](https://mathoverflow.net/questions/7709/splitting-a-space-into-positive-and-negative-parts).
Is there a real (non-trivial) vector space $V$, endowed with a nondegenerate symmetric bilinear pairing $\langle-,-\rangle : V^2 \to \mathbb{R}$, satisfying the following property:
for each function $f \in \mathbb{R}^\mathbb{R}$ with $f(0) = 0$ there is some $F \in V^V$ such that $\langle Fx - Fy, Fx - Fy\rangle = f( \langle x - y, x - y\rangle )$ ($x,y \in V$)?
I'll bet no Krein space can do this.
Somewhat vaguely, the (generally nonlinear) mapping $F$ may be viewed as an "internal realization" of the function $f$.
P.S. By "non-trivial" I mean $V \ne {0}$.
P.P.S. $A^B$ means $\lbrace u | u: B \to A \rbrace$, so that $V^2$ is the Cartesian square (of $V$). Just to avoid any confusion (I hope).
| https://mathoverflow.net/users/2508 | The "ultimate" indefinite inner product space | I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.
### Version 1
I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V\_0 = \mathbb{R}$ with its basis vector $1$. Then let $V\_{n+1}$ be the direct sum of $V\_n$ and the vector space $W\_n$ of formal linear combinations of elements of $V\_n \setminus V\_{n-1}$, where in this formula $V\_{-1} = \emptyset$. If $x \in V\_n \setminus V\_{n-1}$, let $[x]$ denote the corresponding element in $W\_n \subset V\_{n+1}$. Let $V$ be the union of all $V\_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V\_n$.
To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation
$$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$
with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V\_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.
### Version 2
Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W\_n$ be the vector space of formal linear combinations of elements of $(V\_n \setminus V\_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V\_{n+1} = V\_n \oplus W\_n$. In this case, $W\_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.
As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before,
$$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$
with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.
### Version 3
Ady reminds me that the second version still misses the condition that the bilinear form on $V$ should be non-degenerate. I think that the same trick works a third time: We can just enlarge $V$ to also guarantee this condition. This time let $W\_n$ be as in the second version, and let
$$V\_{n+1} = V\_n \oplus W\_n \oplus V\_n^\*,$$
where $V\_n^\*$ is the (algebraic) dual vector space to $V\_n$.
Define the bilinear form on $V\_n \oplus W\_n$ as in version 2, and define $F$ as in version 2. The bilinear form on $V\_n^\*$ is unrestricted, and so is the bilinear pairing between $V\_n^\*$ and $W\_n$. Finally the bilinear pairing between $V\_n^\*$ and $V\_n$ should be the canonical pairing $\langle \phi, x \rangle = \phi(x)$. This guarantees that for every vector $x \in V\_n$, there exists $y \in V\_{n+1}$ such that $\langle y,x \rangle = 1$.
Every version of the construction is cheap in the sense that the image of $F$ is a linearly independent set. Moreover, in the second and third versions, the image of $F$ is far from a basis. My feeling is that it is difficult to ask for much better than that in a universal construction.
| 5 | https://mathoverflow.net/users/1450 | 9650 | 6,590 |
https://mathoverflow.net/questions/9641 | 37 | A while ago I heard of a nice characterization of compactness but I have never seen a written source of it, so I'm starting to doubt it. I'm looking for a reference, or counterexample, for the following . Let $X$ be a Hausdorff topological space. Then, $X$ is compact if and only if $X^{\kappa}$ is Lindelöf for any cardinal $\kappa$.
If the above is indeed a fact, can one restrict the class of $\kappa$'s for which the characterization is still valid?
Note: Here I'm thinking under ZFC.
| https://mathoverflow.net/users/2089 | How far is Lindelöf from compactness? | The answer is **Yes**.
Theorem. The following are equivalent for any Hausdorff
space $X$.
1. $X$ is compact.
2. $X^\kappa$ is Lindelöf for any cardinal
$\kappa$.
3. $X^{\omega\_1}$ is Lindelöf.
Proof. The forward implications are easy, using Tychonoff
for 1 implies 2, since if $X$ is compact, then
$X^\kappa$ is compact and hence Lindelöf.
So suppose that we have a space $X$ that is not compact, but
$X^{\omega\_1}$ is Lindelöf. It
follows that $X$ is Lindelöf. Thus, there is a countable
cover having no finite subcover. From this, we may
construct a strictly increasing sequence of open sets
$U\_0 \subset U\_1 \subset \dots U\_n \dots$
with the union $\bigcup\lbrace U\_n \; | \; n \in \omega \rbrace = X$.
For each $J \subset \omega\_1$ of size $n$, let $U\_J$ be
the set $\lbrace s \in X^{\omega\_1} \; | \; s(\alpha) \in U\_n$ for each $\alpha \in J \rbrace$. As the size of $J$ increases, the set $U\_J$ allows more freedom on the
coordinates in $J$, but restricts more coordinates. If $J$ has
size $n$, let us call $U\_J$ an open $n$-box, since it
restricts the sequences on $n$ coordinates. Let $F$ be the
family of all such $U\_J$ for all finite $J \subset \omega\_1$
This $F$ is a cover of $X^{\omega\_1}$. To
see this, consider any point $s \in X^{\omega\_1}$. For each $\alpha \in
\omega\_1$, there is some $n$ with $s(\alpha) \in
U\_n$. Since $\omega\_1$ is uncountable,
there must be some value of $n$ that is repeated unboundedly
often, in particular, some $n$ occurs at least $n$ times. Let $J$
be the coordinates where this $n$ appears. Thus, $s$ is in
$U\_J$. So $F$ is a cover.
Since $X^{\omega\_1}$ is Lindelöf,
there must be a countable subcover $F\_0$. Let $J^\*$ be
the union of all the finite $J$ that appear in the
$U\_J$ in this subcover. So $J^\*$ is a countable subset
of $\omega\_1$. Note that $J^\*$ cannot be finite,
since then the sizes of the $J$ appearing in $F\_0$
would be bounded and it could not cover
$X^{\omega\_1}$. We may rearrange indices
and assume without loss of generality that $J^\*=\omega$ is
the first $\omega$ many coordinates. So $F\_0$ is
really a cover of $X^\omega$, by ignoring the
other coordinates.
But this is impossible. Define a sequence $s \in
X^{\omega\_1}$ by choosing $s(n)$ to be
outside $U\_{n+1}$, and otherwise arbitrary. Note that
$s$ is in $U\_n$ in fewer than $n$ coordinates below
$\omega$, and so $s$ is not in any $n$-box with $J \subset \omega$, since any such box has $n$ values in $U\_n$.
Thus, $s$ is not in any set in $F\_0$, so it is not a
cover. QED
In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$.
| 41 | https://mathoverflow.net/users/1946 | 9651 | 6,591 |
https://mathoverflow.net/questions/9652 | 25 | Is every curve over $\mathbf{C}$ birational to a smooth affine plane curve?
Bonnie Huggins asked me this question back in 2003, but neither I nor the few people I passed it on to were able to answer it. It is true at least up to genus 5.
| https://mathoverflow.net/users/2757 | Is every curve birational to a smooth affine plane curve? | Yes. Here is a proof.
It is classical that every curve is birational to a smooth one which in turn is birational to a closed curve $X$ in $\mathbb{C}^2$ with atmost double points. Now my strategy is to choose coordinates such that by an automorphism of $\mathbb{C}^2$ all the singular points lie on the $y$-axis avoiding the origin. Now the map $(x,y)\rightarrow(x,xy)$ from $\mathbb{C}^2$ to itself will do the trick of embedding the smooth part of $X$ in a closed manner. Below are the details.
The only thing we need to show is that the smooth locus of a closed curve $X\in\mathbb{C}^2$ with only double points can again be embedded in the plane as a closed curve.
**Step 1.** Let $S$ be the set of singular points of $X$. Choose coordinates on $\mathbb{C}^2$ such that the projection of $X$ onto both the axes gives embeddings of $S$. Call the projection of $S$ on the $y$-axis as $S'$. By sliding the $x$-axis a little bit we can make sure that $S'$ doesn't contain the origin of the plane. Now I claim that there is an automorphism of $\mathbb{C}^2$ which takes $S$ to $S'$. This is easy to construct by a Chinese remainder kind of argument: There is an isomorphism of the coordinate rings of $S$ and $S'$ and we need to lift this to an isomorphism of $\mathbb{C}[x,y]$. I will illustrate with an example where #$\{S\}=3$. Let $(a\_i,b\_i)$ be the points in $S$. Then there exists a function $h(y)$ such that $h|S'=x|S$ as functions restricted to the sets $S$ and $S'$. Here is one recipe: $h(y)=c\_1(\frac{y}{b\_2}-b\_2)(\frac{y}{b\_3}-b\_3)(y-b\_1+1)+\dots$ where $c\_1=a\_1(\frac{b\_1}{b\_2}-b\_2)^{-1}(\frac{b\_1}{b\_3}-b\_3)^{-1}$ etc.
Look at the map $\phi:(x,y)\rightarrow(x-h(y),y)$ on $\mathbb{C}^2$. It is clearly an automorphism and takes the set $S$ to $S'$.
**Step 2.** Now consider the map $\psi:(x,y)\rightarrow(x,xy)$ from the affine plane to itself. It is an easy check that $\psi^{-1}\circ\phi:X-S\rightarrow\mathbb{C}^2$ is a closed embedding.
| 22 | https://mathoverflow.net/users/2533 | 9662 | 6,598 |
https://mathoverflow.net/questions/9557 | 1 | Where can I find graduate level, thorough, parameter estimation/ estimation theory material on the web?
| https://mathoverflow.net/users/2705 | Is there a text on estimation theory online? | I was referred to this text:
Hogg/Craig, Introduction to mathematical statistics. Prentice-Hall
After browsing through a bit I found it to be not so suitable and often garbled.
**UPDATE**
And here is one which fit my needs better:
Kay S.M. Fundamentals of statistical signal processing: estimation theory
| 1 | https://mathoverflow.net/users/2705 | 9666 | 6,602 |
https://mathoverflow.net/questions/9661 | 56 | The "Motivation" section is a cute story, and may be skipped; the "Definitions" section establishes notation and background results; my question is in "My Question", and in brief in the title. Some of my statements go wrong in non-zero characteristic, but I don't know that story well enough, so you are welcome to point them out, but this is my characteristic-zero disclaimer.
### Motivation
In his 1972 talk "Missed Opportunities" (MR0522147), F. Dyson tells the following story of how mathematicians could have invented special and much of general relativity long before the physicists did. Following the physicists, I will talk about Lie groups, but I really mean Lie algebras, or maybe connected Lie groups, ...
The physics of Galileo and Newton is invariant under the action of the *Galileo Group* (indeed, this is the largest group leaving classical physics invariant and fixing a point), which is the group $G\_\infty = \text{SO}(3) \ltimes \mathbb R^3 \subseteq \text{GL}(4)$, where $\text{SO}(3)$ acts as rotations of space, fixing the time axis, and $\mathbb R^3$ are the nonrelativistic boosts $\vec x \mapsto t\vec u$, $t\mapsto t$. This group is not semisimple. But it is the limit as $c\to \infty$ of the *Lorentz Group* $G\_c = \text{SO}(3,1)$, generated by the same $\text{SO}(3)$ part but the boosts are now
$$ t \mapsto \frac{t + c^{-2}\vec u \cdot \vec x}{\sqrt{1 - c^{-2}u^2}} \quad \quad \vec x \mapsto \frac{\vec x + t\vec u}{\sqrt{1 - c^{-2}u^2}} $$
Since semisimple groups are easier to deal with than nonsemisimple ones, by Occam's Razor we should prefer the Lorentz group. Fortunately, Maxwell's equations are not invariant under $G\_\infty$, but rather under $G\_c$, where $c^{-2}$ is the product of the electric permititivity and magnetic permeability of free space (each of which is directly measurable, giving the first accurate measurement of the speed of light). Actually, from this perspective, $c^{-2}$ is the fundamental number, and we should really think of the Galileo Group as a limit to $0$, not $\infty$, of something.
But of course from this perspective we should go further. Actual physics is invariant under more than just the Lorentz group, which is the group that physics physics and a point. So Special Relativity is invariant under the *Poincare Group* $P = G\_c \ltimes \mathbb R^{3+1}$. Again this is not semisimple. It is the limit as $r \to \infty$ of the *DeSitter Group* $D\_r$, which in modern language "is the invariance group of an empty expanding universe whose radius of curvature $R$ is a linear function of time" (Dyson), so that $R = rt$ in absolute time units. Hubble measured the expansion of the universe in the first half of the twentieth century.
Anyway, I'm curious to know if it's true that *every* Lie group is a limit of a semisimple one: how typical are these physics examples? To make this more precise, I'll switch to Lie algebras.
### Definitions
Let $V$ be an $n$-dimensional vector space. If you like, pick a basis $e\_1,\dots,e\_n$ of $V$, and adopt Einstein's repeated index notation, so that given $n$-tuples $a^1,\dots,a^n$ and $b\_1,\dots,b\_n$, then $a^ib\_i = \sum\_{i=1}^n a^ib\_i$, and if $v \in V$, we define the numbers $v^i$ by $v = v^ie\_i$; better, work in Penrose's index notation. Anyway, a *Lie algebra structure* on $V$ is a map $\Gamma: V \otimes V \to V$ satisfying two conditions, one homogeneous linear in (the matrix coefficients) of $\Gamma$ and the other homogeneous quadratic:
$$ \Gamma^k\_{ij} + \Gamma^k\_{ji} = 0 \quad\text{and}\quad \Gamma^l\_{im}\Gamma^m\_{jk} + \Gamma^l\_{jm}\Gamma^m\_{ki} + \Gamma^l\_{km}\Gamma^m\_{ij} = 0 $$
Thus the space of Lie algebra structures on $V$ is an algebraic variety in $V \otimes V^{\*} \otimes V^{\*}$, where $V^{\*}$ is the dual space to $V$.
If $\Gamma$ is a Lie algebra structure on $V$, the corresponding *Killing form* $\beta$ is the symmetric bilinear pairing $\beta\_{ij} = \Gamma^k\_{im}\Gamma^m\_{jk}$. Then $\Gamma$ is *semisimple* if and only if $\beta$ is nondegenerate. Nondegeneracy is a Zariski-open condition on bilinear forms, since $\beta$ is degenerate if and only if a certain homogeneous-of-degree-$n$ expression in $\beta$ vanishes (namely $\sum\_{\sigma \in S\_n} (-1)^\sigma \prod\_{k=1}^n \beta\_{i\_k,j\_{\sigma(k)}} = 0$ as a map out of $V^{\otimes n} \otimes V^{\otimes n}$, where $S\_n$ is the symmetric group on $n$ objects and $(-1)^\sigma$ is the "sign" character of $S\_n$). Since $\beta$ is expressed algebraically in terms of $\Gamma$, semisimplicity is a Zariski-open condition on the variety of Lie algebra structures on a given vector space.
Incidentally, Cartan classified all semisimple Lie algebras (at least over $\mathbb C$ and $\mathbb R$) up to isomorphism, and the classification is discrete. So any two semisimple Lie algebras in the same connected component of the space of semisimple structures are isomorphic.
### My Question
Is the space of semisimple Lie algebra structures on $V$ dense in the space of all Lie algebra structures on $V$? (I.e. if $\Gamma$ is a Lie algebra structure on $V$ and $U \ni \Gamma$ is an open set of Lie algebra structures, does it necessarily contain a semisimple one?) This is really two questions. One is whether it is Zariski-dense. But we can also work over other fields, e.g. $\mathbb R$ or $\mathbb C$, which have topologies of their own. Is the space of semisimple Lie algebra structures on a real vector space $V$ dense with respect to the usual real topology?
(The answer is no when $\dim V = 1$, as then the only Lie algebra structure is the abelian one $\Gamma = 0$, which is not semisimple, and it is no when $\dim V = 2$, as there are nontrivial two-dimensional Lie algebras but no semisimple ones. So I should ask my question for higher-dimensional things.)
**Edit:** I have posted the rest of these as [this follow-up question](https://mathoverflow.net/questions/9719/what-is-the-zariski-closure-of-the-space-of-semisimple-lie-algebras).
If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?
A related question is whether given a nonsemisimple Lie algebra structure, are all its nearby semisimple neighbors isomorphic? Of course the answer is no: the abelian Lie algebra structure $\Gamma = 0$ is near every Lie algebra but in general there are nonisomoprhic semisimple Lie algebras of the same dimension, and more generally we could always split $V = V\_1 \oplus V\_2$, and put a semisimple Lie algebra structure on $G\_1$ and a trivial one on $V\_2$. So the converse question: are there any nonsemisimple Lie algebras so that all their semisimple deformations are isomorphic? Yes, e.g. anyone on the three-dimensional vector space over $\mathbb C$. So: is there a (computationally useful) characterization of those that are?
If the answer to all my questions are "yes", then it's probably been done somewhere, so a complete response could consist of a good link. The further-further question is to what extent one can deform representations, but that's probably pushing it.
| https://mathoverflow.net/users/78 | Is "semisimple" a dense condition among Lie algebras? | The answer to the question in the title is "no". Semisimplicity is an open condition; however, it is not a dense open condition. Indeed, the variety of Lie algebras is reducible. There is one equation which nonsemisimple and only nonsemisimple Lie algebra structures satisfy, namely, that the Killing form Tr(ad(x)ad(y)) is degenerate, just as stated in the question. But there is also a system of equations which all semisimple Lie algebra structures satisfy, as do also all reductive and nilpotent Lie algebra structures, but solvable Lie algebra structures in general don't. These are the unimodularity equations Tr(ad(x))=0 for all x in the Lie algebra. These mean that the top exterior power of the adjoint representation is a trivial representation of the Lie algebra, which is obvious for any Lie algebra that coincides with its commutator subalgebra. But the nonabelian 2-dimensional Lie algebra is not unimodular. Hence in any dimension n, the direct sum of the nonabelian 2-dimensional Lie algebra with the abelian (n-2)-dimensional Lie algebra does not belong to the Zariski closure of semisimple Lie algebras.
| 62 | https://mathoverflow.net/users/2106 | 9668 | 6,603 |
https://mathoverflow.net/questions/9676 | 17 | Hello,
'ordinary' Stiefel-Whitney classes are elements of the singular cohomology ring and are constructed using the Thom isomorphism and Steenrod squares. So I think they should exist for any (generalized) multiplicative cohomology theory for which the Thom isomorphism and cohomology operations like the Steenrod squares exist. If I am not wrong, I would be really happy about some references on this.
Thanks in advance (and merry christmas)
Jonas
| https://mathoverflow.net/users/2699 | Characteristic classes in generalized cohomology theories? | Stiefel-Whitney classes exist for any real-oriented cohomology theory. This is a (multiplicative) cohomology theory E equipped with an isomorphism
$E^\* ( \mathbb{R} P^{\infty} ) \cong E^\*(pt) [[x]]$
The two most well known examples are ordinary cohomology (i.e. singular cohomology) **with $\mathbb{Z}/2$-coefficients** and also unoriented bordism theory MO (this is the Thom spectrum MO, not "MathOverflow"). Note that such an isomorphism might not exist. For example it doesn't exist for ordinary Z cohomology, nor does it exist for K-theory.
The choice of this generator x is the first Stiefel-Whitney class and the other classes can be constructed using the splitting principle. For such a theory you will have Thom isomorphisms for all real bundles (not necessarily oriented).
I believe this is all explained in Switzer's book "Algebraic Topology", but I'm not sure. The course I learned it from didn't have a text book. This is often standard material for a second semester of graduate level algebraic topology, at least it was at UC Berkeley.
| 13 | https://mathoverflow.net/users/184 | 9679 | 6,611 |
https://mathoverflow.net/questions/9647 | 3 | I'm interested generally in discrete optimization problems formulated as 0-1 integer programs; essentially, anything of the form
$$\Phi = \max\_{\mathbf{x} \in \left\{0,1\right\} ^N} f(\mathbf{x})$$
My question is this: suppose the original problem is solvable in polynomial time. Now, add a constraint that $x\_i = 0$ or $x\_i = 1$:
$$\Phi\_{x\_i;j} = \max\_{\mathbf{x} \in \left\{0,1\right\} ^N, x\_i=j} f(\mathbf{x})$$
Can you give me an example problem (preferably a moderately well-known combinatorial optimization problem) where $\Phi\_{x\_i;j}$ can no longer be found in polynomial time? Alternatively, is there an argument to be made that no such example exists?
Edit: clearly there are cases where a variable can switch between hard and easy problems, so examples will exist. I'm looking for a case that isn't "contrived" in this sense--preferably a well-known combinatorial problem that becomes harder when you condition on a partial solution. Is there some characteristic of functions/problems that describes whether they get harder or easier to solve as you condition on more variable assignments?
| https://mathoverflow.net/users/2785 | Hardness of combinatorial optimization after adding one constraint | Okay, here's a less contrived example. While minimal edge coverings can be found in polynomial time, finding a minimal *hyperedge* covering in general (equivalently, set covering) is NP-hard. On the other hand, finding such a covering when one of the hyperedges spans *all* vertices on the graph is easy: you just use that edge.
So, given an arbitrary hypergraph, attach a new hyperedge to every vertex and look for a minimal cover. This can be done quickly. But constrain yourself to not using that edge, and you're back to the original NP-hard problem.
| 2 | https://mathoverflow.net/users/1060 | 9681 | 6,613 |
https://mathoverflow.net/questions/9660 | 2 | I have a monotonic polynomial recurrence of the following form:
c\_n = 1-p + p\*(c\_n-1)^2
This comes from the probability that a specific branching process (Galton-Watson) will be extinct before the nth generation. It's generalization is:
c\_n = 1-p\_1-p\_2-...-p\_n + p\_1(c\_n-1) + p\_2(c\_n-1)^2 + ... + p\_n(c\_n-1)^n
Where p\_1\*1+p\_2\*2+...+p\_n\*n < 1, and c\_1 << 1;
I know that polynomial recurrences have no general solution, but the monotonic case seems like it should be MUCH easier. A basic approximation for my specific c\_n is easily done:
d\_n = 1-c\_n =1- (1-p + p\*(1-c\_n-1)^2)
d\_n = 2p\*d\_n-1 - p\*(d\_n-1)^2 < 2p\*(d\_n-1) = (2p)^(n-1)\*(d\_1);
A cobweb plot also gives strong indication of the behavior.
My problem is, however, that although asymptotic behavior is easy to figure out, I need to be able to determine, with a fair degree of accuracy, d\_i for arbitrary i.
Is there any information out there on monotonic recurrences of polynomial form?
Update:
Any sort of approximation that's more general and useful than mine is also useful to me.
Also, I'm only interested in real numbers. For instance, d\_1 in the specific case I'm looking at is p, where p is < 0.5. Looking at cobweb plot with y = x and y = 2p\*x - p\*x^2 = px(2 - x), it is clear that the recurrence is monotonic and has no chaotic behavior. Is this not a strong enough of a condition to give it a (somewhat) clean form? If so, why not?
| https://mathoverflow.net/users/942 | Closed forms for Monotonic polynomial recurrences? | A lot depends of what you mean by "fair accuracy" and on what exactly you are going to do with your formula. If a 30% upside error in each $d\_n$ is tolerable, you can do the following.
We look at the recursion $d\_{n+1}=qd\_n(1-d\_n)$ with $0<q=2p<1$ starting with some $d\_1\in[0,1]$. It'll be convenient to do the first step separately, so we have $d\_2=q d\_1(1-d\_1)$. The reason is that $q^{-1}d\_2\in[0,1/4]$. Now, denote $b\_n=q^{-(n-1)}d\_n$ and rewrite the recurrence as $b\_{n+1}^{-1}=b\_n^{-1}+q^{n-1}\frac{b\_n}{b\_{n+1}}$. Note now that the ratio of each term to the next is between $1$ and $4/3$ and tends to $1$, so, replacing it by $1$, we get $b\_n^{-1}\approx b\_2^{-1}+\frac{q-q^{n-1}}{1-q}$ with accuracy 30% and being sure that it is an underestimate. Putting all this stuff together, we get
$$
d\_n\approx q^{n-1}\left[\frac 1{d\_1(1-d\_1)}+\frac{q-q^{n-1}}{1-q}\right]^{-1}
$$
for $n\ge 2$.
Note also that 30% is a very rough estimate for the accuracy. In practice, I've never seen it being worse that 6% (though I haven't done too many simulations).
P.S. The estimate
$$
d\_n\approx q^{n-1}\left[\frac 1{d\_1(1-d\_1)}+\frac{q-q^{n-1}}{1-q}
+\log\left(1+d\_1(1-d\_1)\frac{q^2-q^{2(n-1)}}{1-q^2}\right)\right]^{-1}
$$
is even better (3% accuracy for small $n$ and about 0.5% accuracy for large $n$) but noticeably uglier. As several people have already mentioned, there is no exact formula, so the higher precision you want, the longer and uglier the approximation gets.
P.P.S. The main idea behind the second approximation is that if $B\_n=b\_{n}^{-1}$, then we make an error $q^{n-1}\frac{B\_{n+1}-B\_n}{B\_n}\approx \frac{q^{2(n-1)}}{B\_n}$ during each step that led us to the first approximation. To compensate, we would like to take the partial sums of that series. We also know that $B\_n\approx \frac 1{d(1-d)}+q+q^2+\dots+q^{n-2}$. Unfortunately, we cannot sum the resulting series nicely. But if we double all the exponents in the approximate formula for $B\_n$ (which will lead only to a small error when $q\approx 1$), then we'll get the Riemann sum type expression, which we can replace by the corresponding integral. Clearly, it may overshoot on the long run but to my own surprise, it not only corrects the first few terms nicely, but also corrects the asymptotics giving an error below 0.5% for large $n$ in the entire range of parameters (well, at least that is so for all values I looked at and I tried quite a few). Why it works this way remains a mystery, but it does. It can be shown rigorously that the total relative overshot coming from all steps after $n$ is at most $n^{-1}$ and I've never seen the overshots of more than 0.1% up to $n=300$.
| 3 | https://mathoverflow.net/users/1131 | 9686 | 6,617 |
https://mathoverflow.net/questions/9688 | 2 | Following is an argument given by Hempel where I am unable to understand his comment about choosing a loop close enough to a surface. Can somebody please elucidate this:
**Lemma:** If $F$ is a compact connected surface properly embedded in a $3$-manifold $M$ and if $image(i\_\*:H\_1(F;Z/2Z)\rightarrow{H\_1(M;Z/2Z)})=0$, then $F$ is 2-sided in $M$.
**Proof:** By regular neighborhood theory, it suffices to show that $F$ separates some connected neighborhood of $F$. If this is not the case then there is a loop $J\subset{M}$ such that $J\cap{F}$ is a single point, with transverse intersection. **We may choose $J$ close enough to $F$** so that $J$ is homologous to zero (mod 2) in $M$. This contradicts homological invariance (mod 2) of intersection numbers. QED.
**Doubt:** Everything else is clear to me except the bold part in the proof. I don't think we will be able to bring $J$ close to $F$ unless it already bounds a disc. So I can see the proof only in the simply connected $M$ case where after suitably perturbing the loop and using the loop theorem to bound, I can apply an ambient isotopy to bring the disc itself closer to $F$. But how do we see this for the non-simply connected $M$? A complicated loop in $M$ might go all around it.
**Reference:** Lemma 2.1, Chapter 2 (Heegard Splittings), 3-manifolds(book) by Hempel.
| https://mathoverflow.net/users/2533 | Can you explain a step in a proof about 2-sided surfaces in 3-manifolds? | We want to prove that in the case F is not one-sided, we may replace J by a curve J' that is contained in a small neighborhood of F and interesects F in the same way as J. By assumtion F is one sided. Consider the boundary B of a small neighborhood N of $F$. Since F is one-sided, B is connected. Now, conisder the intersection of J with B. There are even number of intersections, since B is the boundary. So you can throw the part of J that does not belong to the neighbohood N an close it to a connected curve J' by segments in B (we assumed that B is connected). This explanes the words written in bold.
| 4 | https://mathoverflow.net/users/943 | 9690 | 6,620 |
https://mathoverflow.net/questions/9667 | 69 | I am preparing to teach a short course on "applied model theory" at UGA this summer. To draw people in, I am looking to create a BIG LIST of results in mathematics that have nice proofs using model theory. (I do not require that model theory be the first or only proof of the result in question.)
I will begin with some examples of my own (the attribution is for the model-theoretic proof, not the result itself).
1. An injective regular map from a complex variety to itself is surjective (Ax).
2. The projection of a constructible set is constructible (Tarski).
3. Solution of Hilbert's 17th problem (Tarski?).
4. p-adic fields are "almost C\_2" (Ax-Kochen).
5. "Almost" every rationally connected variety over Q\_p^{unr} has a rational point (Duesler-Knecht).
6. Mordell-Lang in positive characteristic (Hrushovski).
7. Nonstandard analysis (Robinson).
[But better would be: a particular result in analysis which has a snappy nonstandard proof.]
**Added**: The course was given in July of 2010. So far as I am concerned, it went well. If you are interested, the notes are available at
[http://alpha.math.uga.edu/~pete/MATH8900.html](http://alpha.math.uga.edu/%7Epete/MATH8900.html)
Thanks to everyone who answered the question. I enjoyed and learned from all of the answers, even though (unsurprisingly) many of them could not be included in this introductory half-course. I am still interested in hearing about snappy applications of model theory, so further answers are most welcome.
| https://mathoverflow.net/users/1149 | What are some results in mathematics that have snappy proofs using model theory? | Hilbert's Nullstellensatz is a consequence of the model completeness of algebraically closed fields.
Edit: I don't have a reference, but I can sketch the proof. Suppose you have some polynomial equations that don't have a solution over ${\mathbb C}$. Extend ${\mathbb C}$ by a formal solution, and then algebraically close to get a field $K$. The field $K$ obviously contains a solution, but by model completeness of algebraically closed fields, a first-order statement is true in an algebraically closed field only if it is true in every algebraically closed subfield that contains all its parameters. The existence of a solution to a finite set of polynomial equations is a first-order statement (whose parameters are the coefficients) and ${\mathbb C}$ is algebraically closed. QED.
| 35 | https://mathoverflow.net/users/935 | 9693 | 6,622 |
https://mathoverflow.net/questions/9531 | 5 | I have the quadratic integer program over $\mathbb{Z}^n$
$\displaystyle\min\_{z \in \mathbb{Z}^n} \Phi (z) = \frac{1}{2} z^T Q z - r^T z + s$
subject to $G z = h$, and $z\_i \in \{0,1,2,\dots, b\_i\}$ for all $i \in \{1,2,\dots,n\}$, where $Q$ is symmetric positive-definite. Moreover, $G, h$ are integer-valued and, thus, I suppose that $\{z \in \mathbb{Z}^n : G z = h\}$ defines a sublattice of $\mathbb{Z}^n$.
Let us suppose we solve the relaxed quadratic program over $\mathbb{R}^n$
$\displaystyle\min\_{x \in \mathbb{R}^n} \Psi (x) = \frac{1}{2} x^T Q x - r^T x + s$
subject to $G x = h$, and $0 \leq x\_i \leq b\_i$ for all $i \in \{1,2,\dots,n\}$. Let $x\_{opt} \in \mathbb{R}^n$ be the minimizer of $\Psi$. We can define an $n$-cube (in $\mathbb{Z}^n$) containing $x\_{opt}$ by taking the floor/ceil of each component of $x\_{opt}$. We intersect this $n$-cube with the integer sublattice defined by $G z = h$, and evaluate $\Phi$ at all points of this intersection. Let $z^{\ast}$ be the point in such intersection that minimizes $\Phi$. Let $z\_{opt}$ be the minimizer of the original quadratic integer program. Can we say that $z\_{opt} = z^\*$? I know nothing of integer programming, so I preemptively apologize if my question is silly / elementary...
In other words, can one solve a quadratic integer program by solving a relaxed quadratic real program, and then searching in the neighborhood of the real solution? This seems to work for $n = 1$ and $n = 2$... but it also seems too good to be true in general. If in general $z\_{opt} \neq z^{\ast}$, can we (at least) quantify how sub-optimal $z^{\ast}$ is?
Any feedback will be most welcome!
| https://mathoverflow.net/users/2741 | On Quadratic Integer Programming | The relaxed quadratic programming problem is a red herring. It is true that quadratic programming over $\mathbb{R}$ with linear inequalities can be solved in practice, for one reason because it is a special case of convex programming. But in the stated question, the inequality $0 \le x\_i \le b\_i$ came from nowhere. The correct relaxation is even simpler: You should just minimize $\Phi(z)$ over all of $\mathbb{R}^n$, and the minimum is directly at $z\_0 = Q^{-1}r$.
After that, Mitch is roughly correct. The question as stated is exactly the closest vector problem, which is related to the shortest vector problem that Mitch mentions. The constant $s$ is not important. The question is to find the integer lattice point $z$ which is the closest to $z\_0$ in the metric defined by $Q$. If you like, you can change distance to Euclidean distance, and change the lattice from the standard integer lattice to something else, by applying the operator $Q^{-1/2}$. $L = Q^{-1/2}(\mathbb{Z}^n)$ is a certain lattice, and you are looking for the point which is the closest in Euclidean distance to $Q^{-1/2}(z\_0)$.
In any fixed dimension $n$, the closest and shortest vector problems can be solved in polynomial time. There are various lattice reduction algorithms that only search polynomially many points.
If the dimension $n$ is a parameter, then the situation is very different. For many purposes, people are happy with just a close vector or a short vector, not necessarily the closest or shortest one. The problem varies greatly in difficulty depending on how close is good enough, or equivalently whether there are few lattice points that stick out as much closer than all of the others. Close vector is intuitively harder than short vector, but there is a theoretical result that they are roughly equivalent in difficulty.
Taking the strictest possible requirements, finding a close vector is NP-hard. There are intermediate levels of closeness, given by some tolerance that grows with $n$, that seem hard but are probably not NP-hard. Other levels of closeness can be done in polynomial time. There are lots of papers on the these two problems, and the [Wikipedia page](http://en.wikipedia.org/wiki/Lattice_problem) that Mitch mentions is a pretty good review: The GapCVP section addresses the approximate versions of the question that I mention briefly here.
One weakness of the Wikipedia page is that it has more to say about hardness than practical algorithms. But it does mention two important algorithms: Lenstra-Lenstra-Lovasz and Ajtai-Kumar-Sivakumar.
| 9 | https://mathoverflow.net/users/1450 | 9701 | 6,629 |
https://mathoverflow.net/questions/9704 | 2 | Is there an English translation of Kuratowski's proof about planar graphs?
| https://mathoverflow.net/users/1662 | Is there an English translation of Kuratowski's theorem on forbidden minors of planar graphs? | In case you are asking for the original paper "Sur le problème des courbes gauches en Topologie" by Kuratowski where he first proves his characterization of planar graphs, then a translation by J.Jaworowski can be found in "Graph Theory, Łagów", 1981, M. Borowiecki, J. W. Kennedy and M. M. Sysło. It is the proceedings of a conference held in Łagów, dedicated to the memory of K.Kuratowski.
| 13 | https://mathoverflow.net/users/2384 | 9706 | 6,631 |
https://mathoverflow.net/questions/9708 | 55 | Has the solution of the Poincaré Conjecture helped science to figure out the shape of the universe?
| https://mathoverflow.net/users/1172 | Poincaré Conjecture and the Shape of the Universe | In Einstein's theory of General Relativity, the universe is a 4-manifold that might well be fibered by 3-dimensional time slices. If a particular spacetime that doesn't have such a fibration, then it is difficult to construct a causal model of the laws of physics within it. (Even if you don't see an a priori argument for causality, without it, it is difficult to construct enough solutions to make meaningful predictions.) There isn't usually a geometrically distinguished fibration, but if you have enough symmetry or even local symmetry, the symmetry can select one. An approximate symmetry can also be enough for an approximately canonical fibration. Once you have all of that, the topology of spacelike slices of the universe is not at all a naive or risible question, at least not until you see more physics that might demote the question. The narrower question of whether the Poincaré Conjecture is relevant is more wishful and you could call it naive, but let's take the question of relating 3-manifold topology in general to cosmology.
The [cosmic microwave background](http://en.wikipedia.org/wiki/Cosmic_microwave_background), discovered in the 1964 by Penzias and Wilson, shows that the universe is very nearly isotropic at our location. (The deviation is of order $10^{-5}$ and it was only announced in 1992 after 2 years of data from the [COBE telescope](http://en.wikipedia.org/wiki/Cosmic_Background_Explorer).) If you accept the Copernican principle that Earth isn't at a special point in space, it means that there is an approximately canonical fibration by time slices, and that the universe, at least approximately and locally, has one of the three isotropic Thurston geometries, $E^3$, $S^3$, or $H^3$. The Penzias-Wilson result makes it a really good question to ask whether the universe is a 3-manifold with some isotropic geometry and some fundamental group. I have heard of the early discussion of this question was so naive that some astronomers only talked about a 3-torus. They figured that if there were other choices from topology, they could think about them later. Notice that already, the Poincaré conjecture would have been more relevant to cosmology if it had been false!
The topologist who has done the most work on the question is Jeff Weeks. He coauthored a [respected paper in cosmology](http://arxiv.org/abs/astro-ph/0310253) and wrote an interesting [article in the AMS Notices](http://e-math1.ams.org/notices/200406/fea-weeks.pdf) that promoted the Poincaré dodecahedral space as a possible topology for the universe. But after he wrote that article...
There indeed is other physics that does demote the 3-manifold question, and that is [inflationary cosmology](http://en.wikipedia.org/wiki/Inflationary_theory). The inflation theory posits that the truthful quantum field theory has a vaguely stable high-energy phase, which has such high energy density that the solution to the GR equations looks completely different. In the inflationary solution, hot regions of the universe expand by a factor of $e$ in something like $10^{-36}$ seconds. The different variations of the model posit anywhere from 60 to thousands of factors of $e$, or "$e$-folds". Patches of the hot universe also cool down, including the one that we live in. In fact every spot is constantly cooling down, but cooling is still overwhelmed by expansion. Instead of tacitly accepting certain observed features of the visible universe, for instance that it is approximately isotropic, inflation explains them. It also predicts that the visible universe is approximately flat and non-repeating, because macroscopic curvature and topology have been stretched into oblivion, and that observable anisotropies are stretch marks from the expansion. The stretch marks would have certain characteristic statistics in order to fit inflation. On the other hand, in the inflationary hot soup that we would never see directly, the rationale for canonical time slices is gone, and the universe would be some 4-manifold or even some fractal or quantum generalization of a 4-manifold.
The number of $e$-folds is not known and even the inflaton field (the sector of quantum field theory that governed inflation) is not known, but most or all models of inflation predict the same basic features. And the news from the successor to COBE, called [WMAP](http://en.wikipedia.org/wiki/WMAP), is that the visible universe is flat to 2% or so, and the anistropy statistically matches stretch marks. There is not enough to distinguish most of the models of inflation. There is not enough to establish inflation in the same sense that the germ theory of disease or the the heliocentric theory are established. What is true is that inflation has made experimental predictions that have been confirmed.
After all that news, the old idea that the universe is a visibly periodic 3-manifold is considered a long shot. WMAP didn't see any obvious periodicity, even though Weeks et al were optimistic based on its first year of data. But I was told by a cosmologist that periodicity should still be taken seriously as an alternative cosmological model, if possibly as a devil's advocate. A theory is incomplete science if it is both hard to prove, and if every alternative is laughed out of the room. In arguing for inflation, cosmologists would also like to have something to argue against. In the opinion of the cosmologist that I talked to some years ago, the model of a 3-manifold with a fundamental group, developed by Weeks et al, is as good at that as any proposal.
---
José makes the important point that, in testing whether the universe has a visible fundamental group, you wouldn't necessarily look for direct periodicity represented by non-contractible geodesics. Instead, you could use harmonic analysis, using a suitable available Laplace operator, and this is what used by Luminet, Weeks, Riazuelo, Lehoucq and Uzan. I also that I have not heard of any direct use of homotopy of paths in astronomy, but actually the direct geometry of geodesics does sometimes play an important role. For instance, look closely at [this photograph of galaxy cluster Abell 1689](http://apod.nasa.gov/apod/image/0301/abell1689_hstacs_full.jpg). You can see that there is a strong gravitational lens just left of the center, between the telescope and the dimmer, slivered galaxies. Maybe no analysis of the cosmic microwave background would be geometry-only, but geometry would modify the apparent texture of the background, and I think that that is part of the argument from the data that the visible universe is approximately flat. Who is to say whether a hypothetical periodicity would be seen with geodesics, harmonic expansion, or in some other way.
Part of Gromov's point seems fair. I think it is true that you can always expand the scale of proposed periodicity to say that you haven't yet seen it, or that the data only just starts to show it. Before they saw anisotropy with COBE, that kept getting pushed back too. The deeper problem is that the 3-manifold topology of the universe does not address as many issues in cosmology, either theoretical or experimental, as inflation theory does.
| 67 | https://mathoverflow.net/users/1450 | 9717 | 6,639 |
https://mathoverflow.net/questions/9675 | 1 | Sorry for my precedent tentative, I was a little hasty:
Ok, I think I'd better put the original problem:
I have an action of three fields: $A$ which is the spin-connection, $B$ an skew-symmetric 2-form and $\Phi$ which is traceless ~~and skew-symmetric~~ scalar field. These fields take their values on some algebra, index their components in this algebra by $i,j,k,... = 1,2,3$
I want to implement a certain condition on B by using equations of motion of $\Phi$, the action is:
$S=\int (B\_i \wedge F^i + \Lambda B\_i \wedge B^i + \Phi\_{ij} B^i \wedge B^j) $
Now for me equations of motions are simply:
$B^i \wedge B^j=0$
perhaps with the condition that all diagonal elements are equal (as jc showed) ~~but this is automatically satisfied for a skew-symmetric matrix (here $B^i \wedge B^j$)~~.
But in all papers I find:
$B^i \wedge B^j - \frac{1}{3}\delta^{ij}B\_k\wedge B^k = 0$
So I see that they all took the traceless part of the matrix representing equations of motion, necessarily it has a relation with the traceless character of $\Phi$ but I do not see which one.
~~In addition, this expression is not antisymmetric in $i,j$~~.
Would anyone have an idea?
| https://mathoverflow.net/users/2597 | Extremum under variations of a traceless matrix | If $B^i$ are 2-forms, then $B^i \wedge B^j$ is symmetric, not skewsymmetric. Since $\Phi\_{ij}$ is traceless, only the traceless part of $B^i \wedge B^j$ that couples to $\Phi$. So I see nothing wrong with the equation you find in the papers.
The reason you take $\Phi$ to be traceless is that the trace is already contained in the second term in the action.
---
**Edit** (in response to the comment thread below)
Let me give more details. Let $M^{ij} := B^i \wedge B^j$. We can think of $M$ as a matrix with 4-forms as entries. The space of even forms is a commutative algebra, so we can work with $M$ as if it were a real or complex matrix, say. In particular, we can take its trace (which will be a 4-form): $T = \delta\_{ij} M^{ij}$, where as in the question the Einstein summation convention is in force. We can then decompose $M$ into a traceless part we shall call $M\_0$ and a part containing the trace:
$$M^{ij} = M\_0^{ij} + \frac{1}{N} T \delta^{ij},$$
where I assume that $M$ is an $N\times N$ matrix. If you take the trace of this equation, you find that $M\_0$ is indeed traceless. Its explicit form is given by solving that equation for $M\_0$, but we do not need it.
Now let $\Phi\_{ij}$ be a symmetric traceless matrix. This means that $\delta^{ij} T\_{ij} = 0$. Contracting with $M$ we find
$$\Phi\_{ij} M^{ij} = \Phi\_{ij} M\_0^{ij}.$$
In other words, $\Phi$ never sees the trace of $M$ and hence if you have a lagrange multiplier term in an action functional of the form
$$\int \Phi\_{ij} M^{ij}$$
this is really equal to
$$\int \Phi\_{ij} M\_0^{ij}$$
and hence the resulting Euler-Lagrange equation is $M\_0^{ij} = 0$.
| 1 | https://mathoverflow.net/users/394 | 9722 | 6,642 |
https://mathoverflow.net/questions/9721 | 11 | The number ${n \choose k}$ of $k$-element subsets of an $n$-element set and the number $\left( {n \choose k} \right)$ of $k$-element **multisets** of an $n$-element set satisfy the reciprocity formula
$\displaystyle {-n \choose k} = (-1)^k \left( {n \choose k} \right)$
when extended to negative integer indices, for example by applying the usual recurrence relations to all integers. There's an interesting way to think about the "negative cardinalities" involved here using Euler characteristic, which is due to Schanuel; see, for example, [this paper of Jim Propp](http://arxiv.org/abs/math/0203289). Another (related?) way to think about this relationship is in terms of the symmetric and exterior algebras; see, for example, [this blog post](http://qchu.wordpress.com/2009/11/06/set-multiset-duality-and-supervector-spaces/).
The number $S(n, k)$ of $k$-block partitions of a set with $n$ elements and the number $c(n, k)$ of permutations of a set with $n$ elements with $k$ cycles satisfy a well-known inverse matrix relationship, but they also satisfy the reciprocity formula
$c(n, k) = S(-k, -n)$
when extended to negative integer indices, again by applying the usual recurrence relations.
**Question:** Are there any known highbrow interpretations of this reciprocity formula?
| https://mathoverflow.net/users/290 | Highbrow interpretations of Stirling number reciprocity | Supplementary Exercise 3.2(d,e) on page 313 of my book *Enumerative Combinatorics*, vol. 1, second printing, shows that this Stirling number reciprocity is a special case of the reciprocity theorem for order polynomials (Exercise 3.61(a)). Thus it is related to a lot of "highbrow" math, such as the reciprocity between a Cohen-Macaulay ring and its canonical module. (For the basic properties of canonical modules, see Section I.12 of my other book *Combinatorics and Commutative Algebra,* second ed.)
| 23 | https://mathoverflow.net/users/2807 | 9724 | 6,644 |
https://mathoverflow.net/questions/9127 | 2 | Hi, i'm looking to get into nonparametric bayesian techniques but I'm having problem understanding what's going on in the definition of the Dirichlet process or how it works. So what does P ~ DP(α\*P0) mean?
What does a distribution P looks like? Is the samples being used, Xi ~ P?
| https://mathoverflow.net/users/2633 | How does the Dirichlet process work? | I'm a fan of Yee Whye Teh's tutorials, listed under "Short Courses" here:
<http://www.gatsby.ucl.ac.uk/~ywteh/teaching/teaching.html>
You can also watch the video on videolectures if you want an explanation to accompany the slides.
| 2 | https://mathoverflow.net/users/2785 | 9730 | 6,649 |
https://mathoverflow.net/questions/9738 | 5 | A not necessarily commutative algebra A (over C, say) is called formally smooth (or quasi-free) if, given any map $f:A \to B/I$, where $I \subset B$ is a nilpotent ideal, there is a lifting $F:A \to B$ that commutes with the projection. (The reason for the terminology is that if we restrict to the category of finitely generated commutative algebras, this condition is equivalent to Spec(A) being smooth. For more info see the paper "algebra extensions and nonsingularity" by Cuntz and Quillen, 1995.) It isn't hard to see that if A is formally smooth, then the representation varieties $Rep\_\mathbb{C}(A,V)$ are smooth (V is finite dimensional). Does anyone know of an example of an algebra that is not formally smooth, but whose representation varieties are smooth?
One almost-answer is the Weyl algebra $A = \mathbb{C}\langle x,y\rangle/(xy - yx = 1)$. This isn't formally smooth, but its representation varieties are all empty. (To see this, take the trace of $xy - yx = 1$ to get $0 = n$.) This doesn't seem like it should count as answer, does anyone know a better one?
| https://mathoverflow.net/users/2669 | Non-smooth algebra with smooth representation variety | Take any semi-simple lie algebra g and consider its enveloping algebra U(g). As all finite dimensional representations are semi-simple, every representation variety rep\_n U(g) is a finite union of orbits, whence smooth. No such U(g) is formally smooth.
| 11 | https://mathoverflow.net/users/2275 | 9744 | 6,657 |
https://mathoverflow.net/questions/9736 | 8 | Let Q be a finite quiver without loops. Then its global dimension is 1 if it contains at least one arrow.
I'm trying to get some intuition about how much the global dimension can grow when we quotient by some homogeneous ideal of relations I. In general, if Q is acyclic (is this necessary?), then the global dimension is bounded by the number of vertices, but I want something that uses information from I. For simplicity, let's assume that there is at most 1 edge between any two vertices. If I add the relation that a single path of length 2 is 0, then the global dimension goes up to 2, and the same is true if 2 is replaced by any r>2 (right?). I can get higher global dimensions by the following: take some consecutive arrows $a\_1, a\_2, \dots, a\_n$, and require that each path of length 2 $a\_{i+1} a\_i$ is 0, then the global dimension goes up to n-1.
The way I am trying to picture this is by thinking of projective modules as flowing water which gets stopped by some rock placed where the relations are, and seeing how many times the flow needs to restart before it can reach the end (I don't know if this is a useful comment.)
Anyway, here is my question: is there some simple way to bound the global dimension of Q/I assuming that Q is acyclic and no multiple edges between any two vertices? In this case, we're only allowed to say that certain paths are 0, so I am suspecting this has something to do with "number of overlaps." My guess would be something like, define an overlap to be when an initial segment of one path coincides with an ending segment of another path, and then the global dimension should be less than or equal to number of overlaps + 2.
| https://mathoverflow.net/users/321 | Global dimenson of quivers with relations | If you are only considering *monomial* algebras (that is, if you are generating the ideal I by paths) then your intuition about overlaps is correct, once you see which overlaps you need to consider. There is a paper by Bardzell (The alternating behaviour of monomial algebras) where he constructs explicitely a projective resolution of the quotient algebra as a bimodule over itself (whose length bounds the gldim of the algebra) which is constructed precisely by considering overlaps.
By the way, if the graph is not acyclic, then the global dimension can very well be infinite. The simplest example is a quiver with one vertex and a loop, and the ideal geberated by the square of the loop.
**Later:** Let me be more explicit about what I meant by "once you see which overlaps you need to consider"... Consider the quiver $Q$ which is an oriented path with 15 arrows, and let $I$ be ideal generated by all paths in $Q$ of length 8. There are then 8 minimal relations, they all overlap, but if you work through the construction of minimal projective resolutions of the simple modules of $kQ/I$ you'll see that most of those overlaps do not matter, and that the global dimension is $3$ in this case. You can play this game with longer paths, as long as you divide by not too short relations.
It is not too hard to single out precisely which are the overlaps that *do* matter when the quiver is a path. The general case is not horribly more complicated, yet it always manages to confuse me.
| 15 | https://mathoverflow.net/users/1409 | 9747 | 6,659 |
https://mathoverflow.net/questions/9733 | 34 | I have heard during some seminar talks that there are applications of the theory of
matrix factorizations in string theory. A quick search shows mostly papers written by physicists. Are there any survey type papers aimed at an algebraic audience on this topic, especially with current state/open questions motivated by physics?
Background: A matrix factorization of an element $x$ in a ring $R$ is a pair of square matrices $A,B$ of size $n$ such that $AB=BA=xId\_n$. For more see, for example, Section 3 of
[this paper](http://www.leuschke.org/uploads/MCMs_on_detX-arXiv.pdf) .
| https://mathoverflow.net/users/2083 | Matrix factorizations and physics | Indeed matrix factorizations come up in string theory. I don't know if there are good survey articles on this stuff, but here is what I can say about it. There might be an outline in the big Mirror Symmetry book by Hori-Katz-Klemm-etc., but I am not sure.
When we are considering the B-model of a manifold, for example a compact Calabi-Yau, the D-branes (boundary states of open strings) are given by coherent sheaves on the manifold (or to be more precise, objects of the derived category of coherent sheaves). Matrix factorizations come up in a different situation, namely, they are the D-branes in the B-model of a *Landau-Ginzburg model*. Mathematically, a Landau-Ginzburg model is just a manifold (or variety) $X$, typically non-compact, plus the data of a holomorphic function $W: X \to \mathbb{C}$ called the superpotential. In this general situation, a matrix factorization is defined to be a pair of coherent sheaves $P\_0, P\_1$ with maps $d : P\_0 \to P\_1$, $d : P\_1 \to P\_0$ such that $d^2 = W$. I guess you could call this a "twisted (or maybe it's 'curved'? I forget the terminology) 2-periodic complex of coherent sheaves". When $X = \text{Spec}R$ is affine, and when the coherent sheaves are free $R$-modules, this is the same as the definition that you gave.
The relationship between matrix factorization categories and derived categories of coherent sheaves was worked out by Orlov: <http://arxiv.org/abs/math/0503630> <http://arxiv.org/abs/math/0503632> <http://arxiv.org/abs/math/0302304>
I believe that the suggestion to look at matrix factorizations was first proposed by Kontsevich. I think the first paper that explained Kontsevich's proposal was this paper by Kapustin-Li: <http://arXiv.org/abs/hep-th/0210296v2>
There are some interesting recent papers regarding the relationship between the open-string B-model of a Landau Ginzburg model (which is, again, mathematically given by the matrix factorizations category) and the closed-string B-model, which I haven't described, but an important ingredient is the Hochschild (co)homology of the matrix factorizations category. Take a look at Katzarkov-Kontsevich-Pantev <http://arxiv.org/abs/0806.0107> section 3.2. There is a paper of Tobias Dyckerhoff <http://arxiv.org/abs/0904.4713> and a paper of Ed Segal <http://arxiv.org/abs/0904.1339> which work out in particular the Hochschild (co)homology of some matrix factorization categories. The answer is it's the Jacobian ring of the superpotential. This is the correct answer in terms of physics: the Jacobian ring is the closed state space of the theory.
Katzarkov-Kontsevich-Pantev also has some interesting stuff about viewing matrix factorization categories as "non-commutative spaces" or "non-commutative schemes".
Edit 1: I forgot to mention: Kontsevich's original homological mirror symmetry conjecture stated that the Fukaya category of a Calabi-Yau is equivalent to the derived category of coherent sheaves of the mirror Calabi-Yau. Homological mirror symmetry has since been generalized to non-Calabi-Yaus. The rough expectation is that given any compact symplectic manifold, there is a mirror Landau-Ginzburg model such that, among other things, the Fukaya category of the symplectic manifold should be equivalent to the matrix factorizations category of the Landau-Ginzburg model. For example, if your symplectic manifold is $\mathbb{CP}^n$, the mirror Landau-Ginzburg model is given by the function $x\_1+\cdots+x\_n + \frac{1}{x\_1\cdots x\_n}$ on $(\mathbb{C}^\ast)^n$. This is sometimes referred to as the Hori-Vafa mirror <http://arxiv.org/abs/hep-th/0002222>
I think that various experts probably know how to prove this form of homological mirror symmetry, at least when the symplectic manifold is, for example, a toric manifold or toric Fano manifold, but it seems that very little of this has been published. There may be some hints in this direction in Fukaya-Ohta-Oh-Ono <http://arxiv.org/abs/0802.1703> <http://arxiv.org/abs/0810.5654>, but I'm not sure. There is an exposition of the case of $\mathbb{CP}^1$ in this paper of Matthew Ballard <http://arxiv.org/abs/0801.2014> -- this case is already non-trivial and very interesting, and the answer is very nice: the categories in this case are equivalent to the derived category of modules over a Clifford algebra. I quite like Ballard's paper; you might be interested in taking a look at it anyway.
Edit 2: Seidel also has a proof of this form of homological mirror symmetry for the case of the genus two curve. Here is the paper <http://arxiv.org/abs/0812.1171> and here is a video <http://www.maths.ed.ac.uk/~aar/atiyah80.htm> of a talk he gave on this stuff at the Atiyah 80 conference.
| 28 | https://mathoverflow.net/users/83 | 9748 | 6,660 |
https://mathoverflow.net/questions/9751 | 23 | As suggested by Poonen in a comment to an answer of [his question](https://mathoverflow.net/questions/9652/is-every-curve-birational-to-a-smooth-affine-plane-curve/9662#9662) about the birationality of any curve with a smooth affine plane curve we ask the following questions:
Q) Is it true that every smooth affine curve is isomorphic to a smooth affine plane curve?
(a) In particular, given a smooth affine plane curve $X$ with an arbitrary Zariski open set $U$ in it, can one give a closed embedding of $U$ in the plane again?
(b) An extremely interesting special case of (Q) above: Suppose $X$ is a singular plane algebraic curve with $X\_{sm}$ the smooth locus. Can one give a closed embedding of $X\_{sm}$ in the plane?
All varieties in question are over $\mathbb{C}$.
---
**UPDATE**:
Bloch, Murthy and Szpiro have already proven in their paper ["Zero cycles and the number of generators of an ideal"](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=223540&vfpref=html&r=7&mx-pid=1044346) , a much more general result (see Theorem 5.7, op.cit), that every reduced and irreducible prjective variety has an affine open set which is a hypersurface. This settles the above question birationally, in particular.
The authors give a very short and beautiful alternate proof of their result by M.V. Nori which I include here for its brevity and for anyone who may not have access to the paper:
**Proof**: Suppose $X$ is an integral projective variety of dimension $d$. By a generic projection, easily reduce to the case of a (possibly singular) integral hypersurface $X$ of $\mathbb{A}^{d+1}$. Suppose the coordinate ring of $X$ is $A=\mathbb{C}[x\_1,\dots,x\_{d+1}]$ and its defining equation is $F=\Sigma\_0^m{f\_i}x\_{d+1}^{i}=0$ with $f\_0\neq{0}$. For some element $h$ in $J\cap\mathbb{C}[x\_1,\dots,x\_d]$, where $J$ defines the singular locus of $X$, put $x\_{d+1}'=x\_{d+1}/(hf\_0^2)$ in $F=0$ to observe that $1/(hf\_0)\in\mathbb{C}[x\_1,\dots,x\_{d+1}']$ and $A\_{hf\_0}=\mathbb{C}[x\_1,\dots,x\_{d+1}']$. Clearly $\rm{Spec}\ {A\_{hf\_0}}$ admits a closed immersion in $\mathbb{A}^{d+1}$.
---
However, the above authors also prove in their Theorem 5.8 that there exist affine varieties of any dimension, which are not hypersurfaces. This answers our question in negative. This was also known to Sathaye for curves, see [On planar curves](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=154880&vfpref=html&r=19&mx-pid=466148). He gives a nice example of a double cover of a punctured elliptic curve, ramified at 9 points and also at the point at infinity. This curve cannot be embedded in $\mathbb{A}^2$. Sathaye uses the value semigroup at the only point at infinity to prove this. His example has trivial canonical divisor. So it answers Poonen's question in the comments below, negatively.
In short, $K=0$ for an affine curve is necessary but not sufficient for the curve to be planar, however one should note that $K=0$ is necessary and sufficient for an affine curve to be a complete intersection.
| https://mathoverflow.net/users/2533 | Is every smooth affine curve isomorphic to a smooth affine plane curve? | You can try this:
<http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.52.6348>
| 14 | https://mathoverflow.net/users/2083 | 9752 | 6,662 |
https://mathoverflow.net/questions/9749 | 9 | Suppose H is a subgroup of a finite group G. Can the group of all automorphisms of H that extend to G can be characterized somehow? What condition on the group extension would guarantee that any automorphism of H can be extended to G?
| https://mathoverflow.net/users/2805 | Characterising extendable automorphisms | An abstract answer to the question for all groups is given in the papers below. I have not followed the field in recent years. There may be other papers specific to finite groups.
[Charles Wells, Automorphisms of Group Extensions, 1970.](http://www.cwru.edu/artsci/math/wells/pub/pdf/AGEPackage.pdf)
Kung Wei Yang
Isomorphisms of group extensions.
Pacific J. Math. Volume 50, Number 1 (1974), 299-304.
D.J.S. Robinson, Applications of cohomology to the theory of groups, Groups – St. Andrews 1981, London Math. Soc. Lecture Notes vol. 71 (1982), pp. 46–80.
Jin Ping, Automorphisms of groups
Journal of Algebra
Volume 312, Issue 2, 15 June 2007, Pages 562-569
| 5 | https://mathoverflow.net/users/342 | 9776 | 6,678 |
https://mathoverflow.net/questions/8606 | 41 | The answers to [this question](https://mathoverflow.net/questions/6200/what-is-to-quantize-something) do a good job of exploring, at a heuristic level, what "quantization" should be. From my perspective, quantization involves replacing a (commutative) Poisson algebra by some related noncommutative associative algebra. Poisson algebras arise naturally especially as algebras of functions in geometry and physics. Noncommutative algebras arise naturally as algebras of operators on linear spaces.
I've often heard it said that "quantization is not a functor". I'm wondering what a precise statement of this is.
For example, I could imagine statements of the following form.
1. There is no functor from the category of Poisson manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
2. There is no functor from the category of symplectic manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
3. Recall that for any smooth manifold, its cotangent bundle is naturally symplectic. There is no functor from the category of smooth manifolds to the category of associative algebras that quantizes the cotangent bundle.
4. Recall that the dual to the universal enveloping algebra of a Lie bialgebra is naturally Poisson Hopf. There is no functor from the category of Lie bialgebras to the category of Hopf algebras satisfying some nice property.
Actually, 4. is false. Indeed, Etingof and Khazdan constructed a functor from bialgebras to Hopf algebras satisfying a host of properties, and Enriquez classified all the ones with nice properties. Note that Kontsevich does give a quantization of any Poisson manifold, but perhaps his isn't functorial?
| https://mathoverflow.net/users/78 | What does "quantization is not a functor" really mean? | Here is one precise statement of how quantization is not a functor:
5) There is no functor from the classical category $\mathcal C$ of Poisson manifolds and Poisson maps to the quantum category $\mathcal Q$ of Hilbert spaces and unitary operators that is consistent with the cotangent bundle/$\frac12$-density relation (explained below).
The result is due to Van Hove, in "Sur le probleme des relations entre les transformations unitaires de la mecanique quantique et les transformations canoniques de la mecaniques classique." This is an old paper and I can't find a link for it, but the reference I found it in is Weinstein's "Lectures on Symplectic Manifolds."
By "cotangent bundle/$\frac12$-density relation" I mean the following: if $\mathcal M$ is the category of smooth manifolds and diffeomorphisms, we have a cotangent functor $\mathcal M \to \mathcal C$. This assigns to each manifold its cotangent bundle with the canonical symplectic structure, and to each diffeomorphism the induced symplectomorphism of cotangent bundles.
We also have a natural functor $\mathcal M \to \mathcal Q$. For any smooth manifold $X$ consider the bundle of complex $\frac12$-densities on $X$. (What is the bundle of complex $s$-densities? Well, the fiber over a point $x \in X$ is the set of functions $\delta\_x: \bigwedge^{top} T\_xX \to \mathbb{C}$ such that $\delta(cv) = |c|^{s}\delta(v)$.) If $\delta^1$ and $\delta^2$ are smooth compactly-supported $\frac12$-densities, their pointwise product $\delta^1 \bar{\delta^2}$ is a compactly supported 1-density which we can integrate to get a complex number. This turns the space of all such sections into a pre-Hilbert space, the completion of which is what our functor assigns to the manifold $X$. As we would hope for, the canonical nature of the construction lets us assign unitary operators between Hilbert spaces to diffeomorphisms between smooth manifolds, hence is functorial.
(Note: If we choose a volume form on $X$, the above procedure produces something isomorphic with the space of $L^2$ functions on $X$ with respect to this form, but to get something functorial we want a canonical construction.)
From this pair of functors $\mathcal M \to \mathcal C$ and $\mathcal M \to \mathcal Q$ we get a product functor $\mathcal M \to \mathcal{C} \times \mathcal{Q}$. The image of this functor is a subcategory of $\mathcal C \times \mathcal Q$ which we will call the "cotangent bundle/$\frac12$-density relation." (The word relation is meant in the same sense that an ordinary relation between two sets is a subset of their product).
Now we can clarify just what is meant by our original statement: there is no functor $\mathcal C \to \mathcal Q$ whose graph contains the cotangent bundle/$\frac12$-density relation. The reasons why this is a desirable condition come from physics and are beyond me, but roughly speaking I think the point is that there exists a good idea of what a quantization functor is supposed to do to cotangent bundles.
| 20 | https://mathoverflow.net/users/361 | 9788 | 6,687 |
https://mathoverflow.net/questions/9778 | 42 | There are two great first examples of complete discrete valuation ring with residue field $\mathbb{F}\_p = \mathbb{Z}/p$: The $p$-adic integers $\mathbb{Z}\_p$, and the ring of formal power series $(\mathbb{Z}/p)[[x]]$. Any complete DVR over $\mathbb{Z}/p$ is a ring structure on left-infinite strings of digits in $\mathbb{Z}/p$. The difference between these two examples is that in the $p$-adic integers, you add the strings of digits with carries. (In any such ring, you can say that a $1$ in the $j$th place times a 1 in the $k$th place is a 1 in the $(j+k)$th place.) At one point I realized that these two examples are not everything: You can also add with carries but move the carry $k$ places to the left instead of one place to the left. The ring that you get can be described as $\mathbb{Z}\_p[p^{1/k}]$, or as the $x$-adic completion of $\mathbb{Z}[x]/(x^k - p)$. This sequence has the interesting feature that the terms are made from $\mathbb{Z}\_p$ and have characteristic $0$, but the ring structure converges topologically to $(\mathbb{Z}/p)[[x]]$, which has characteristic $p$.
(**Edit:** Per Mariano's answer, I have in mind a discrete valuation in the old-fashioned sense of taking values in $\mathbb{Z}$, not $\mathbb{Z}^n$.)
I learned from Jonathan Wise in [a question on mathoverflow](https://mathoverflow.net/questions/7840/) that these examples are still not everything. If $p$ is odd and $\lambda$ is a non-quadratic residue, then the $x$-adic completion of $\mathbb{Z}[x]/(x^2-\lambda p)$ is a different example. You can also call it $\mathbb{Z}\_p[\sqrt{\lambda p}]$.
So my question is, is there is a classification or a reasonable moduli space of complete DVRs with residue field $\mathbb{Z}/p$? Or whose residue field is any given finite field? Or if not a classification, an indexed family that includes every example at least once?
I suppose that the question must be related to the Galois theory of $\mathbb{Q}\_p$; maybe the best answer would be a relevant sketch of that theory. But part of my interest is in continuous families of DVR structures on the Cantor set of strings of digits in base $p$.
---
As question authors often say in mathoverflow, I learned stuff from many of the answers and it felt incomplete to only accept one of them. I upvoted several others, though. Following Kevin, the set of pairs $(R,\pi)$, where $R$ is a CDVR and $\pi$ is a uniformizer, is an explicit indexed family that includes every example (of $R$) at least once. The mixed characteristic cases are parametrized by Eisenstein polynomials, and there is only one same-characteristic choice of $R$.
One side issue that was not addressed as much is continuous families of CDVRs. To explain that concern a little better, all CDVRs with finite residue field are homeomorphic (to a Cantor set). For any given finite residue field, there are many homeomorphisms that commute with the valuation. So you could ask what a continuous family of CDVRs can look like, where both the addition and multiplication laws can vary, but the topology and the valuation stay the same. Now that I have been told what the CDVRs are, I suppose that not all that much can happen. It seems key to look at $\nu(p)$, the valuation of the integer element $p$, in a sequence of such structures (say). If $\nu(p) \to \infty$, then the CDVRs have to converge to $k[[x]]$. Otherwise $\nu(p)$ is eventually constant; it eventually equals some $n$. Then it looks like you can convert the convergent sequence of CDVRs to a convergent sequence of Eisenstein polynomials of degree $n$.
| https://mathoverflow.net/users/1450 | Complete discrete valuation rings with residue field ℤ/p | Greg, I want to say some basic things, but people are giving quite "high-brow" answers and what I want to say is a bit too big to fit into a comment. So let me leave an "answer" which is not really an answer but which is basically background on some other answers.
So firstly there is this amazing construction of Witt vectors, which takes as input a finite field $k$ of characteristic p (well, it can take a lot more than that but let me stick with finite fields of characteristic p) and spits out a canonical complete DVR $W(k)$ with residue field $k$ and uniformiser $p$. If you feed in $Z/pZ$ it spits out $Z\_p$ and if you feed in, say, the field $Z/5Z[\sqrt{2}]$ with 25 elements it spits out something isomorphic to $Z\_5[\sqrt{2}]$, and so on. Witt vectors are just a way of formalising the "carry" business---it gets a bit trickier when the residue field isn't Z/pZ. Turns out that any complete DVR $R$ with fraction field of characteristic 0 and residue field $k$ is canonically and uniquely a $W(k)$-algebra. So that's pretty cool. Furthermore, $R$ will be finite and free over $W(k)$, and $R$ is the "ring of integers" (this makes sense) in its fraction field $K$, which is a finite extension of the fraction field of $W(k)$, something which in turn will be finite over $Q\_p$ of degree $d$ if $[k:Z/pZ]=d$. [Note in particular that "fraction field of Witt vectors" gives us a whole bunch of extension of $Q\_p$---the so-called "unramified" ones---one for each finite extension of $Z/pZ$.]
Conversely, given an arbitrary finite field extension $K$ of $Q\_p$, the ring of integers of $K$ (that is, the elements in $K$ satisfying a monic polynomial with coefficients in $Z\_p$) will be a complete DVR with residue field some finite field $k$, and then the field of fractions of $W(k)$ embeds into $K$. This subfield of $K$---the "maximal unramified subextension of $K$" is canonical and intrinsic. An extension $K$ of $Q\_p$ is "totally ramified" if the residue field of $R$ is $Z/pZ$. A FABULOUS place to read about this stuff is Serre's book "local fields". Everything there, with complete canonical proofs.
So one aspect of your question is whether there is a moduli space of totally ramified extensions of $Q\_p$. I wouldn't rule such a gadget out but I've not seen one. This might be just a question in algebra. If $K$ is a finite totally ramified extension of $Q\_p$ then $K=Q\_p(\pi)$ with $\pi$ a uniformiser of the integers of $K$, and $\pi$ will satisfy a degree $d$ equation if $d=[K:Q\_p]$. Furthermore this degree $d$ equation will be monic, with coefficients in $Z\_p$, and the constant term will be $p$ times a unit. Conversely any such equation will give a totally ramified extension of $Q\_p$ (they will all be irreducible by Eisenstein's criterion).
So now we have a list of all totally ramified extensions of $Q\_p$, and hence all complete DVRs with residue characteristic $Z/pZ$ but generic characteristic zero, because we just list all polynomials of this form. Unfortunately each $R$ is in our list infinitely often. So to make the $R$s the $Z/pZ$-points of a moduli space we need to quotient out the set of degree $d$ Eisenstein polynomials by the relation "the extension of $Q\_p$ generated by these polynomials are the same". A map $Q\_p(\pi\_1)\to Q\_p(\pi\_2)$ is just another polynomial so it looks to me like there is hope that this can be done, but I've not done it.
Finally, everything I said above has a natural generalisation to any finite field, not just $Z/pZ$.
When people talk about class field theory or Lubin-Tate groups above, what they're saying is that there are certain totally ramified extensions of $Q\_p$, namely those which are Galois over $Q\_p$ with abelian Galois group, which can be constructed explicitly using other techniques (like formal groups or Artin maps or whatever), and these constructions generalise to give all abelian extensions of an arbitrary finite extension of $Q\_p$. However if you're looking for general moduli spaces then it seems to me that these notions might not be of too much use to you because they don't give all extensions, just abelian ones.
There. So really that was just a comment but it was visibly too large. Hopefully someone will now quotient out the Eisenstein polynomials by an equivalence relation thus giving you your moduli space, because that seems to me to be the heart of your question.
| 29 | https://mathoverflow.net/users/1384 | 9789 | 6,688 |
https://mathoverflow.net/questions/9746 | 22 | Who was the first person who solved the problem of extending the factorial to non-integer arguments?
Detlef Gronau writes [1]: "As a matter of fact, it was Daniel Bernoulli who gave in 1729 the first representation of an interpolating function of the factorials in form of an infinite product, later known as gamma function."
On the other hand many other places say it was Leonhard Euler. Will the real inventor please stand up?
[1] "Why is the gamma function so as it is?" by Detlef Gronau, Teaching Mathematics and Computer Science, 1/1 (2003), 43-53.
| https://mathoverflow.net/users/2797 | Who invented the gamma function? | I don't have a complete answer. As you say, many sources say that Euler did it, but Gronau gives compelling reason to doubt this. The best source I have found for this issue is ["The early history of the factorial function" by Dutka](https://doi.org/10.1007/BF00389433), and for what it's worth I am convinced that Gronau's assessment is a fair one.
First, I'll summarize the usual story. Kline discusses this in chapter 19, section 5 of *Mathematical Thought from Ancient to Modern Times* (which falls in volume 2 of the paperback printing), and a more thorough source is Davis's article "[Leonhard Euler's Integral: A Historical Profile of the Gamma Function](https://www.jstor.org/stable/2309786)". There is agreement in these sources that Euler solved the problem after unsuccessful attempts by Stirling, D. Bernoulli, and Goldbach, and that the first record of Euler's solution appears in outline form in a 1729 letter from Euler to Goldbach. This was expanded in subsequent letters and written up in the [article](http://math.dartmouth.edu/%7Eeuler/pages/E019.html) to which Kristal Cantwell links (apparently the article was written in 1729 but not published until 1738). Euler's letters to Goldbach start on the third page of [this pdf](http://math.dartmouth.edu/%7Eeuler/correspondence/fuss/goldbach1-59).
However, Gronau cites a letter by Bernoulli that was written a few days before Euler's and that contains at least a partial solution, possibly contradicting Kline and Davis. Dutka's paper goes into more detail and also claims that Euler's work was influenced by Bernoulli's earlier solution. I could only speculate on what led to the confusion among other authors, and I won't do so here. Perhaps it should be mentioned here (as is done by Gronau and Dutka) that Euler did much more than Bernoulli. For instance, Euler gave the first integral representations of the gamma function.
**Edit:** Because this answer is accepted and yet incomplete, I want to direct attention to Bruce Arnold's answer below. It contains a link to a copy of the too often neglected letter of D. Bernoulli cited by Gronau and Dutka.
| 12 | https://mathoverflow.net/users/1119 | 9790 | 6,689 |
https://mathoverflow.net/questions/9793 | 20 | My grandfather had a PhD in math. When he died, he left a lot of math textbooks, which I took. These include things like Van der Waerden's 2-volume algebra set from the 1970s,
"Studies in Global Geometry and Analysis" by Shiing-Shen Chern, a series called "Mathematics: it's content, methods, and meaning," and many more.
I'm keeping about 20 of them, but there are 103 which I don't want to keep, but which I don't know what to do with. I obviously don't want to throw them away, and I don't really know what will happen to them if I donate them to the giant used-books depository in downtown Baltimore (called "the book thing," where people drop off and pick up used books for free). I'd like to donate them to some math collector or math library. But maybe there are just too many used antique math books floating around.
RECAP: I have 103 antique used math books which I cannot keep. Do you have a suggestion for what to do with them?
Thanks,
David
| https://mathoverflow.net/users/2811 | What to do with antique math books? | David, Older mathematics books can be surprisingly rare.
An option is to sell them on Advanced Book Exchange (abe.com).
I would be happy to help you triage your books. I did this once for the daughter of a philosopher who had a large mathematics book collection. It did not take long on the telephone.
Dan
| 12 | https://mathoverflow.net/users/2813 | 9797 | 6,693 |
https://mathoverflow.net/questions/9792 | 5 | First: Is there a precise meaning to the term "model for (oo,n)-categories"? A related question, which might actually be the same question, is: what exactly do we want to get out of (oo,n)-category theory for general n? Whatever definition of (oo,n)-categories we use, what are the desired things it should satisfy? What should the main examples be, for general n? I know that [bordism categories](http://www.math.harvard.edu/~lurie/papers/cobordism.pdf) should be examples. What else? Actually, aside from the cobordism hypothesis and other TQFT-y things I don't really know what the motivations are for (oo,n)-category theory for general n (or at least n bigger than 1), so I hope that people can say some words about that as well. (For n=1, there seems to be a lot of motivation, see for example [this](https://mathoverflow.net/questions/2185/how-to-think-about-model-categories) or [this](https://mathoverflow.net/questions/8663/infinity-1-categories-directly-from-model-categories) or [this](https://mathoverflow.net/questions/4689/stable-presentable-categories-as-module-categories) or [this](https://mathoverflow.net/questions/815/triangulated-vs-dg-a-infinity/) or ...)
Second: Presently, what are the models that we have for (oo,n)-categories? Which models have been proven to be equivalent? Of course, there's already a lot about this on the nLab:
[(oo,1)-categories](http://ncatlab.org/nlab/show/(infinity,1)-category)
[(oo,2)-categories](http://ncatlab.org/nlab/show/(infinity,2)-category)
[(oo,n)-categories](http://ncatlab.org/nlab/show/(infinity,n)-category)
I'm mainly just curious about the current status on (oo,n)-categories for general n. Aside from n-fold complete Segal spaces, are there other definitions/"models"?
| https://mathoverflow.net/users/83 | Models for, and motivation for, (oo,n)-categories for general n | For me a "model of (∞,n)-categories" is something (e.g., a model category) from which one can extract "the" (∞,1)-category of (∞,n)-categories. One could make this more precise by choosing a preferred definition of (∞,n)-categories and asking for things equivalent to it. Of course it's currently less clear than for, say, models of spaces or spectra that there really is a unique "correct" (∞,1)-category which really is equivalent to everything we hope it's equivalent to. And indeed already for n = 1 there are related but distinct useful notions of *category* as I write [here](https://mathoverflow.net/questions/8714/what-are-natural-transformations-in-1-categories/8959#8959) which manifest themselves in homotopy theory as Segal spaces and complete Segal spaces. But I think most everyone expects that there's a natural notion of (∞,n)-category up to n-categorical equivalence which is the right analogue of n-category (whatever that means).
The easiest examples are of course in the case n = 2, where we have (∞,2)-categorical analogues of the usual examples of 2-categories, for instance the (∞,2)-category of A∞ ring spectra and bimodules, or the (∞,2)-category of (∞,1)-categories, or presentable (∞,1)-categories, or stable (∞,1)-categories, ... For instance if you wanted to understand the relationship between (∞,1)-categories and their stabilizations—(∞,1)-categories form an (∞,2)-category, and stable ones form some kind of subcategory, and you might ask whether there is something like an adjoint to the inclusion—there isn't quite, but maybe there's an adjunction if we view these (∞,2)-categories as objects of some other (∞,3)-category. So (∞,n)-categories do arise naturally in the study of (∞,k)-categories for k < n. These examples are in a sense "algebraic" objects, as opposed to bordism categories, which turn out to be algebraic too, in a sense, but *a priori* are given by geometric constructions.
As for models for (∞,n)-categories: Besides the iterated complete Segal space model we have Charles Rezk's Θn-spaces, and I think simplicial strict n-categories are also supposed to give the right notion. There's the "complicial sets" model, which seems to me to be more conjectural. I would also like to hear about results about equivalence of these models—as far as I know none have been written down yet, except in the case n = 2.
| 5 | https://mathoverflow.net/users/126667 | 9802 | 6,697 |
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