source_id
int64 1
4.64M
| question
stringlengths 0
28.4k
| response
stringlengths 0
28.8k
| metadata
dict |
|---|---|---|---|
53,988 |
The mathematical definition of a vertex algebra can be found here: http://en.wikipedia.org/wiki/Vertex_operator_algebra Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean. To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them a priori (i.e. not "they come from physics" nor "you can prove moonshine with them"). As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are: What is the basic physical phenomenon/problem/question that vertex operators model? What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra? Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas? Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively? Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects?
|
Vertex algebras precisely model the structure of "holomorphic one-dimensional algebra" -- in other words, the algebraic structure that you get if you try to formalize the idea of operators (elements of your algebra) living at points of a Riemann surface, and get multiplied when you collide. Our geometric understanding of how to formalize this idea has I think improved dramatically over the years with crucial steps being given by the point of view of "factorization algebras" by Beilinson and Drinfeld, which is explained (among other places :-) ) in the last chapter of my book with Edward Frenkel, "Vertex algebras and algebraic curves" (second edition only). This formalism gives a great way to understand the algebraic structure of local operators in general quantum field theories -- as is seen in the recent work of Kevin Costello -- or in topological field theory, where it appears eg in the work of Jacob Lurie (in particular the notion of "topological chiral homology"). In fact I now think the best way to understand a vertex algebra is to first really understand its topological analog, the structure of local operators in 2d topological field theory.
If you check out any article about topological field theory it will explain that in a 2d TFT, we assign a vector space to the circle, it obtains a multiplication given by the pair of pants, and this multiplication is commutative and associative (and in fact a commutative Frobenius algebra, but I'll ignore that aspect here). It's very helpful to picture the pair of pants not traditionally but as a big disc with two small discs cut out -- that way you can see the commutativity easily, and also that if you think of those discs as small (after all everything is topologically invariant) you realize you're really describing operators labeled by points (local operators in physics, which we insert around a point) and the multiplication is given by their collision (ie zoom out the picture, the two small discs blend and look like one disc, so you've started with two operators and gotten a third). Now you say, come on, commutative algebras are SO much simpler than vertex algebras, how is this a useful toy model?
well think about where else you've seen the same picture -- more precisely let's change the discs to squares. Then you realize this is precisely the picture given in any topology book as to why $\pi_2$ of a space is commutative (move the squares around each other). So you get a great model for a 2d TFT by thinking about some pointed topological space X.. to every disc I'll assign maps from that disc to X which send the boundary to the basepoint (ie the double based loops of X), and multiplication is composition of loops -- i.e. $\Omega^2 X$ has a multiplication which is homotopy commutative (on homotopy groups you get the abelian group structure of $\pi_2$). In homotopy theory this algebraic structure on two-fold loops is called an $E_2$ structure. My claim is thinking about $E_2$ algebras is a wonderful toy model for vertex algebras that captures all the key structures. If we think of just the mildest generalization of our TFT story, and assign a GRADED vector space to the circle, and keep track of homotopies (ie think of passing from $\Omega^2 X$ to its chains) we find not just a commutative multiplication of degree zero, but a Lie bracket of degree one, coming from $H^1$ of the space of pairs of discs inside a bigger disc (ie from taking a "residue" as one operator circles another). This is in fact what's called a Gerstenhaber algebra (aka $E_2$ graded vector space). Now all of a sudden you see why people say you can think of vertex algebras as analogs of either commutative or Lie algebra (they have a "Jacobi identity") -- -the same structure is there already in topological field theory, where we require everything in sight to depend only on the topology of our surfaces, not the more subtle conformal geometry. Anyway this is getting long - to summarize, a vertex algebra is the holomorphic refinement of an $E_2$ algebra, aka a "vector space with the algebraic structure inherent in a double loop space", where we allow holomorphic (rather than locally constant or up-to-homotopy) dependence on coordinates. AND we get perhaps the most important example of a vertex algebra--- take $X$ in the above story to be $BG$, the classifying space of a group $G$. Then $\Omega^2 X=\Omega G$ is the "affine Grassmannian" for $G$, which we now realize "is" a vertex algebra.. by linearizing this space (taking delta functions supported at the identity) we recover the Kac-Moody vertex algebra (as is explained again in my book with Frenkel).
|
{
"source": [
"https://mathoverflow.net/questions/53988",
"https://mathoverflow.net",
"https://mathoverflow.net/users/332/"
]
}
|
53,999 |
I can't seem to find any work on the following question: Can every (closed, of finite type) Riemann surface $S$ be realized as an embedded (or even immersed) smooth surface in Euclidean $3$-space, where by realized, I mean that the induced conformal structure on $T \subset \mathbb{E}^3$ is the conformal structure on $S?$ The answer is obviously yes when $S$ is of genus $0,$ but that's where the obvious statements end -- I don't know the answer for $g=1.$
|
The embedding question has been answered in the affirmative by Adriano Garsia in 1961. Here is the link to MR. In fact, he has shortly after shown that the embedding might be chosen with real algebraic image.
|
{
"source": [
"https://mathoverflow.net/questions/53999",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11142/"
]
}
|
54,122 |
I hope this problem is not considered too "elementary" for MO. It concerns a formula that I have always found fascinating. For, at first glance, it appears completely "obvious", while on closer examination it does not even seem well-defined. The formula is the one that I was given as the definition of the cross-product in $\mathbb R^3 $ when I was first introduced to that concept: $$
B \times C :=
\det
\begin{vmatrix}
{\mathbf i } & {\mathbf j } & {\mathbf k } \\\\
B_1 & B_2 & B_3 \\\\
C_1 & C_2 & C_3\\\\
\end{vmatrix}
$$
On the one hand, if one expands this by minors of the first row, the result is clearly correct---and to this day this is the only way I can recall the formula for the components of the cross-product when I need it. But, on the other hand, the determinant of an $n \times n$ matrix whose elements are a mixture of scalars and vectors is undefined. Just think what happens if you interchange one element of the first row with the element just below it. In fact, as usually understood, for a determinant of a matrix to be well-defined, its elements should all belong to a commutative ring. But then again (on the third hand :-) if we take the dot product of both sides of the formula with a third vector, $A$, we seem to get: $$
A \cdot B \times C =
A \cdot
\det
\begin{vmatrix}
{\mathbf i } & {\mathbf j } & {\mathbf k } \\\\
B_1 & B_2 & B_3 \\\\
C_1 & C_2 & C_3\\\\
\end{vmatrix}
=
\det
\begin{vmatrix}
A_1 & A_2 & A_3 \\\\
B_1 & B_2 & B_3 \\\\
C_1 & C_2 & C_3\\\\
\end{vmatrix}
$$
and of course the left and right hand sides are well-known formulas for the (signed) volume of the parallelepiped spanned by the three vectors, $A, B, C$. Moreover, the validity of the latter formula for all choices of $A$ indicates that the original formula is "correct". So, my question is this: Is there a rigorous way of defining the original determinant so that all of the above becomes meaningful and correct?
|
But there is a commutative ring available, along the lines of what Mariano says. If $k$ is a field and $V$ is a vector space, then $k \oplus V$ is a commutative ring by the rule that a scalar times a scalar, or a scalar times a vector, or a vector times a scalar, are all what you think they are. The only missing part is a vector times a vector, and you can just set that to zero. The dot product is then a special bilinear form on the algebra. In the formalism, I think that everything that you wrote makes sense. Theo says in a comment that "even better", one should work over $\Lambda^*(V)$, the exterior algebra over $V$. The motivation is that this algebra is supercommutative. I considered mentioning this solution, and supposed that I really should have, because it arises in important formulas. For example, the Gauss formula for the linking number between two knots $K_1, K_2 \subseteq \mathbb{R}^3$ is:
$$\mathrm{lk}(K_1,K_2) = \int_{K_1 \times K_2} \frac{\det \begin{bmatrix} \vec{x} - \vec{y} \\ d\vec{x} \\ d\vec{y} \end{bmatrix}}{4\pi |\vec{x} - \vec{y}|^3}$$
$$= \int_{K_1 \times K_2}
\frac{\det \begin{bmatrix} x_1 - y_1 & x_2 - y_2 & x_3 - y_3 \\
dx_1 & dx_2 & dx_3 \\ dy_1 & dy_2 & dy_3 \end{bmatrix}}{4\pi |\vec{x} - \vec{y}|^3}.$$
The right way to write and interpret this formula is indeed as a determinant in the exterior algebra of differential forms. For one reason, it makes it easy to generalize Gauss' formula to higher dimensions. However, supercommutative is not the same as commutative, and this type of determinant has fewer properties than a determinant over a commutative ring. And different properties. Such a determinant has a broken symmetry: you get a different answer if you order the factors in each term by rows than by columns. (I am using row ordering.) Indeed, the row-ordered determinant can be non-zero even if it has repeated rows. To give two examples, the determinant in the generalized Gauss formula has repeated rows, and the standard volume form in $\mathbb{R}^n$ is
$$\omega = \frac{\det ( d\vec{x}, d\vec{x}, \ldots, d\vec{x} )}{n!}.$$ Happily, for Dick's question, you can truncate the exterior algebra at degree 1, which is exactly what I did. This truncation is both supercommutative and commutative.
|
{
"source": [
"https://mathoverflow.net/questions/54122",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7311/"
]
}
|
54,193 |
Consider graphs on $n$ nodes. I am trying to find a graph $G$ that contains all $n$-node trees as sub-graphs but contains as few edges as possible. The complete graph $K_n$ suffices, but can we get by with fewer edges? Maybe $O(n)$ edges? (This problem arose in the context of circuit design, where edges in $G$ correspond to wires in a chip layout.)
|
See Chung and Graham, On Universal Graphs for Spanning Trees . They prove that the number of edges is $\Theta(n\log n)$.
|
{
"source": [
"https://mathoverflow.net/questions/54193",
"https://mathoverflow.net",
"https://mathoverflow.net/users/8938/"
]
}
|
54,197 |
The Hodge Conjecture states that every Hodge class of a non singular projective variety over $\mathbf{C}$ is a rational linear combination of cohomology classes of algebraic cycles: Even though I'm able to understand what it says, and at first glance I do find it a very nice assertion, I cannot grasp yet why it is so relevant as to be considered one of the biggest open problems in algebraic geometry. Which are its implications? Going a bit further, what about the Tate conjecture?
|
Let $K$ be one of the following fields: the complex numbers, a finite field, a number field (and we could amalgamate the last two into the more general case of a field finitely generated overs its prime subfield). In each case we can consider the category of smooth projective varieties over $K$, with morphisms being correspondences [added: modulo the relation of cohomological equivalence;
see David Speyer's comment below for an elaboration on this point]. (I am really thinking of the category of pure motives, but there is no particular need to invoke that word.) In each case we also have a natural abelian (in fact Tannakian) category in play: in the complex case, the category of pure Hodge structures, and in the other cases, the category of $\ell$-adic representations of $G_K$ (the absolute Galois group of $K$) (for some prime $\ell$, prime to the characteristic of $K$ in the case when $K$ is a finite field). Now taking cohomology gives a functor from the category of smooth projective varieties to this latter category (via Hodge theory in the complex case, and the theory of etale cohomology in the other cases). The Hodge conjecture (in the complex case) and the Tate conjecture (in the other cases) then says that this functor is fully faithful. The consequence (if the conjecture is true) is that we can construct correspondences between varieties simply by making computations in some much more linear category. Let me give an example of how this can be applied in number theory: Fix two primes $p$ and $q$, and let $\mathcal O_D$ be a maximal order in the quaternion algebra $D$ over $\mathbb Q$ ramified at $p$ and $q$, and split everywhere else (including at infinity).
Let $\mathcal O_D^1$ denote the multiplicative group of norm one elements in $\mathcal O_D$. Since $D \otimes_{\mathbb Q} \mathbb R \cong M_2(\mathbb R)$, we may regard $\mathcal O_D^1$ as a discrete subgroup of $SL_2(\mathbb R)$, and form the quotient $X := \mathcal O_D^1\backslash \mathcal H$ (where $\mathcal H$ is the upper half-plane). We may also consider the usual congruence subgroup $\Gamma_0(pq)$ consisting of matrices in $SL_2(\mathbb Z)$ that are upper triangular modulo $pq$, and form $X_0(pq)$, the compacitifcation of $\Gamma_0(pq)\backslash \mathcal H$. Now the theory of modular and automorphic forms and their associated Galois representations show that $X$ and $X_0(pq)$ are both naturally curves over $\mathbb Q$, and that there
is an embedding of Galois representations $H^1(X) \to H^1(X_0(pq))$. Thus the Tate conjecture predicts that there is a correspondence between $X$ and $X_0(pq)$ inducing
this embedding. Passing to Hodge structure, we would then find that the periods of holomorphic one-forms on $X$ should be among the periods of holomorphic one-forms on $X_0(pq)$. Now the theory of $L$-functions shows that the periods of holomorphic one-forms on $X_0(pq)$ in certain cases compute special values of $L$-functions attached to modular forms on $\Gamma_0(pq)$. So putting this altogether, we would find that (given the Tate conjecture) we could compute special values of $L$-functions for certain modular forms by taking period integrals on the curve $X$. In certain respects $X$ is better behaved than $X_0(pq)$, and so this is an important technique in investigating the arithmetic of $L$-functions. Now it happens that in this case the Tate conjectures is a theorem (of Faltings), and so
the above argument is actually correct and complete. But there are infinitely many other analogous situations in the theory of Shimura varieties where the Tate conjecture is not yet known, but where one would like to know it. There are also similar arguments where one uses the Hodge conjecture to move from information about periods to information about Galois representations and arithmetic. I should also say that there are lots of partial results along the above lines in these cases; there are many inventive techniques that people have introduced for getting around the Hodge and Tate conjectures. (Deligne's use of absolute Hodge cycles, applications of theta corresondences by Shimura, Harris, Kudla, and others, ... .) But the Hodge and Tate
conjectures stand as fundamental guiding principles telling us what should be true, and which we would dearly like to see proved. Summary: algebraic cycles are very rich objects, which straddle two worlds, the world of period integrals and the world of Galois representations. Thus, if the Hodge and Tate conjectures are true, we know that there are profound connections between those two worlds: we can pass information from one to the other through the medium of algebraic cycles. If we had these conjectures available, it would be an incredible enrichment of our understanding of these worlds; as it is, people expend a lot of effort to find ways to pass between the two worlds in the way the Hodge and Tate conjectures predict should be possible. Another example, added following the request of the OP to provide an example just involving complex geometry: If $X$ is a K3 surface, then from
the Hodge structure on the primitive part of $H^2(X,\mathbb C)$, one can construct an
abelian variety, the so-called Kuga--Satake abelian variety associated to $X$. The
construction is made in terms of Hodge structures. One expects that in fact there should
be a link between $X$ and its associated Kuga--Satake abelian variety provided by some correspondence, but this is not known in general. It would be implied by the Hodge conjecture. (This constructions is discussed in this answer and in the accompanying comments.)
|
{
"source": [
"https://mathoverflow.net/questions/54197",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12420/"
]
}
|
54,221 |
Let $j$ be the Klein $j$-invariant (from the theory of modular functions). Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for
any rational function $f$?
|
No. Conceptually, the reason is that $j'(z)$ is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight $2$, so it cannot be expressed in terms of $j$ (weakly holomorphic modular form of weight $0$) and $z$ (not anywhere near being a modular form). For a rigorous proof: Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$. Suppose that the $j$ invariant did satisfy a differential equation of your form. Then we'd have $f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions $z$ and $j(z)$ are algebraically independent (this is just saying that $j(z)$ is a transcendental function). Hence the underlying two-variable rational function $f(x, y)$ satisfies $f(x, y) = f(x, y+1)$. This then easily implies that $f(x, y)$ must be independent of $y$, e.g. $f(x, y) = g(x)$ for some rational function $g$. So our original differential equation must actually take the form $j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight $2$ while the right hand side has weight 0, and a nonzero modular form has a unique weight, so this is impossible.
|
{
"source": [
"https://mathoverflow.net/questions/54221",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12669/"
]
}
|
54,232 |
Writing a book from the beginning to the end is (so I heard) a very hard process. Planning a book is easier. This question is dual in a sense to the question " Books you would like to read (if somebody would just write them )". It is about a book that you feel you would like to write (if you just have the time). A book that will describe a topic not yet properly discribed or give a new angle to a subject that you can contribute. The question is meant to refer to realistic or semi-realistic projects (related to mathematics). Answers about book projects based on existing survey articles or lecture notes can be especially useful. Of course, If you had some progress in writing a book mentioned here please please update your answer!
|
Book : The Differential Topology of Loop Spaces Why : Because they are one of the first examples of spaces that are almost, but not quite, entirely unlike manifolds. They are relatively straightforward spaces which can be fairly conceptually grasped, but still contain enough intricacies to reveal some of the important differences between finite and infinite dimensions (though perhaps I should say between manifolds modelled on Banach spaces and more general manifolds). A book on their differential topology would thus be a gentle introduction to the topic than is (as far as I'm aware) currently available (in particular, although just about everything I'd want to say is covered in Kriegl and Michor's works, it's in such a context and with such generality that "daunting" doesn't quite cut the mustard). Who For : Me, 10 years ago. That is, I'd try to write the book I wish I'd had when starting out in infinite dimensional differential topology so I wouldn't have made all the mistakes that I made. Why Me : Because I work in that area and I think I've made just about every wrong assumption about loop spaces possible so I know lots of the traps for unwary differential topologists venturing out into the miasma that is infinite dimensional topology. Will I Ever Actually Write It : Maybe, maybe not (vote for this answer if you want me to!). I made a start by writing up some seminar notes. I've started transferring them in to the nLab (but in the process I've been generalising them which slightly goes against the purpose of the project as I described it above). I'd certainly like to write it, if only to convince myself that I no longer have all those false assumptions, but whether or not I ever actually do it ... (hey, I've an idea, maybe all the time I put into MO and meta.MO could be reallocated to book-writing. Then it'll be finished next week.). Update: 2019-01-07 Due to changes in circumstances, I am extremely unlikely to spontaneously develop the above-mentioned notes into a book. Should anyone be in a position to say to me "If you did polish those notes into a book we'd definitely publish it" feel free to get in touch.
|
{
"source": [
"https://mathoverflow.net/questions/54232",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1532/"
]
}
|
54,239 |
I want to start out by giving two examples: 1) Graham's problem is to decide whether a given edge-coloring (with two colors) of the complete graph on vertices $\lbrace-1,+1\rbrace^n$ contains a planar $K_4$ colored with just one color. Graham's result is that such a $K_4$ exists provided $n$ is large enough, larger than some integer $N$. Recently, I learned that the integer $N$ determined by Graham's problem is known to lie between $13$ and $F^{7}(12)$, which is called Graham's number , a number which beats any imagination and according to Wikipedia is practically incomprehensible . Some consider it to be the largest number which was ever used in a serious mathematical argument. One way of looking at this is the following: Graham's problem of deciding whether for given $n$ and given coloring such a planar $K_4$ exists takes constant time . Indeed, if $n$ is small you have to look, if $n$ is large you are done. However, this takes polynomial time (check all four-tuples of vertices) for all practical purposes (assuming that $N$ is close to its upper bound). I am sure someone can now cook up an algorithm which needs exponential time for all practical purposes but constant or polynomial time in general. 2) I also learned that the best proof of Szemerédi's regularity lemma yields a bound on a certain integer $n(\varepsilon)$ which is the $\log(1/\varepsilon^5)$́-iteration of the exponential function applied to $1$. This bound seems ridiculous in the sense that it does not even allow for interesting applications of this result (say with $\varepsilon=10^{-6}$) to networks like the internet, neural networks or even anything practically thinkable. At this point, this is only an upper bound, but Timothy Gowers showed that $\log(1/\varepsilon)$-iterations are necessary. Again, it seems that one could cook up reasonable algorithmic problems which have solutions which are polynomial time but practically useless. Maybe one can do better in concrete cases, but this then needs additional input. Coming closer to the question, what if finally $P=NP$ holds, but the proof involves something like the existence of a solution to Graham's problem or an application of Szemerédi's regularity lemma, so that finally the bounds of the polynomial time algorithm are for some reason so poor that nobody even wants to construct it explicitly. Maybe the bounds are exponential for all practical purposes, but still polynomial. I often heard the argument that once a polynomial solution for a reasonable problem is found, further research has also produced practicable polynomial time algorithms. At least for Graham's problem this seems to fail miserably so far. Question: Is there any theoretical evidence for this? Now, maybe a bit provocative: Question: Why do we think that $P\neq NP$ is necessarily important? I know that $P$ vs. $NP$ is important for theoretical and conceptional reasons, but what if finally $P=NP$ holds but no effective proof can be found. I guess this wouldn't change much. EDIT: Just to be clear: I do not dismiss complexity theory at all and I can appreciate theoretical results, even if they are of no practical use.
|
The $P \ne NP$ problem is the best way we know to formulate the belief (which was expressed even before the problem was formally stated) that certain specific algorithmic problems (such as finding a Hamiltonian cycle in a graph) requires exponential number of steps as a function of their description. The formulation is based on the important notion of a nondeterministic algorithm. The conjecture that P is not equal to NP is the basis of a mathematical theory of complexity which is remarkably beautiful. It is likely that a proof for the conjecture will lead to better understanding of the limitations of computation which will go beyond the conjecture itself. (Unfortunately we are very far from such a proof.) A counterexample (which is not expected) may lead to a major change of our reality and not just our understanding of it. (I suppose this is one of the main reasons in believing the conjecture.) Your concern is that maybe, because of the asymptotic nature of the conjecture, we can question its real relevance. Namely problems we regard as intractable may still be even if P=NP because of huge constants in the polynomial involved, or perhaps even if NP not equal P there can be efficient algorithm to relevant cases of intractable problems. These are serious concerns that are often raised and sometimes there are efforts to study them scientifically. Overall, it looks that these asymptotic concerns are secondary compared to the problem itself. Moreover, there are many examples that the asymptotic behavior and practical behavior are similar beyond what the theory dictates (but a few examples also in the other direction). There are other related concerns regarding the relevance for the $P \ne NP$ problem. Is the worst case analysis relevant to practical problems? Is the hardness apply when we are interested in approximate solution rather than in exact solution? Both these questions (like the asymptotic issue) can be regarded as secondary in importance compared to the $NP \ne P$ problem itself but otherwise very important. Indeed both hardness of approximation and average case analysis are very central research topics. (A side remark, there is something unnatural about the way Graham's numbers are used in the question. You are dealing with a decision problem where the answer is always yes when n is sufficiently large. And there is a simple polynomial time algorithm as a function of n to decide the answer for a every given n. So the relevance of this particular example (which is interesting) to the question asked about NP complete problems (which is also interesting) is not so strong.) Finally, regarding relevance. Andreas raised the interesting possibility that NP=P but the constant involved are so huge that the polynomial algorithm for solving NP complete problems is utterly not practical. A related interesting possibility is that there even exists a practical polynomial time algorithm for NP complete problems but that finding this algorithm itself is computationally intractable. As I said, both these possibilities are considered unplausible. Even if true they will not harm the relevance of the NP=P problem in making the central distinction between intractable and tractable problems. In order to make these concerns interesting one should come up with a theoretical framework to study them, and then study them fruitfully. (As I mentioned above, this was done for related concerns regarding approximation and regarding average case behavior.)
|
{
"source": [
"https://mathoverflow.net/questions/54239",
"https://mathoverflow.net",
"https://mathoverflow.net/users/8176/"
]
}
|
54,252 |
The standard Reuleaux triangle is not smooth, but the three
points of tangential discontinuity can be smoothed as
in the figure below (left), from the Wikipedia article .
However, it is unclear (to me) from this diagram
whether the curve is $C^2$ or $C^\infty$ . Meissner’s tetrahedron is a 3D body of constant width,
but it is not smooth, as is evident in the right figure
below. My question is: Are there $C^\infty$ constant-width bodies in $\mathbb{R}^d$ (other than the spheres)? The image of Meissner’s tetrahedron above is taken
from the impressive work of
Thomas Lachand–Robert and Edouard Oudet,
"Bodies of constant width in arbitrary dimension"
( Math. Nachr. 280, No. 7, 740-750 (2007); pre-publication PDF here ). Here is a link to Wayback Machine .) I suspect the answer to my question is known, in which case a reference would suffice. Thanks! Addendum. Thanks to the knowledgeable (and rapid!) answers by Gerry, Anton, and Andrey,
my question is completely answered—I am grateful!!
|
Fillmore showed that there are sets of constant width in $\mathbb R^d$ with analytic boundaries which have a trivial symmetry group (so these are very different from spheres;
see "Symmetries of surfaces of constant width" , J. Differential Geom. , Vol 3, (1969), pp. 103-110). Moreover, the set of bodies of constant width with analytic boundaries is dense in the space of all convex bodies of constant width in $\mathbb R^d$ with respect to the Hausdorff metric (see e.g. "Smooth approximation of convex bodies" by Schneider).
|
{
"source": [
"https://mathoverflow.net/questions/54252",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6094/"
]
}
|
54,333 |
In G. F. Simmons' Differential Equations book (p.141) , the following claim is made:
“... As a matter of fact, there is no known type of second order linear differential equation- apart from those with constant coefficients, and those reducible to these by changes of the independent variable--which can bee solved in terms of elementary functions.” About 36 years have now passed since this statement made its published appearance. Is the remark still true or was it false even before? My motivation is the analogous theory of integrals of finite combinations of elementary functions where you know certain large and important classes of functions whose integral is expressible as an elementary function( or as a finite combination of such). As we know, there is a somewhat complete classification as to which integral of finite combination of elementary function of one variable is expressible as a finite combination of elementary functions.(References are Piskunov's book and this ). So it seemed to me that it is natural to care whether such theorems existed for differential equations. Thanks.
|
This claim is not valid. The big breakthrough was the paper below: MR0839134 (88c:12011)
Kovacic, Jerald J.
An algorithm for solving second order linear homogeneous differential equations.
J. Symbolic Comput. 2 (1986), no. 1, 3–43.
12H05 (34A30) Later, M. Petkovsek extended the ideas to second order difference equations.
|
{
"source": [
"https://mathoverflow.net/questions/54333",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5627/"
]
}
|
54,343 |
There are different conventions for defininig the wedge product $\wedge$. In Kobayashi-Nomizu, there is $\alpha\wedge\beta:=Alt(\alpha\otimes\beta)$,
in Spivak, we find $\alpha\wedge\beta:=\frac{(k+l)!}{k!l!}Alt(\alpha\otimes\beta)$,
where $\alpha$ and $\beta$ are any forms of degree $k$ and $l$ respectively, and $Alt(\cdot)$ take the alternating part of the tensor. But, is there a rationale to prefer one of them among the others? If not, what do you prefer? and for what reason?
|
I think a lot of people run into this issue. The way I think about it is the following: Take your finite-dimensional vector space $V$ and form its tensor algebra $T(V)$. Define $\mathcal{J}$ to be the 2-sided ideal in $T(V)$ generated by elements of the form $v \otimes v$, and then define the exterior algebra to be $\Lambda(V) = T(V) / \mathcal{J}$. This exhibits the exterior algebra as a quotient of the tensor algebra. The different conventions you see for the wedge product arise from different embeddings of the exterior algebra into the tensor algebra. Define on $V^{\otimes n}$ the map
$$ A_n (v_1 \otimes \dots \otimes v_n) = \frac{1}{n!} \sum_{\pi \in S_n} sgn(\pi) v_{\pi(1)} \otimes \dots \otimes v_{\pi(n)},
$$
(or possibly with $\pi^{-1}$ instead of $\pi$, although I guess it doesn't matter)
and then define on the tensor algebra the map
$$A = \bigoplus_{n=0}^{\infty} A_n.$$
Then you can show easily that $A_n^2 = A_n$ for all $n$, so that $A$ is a projection. The point is that $\mathcal{J} = \mathrm{ker} (A)$, so that you can identify the quotient $\Lambda(V)$ with $\mathrm{im} A$, i.e. we have now embedded the exterior algebra as a subspace of the tensor algebra. This is where the two conventions differ. I have defined $A_n$ with a $\frac{1}{n!}$ in front, but some don't do so. Of course, this doesn't change the kernel of the map, but it does change the embedding of the exterior algebra into the tensor algebra. The important point is that $A$ is not an algebra map of $T(V)$ to itself, so the embedding $\Lambda(V) \to T(V)$ is not an embedding of algebras. Now you ask how to describe the exterior product in terms of the product in $T(V)$. Take $\alpha \in \Lambda^k(V)$ and $\beta \in \Lambda^l(V)$ with representatives $\tilde{\alpha} \in \mathrm{im}(A_k)$ and $\tilde{\beta} \in \mathrm{im}(A_l)$, respectively. Then $A_{k+l}(\tilde{\alpha} \otimes \tilde{\beta})$ is the representative of $\alpha \wedge \beta$ that you're looking for. Essentially, it boils down to whether or not you put the $\frac{1}{n!}$ in front of your alternating map or not.
|
{
"source": [
"https://mathoverflow.net/questions/54343",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12617/"
]
}
|
54,402 |
Could anyone point me to a place where I could find Deligne's letter to Piatetskii-Shapiro from 1973? It is cited for example in Berkovich's "Vanishing cycles for formal schemes II".
|
I have typeset Deligne's letter, and placed the result here: http://www.math.ias.edu/~jaredw/DeligneLetterToPiatetskiShapiro.pdf I have made some minor edits so that the text reads more naturally to a native speaker of English. Also I made a few annotations where I truly believe there is an error in the original. Any other errors are mine. The letter struck me with how much it accomplishes in such a short space. Actually, I would say that the meat of the argument is confined to the final three pages, with the rest there only to establish notation. This letter ought to be required reading for anyone studying automorphic forms in arithmetic!
|
{
"source": [
"https://mathoverflow.net/questions/54402",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10701/"
]
}
|
54,430 |
It can be difficult to learn mathematics on your own from textbooks, and I often wish universities videotaped their mathematics courses and distributed them for free online. Fortunately, some universities do that (albeit to a very limited extent), and I hope we can compile here a list of all the mathematics courses one can view in their entirety online. Please only post videos of entire courses; that is, a speaker giving one lecture introducing a subject to the audience should be off-limits, but a sequence of, say, 30 hour-long videos, each of which is a lecture delivered in a class would be very much on-topic.
|
The lecture videos of Introduction to Abstract Algebra, taught by Benedict Gross at Harvard, can be downloaded here .
|
{
"source": [
"https://mathoverflow.net/questions/54430",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1407/"
]
}
|
54,434 |
As we all know, for a Riemannian manifold $(M,g)$ , there exists a unique torsion-free connection $\nabla_g$ , the Levi-Civita connection, that is compatible with the metric. I was wondering if one can reverse this situation: Given a manifold with $M$ with connection $\nabla$ , when does there exist a Riemannian metric $g$ for which $\nabla$ is the Levi-Civita connection. If this were true for complex projective manifolds it would make me be very happy.
|
Bill and Willie have (of course) given correct answers in terms of the holonomy of the given torsion-free connection $\nabla$ on the $n$-manifold $M$. However, it should be pointed out that, practically, it is almost impossible to compute the holonomy of $\nabla$ directly, since this would require integrating the ODE that define parallel transport with respect to $\nabla$. Even though they are linear ODE, for most connections given explicitly by some functions $\Gamma^i_{jk}$ on a domain, one cannot perform their integration. Although, as Bill pointed out, you cannot always tell from local considerations whether $\nabla$ is a metric connection, you can still get a lot of information locally, and this usually suffices to determine the only possibilities for $g$. The practical tests (carried out essentially by differentiation alone) were of great interest to the early differential geometers, but they don't get much mention in the modern literature. For example, one should start by computing the curvature $R$ of $\nabla$, which is a section of the bundle $T\otimes T^\ast\otimes \Lambda^2(T^\ast)$. (To save typing, I won't write the $M$ for the manifold.) Taking the trace (i.e., contraction) on the first two factors, one gets the $2$-form $tr(R)$. This must vanish identically, or else there cannot be any solutions of $\nabla g = 0$ for which $g$ is nondegenerate. (Geometrically, $\nabla$ induces a connection on $\Lambda^n(T^\ast)$ (i.e., the volume forms on $M$) and $tr(R)$ is the curvature of this connection. If this connection is not flat, then $\nabla$ doesn't have any parallel volume forms, even locally, and hence cannot have any parallel metrics.) To get more stringent conditions, one should treat $g$ as an unknown section of the bundle $S^2(T^\ast)$, pair it with $R$ (i.e., 'lower an index') and symmetrize in the first two factors, giving a bilinear pairing $\langle g, R\rangle$ that is a section of $S^2(T^\ast)\otimes \Lambda^2(T^\ast)$. By the Bianchi identities, the equation $\langle g, R\rangle = 0$ must be satisfied by any solution of $\nabla g = 0$. Notice that these are linear equations on the coefficients of $g$. For most $\nabla$ when $n>2$, this is a highly overdetermined system that has no nonzero solutions and you are done. Even when $n=2$, this is usually $3$ independent equations for $3$ unknowns, and there is no non-zero solution. Often, though, the equations $\langle g, R\rangle = 0$ define a subbundle (at least on a dense open set) of $S^2(T^\ast)$ of which all the solutions of $\nabla g= 0$ must be sections. (As long as $R$ is nonzero, this is a proper subbundle. Of course, when $R=0$, the connection is flat, and the sheaf of solutions of $\nabla g = 0$ has stalks of dimension $n(n{+}1)/2$.) The equations $\nabla g = 0$ for $g$ a section of this subbundle are then overdetermined, and one can proceed to differentiate them and derive further conditions. In practice, when there is a $\nabla$-compatible metric at all, this process spins down rather rapidly to a line bundle of which $g$ must be a section, and one can then compute the only possible $g$ explicitly if one can take a primitive of a closed $1$-form. For example, take the case $n=2$, and assume that $tr(R)\equiv0$ but that $R$ is nonvanishing on some simply-connected open set $U\subset M$. In this case, the equations $\langle g, R\rangle = 0$ have constant rank $2$ over $U$ and hence define a line bundle $L\subset S^*(T^\ast U)$. If $L$ doesn't lie in the cone of definite quadratic forms, then there is no $\nabla$-compatible metric on $U$. Suppose, though, that $L$ has a positive definite section $g_0$ on $U$. Then there will be a positive function $f$ on $U$, unique up to constant multiples, so that the volume form of $g = f\ g_0$ is $\nabla$-parallel. (And $f$ can be found by solving an equation of the form $d(\log f) = \phi$, where $\phi$ is a closed $1$-form on $U$ computable explicitly from $\nabla$ and $g_0$. This is the only integration required, and even this integration can be avoided if all you want to do is test whether $g$ exists, rather than finding it explicitly.) If this $g$ doesn't satisfy $\nabla g = 0$, then there is no $\nabla$-compatible metric. If it does, you are done (at least on $U$). The complications that Bill alludes to come from the cases in which the equations $\langle g, R\rangle = 0$ and/or their higher order consequences (such as $\langle g, \nabla R\rangle = 0$, etc.) don't have constant rank or you have some nontrivial $\pi_1$, so that the sheaf of solutions to $\nabla g = 0$ is either badly behaved locally or doesn't have global sections. Of course, those are important, but, as a practical matter, when you are faced with determining whether a given $\nabla$ is a metric connection, they don't usually arise.
|
{
"source": [
"https://mathoverflow.net/questions/54434",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1648/"
]
}
|
54,513 |
This question is similar to a previous one about "urban legends" , but not the same. It is established that Milnor proved the Fáry-Milnor theorem as an undergraduate at Princeton. For the record, Fáry was a professor in France and proved the result independently. Milnor has a solely authored paper in the Annals that acknowledges Fox. The problem had been posed by Borsuk in 1947. In Milnor's version of the theorem, the infimal total curvature of any smooth knot equals its bridge number. There are many stories to that Milnor thought it was a homework problem or a test question, that he came to class late and solved it on the spot, etc. My question is: Is there any good evidence that when Milnor solved this problem, that he thought it was anything other than an open problem? I am not so interested in answers with "names withheld to spare embarrassment", nor otherwise in digressions or mischievous answers. I am more interested in a convincing citation, either to print or to a named person. (Like Milnor himself, although he has the right not to discuss the matter.) Frankly I think that some versions of this Milnor story are a little bit tasteless. At least in my own mind, I'd like to have it in a more dignified form, if it's true at all. The answer by "none" is very good and at first I accepted it. But when I checked Nasar's book, to my surprise it does not completely put the issue to rest. Nasar says that Milnor, as a freshman, showed the proof to his differential geometry professor Albert Tucker with the request, "Would you be good enough to point out the flaw in this attempt. I'm sure there is one, but can't find it." (Tucker then passed around the proof to Fox and Chern.) It would make no sense for Milnor to say this if he truly thought that it was homework. But Nasar also contradicts this inference with the statement, "The story goes that Milnor mistook the conjecture for a homework assignment." For this statement she cryptically cites "Princeton University Archives". Besides the conflicting evidence, the mistaken-for-homework story as attributed to Milnor is also suspiciously similar to the confirmed story about Dantzig. Update: I sent e-mail to Milnor and I got a short reply that begins, "It seems a pity to contradict such a pleasant tall tale; but for my version see..." As I see it, the story that Milnor mistook Borsuk's curvature conjecture for homework has now died three deaths: (1) Milnor says that it's not true. (2) It isn't consistent with either Milnor's or Tucker's account of what happened. (3) It is similar to a true story about Dantzig that has mutated and that has also been pasted onto other mathematicians, such as for instance Fefferman. By 1991, Tucker said that "as a bad joke", he called Borsuk's conjecture "an assignment". But this is 40 years after the fact, when history was already muddied by the mutating story that started with Dantzig. Actually it's not necessarily so bad, if it happened, because it should have been clear from context that it was only a joke. Milnor referred me to a short autobiographical account, "Growing up in the Old Fine Hall". This version of the story says that Tucker first discussed Fenchel's theorem that total curvature of any topological circle is at least $2\pi$, and then stated Borsuk's conjecture; then a few days later Milnor had a draft of a proof. This account emphasizes the mathematics more than the human drama. Even so, it's just not consistent with the "thought it was homework" story. In various sources (many provided in the second answer by "none" below), Tucker gives a consistent account that Milnor first thought that his proof was wrong and asked Tucker to find the mistake. This detail is decidedly absent from Milnor's published accounts. Obviously he asked Tucker to check his proof; that could easily be confused with being sure that there is a mistake. Maybe only two people were there; I would leave it at saying that one of them says so and the other one doesn't say. I re-accepted the first answer given by "none" since it is most of the whole story. Finally the true case of Dantzig. One reason that this story is so well-known, and a reason that it has mutated so much and gotten a little tacky, began with a coincidence on a plane trip to or from California. Dantzig happened to be seated next to mega-Reverend Robert Schuller and told him his account. Schuller relayed it to his flock with some silly exaggerations, and the story made its way into many other churches. Eventually Dantzig's name disappeared from this story spreading mostly among non-mathematicians. It would then be easy to fill in the name of some other mathematician. This history is told in the book "Curses! Broiled Again!" (which apparently got it from the College Mathematics Journal) For the record, in the true version of the story, Dantzig was in graduate school in statistics, it was 1939, he came in late to class, he saw two open problems on the board and thought they were homework. The professor was Neyman, who must have been a bit impenetrable to his students, because he didn't tell Dantzig for six weeks what had really happened.
|
The story is told in some detail in Sylvia Nasar's "A Beautiful Mind", a biography of John Nash, who was a fellow student of Milnor at one point. In that version, Milnor knew that Borsuk's conjecture was an open problem; he wrote up his apparent answer not believing it to be correct, and asked Fox to look it over since he (Milnor) hadn't been able to find the error himself. Fox told him to write up the result for publication; the final result was generalized considerably over the original version. It's pretty likely that Nasar interviewed Milnor (because of his biographical connection with Nash) while writing the book, so her version is probably as good as you'll find. The "came to class late and thought it was a homework problem" story is about George Dantzig and is easy to find on the internet (e.g. Wikipedia or Snopes). It was about some problem in statistics. I think there may have actually been two open problems involved. Sometimes the Dantzig story gets told with SIX open problems. That might be confabulated with Grothendieck's PhD thesis. Dieudonné and Schwartz had written a paper on functional analysis ending with six apparently difficult open problems. They turned these over to Grothendieck saying something like "see if you can make some progress on any of these, and that can be your thesis." Within a few months Grothendieck developed the theory of nuclear spaces, that turned all six problems into trivial calculations, and basically killed off the research direction originally proposed (by completely solving it). That story is from Allyn Jackson's biographical profile of Grothendieck in Notices of the AMS, I think.
|
{
"source": [
"https://mathoverflow.net/questions/54513",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1450/"
]
}
|
54,603 |
I tried reading about Arakelov theory before, but I could never get very far. It seems that this theory draws its motivation from geometric ideas that I'm not very familiar with. What should I read to learn about the geometry that in turn inspired the ideas in Arakelov theory?
|
To get a nice overview of how and why Arakelov theory started you could read the introduction to R. de Jong's Ph.D. thesis on http://www.math.leidenuniv.nl/~rdejong/publications/ I remember that being very helpful to me. To avoid too many complex analytic difficulties you should stick to the case of arithmetic surfaces (i.e. integral regular flat projective 2-dimensional $\mathbf{Z}$ -schemes). The complex analysis involved is all "Riemann surfaces theory". An elementary and thorough treatment of this is given in P. Bruin's master's thesis http://www.math.leidenuniv.nl/~pbruin/ Arakelov theory provides an intersection pairing on an arithmetic surface $X$ . The idea is to add vertical divisors on $X$ above the "points at infinity" on Spec $\mathbf{Z}$ (or Spec $O_K$ ). There will be two contributions: finite and infinite. To get a good understanding of the finite contributions I recommend reading Chapter 8.3 and 9.1 of Q. Liu's book. I remember that after reading these texts the article by Faltings was much more readible to me. I also enjoyed the very nice asterisk by Szpiro on the subject, Séminaire sur les pinceaux de courbes de
genre au moins deux (all in French, last page has an English abstract). Here's some advice on what you shouldn't read when you just start. I wouldn't start immediately reading the papers by Gillet and Soulé (unless you really want too). The complex analysis is very involved. The paper by Bost "Potential theory and Lefschetz theorems for arithmetic surfaces" introduces the most general intersection theory (based upon $L^2_1$ Green functions) and should also be left for later reading in my opinion. To learn Arakelov theory the proofs don't really help me understand the statements for they are based upon moduli space arguments usually (e.g. the proof of the Noether formula). Therefore, I would also recommend you skip most of the proofs on a first reading. What did help is seeing how Arakelov theory gets applied. I recommend the recent book by Couveignes, Edixhoven, et al. available here http://arxiv.org/abs/math/0605244
|
{
"source": [
"https://mathoverflow.net/questions/54603",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5309/"
]
}
|
54,612 |
May I respectfully ask what the minimal background needed to read Wiles' proof of Fermat's Last Theorem is? I'm not an expert on number theory, but out of curiosity I wanted to understand - at a cursory level if possible - the outline of the proof. Thank you to all responders in advance. My background: Junior-year undergraduate in Theoretical Physics.
|
This is a very hard proof to do for an undergraduate but there are books available. Tthe book "Invitation to Fermat Wiles" ( http://www.amazon.com/Invitation-Mathematics-Fermat-Wiles-Yves-Hellegouarch/dp/0123392519 ) is an exposition on the proof written for undergraduates for example.
|
{
"source": [
"https://mathoverflow.net/questions/54612",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12723/"
]
}
|
54,677 |
In how many different ways can k bishops be placed on an nxn chessboard such that no two bishops attack each other? Please try to respond with a formula and explanation.
|
Call this number $B_k(n)$. For fixed $k$ it is known that $B_k(n)$ has the form
$P_k(n)+(-1)^nQ_k(n)$, where $P_k$ and $Q_k$ are polynomials. These polynomials have been computed for (at least) $k\leq 6$. We also have (unsurprisingly) the asymptotic formula
$B_k(n)\sim n^{2k}/k!$.
For further information see http://www.math.binghamton.edu/zaslav/Tpapers/bishops.slides.20100729.pdf . Update. I learned from Tom Zaslavsky that an explicit formula for $B_k(n)$ as a triple sum was given by C. E. Arshon in 1936.
|
{
"source": [
"https://mathoverflow.net/questions/54677",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12803/"
]
}
|
54,735 |
I teach elementary number theory and discrete mathematics to students who come with no abstract algebra. I have found proving the key theorem that finite multiplicative subgroups of fields are cyclic a pedagogical speedbump. For example, Serre's proof in A Course in Arithmetic runs a full page, requires introducing Euler's $\phi$ -function, and depends on a counting argument that might seem to beginners too clever or magical for a cornerstone result. I'd like to have a collection of proofs of this fact, to compare their advantages,
to match their viewpoints to my various audiences, to contrast for my students, etc. To get the ball rolling, here's the shortest argument I can think of (and if it's in the literature somewhere I'd love a reference). Induction on the order of the subgroup. So suppose multiplicative
subgroup $G$ of field $F$ has order $n$ . If $n=p^k$ with $p$ prime and $G$ isn't
cyclic, all $p^k$ elements of $G$ satisfy $x^{p^{k-1}}-1=0$ , impossible. If $n=ab$ , $\gcd(a,b)=1$ , then $(\cdot)^a:G\rightarrow G$ has a kernel $A$ of size at most $a$ and a range $B$ of size at most $b$ (since the $y\in B$ satisfy $y^b=1$ ), so $|A|=a$ , $|B|=b$ , and a product $xy$ of cyclic generators $x,y$ for $A,B$ respectively generates $G$ . If you know published proofs distinctly different from either of these, please cite a source. No need to spell out the details, but please mention a key feature to help avoid duplicates. If you have your own favorite approach, please share it.
|
Let $n = |G|$ and let $m$ be the l.c.m. of the orders of the cyclic factors of $G$. Then $x^m = 1$ for all $x \in G$; since we are in a field this equation has at most $m$ roots, which shows that $m \geq n$. It follows that $m = n$ and $G$ is cyclic. Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.
|
{
"source": [
"https://mathoverflow.net/questions/54735",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10909/"
]
}
|
54,759 |
There are several well-known dualization results in category theory, i.e. that such-and-such a well-known category D is isomorphic to the opposite C^{op}. Does anyone know of such a result concerning what the opposite category to Rng, rings (*-monoid on +-group) with their homomorphisms, looks like? I ask (naively I should add) because I'm curious if there's a natural "algebraic" structure on homology dual to that of the ring-structure of cohomology.
|
Here is a too-serious answer to your question, along with answers to a couple questions I think you should be asking: The category you're interested in, as noted by others, is the category of coalgebras / corings, which is emphatically not the opposite category of rings --- but we're going to see exactly what's different between the two in the nicest case. To start things off, here's a definition of a coalgebra so that we're all on the same page: an $R$-coalgebra is an $R$-module $A$ together with morphisms $\Delta: A \to A \otimes_R A$ and $\epsilon: A \to R$ satisfying Coassociativity: $(\Delta \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \Delta) \circ \Delta$, Counitality: $(\epsilon \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \epsilon) \circ \Delta = \operatorname{id}$ after identifying $R \otimes_R A$ and $A \otimes_R R$ with $A$. Coalgebras are familiar objects in algebraic topology, and you've already found the biggest source of them. Suppose you have a cohomology theory $E$ whose coefficient ring is a (graded) field. For any space $X$, the constant map $X \to \mathrm{pt}$ induces a map in homology $E_* X \to E_*$ which serves as the counit for $E_* X$. Toward a comultiplication, we have maps $E_* X \xrightarrow{\Delta} E_* (X \times X) \leftarrow E_* X \otimes_{E_*} E_* X$, but asking for the right-hand map to have an inverse is the same as asking for a Kunneth isomorphism. This only happens under restrictions on $X$ or restrictions on $E$ --- such as when $E_*$ is a field, which is one reason we requested that. Now, let's lay our cards on the table and just announce some dualities we see in front of us: There's your opposite category $\mathsf{Rings}^{op}$, which is dual to rings in the sense that $(\mathsf{Rings}^{op})^{op}$ is equal to $\mathsf{Rings}$. In the language of the comments below, coalgebras are Eckmann-Hilton duals of algebras. This is really a statement about how we produced the definition above: we took the definition for an algebra, and we flipped all the arrows around. There's also the notion of linear algebraic duals: $V^\vee = \operatorname{Hom}_k(V, k)$. To avoid some very serious technicalities, we'll want to work in the nicest, most familiar setting possible: modules of finite rank over a ground ring that's a field. Now, we would like to compare these three ideas. There is another category of interest floating around: the category of $k$-algebras has an associated category of presheaves $\widehat{\mathsf{Algebras}_k}$ $=$ $\operatorname{Functors}(\mathsf{Algebras}_k, \mathsf{Sets})$ which receives a map $\mathsf{Algebras}_k^{op} \to \widehat{\mathsf{Algebras}_k}$ described by the left side of the $\operatorname{Hom}$-functor: $X \mapsto \operatorname{Hom}(X, -)$. This assignment, called the Yoneda embedding, is a functor into a cocomplete category which is an equivalence onto its image and whose image is codense --- these are consequences of the Yoneda lemma. That the target of the Yoneda embedding is cocomplete makes it a much nicer category to play around in, and so it's worth considering what this embedding's use is. I claim there's a relation between the category of $k$-coalgebras and $\widehat{\mathsf{Algebras}_k}$. Again, to make linear algebra behave nicely, we need to encode finiteness restrictions into our setup, and to make that happen we'll turn to "$k$-formal schemes". The classical $\operatorname{Spec}$ construction in algebraic geometry also gives a contravariant functor off the category of $k$-algebras which is an equivalence onto its image. Rather than fussing with what a Zariski spectrum is, since we're just playing around with categories, I will instead take the Yoneda embedding to be my definition of $\operatorname{Spec}$ and the presheaf category to be something dimly, vaguely, sorta like the category of schemes. Representable presheaves (i.e., those in the image of $\operatorname{Spec}$) are called affine schemes. Plenty of constructions from algebraic geometry transfer almost without comment; for instance, defining $\mathbb{A}^1 = \operatorname{Spec} \mathbb{Z}[x]$, we recover the functor $\mathcal{O}(X) = \operatorname{Hom}_{\widehat{\mathsf{Algebras}_k}}(X, \mathbb{A}^1)$, which in the case of an affine $X = \operatorname{Spec}(A)$ gives $\mathcal{O}(\operatorname{Spec} A) \cong A$. A scheme $X$ will be called finite if it is $\operatorname{Spec}$ of an algebra of finite dimension as a $k$-module. These, too, are in ample supply in algebraic topology. If $X$ is a compact pointed space, then the algebra $H^* X$ will be finite in the sense we need. Of course, algebraic topology gets done on more than compact spaces, so we need to broaden our perspective a little bit: we can ask instead that $X$ be compactly generated, so that if $X_\alpha$ denotes the collection of compact subsets of $X$ directed by inclusion, we then have $X = \operatorname{colim} X_\alpha$. In the case that $X$ is a CW-complex, it is sufficient to take $X_\alpha$ to be just the finite subcomplexes of $X$. We might then be interested in the scheme $X_E := \operatorname{colim} \operatorname{Spec} E^* X_\alpha$. Such a scheme which occurs as the colimit of a directed system of finite $k$-schemes is called a $k$-formal scheme. So now we have a suitable notion of a 'finite' scheme that still captures all our interesting (and frequently large) cohomology rings. Here's the comparison I promised early on: the functor taking a formal scheme $\operatorname{colim} \operatorname{Spec} A_i$ to the $k$-coalgebra $\operatorname{colim} A_i^\vee$ is an equivalence of categories. There's almost nothing to say because the sea of definitions we've made take care of most everything, but there is one key lemma: given a $k$-coalgebra, you need to know that you can write it as the colimit of finite $k$-coalgebras. The exact lemma that gets used is: if $E$ is a finite dimensional subspace of a $k$-coalgebra $A$, then there exists a subcoalgebra $F \subseteq A$ which is finite as a $k$-vector space and which contains $E$ (i.e., $E$ can be finitely enlarged so that it becomes closed under comultiplication). If you push around elements a bit you'll see that that's the case (and that this requires working over a ground field). Then, straight after, here's the second big assertion: the assignments $X \mapsto E_* X$ and $X \mapsto X_E$ are equivalent under the above equivalence of categories. Now, this construction is delicate, and the limitations are not meant to be taken lightly! You need a ring structure on the underlying (co)homology theory to even dream of having products, you need Kunneth isomorphisms to make sense of the coalgebra structure, you need a coefficient field to have good linear-algebraic duality, and there are still potential problems with supercommutativity that we haven't addressed. But, when all the stars align and God smiles on us, this is what the coalgebra structure on homology is supposed to mean: it's another presentation of the formal scheme associated to the (perhaps more familiar) ring structure on cohomology. To counterbalance that caveat, that's not to say that this point of view is not immensely useful. Here are some sample applications: All of chromatic homotopy: $\mathbb{C}\mathrm{P}^\infty$ ($ = B\mathrm{U}(1) = \mathrm{Gr}_1$) carries the structure of an $H$-group, and there are a whole sea of cohomology theories $E$, called complex-orientable, for which $\mathbb{C}\mathrm{P}^\infty_E$ is (noncanonically) isomorphic to $\hat{\mathbb{A}}^1 = \operatorname{colim} \operatorname{Spec} E^*[x] / \langle x^n \rangle$, an object which behaves a lot like an infinitesimal one-dimensional Lie group. This "formal Lie group" carries an immense amount of information about the cohomology theory $E$, and the "space" of available formal Lie groups carries an immense amount of information about stable homotopy theory as a whole. There are a variety of partial theorems in the following spirit: if $E$ and $F$ are complex oriented cohomology theories and $F$ is represented by the spaces $F_k$ in the sense that $F^k X = [X, F_k]$, then $\bigoplus_k E_* F_k$ behaves like $\operatorname{Hom}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$. Goerss has shown this when $E = H\mathbb{F}_p$ and $F$ satisfies a certain condition on $F_* \Omega^2 S^3$ (which is also satisfied for a class of complex-oriented spectra called Landweber exact). (Addendum: Goerss spends a lot of time setting up a "super" version of Dieudonne modules, which is meant to address in part issues with supercommutativity ignored/avoided here.) Armed with an ample supply of Kunneth isomorphisms, Ravenel and Wilson, the progenitors of the idea above, computed these coalgebras in the cases where $E$ and $F$ ranged in $K(n)$ (Morava $K$-theory), $E(n)$ (Johnson-Wilson theories), $BP$ (Brown-Peterson theory), and $H\mathbb{Z}/p^j$ (singular theories / Eilenberg-Mac Lane spaces). For instance, one can define the free (supercommutative) algebra on a group object in the category of $k$-formal schemes, and it turns out that $H_*(K(\mathbb{Z}/p^j, q); \mathbb{F}_p)$ is the free alternating algebra in this sense on the formal group scheme $B\mathbb{Z}/p^j_{H\mathbb{F}_p}$. A similar statement can be made for $K(n)_* K(\mathbb{Z}/p^j, q)$. The above ideas have stable versions too: the homology of spectra $E_* F$ is to be thought of as the scheme of isomorphisms $\operatorname{Iso}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$, and so, for instance, the dual of the Steenrod algebra, an object of classical interest, can be thought of as $\mathcal{O}$ of the automorphisms of a particular formal Lie group $\hat{\mathbb{G}}_a$. Relatedly, the statement that the dual Steenrod algebra coacts on the homology coalgebras $H_* X$ is straightened out in the category of formal schemes by saying that $\operatorname{Aut} \hat{\mathbb{G}}_a$ acts on the formal scheme $X_{H\mathbb{F}_2}$. This has also received classical interest, though not in this language: the final part of Thom's thesis on calculations in the real bordism ring amount to showing that this $\operatorname{Aut} \hat{\mathbb{G}}_a$ action on the homology of the real bordism spectrum is free. Finally, this is an example of a broader phenomenon: often the schemes associated to rings have enlightening interpretations in moduli theoretic terms. A great many classical objects in stable homotopy theory lead double lives in this framework as moduli spaces, and the geometry of the moduli space frequently informs us on how the original topological objects behave. Anyway, this is all to say that you should definitely care deeply about the coalgebra structure on homology, as it's one way to get into formal schemes, where everything is interesting and magical and great.
|
{
"source": [
"https://mathoverflow.net/questions/54759",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12823/"
]
}
|
54,775 |
The question is self-explanatory, but I want to make some remarks in order to prevent the responses from going off into undesirable directions. It seems that every few years I hear someone ask this question; it seems to hold a perennial fascination for research mathematicians, just as quests for short proofs do. The trouble is that it has strong urban-legend tendencies: someone will say, "So-and-so's thesis was only $\epsilon$ pages long!" where $\epsilon \ll 1$ . It will often be very difficult to confirm or disconfirm such claims, since Ph.D. theses are often not even published, let alone readily available online. If you Google around for a while, as I did, you will find many dubious leads and can easily waste a lot of time on wild goose chases. Frankly, I'm a bit fed up with this state of affairs. I am therefore asking this question on MO in the hope that doing so will put this old question to rest, or at least establish provable upper bounds. I would therefore request that you set yourself a high standard before replying. Don't post a candidate unless you're sure your facts are correct, and please give some indication why you're so sure. Read the meta discussion before posting. (Note that the meta discussion illustrates that even a MathSciNet citation isn't always totally definitive.) Include information about the content and circumstances of the thesis if you know it, but resist the temptation to gossip or speculate. I'm not making this question community wiki or big-list because it should ideally have a definite answer, though I grant that it's possible that there are some borderline cases out there (perhaps there are theses that were not written in scholarly good faith, or documents that some people would regard as equivalent to a Ph.D. thesis but that others would not, or theses in subjects that are strictly speaking distinct from mathematics but that are arguably indistinguishable from mathematics dissertations). Finally, to anticipate a possible follow-up question, there is a list of short published papers here (search for "Nelson"). Note that the question of the shortest published paper is not as urban-legendy because the facts are easier to verify. I looked up the short papers listed there myself and found them to be quite interesting. So in addition to trying to settle an urban legend, I am hoping that this question will bring to light some interesting and lesser known mathematics.
|
David Rector's thesis ("An Unstable Adams Spectral Sequence", MIT 1966) is 9 pages, according to the record at the MIT library . I haven't seen the actual thesis for many years, but I'm pretty the actual mathematical content takes about 3 pages total, and is largely identical to the published version in Topology (1966, same title, doi link: https://doi.org/10.1016/0040-9383(66)90025-5 ), which is 3 pages plus bibliography. (Dan Kan, his advisor, likes short papers.)
|
{
"source": [
"https://mathoverflow.net/questions/54775",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3106/"
]
}
|
54,851 |
For a finite group G, let |G| denote the order of G and write $D(G) = \sum_{N \triangleleft G} |N|$, the sum of the orders of the normal subgroups. I would like to call G "perfect" if D(G) = 2|G|, since then the cyclic group of order n is perfect if and only if the number n is perfect. But the term "perfect group" is taken , so I'll call such a group immaculate . My question is: Does there exist an immaculate group of odd order? Since the cyclic immaculate groups correspond one-to-one with the perfect numbers, a "no" answer would immediately prove the famous conjecture that there are no odd perfect numbers. However, perhaps someone can easily see that there is a non -cyclic immaculate group of odd order, proving that the answer is "yes". Here's what I know. There are no abelian immaculate groups except for the cyclic ones. ( Edit : more generally, if $D(G) \leq 2|G|$ then every abelian quotient of $G$ is cyclic. Proof: not hard, and given here .) However, there do exist nonabelian immaculate groups, e.g. $S_3 \times C_5$ (of order 30). Derek Holt has computed all the immaculate groups of order less than or equal to 500. Their orders are
$$
6, 12, 28, 30, 56, 360, 364, 380, 496
$$
( Integer Sequence A086792 ). Of these, only 6, 28 and 496 are perfect numbers; the rest correspond to nonabelian immaculate groups. Some nonabelian immaculate groups of larger order are also known, e.g. $A_5 \times C_{15128}$, $A_6 \times C_{366776}$, and, for each even perfect number n, a certain group of order 2n. But these, too, all have even order. Edit : Steve D points out that p-groups can never be immaculate. This also appears as Example 2.3 here ; it follows immediately from Lagrange's Theorem. I should have mentioned this, as it rules out an easy route to a "yes" answer.
|
I did a little computer search and I think I found an example of an odd immaculate group. I searched for groups of the form $G=(C_q \rtimes C_p) \times C_N$ with odd primes $p,q$ such that $p | q-1$ and $N$ an odd integer satisfying $(N,pq)=1$. Using Tom's notations and results, we have \begin{equation*}
\frac{D(G)}{|G|} = \frac{D(C_q \rtimes C_p)}{|C_q \rtimes C_p|} \cdot \frac{D(C_N)}{|C_N|} = \frac{1+q+pq}{pq} \cdot \frac{\sigma(N)}{N}
\end{equation*}
where $\sigma(N)$ denotes the sum of divisors of $N$. We want $\frac{\sigma(N)}{N} = \frac{2pq}{1+q+pq}$. Since the last fraction is irreducible, $N$ has to be of the form $N=(1+q+pq)m$ with $m$ odd. I found the following solution : \begin{equation*}
p=7, \quad q=127, \quad m=393129.
\end{equation*}
This gives the immaculate group $G=(C_{127} \rtimes C_7) \times C_{399812193}$, which has order $|G| = 355433039577 = 3^4 \cdot 7 \cdot 11^2 \cdot 19^2 \cdot 113 \cdot 127$. Edit : here is the beginning of an explanation of "why" $m$ is square in this example ($393129=627^2$). Recall that an integer $n \geq 1$ is a square if and only if its number of divisors is odd (consider the involution $d \mapsto \frac{n}{d}$ on the set of divisors of $n$). If $n$ is odd, then all its divisors are odd, so that $n$ is a square if and only if $\sigma(n)$ is odd. Now consider $N=(1+q+pq)m$ as above. The condition on $\sigma(N)/N$ implies that $\sigma(N)$ is even but not divisible by $4$. If we assume that $1+q+pq$ and $m$ are coprime, then $\sigma(N)=\sigma(1+q+pq) \sigma(m)$, so the reasoning above shows that $1+q+pq$ or $m$ is a square (but not both). If $1+q+pq=\alpha^2$ then $\alpha \equiv \pm 1 \pmod{q}$ so that $\alpha \geq q-1$, which leads to a contradiction. Thus $m$ is a square (it is possible to show further that $1+q+pq$ is a prime times a square). If $1+q+pq$ and $m$ are not coprime, the situation is more intricate (this is what happens in the example I found : we had $\operatorname{gcd}(1+q+pq,m)=9$). Let $m'$ be the largest divisor of $m$ which is relatively prime to $1+q+pq$. Put $m=\lambda m'$. Then $\lambda(1+q+pq)$ or $m$ is a square. I don't see an argument for excluding the first possibility, but at least if $\lambda$ is a square then so is $m$.
|
{
"source": [
"https://mathoverflow.net/questions/54851",
"https://mathoverflow.net",
"https://mathoverflow.net/users/586/"
]
}
|
54,853 |
What advice do you have for giving a talk on a mathematical research paper to people in other fields in science (not physics nor astronomy) but without lot of math background? Thanks.
|
I did a little computer search and I think I found an example of an odd immaculate group. I searched for groups of the form $G=(C_q \rtimes C_p) \times C_N$ with odd primes $p,q$ such that $p | q-1$ and $N$ an odd integer satisfying $(N,pq)=1$. Using Tom's notations and results, we have \begin{equation*}
\frac{D(G)}{|G|} = \frac{D(C_q \rtimes C_p)}{|C_q \rtimes C_p|} \cdot \frac{D(C_N)}{|C_N|} = \frac{1+q+pq}{pq} \cdot \frac{\sigma(N)}{N}
\end{equation*}
where $\sigma(N)$ denotes the sum of divisors of $N$. We want $\frac{\sigma(N)}{N} = \frac{2pq}{1+q+pq}$. Since the last fraction is irreducible, $N$ has to be of the form $N=(1+q+pq)m$ with $m$ odd. I found the following solution : \begin{equation*}
p=7, \quad q=127, \quad m=393129.
\end{equation*}
This gives the immaculate group $G=(C_{127} \rtimes C_7) \times C_{399812193}$, which has order $|G| = 355433039577 = 3^4 \cdot 7 \cdot 11^2 \cdot 19^2 \cdot 113 \cdot 127$. Edit : here is the beginning of an explanation of "why" $m$ is square in this example ($393129=627^2$). Recall that an integer $n \geq 1$ is a square if and only if its number of divisors is odd (consider the involution $d \mapsto \frac{n}{d}$ on the set of divisors of $n$). If $n$ is odd, then all its divisors are odd, so that $n$ is a square if and only if $\sigma(n)$ is odd. Now consider $N=(1+q+pq)m$ as above. The condition on $\sigma(N)/N$ implies that $\sigma(N)$ is even but not divisible by $4$. If we assume that $1+q+pq$ and $m$ are coprime, then $\sigma(N)=\sigma(1+q+pq) \sigma(m)$, so the reasoning above shows that $1+q+pq$ or $m$ is a square (but not both). If $1+q+pq=\alpha^2$ then $\alpha \equiv \pm 1 \pmod{q}$ so that $\alpha \geq q-1$, which leads to a contradiction. Thus $m$ is a square (it is possible to show further that $1+q+pq$ is a prime times a square). If $1+q+pq$ and $m$ are not coprime, the situation is more intricate (this is what happens in the example I found : we had $\operatorname{gcd}(1+q+pq,m)=9$). Let $m'$ be the largest divisor of $m$ which is relatively prime to $1+q+pq$. Put $m=\lambda m'$. Then $\lambda(1+q+pq)$ or $m$ is a square. I don't see an argument for excluding the first possibility, but at least if $\lambda$ is a square then so is $m$.
|
{
"source": [
"https://mathoverflow.net/questions/54853",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12435/"
]
}
|
54,876 |
What is the geometric meaning of Cohen-Macaulay schemes? Of course they are important in duality theory for coherent sheaves, behave in many ways like regular schemes, and are closed under various nice operations. But whereas complete intersections have an obvious geometric meaning, I don't know if this is true for CM schemes. Perhaps we can make somethinkg out of the following theorem: A noetherian ring $R$ is CM iff every ideal $I$ which can be generated by $ht(I)$ many elements is unmixed, i.e. has no embedded associated primes. Also, Eisenbud suggests that Cor. 18.17 in his book "Commutative algebra with a view toward algebraic geometry" reveals some kind of geometric meaning, but perhaps someone can explain this in detail? Every integral curve is CM. Now assume that you are given a surface, given by some homogeneous equations, how do you "see" if it is CM or not? Hailong's answer contains the link to How to think about CM rings? , which is pretty the same question as mine and has already some very good answers. So I appologize for this duplicate. But the answers here reveal some more insights, thanks!
|
[ EDIT: I rewrote the first couple of paragraphs, because I realized a better way to say what I had in mind.] There are many ways to define dimension and some of them give the same answer some of them don't. Depth is a sort of dimension. Perhaps not the most obvious, but one that works well in many situation. In general we count dimension by chains and the main difference between Krull dimension and depth is about the same as the difference between Weil divisors and Cartier divisors. For simplicity assume that we are talking about finite dimensional spaces. Infinite dimension can be dealt with by saying that it contains arbitrary dimensional finite dimensional spaces where we may substitute "Krull dimension" or "depth" in place of "dimension". I usually think of Krull dimension as going from small to large: We start with a (closed) point, embed it into a curve, then to a surface until we get to the maximal dimension. However, for comparing to depth it is probably better to think of it as going from large to small: Take a(n irreducible) Weil divisor, then a(n irreducible) Weil divisor in that and so on until you get to a point. In contrast, when we deal with depth we take Cartier divisors: We start with the space itself (or an irreducible component), then take a(n irreducible) hypersurface, then the intersection of two hypersurfaces (such that it is a "true" hypersurface in each irreducible component this condition corresponds to the "non-zero divisor" provision)=a codimension $2$ complete intersection, and so on until we reach a zero dimensional set. So, I would say that the geometric meaning of Cohen-Macaulay is that it is a space where our intuition about these two notions giving the same number is correct. I would also point out that this does not mean that necessarily all Weil divisors are Cartier, just that one cannot get a longer sequence of subsequent Weil divisors than Cartier divisors. Another, less philosophical explanation is the following:
Cohen-Macaulay means that depth = dimension . $S_n$ means that this is true up to codimension $n$. Then one may try to give geometric meaning to the $S_n$ property and say that CM means that all of those properties hold.
So, $S_1$ --- means the existence of non-zero divisors, i.e., that there exists hypersurfaces that are like the ones we imagine. $S_2$ --- is perhaps the most interesting one, or the one that is the easiest to explain. See this answer to another question where it is explained how it corresponds to the Hartogs property , that is, to the condition that functions defined outside a codimension $2$ set can be extended to the entire space. $S_3$ --- I don't have a similarly nice description of this, but I am sure something could be made up, or some people might even know something nice. One thing is sure. This means that every ("true") hypersurface has the $S_2$ property, which has a geometric meaning as above. [...] So, one could say that $S_n$ means that every ("true") hypersurface has the $S_{n-1}$ property, which we already described. I realize that this description of $S_n$ may not seem satisfactory, but in practice, this is very useful. I would also add that in moduli theory it is actually important to know that some properties are inherited by hypersurfaces (=fibers of morphisms), so saying that hypersurfaces are $S_2$ is actually a good property. More specifically, for example, the total space of a family of stable (resp. normal, $S_2$) varieties is $S_3$ ("is" as in "has to be"). Then one might (as in Shafarevich's conjecture, see Parshin's Theorem, Arakelov's Theorem, Manin's Theorem, Faltings' Theorem) study deformations of these families (say the embedded deformations of a curve in the moduli stack of the corresponding moduli problem). Then the total space of these deformations ought to be $S_4$ on account of their fibers having to be $S_3$ since their fibers have to be $S_2$. This actually explains why it is not entirely bogus to say that $S_4$ means that codimension $2$ complete intersections satisfy the Hartogs property. This actually reminds me another thing that is important about CM. A lot of properties are inherited by general hypersurfaces . The CM property is inherited by all of them. This makes them perfect for inductive proofs. One way to see that a surface is CM is that normal $\Rightarrow$ CM . Of course, the point is that normal is equivalent to $R_1$ and $S_2$, so it always implies $S_2$ and if the dimension is at most $2$, then $S_2$ is the same as CM. If you have a non-normal surface, but it is non-normal only because it has normal crossings in codimension one, then it is CM. You may also try to test directly for the Hartogs property mentioned above: A reduced surface $S$ is Cohen-Macaulay if and only if for any $P\in S$ and any regular function $f$ defined on $U\setminus \{P\}$ for an open set $P\in U\subseteq S$ there exists a regular function $g$ on $U$ such that $g_{|U\setminus \{P\}}=f$.
|
{
"source": [
"https://mathoverflow.net/questions/54876",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2841/"
]
}
|
54,907 |
This question appeared in my answer to this question, but it seems to be interesting in itself. Let $G$ be an infinite finitely generated group, $\epsilon\gt 0$. Is there a finite subset $S\subset G$ such that every subset of $S$ with at least $\epsilon|S|$ elements generates $G$? If the answer is "yes", it should have a trivial proof by Gromov's thesis (every property of all finitely generated groups is either false or trivial). Update. In view of Stephen's answer and Kevin's comment below, perhaps a more correct question is this: Is it true that if we represent an infinite group $G$ as a union of a finite number of subsets, then one of these subsets generates a finite index subgroup of $G$? Compare with Adreas Thom's question .
|
This is false for the infinite dihedral group $\langle a,b\mid b^2=1, ba=a^{-1}b\rangle$. No set $S$ works for $\epsilon\le1/3$, because there is always a subset with $\lceil{\epsilon|S|}\rceil$ elements that lies entirely in $\{a^n\mid n\in\mathbb{Z}\}$, $\{a^{2n}b\mid n\in\mathbb{Z}\}$ or $\{a^{2n+1}b\mid n\in\mathbb{Z}\}$, and none of these three sets generates the whole group.
|
{
"source": [
"https://mathoverflow.net/questions/54907",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
54,923 |
I'm looking for a measure-theoretic analogue to the disjoint union topology, or for work on the $\sigma$-algebra generated by canonical injections. More formally: For an indexed family of sets $\{A_i\}_{i \in I}$, define $\psi_i : A_i \to A$, $a \mapsto (a,i)$ (the canonical injections), where $A = \bigcup_{i \in I} (A_i \times \{i\})$. Then the disjoint union $\sigma$-algebra $\mathcal{A}$ is the finest $\sigma$-algebra on $A$ such that for every $i \in I$, $\psi_i$ is $\mathcal{A}_i$-$\mathcal{A}$-measurable. There are so many analogies between measure theory and topology that I've been surprised at how difficult it's been for me to find anything on this. I'd appreciate references to any related ideas as well.
|
This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward. From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces. But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $S \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.) One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets. I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?
|
{
"source": [
"https://mathoverflow.net/questions/54923",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10828/"
]
}
|
55,042 |
I have a probably stupid question on schemes ...
Let $S$ be a scheme, and let $A = \mathsf{Aut}(S)$ be its automorphism group. Does $A$ carry
a scheme structure itself, that is, can one see $A$ as a group scheme ?
Thanks !
|
The answer is yes when the scheme is flat and projective over the base. This follows from the existence of the Hom scheme, which in turn is proven via the existence of the Hilbert scheme. A readable reference is Nitin Nitsure's part of the book Fundamental algebraic geometry . In particular Theorem 5.23 in his notes states that when S is noetherian, X projective and flat over S , and Y quasi-projective over S , then there exists a Hom-scheme parametrizing morphisms from X to Y over S . (The precise definition of the functor that this scheme represents is given in the text). The automorphism group scheme of X is then an open subscheme of $\mathrm{Hom}_S(X,X)$. Addendum : The proof that Isom is open in Hom is very similar to the proof that Hom is open in Hilb. The map from the Hom scheme to $\mathrm{Hilb}_{X \times_S Y / S}$ is given by associating to a morphism $f : X \to Y$ its graph. Now the image consists of those $Z \subset X \times_S Y$ such that projection onto X induces an isomorphism $Z \cong X$, and the crucial part of the proof is showing that this is an open condition. (In Nitsure's notes this is Theorem 5.22.(b).) But then the condition that Z maps isomorphically onto Y is also open, and this is exactly the condition that defines the Isom scheme inside the Hom scheme.
|
{
"source": [
"https://mathoverflow.net/questions/55042",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12884/"
]
}
|
55,085 |
I am wondering, historically, when has a new proof of an old theorem been particularly fruitful. A few examples I have in mind (all number theoretic) are: First example is classical... which is Euler's proof of Euclid's theorem which asserts that there exist infinitely many primes. Here is when the factorization $\displaystyle \prod_p (1-p^{-s})^{-1} = \sum_{n=1}^\infty \frac{1}{n^s}$ was first introduced, leading of course to what is now known as the Riemann Hypothesis. Second example is when Hardy and Littlewood gave an alternative proof of Waring's problem, which was done by Hilbert earlier. Their proof introduced what is now known as the Hardy-Littlewood Circle Method and gave an exact asymptotic for the Waring bases, which is stronger than Hilbert's result which only asserted that every sufficiently large positive integer can be written as the sum of a bounded number of $k$th powers. Later on the Hardy-Littlewood method proved very fruitful in other results, namely Vinogradov's Theorem asserting that every sufficiently large odd positive integer can be written as the sum of three primes. Third example is Tim Gowers' alternate proof to Szemerédi's Theorem asserting that every subset of the positive integers with positive upper density contains arbitrarily long arithmetic progressions. This advance, namely the introduction of Gowers uniformity norms, led eventually to the Green-Tao Theorem proving the existence of arbitrarily long arithmetic progressions in the primes. So I am wondering if there exist other incidences (number theory related or not) where a new proof really gave legitimate new insights, perhaps even a proof of a (major) new result. Edit: I am primarily interested in examples where a new proof sparked off a new direction in research. This is best supported by having a major new theorem proved using techniques inspired by the new proof. An example of something that I am not interested in is something like Donald Newman's proof of the prime number theorem, which while elegant and 'natural' as he puts it, has seen limited generalization to other areas and one is hard pressed to apply the same technique to other problems.
|
Here are a few examples from the 19th century. Unsolvability of the quintic equation . Abel (1826) proved this by algebraic
ingenuity, but without clarifying the concepts involved. Galois (1830) gave a
proof that introduced the concepts of group, normal subgroup, and solvability
(of groups), thus laying the foundations of group theory and Galois theory. Double periodicity of elliptic functions . Abel and Jacobi established this
(1820s) mainly by computation. Riemann (1850s) put elliptic functions on
a clear conceptual basis by showing that the underlying elliptic curve is a
torus, and that the periods correspond to independent loops on the torus. Riemann-Roch theorem . Riemann (1857) discovered this theorem using
Riemann surfaces, but applying physical intuition (the "Dirichlet principle").
This principle was not made rigorous until 1901. In the meantime, Dedekind
and Weber (1882) gave the first rigorous and complete proof of Riemann-Roch,
by reconstructing the theory of Riemann surfaces algebraically. In the process
they paved the way for modern algebraic geometry.
|
{
"source": [
"https://mathoverflow.net/questions/55085",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10898/"
]
}
|
55,214 |
The first place where the amenability problem for Thompson's group $F$ appears in the literature is, I believe, 1980 in a problems article by Ross Geoghegan. I have heard, however, vague comments to the effect that the problem was considered by other people before this. Does anyone have any knowledge about the existence of this problem prior to 1980? Edit: Following Mark's advice offline, I wrote Richard Thompson to verify the details of Mark's answer. He did confirm that he considered the problem. He first observed that his group $F$ did not contain $\mathbb{F}_2$. He then discovered the material on amenability in Hewitt and Ross's text on abstract harmonic analysis. He then observed that $F$ was not elementarily amenable. This occurred sometime prior to his 1973 visit to University of Illinois at Urbana-Champaign to visit Day. He did not, however, attend Greenleaf's series of lectures in 1967. Instead he read Greenleaf's 1969 text "Invariant Means on Topological Groups and Their Applications." It seems that the best date to attach to Thompson's consideration of the question is 1973. Of course Thompson never published his observations and they were not widely circulated. His observations mentioned above were rediscovered by others in the 1980s (the question itself by Geoghegan). I have invited Richard to post an answer, in which case I will delete this edit.
|
Richard Thompson visited me at Princeton several times I believe in the mid '70's, gave me copies of some handwritten notes about his groups (which I shared with a few people), and raised the amenability question, which we discussed a bit. I don't think Thompson's groups were very widely known at the time.
|
{
"source": [
"https://mathoverflow.net/questions/55214",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10774/"
]
}
|
55,244 |
On the Wikipedia page 1 about algebraic varieties https://en.wikipedia.org/wiki/Algebraic_variety , a sentence reads as follows: [[A more significant modification is to allow nilpotents in the sheaf of rings. A nilpotent in a field must be 0: these if allowed in coordinate rings aren't seen as coordinate functions. From the categorical point of view, nilpotents must be allowed, in order to have finite limits of varieties (to get fiber products).]] So I am wondering if there is an intuitive example to get non-reduced 'schemes' from reduced 'algebraic varieties' (probably by taking fiber product or alike)? 1 Link to a revision from February 2011 .
|
I'm a bit confused by the quoted wikipedia entry, because the category of reduced rings also has coproducts (take the tensor product, and then pass to the quotient by the nilradical), and hence the category of reduced schemes, the category of varieties over a field, and so on, all admit fibre products. [Added: See Jim's Borgers series of comments below for a discussion of why, nevertheless, there may be a purely categorical description of the sense in which constructions in the category of reduced rings can be "wrong", while constructions in the category of all rings are the right ones.] So the answer to the question of why we need nilpotents is not that it is necessary for the existence of fibre products. Grothendieck introduced nilpotents for many reasons, a number of which are discussed in the other answers: to get correct counting in degenerate situations, it is typically necessary to allow nilpotents; they are also the bedrock of deformation theory and other applications of analytic ideas in algebraic geometry. It might be helpful to recall another motivation, which forms a significant part of Grothendieck's overall strategy for studying algebraic geometry: Suppose that we want to prove a property about a morphism $f: X \to S$. A typical approach is to first show that it
is a local property, in some sense, so that we can reduce to the case when Spec $\mathcal O_{S,s}$
for some point $s \in S$, and hence assume that $S$ is local; and then to use a flat descent argument to pass from $\mathcal O_{S,s}$ to its completion, and thus assume that $S$ is the Spec of a complete local ring. We then write this complete local ring as the projective limit of the quotients by its maximal ideals, and so reduce to the case when $S$ is the Spec of an Artinian local ring. Since such a Spec has a single point, we can then hope to reduce to checking our property on the fibre over this one point, which reduces us to the case when $S$ is the Spec of a field. This is a powerful method, which absolutely requires us to be able to do geometry over an Artinian ring (and hence requires us to allow nilpotents). It comes up in lots of places, e.g. in establishing basic properties of abelian schemes, by reducing to the abelian variety case. See the anwers to this question for some examples.
|
{
"source": [
"https://mathoverflow.net/questions/55244",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1992/"
]
}
|
55,288 |
In the days before [W, TW, BCDT], how did people show that specific elliptic curves over $\mathbb{Q}$ were modular? For instance, I was reading through a paper of Buhler, Gross and Zagier from 1985 on the curve 5077a, and they say that modularity can be checked by a finite computation in the 422-dimensional space of cuspforms of weight 2 and level 5077 (and remark at the end that Serre and Mestre have checked it). A google search brought up the name "Faltings-Serre method": was this the technique of choice? Also, are there any good references for it?
|
They explicitly computed quotients of $X_0(N)$ and identified them with elliptic curves.
Suppose you can compute the space $S_2(\Gamma_0(N),\mathbf{C})$ of modular forms. An (isogeny class of) elliptic curves of conductor $N$ corresponds (by modularity) to a normalized new Hecke eigenform with coefficients in $\mathbf{Z}$. Given such an $f$, one can compute the periods of $f$, which allows one to write down a Weierstrass equation for $f$. If one can do this in such a way to guarantee that the coefficients of the Weierstrass equation are integers , this allows one to computationally determine exactly the modular elliptic curves of conductor $N$. All this is very well explained in Cremona's book, which is available free online: http://www.warwick.ac.uk/~masgaj/book/amec.html The particular method referred to in the paper of [BGZ] was presumably the "method of graphs", see for example here: http://modular.math.washington.edu/msri06/refs/mestre-method-of-graphs/mestre-en.pdf The Faltings-Serre method is slightly different; it allows you to determine, given two Galois representations $\rho_1$ and $\rho_2$ from
$G_{K}$ to (say) $\mathrm{GL}_n(\mathbf{Z}_p)$ such that: $\overline{\rho}_1 \simeq \overline{\rho}_2$, $\rho_1$ and $\rho_2$ are both unramified outside some finite (given) set of places $S$ of $K$. whether $\rho_1 \simeq \rho_2$. You could use it to prove that an elliptic curve $E/\mathbf{Q}$ is modular by comparing the Galois representation attached to the $p$-adic Tate module of $E$ and the Galois representation attached to the conjecturally corresponding eigenform of level $N$ (EDIT: this works because the Galois representation determines the isogeny class of $E$ by Faltings' proof of the Tate conjecture for abelian varieties). However, this wouldn't be so efficient. However, there are other situations (for example, elliptic curves over other number fields) where the corresponding automophic form is not enough to compute the periods. In this case, the Faltings-Serre method can be used to prove modularity. Taylor uses this to prove that a certain elliptic curve over $\mathbf{Q}(\sqrt{-3})$ is modular.
|
{
"source": [
"https://mathoverflow.net/questions/55288",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2698/"
]
}
|
55,397 |
Is the number $\sum_{n=1}^\infty \frac{1}{2^{n^2}}$ known to be transcendental? Is there a survey with up-to-date transcendence results?
|
I have checked with Introduction to Algebraic Independence Theory , where it is mentioned in the preface (p. V) that D. Bertrand and independently D. Duverney, Ke. Nishioka, Ku. Nishioka, I. Shiokawa (DNNS) deduced results on algebraic independence of the values of theta-functions at algebraic points and in particular derived the transcendence of the sums $\sum_{n=1}^\infty q^{n^2}$ for any algebraic $q$ satisfying $0 < |q| < 1$ . The precise references are not given but a little googling turned up the paper by D. Bertrand, Theta Functions and Transcendence , The Ramanujan Journal , Vol. 1 (1997), pp. 339-350, which seems to be relevant. The second reference is DNNS, Transcendence of Jacobi's theta series , Proc. Japan Acad. Ser. A Math. Sci. , Vol. 72 (1996), pp. 202-203.
|
{
"source": [
"https://mathoverflow.net/questions/55397",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2631/"
]
}
|
55,458 |
I am very interested in proofs. I have taken an undergraduate course
called "Logic and Set Theory" which I found very interesting, but ultimately
unsatisfying. My biggest disappointment has to do with the language in which
proofs are expressed. It seems to me that we have all of the symbols necessary
to express a proof in "pure math". By which I mean, only using symbols and a
few specialized words (iff, let, ...). And yet most proofs that I have seen are
just walls of English text, interpolated with mathematical symbols. When I read a complex proof, I find myself needing to transcribe it into pure
symbols before I have any chance at understanding it. I have talked to a
professor about this, and he informed me that my "pure math" proofs were
actually considered informal, and not proper proofs at all! He seemed skeptical
that anyone would actually prefer symbols to English. I have searched Wikipedia and Google for more information, and I see that there
is something called a "Formal Proof" (although I have heard this term used in
other situations, and so I am not quite sure it means what I think it means)
which uses a computer to verify a proof written in a special programing
language. As fascinating as that is, it seems to be a step further than what I
am looking for. Is there a well known method for writing and sharing proofs of mathematical
statements that uses only mathematical symbols and is not a full blown
programming language? And if not, why is this considered "taboo" or "informal"? Thanks, --jc EDIT:
I guess this turned out to not be a real question? Strange, I checked, it definitely ends in a question mark. Thanks everyone for the help, advice, and links. I appreciate your input.
|
The question becomes interesting when it is interpreted as a technical question about the extent to which we can have a semi-formal language somehow in-between the truly formal proofs, which are largely unreadable by humans, and the informal proofs used by professional mathematicians. In fact, there has been some truly interesting work on this topic. In particular, the Naproche proof system implements this semi-formal language idea. See also this article describing the system and try out the web interface examples ). The idea of Naproche (for Natural language Proof Checking) is to focus precisely on the layer of proof detail that exists between the fully formal proofs that can be checked by computer and the fully natural language proofs used by humans. When using Naproche, one creates proofs in a controlled natural language, a semi-formal natural-seeming language, while under the hood the system converts the semi-formal proof to an unseen fully formal proof, which is proof-checked by one of the standard formal proof-checkers. The effect is that by using the semi-formal language, one guides Naproche to a formal proof which can then be verified. Thus, one gains the value of the verified formal proof, without needing ever to explicitly consider the formal proof object. Furthermore, because the syntax of the controlled natural language uses TeX formalisms, the semi-formal proofs and theorem can be automatically typeset in an appealing way. I encourage everyone to go try out the web interface examples , which includes Naproche semi-formal (but fully verified) proofs of
elementary results in group theory, set theory, and a chunk of Landau's text. Here is an example of Naproche text, and you may also consult the pdf output here . This text entered verbatim results in the formal proof object , which is verified as correct.
(The pdf and proof object are temporary files, but can be generated by clicking on "create pdf" or "Logical check" at the web interface.) Axiom.
There is no $y$ such that $y \in \emptyset$.
Axiom.
For all $x$ it is not the case that $x \in x$.
Define $x$ to be transitive if and only if
for all $u$, $v$, if $u \in v$ and $v \in x$
then $u\in x$. Define $x$ to be an ordinal
if and only if $x$ is transitive and for all
$y$, if $y \in x$ then $y$ is transitive.
Theorem.
$\emptyset$ is an ordinal.
Proof.
Consider $u \in v$ and $v \in \emptyset$.
Then there is an $x$ such that $x \in \emptyset$.
Contradiction. Thus $\emptyset$ is transitive.
Consider $y \in \emptyset$. Then there is an
$x$ such that $x \in \emptyset$. Contradiction.
Thus for all $y$, if $y \in \emptyset$ then $y$
is transitive. Hence $\emptyset$ is an ordinal.
Qed.
Theorem.
For all $x$, $y$, if $x \in y$ and $y$ is an
ordinal then $x$ is an ordinal.
Proof.
Suppose $x \in y$ and $y$ is an ordinal. Then
for all $v$, if $v \in y$ then $v$ is transitive.
Hence $x$ is transitive. Assume that $u \in x$.
Then $u \in y$, i.e. $u$ is transitive. Thus $x$
is an ordinal.
Qed.
Theorem: There is no $x$ such that for all $u$,
$u \in x$ iff $u$ is an ordinal.
Proof.
Assume for a contradiction that there is an $x$
such that for all $u$, $u \in x$ iff $u$ is an ordinal.
Lemma: $x$ is an ordinal.
Proof:
Let $u \in v$ and $v \in x$. Then $v$ is an ordinal,
i.e. $u$ is an ordinal, i.e. $u \in x$. Thus $x$ is
transitive. Let $v \in x$. Then $v$ is an ordinal,
i.e. $v$ is transitive. Thus $x$ is an ordinal. Qed.
Then $x \in x$. Contradiction. Qed.
|
{
"source": [
"https://mathoverflow.net/questions/55458",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2377/"
]
}
|
55,526 |
I know the definition of $K_X$ on a normal, singular variety, but I don't have a good set of examples in my mind. What's an example of a variety where $K_X$ is $\mathbb Q$-Cartier but not Cartier? Are there any conditions under which an adjunction formula lets me compute the canonical class of a singular divisor?
|
Note: I added an addendum below in response to quinque 's comment and the subsequent discussion of the issue (s)he raised on math.stackexchange (see the link in quinque 's comment). An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not. I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not :
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that the square of the exceptional divisor of the blow up of $\mathbb A^6$ is $-1$-times the hyperplane in $\mathbb P^5$ which restricts to a conic on $Y$. Now write $$K_Y\sim_{\mathbb Q} \pi^*K_X + aE \tag{$\star$}$$ and use the adjunction formula ($Y$ is smooth!) to get
$$
(a+1)E^2\sim K_E=K_{\mathbb P^2} \sim -3L.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier. Addendum ( a.k.a. Intermezzo ): quinque raised the interesting point that $\dfrac 12 E$ is actually $\mathbb Q$-linearly equivalent to a Cartier divisor, so (s)he was worried that then the above does not prove that $K_X$ is not Cartier. It actually does, but this points to an interesting consequence, namely that this means that $\pi^*K_X$ is actually numerically equivalent to a Cartier divisor, so using only intersection numbers one will not be able to prove that it is not Cartier. Anyway, here is why the above implies that $K_X$ is not Cartier: Suppose it is. That means that $\omega_X$ is the trivial line bundle in a neighborhood of the singular point. Consider the pull-back of a local generator. On $Y\setminus E$ this will generate $\omega_Y$ and hence the corresponding (integral!) divisor $\pi^*K_X$ is equal (not just linearly equivalent!!) to $K_Y+bE$ for some $b\in \mathbb Z$. Then the same calculation as above implies that $b=-a=-\dfrac 12$ which is a contradiction. The point is that if $K_X$ is Cartier, then ($\star$) holds with linear equivalence, and in fact with equality, instead of $\mathbb Q$-linear equivalence and hence $a$ has to be an integer in that case. Also interesting to note that the same construction does not give a desired example in dimension $2$: The quotient $\mathbb A^2/(x,y)\sim(-x,-y)$ is a cone over a conic which is a surface in $\mathbb P^3$. In particular it is Gorenstein and hence $K_X$ is Cartier. On the other hand, one gets a $2$-dimensional example by $\mathbb A^2/\mu_3$ where $\mu_3$ acts by multiplication by a primitive third root of unity. The proof is essentially the same as above. It is relatively easy to see (just as above) that this is the same as the cone over a twisted cubic. As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grothendieck Duality then.
|
{
"source": [
"https://mathoverflow.net/questions/55526",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12992/"
]
}
|
55,718 |
While preparing a course in complex analysis, I stumbled over a remark in Dudziak's book on removable sets, namely that any totally disconnected $K \subset\subset {\mathbb C}$ must have a connected complement; a remark, "that verifying the reader may find one of those exercises in 'mere' point-set topology that is a wee bit frustrating". Out of curiosity I spent an evening with this question. The assertion turned out to be a simple consequence of Theorem 14.2 ("If $x$ and $y$ are separated by the closed set $F$ in the open or closed plane they are separated by a component of $F$.") in an old book "Elements of the Topology of Plane Sets of Points" by M.H.A. Newman, Cambridge 1951. There, the proof is based on an lemma by Alexander (used in his proof of the Jordan-Brouwer separation theorem; Trans. AMS 23, 333-349, 1922), stating that, for disjoint closed sets $F_1$ and $F_2$ in the plane, two points which are connected in the complement of $F_1$ and in the complement of $F_2$ are connected in the complement of $F_1 \cup F_2$. This lemma fails for more general surfaces (Newman gives a counterexample for the torus) and is proved by homological methods. So, here are my questions: (a) is there a more modern reference to these kind of results (Newman's book uses quite an ideosyncratic terminology and notation); (b) is there a simpler proof not using Alexander's lemma (or similarily deep results); (c) how about the connectedness of the complement of a totally disconnected closed set in surfaces (or topological spaces) more general than the plane?
|
A proof of the statement using knowledge that has withstood the test of time is to
cite the Alexander duality theorem , for which you can find multiple modern sources. The relevant form of Alexander duality states that the if $X$ is a compact subset of $S^n$ , then the reduced Čech cohomology of $X$ is isomorphic to the reduced homology of its complement in the complementary dimension less 1, dimension $q \leftrightarrow n-q-1$ . So the reduced $H_0$ of $\mathbb R^2 \setminus X $ , which measures
disconnectedness and is equal to that of $S^2 \setminus X$ , is isomorphic to the
1-dimensional Čech cohomology of $X$ ; all but 0-dimensional Čech cohomology is trivial
for any totally disconnected space, since there are fine covers with 0-dimensional
nerve. The Čech cohomology of $X$ is the direct limit of the cohomology of the
nerve of coverings; the nerve is the simplicial complex whose simplices assert that the elements of the cover indexed by its vertices have a non-empty intersection. Using
extension theorems, there is a continuous map from a topological space to the nerve of an open cover. What this translates to for any $X$ such that $\mathbb{R}^2 \setminus X$ is disconnected,
you can use that fact to define a map from the nerve of any cover of $X$ to $S^1$ which is not null-homotopic, and even when you refine the cover and pull the map back to
the refinement, it's still not nullhomotopic. That is inconsistent with $X$ being
totally disconnected. Intuitively, the Alexander theorem detects linking between $X$ and its complement.
Every $q$ -cycle $Z^q \subset S^n \setminus X$ , when $q \ne 0, n$ is the boundary of a $q+1$ -chain $C^{q+1}$ in $S^n$ , so the nontriviality of $Z^q$ is captured by the way $C^{q+1}$ intersects $X$ . Alexander duality is a very natural way to formalize this
idea. Ordinary homology is good enough for the complement of $X$ : since it is open,
it is locally contractible, and the brand of homology or cohomology is not an issue. The same reasoning shows that the complement of any totally disconnected subset of
any
manifold of dimension $n \ge 2$ is connected.
|
{
"source": [
"https://mathoverflow.net/questions/55718",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13034/"
]
}
|
55,726 |
When looking definition, and theorems related to Properly discontinuous action of a group $G$ on a topological space $X$, it is different in different books (Topology and Geometry-Bredon, Complex Functions-Jones, Three Dimensional Geometry and Topology- Thurston). Therefore, it will be clarified, if we write these definitions separately, and see which is stronger or which are equivalent? I will name them as "Type A", "Type B"..) Let $X$ be a topological space and $G$ be a group acting on $X$. Definition 1: The action is of " Type A " if the map $G \times X \rightarrow X \times X$, given by $(g,x)\mapsto (x,g.x)$ is proper, i.e. inverse image of any compact set under this map is compact. Definition 2: The action is of " Type B" if for any compact set $K\subseteq X$, $K\cap g.K=\phi$ for all but finitely many $g\in G$. Definition 3: The action is of " Type C " if for each $x\in X$ has an open neighbourhood $U$ such that $g.U\cap U=\phi$ for all but finitely many $g\in G$. Definition 4: The action is of " Type D " if for each $x\in X$, there exist an open neighbourhood $U$ of $x$, such that $g.U\cap U\neq \phi$ for $g\in G$ implies $g.x=x$. Definition 5: The action is of " Type E " if each $x\in X$ has a neighbourhood $U$ such that the set {$ g\in G \colon g.x\in U $} is finite. Q.1 Which type of actions imply which other type of action? Q.2 If $X$ is Hausdorff, then under which type action, the quotient $X/G$ is Hausdorff? (These are required, when studying action of a group on a compact Riemann surface, its quotient, whether quotient map is branched or unbranched, etc,) (This question may be not applicable to post for MO; but when reading a paper related to enumeration of equivalent coverings of a space, with given (finite) transformation group, I came across this notion, and when looked into details, the different definitions puzzled.)
|
Below locally compact spaces are assumed to be Hausdorff . The following is essentially a distillate of results from Bourbaki's Topologie Générale , Chapitres II and III. Definition. A continuous function $f: X \to Y$ is called proper if $f$ maps closed sets to closed sets and $f^{-1}(K)$ is compact for all compact $K \subset Y$. Remark. If $X$ is Hausdorff and $Y$ is locally compact then a continuous function $f: X \to Y$ is proper if and only if $f^{-1}(K)$ is compact for all compact $K \subset Y$. Moreover, $X$ must be locally compact. To see this, cover $Y$ with open and relatively compact sets $U_{\alpha}$. Then $f^{-1}(U_{\alpha})$ is an open covering of $X$ by relatively compact sets, hence $X$ is locally compact. If $F \subset X$ is closed then $f(F)$ is closed. Indeed, if $(y_{n}) \subset f(F)$ is a net converging to $y$, then we may assume that all $y_{n}$ are in a compact neighborhood $K$ of $y$. Pick a pre-image $x_{n}$ of each $y_{n} \in f^{-1}(K)$, which is compact by assumption. If $x_{i} \to x \in f^{-1}(K)$ is a convergent subnet of $(x_{n})$ then $(f(x_{i}))$ is a subnet of $(y_{n})$, hence $f(x) = y$ by continuity and thus $y \in f(F)$. Remark. In the definition of properness it would suffice to require that $f$ is closed and $f^{-1}(y)$ is compact for all $y \in Y$, but the definition above is good enough for the present purposes. Definition. Let $G$ be a topological group acting continuously on a topological space $X$. The action is called proper if the map $\rho: G \times X \to X \times X$ given by $(g,x) \mapsto (x,gx)$ is proper. Proposition. If $G$ acts properly on $X$ then $X/G$ is Hausdorff. In particular, each orbit $Gx$ is closed. The stabilizer $G_{x}$ of each point is compact and the map $G/G_{x} \to Gx$ is a homeomorphism. Moreover, if $G$ is Hausdorff then so is $X$. Proof. Indeed, the orbit equivalence relation is the image of $\rho$, hence it is closed. Since the projection $X \to X/G$ is open, this implies that $X/G$ is Hausdorff. Since the pre-image of the point $[x]$ in $X/G$ is its orbit $Gx$, we see that orbits are closed. The stabilizer $G_{x}$ of a point $x$ is the projection of $\rho^{-1}(x,x)$ to $G$, hence it is compact. The map $G/G_{x} \to Gx$ is proper and $1$-to-$1$, hence a homeomorphism. Finally, if $G$ is Hausdorff, then $\{e\} \times X \subset G \times X$ is closed and therefore the diagonal $\Delta_{X} = \rho(\{e\} \times X)$ of $X \times X$ is closed, hence $X$ is Hausdorff. Exercise. Let $G$ be a Hausdorff topological group acting properly on a locally compact space $X$. Then $G$ and $X/G$ are both locally compact. If $X$ is compact Hausdorff then so are $G$ and $X/G$. Replace finite by compact in Type A and Type B. Then we have the following implications for a continuous action: Proper $\Longrightarrow$ Type A, the converse holds if both $G$ and $X$ are locally compact. Type A $\Longrightarrow$ Type B. Let $K \subset X$ be compact. Then $K \times K \subset X \times X$ is compact. Thus, if the action is of type A, then $\rho^{-1}(K \times K) = \{(g,x) \in G \times X\,:\,(x,gx) \in K \times K\} \subset G \times X$ is compact. The projection of this set to $G$ is compact and consists precisely of the $g \in G$ for which $K \cap gK \neq \emptyset$. Type B $\Longrightarrow$ Type A if $X$ is Hausdorff. We have to show that $\rho^{-1}(L)$ is compact for every compact $L \subset X \times X$. Let $K$ be the union of the two projections of $L$. Then $(g,x) \in \rho^{-1}(K \times K)$ is equivalent to $x \in K \cap gK$. Since $\rho^{-1}(K \times K)$ is compact and $\rho^{-1}(L)$ is a closed subset of $\rho^{-1}(K \times K)$, we have that $\rho^{-1}(L)$ is compact. Corollary. If $G$ and $X$ are locally compact, properness, Type A and Type B are all equivalent. Let me now show that in the locally compact setting properness is equivalent to a refinement of Type C: Proposition. Let $G$ and $X$ be locally compact and assume that $G$ acts continuously on $X$. The following are equivalent: The action is proper. For all $x,y \in X$ there are open neighborhoods $U_{x}, U_{y} \subset X$ of $x$ and $y$ such that $C = \{g \in G\,:\,gU_x \cap U_{y} \neq \emptyset \}$ is relatively compact. Proof. $1.$ implies $2.$ Let
$K_{x}$ and $K_{y}$
be compact neighborhoods of $x$ and $y$. Then the set $\rho^{-1}(K_{x} \times K_{y})$ is compact and its projection to $G$ contains $C$ and is compact. Now let $U_{x}$ and $U_{y}$ be the interiors of $K_{x}$ and $K_{y}$. $2.$ implies $1$. Let $K \subset X \times X$ be compact. We want to show that $\rho^{-1}(K)$ is compact as well. Let $(g_{n},x_{n})$ be a universal net in $\rho^{-1}(K)$. Then $(x_{n},g_{n}x_{n})$ is a universal net in $K$ and hence converges to some $(x,y) \in K$. Let $U_{x}, U_{y}$ and $C$ be as in $2.$. Then $(x_{n},g_{n}x_{n}) \in U_{x} \times U_{y}$ eventually and thus also $(g_{n}) \subset C$ eventually. Since $(g_{n})$ is universal and $C$ is relatively compact, $(g_{n})$ converges to some $g \in G$. Hence $(g_{n},x_{n})$ converges to $(g,x) \in \rho^{-1}(K)$. Example. To see that Type C is weaker than properness, consider $A = \begin{pmatrix} 2 & 0 \\ 0 & 2^{-1} \end{pmatrix}$ and the action of $\mathbb{Z}$ on $\mathbb{R}^{2} \smallsetminus \{0\}$ given by $n \cdot x = A^{n} x$. For instance for $x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $y = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and all neighborhoods $U_{x} \ni x$ and $U_{y} \ni y$ the set $\{n \in \mathbb{Z}\,:\, U_{x} \cap n \cdot U_{y} \neq \emptyset \}$ is infinite. Thus this action isn't proper. On the other hand, it is easy to see that it is of Type C. Remark. The previous example shows that properness of an action is not a local property. Exercise. If the action of a locally compact group $G$ on a locally compact space $X$ is of type C and $X/G$ is Hausdorff then it is proper. To finish this discussion, it is evident that an action of type C is also of type E, hence type E is also weaker than properness. Finally, a trivial action is of type D, hence this property has nothing to do with properness. Here are some references: I've followed Bourbaki, Topologie Générale, Ch. III, in terminology, and the proofs I've given are variants of Bourbaki's. I happen to like Koszul's Lectures on groups of transformations . If you're looking for a more pedestrian approach, you can find the most important facts in Lee's Introduction to topological manifolds .
|
{
"source": [
"https://mathoverflow.net/questions/55726",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12484/"
]
}
|
55,769 |
Infinite dimensional constructions, such as spaces of diffeomorphisms, spectra, spaces of paths, and spaces of connections, appear all over topology. I rather like them, because they sometimes help me to develop a good mental picture of what is going on. Emmanuel Farjoun, in the first lecture of an Algebraic Topology course about a decade ago, described "becoming comfortable with the idea of infinite dimensional manifolds" as being "one of the main conceptual advances in topology in the latter half of the 20th century". But, as I realized in a discussion yesterday, I don't understand whether infinite dimensional spaces are needed , or whether they are merely an intuitive crutch. Are there situations in which a significant finite dimensional result strictly requires an infinite dimensional construction in order to prove it or in order to properly understand it? If the answer is no, then at least are there important finite-dimensional theorems for which infinite-dimensional proofs are "clearly" the easiest and the most natural? (for some conceptual reason which you can explain; not just "a finite-dimensional proof isn't known yet"). A closely related question is this , although its focus is somewhat different, and none of the answers there apply here; however Andrew Stacey's answer which argues that infinite-dimensional constructions are usually not strictly necessary, is relevant. Edit : Your favourite finite-dimensional result proved by infinite-dimensional means answers this question only if you can explain to me why I should not expect a finite-dimensional proof to exist or to be anywhere near as "good".
|
Any compact Lie group admits an embedding into a large linear group $GL_n (\mathbb{C})$. I think it is no question that this is a genuinely finite-dimensional and important statement. The proof is via the Peter-Weyl theorem; essentially one has to show that there are enough finite-dimensional representations. How is this done? There is an obvious faithful representation on the Hilbert space $L^2 (G)$. One constructs a compact, $G$-equivariant self-adjoint and injective operator $F$ on $L^2 (G)$. By the spectral theorem, the sum of the eigenspaces of $F$ is dense in $L^2 (G)$, and they are finite-dimensional, so here is a load of finite-dimensional representations! Of course the completeness of the Hilbert space is crucial. The curious thing is that the role of the Hilbert space theory is to cut down infinite dimensional things down to finite dimensions. The appearance of infinite-dimensional spaces in complex geometry (see Georges answer), Hodge theory, elliptic PDE etc. has of course a similar flavour. On the other hand, it seems that in algebraic topology, most infinite-dimensional spaces can be kicked out by finite-dimensional approximation, of course at the price of making arguments more cumbersome. The algebraic topologist's infinite dimensional spaces are of the $\mathbb{R}^{\infty}$-type, not of the $\ell^2$-type. Completeness does not play a big role in algebraic topology.
|
{
"source": [
"https://mathoverflow.net/questions/55769",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2051/"
]
}
|
55,801 |
One possibility is the integration by substitution formula. But this doesn't explain why $\int_1^{\infty}x^{-3}dx$ is easy to evaluate (and rational) whilst $\sum_1^{\infty}n^{-3}$ is unknown (and seemingly not a simple combination of well known numbers). It is true that $x^{-3}$ has a simple inverse derivative whilst $n^{-3}$ does not have a known simple inverse difference, but is there more to it than that?
|
Well, some things are easier in one place, and other things are easier in the other. The many assertions in answers that integrals of $x^n$ are easier than the corresponding sums is misleading. In the continuous world, $x^n$ is a very natural object. In the discrete world, however, the natural object is the falling factorial: $x^{\underline{n}}$ is defined as $x(x-1)\dots(x-n+1)$. Compute the derivative of $x^{\underline{n}}$ and you get a mess. Compute the forward difference and you get the beautifully simple $f(x+1)-f(x) = (x+1)^{\underline{n}} - x^{\underline{n}} = n x^{\underline{n-1}}$. That is,
$$\sum_{i=1}^n i^9= \frac{1}{20} n^2 (n+1)^2 \left(n^2+n-1\right) \left(2 n^4+4 n^3-n^2-3 n+3\right)$$
is a little messy. Certainly messier than $\int x^9 dx$. On the other hand, $$\sum_{i=1}^n i^{\underline{9}} = \frac1{10} (n+1)^{\underline{10}}.$$
while the integral $\int x^{\underline{9}}dx$ is bad enough that I dare you to work it out by hand. The truth here is that the discrete world is often easier (and more fundamental, too), but history has prejudiced us to be more familiar with the continuous world. Zeilberger (who else?) has written passionately about this, and if you know of other sources for discrete calculus (online or off) please add them as comments to this answer!
|
{
"source": [
"https://mathoverflow.net/questions/55801",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4692/"
]
}
|
55,906 |
Are there any good detailed historical sources about development of connections on vector/principal bundles over the last 100 years? The best source I am aware of is Michael Spivak's 5 volume opus, but this is not detailed enough for the project I have in mind (I am intending to set this as a topics essay for my first year graduate differential geometry class, and I want to make sure I know enough myself first!).
|
Here is a rough historical overview: 1900: Ricci and his student Levi-Civita introduce the concept of a "tensor" in MR1511109 Ricci, G.; Levi-Civita, T. Méthodes de calcul différentiel absolu et leurs applications. (French) Math. Ann. 54 (1900), no. 1-2, 125--201. There they define an operation called "covariant differentiation" of a tensor which generalizes the usual differentition to curvelinear coordinates; the correction term is given by the Christoffel symbols. By that time there is no abstract notion of covariant derivative of a section or whatsoever nor the differentiation is considered together with the associated parallel transport. For more about this I would refer you to the book by Morris Kline on the history of mathematics. There is also an edited edition of this article available (Editor is Hermann who was already mentioned by Deane Yang). Whitney used the word "tensor" in connection with a group theoretic construction in 1938 (he obviuosly was influenced by what he knew about tensors; see end of the paper where he speaks about parallel transport which gives an impression on how people thought about thing like that in the 1930's). Later this was generalized to modules by Cartan and Dieudonné (see Weibel's history on homological algebra). Meanwhile Einstein and his friend Grossmann used this "absolute differential calculus" (as it is also called) to give a mathematical footing to general relativity. Later Einstein had a lot of correspondence with Levi-Civita as well as Cartan on topics like parallel transport which can be found in their scientific correspondence. 1917: Levi-Civita, T. Nozione di parallelismo in una varietà qualunque e conseguente specificazione geometrica della curvatura riemanniana. (Italian)
Palermo Rend. 42, 172-205 (1917). Here Levi-Civita gives, following an indication of his teacher Ricci, a geometric interpretation of the covariant differentiation by means of a associated parallel transport. I only know of the italian version of this article but there is book of Levi-Civita (from the 1930s?) on tensor analysis where he gives a more or less plausible derivation of this concept; though I found it hard to follow in all details. Felix Klein by the way was not very happy with the derivation given by Levi-Civita and in some book (currently I can not recall the title but it is about geometry and also available in English) he derived the whole thing from a physical experiment with some peculiar machine which can be used to detect curvature. I think the books title must be something like "Higher Geometry". The idea seems to be due to Radon (1918); but seemingly published nowhere else. Anyhow the review by Blaschke (one of the leading person in differential geometry at this time) for the Zentralblatt is very illuminating (almost a prophecy). between 1917 and 1920 (third German edition is published in 1919) Weyl published "Raum, Zeit, Materie" ("Space-time-matter"; available here: http://www.archive.org/details/spacetimematter00weyluoft ). There he uses the Christoffel symbols and their transformation behaviour to define "connection". Around the same time as Levi-Civita, Schouten and Hesse made similar observations. This can be found in an easy to find article by acclaimed German mathematics historian Karin Reich. Unfortunately, this article is in German. Nevertheless you can see pictures of models of surfaces with parallel transport drawn on them, that Schouten produced. Later Weyl in connection with his physical studies introduced the concept of a Weyl connection (whose most peculiar property is that it does not preserve length); this was already mentioned in the comments. More can be found in Chapter I of this brilliant book (again German would be required): http://books.google.de/books?id=oZLiqDQGnjgC&printsec=frontcover&dq=Scholz+and+Weyl&hl=de&ei=IcVfTeSULILXsgaRz422CA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwAA#v=onepage&q&f=false Around 1923 Élie Cartan starts to study connections (his so-called projective connections) More on this can be found in the splendid article "Vector bundles and connections in physics and mathematics: some historical remarks" by Varadarajan. Somewhere behind 1940 Norman Steenrod introduces the concept of a fibre bundle and
around 1950 Charles Ehresmann introduces the concept of a connection on a fibre resp. principal fibre bundle. His article is very readable and a valuable historical source.
|
{
"source": [
"https://mathoverflow.net/questions/55906",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13074/"
]
}
|
55,913 |
One always sees the definition of a contact manifold $(X,\xi)$ as an odd dimensional manifold with a hyperplane distribution $\xi$ which can locally be expressed as $\xi = \ker \alpha$ for a 1-form $\alpha$. But in fact, it seems that in every example I know of, one always assumes that $\xi$ is cooriented , and hence we can write $\xi = \ker \alpha$ globally . Is there a reason (other than historical) as to why coorientation wasn't built in automatically in the definition of a contact manifold? It seems strange that this isn't required in the definition.
|
Here is a rough historical overview: 1900: Ricci and his student Levi-Civita introduce the concept of a "tensor" in MR1511109 Ricci, G.; Levi-Civita, T. Méthodes de calcul différentiel absolu et leurs applications. (French) Math. Ann. 54 (1900), no. 1-2, 125--201. There they define an operation called "covariant differentiation" of a tensor which generalizes the usual differentition to curvelinear coordinates; the correction term is given by the Christoffel symbols. By that time there is no abstract notion of covariant derivative of a section or whatsoever nor the differentiation is considered together with the associated parallel transport. For more about this I would refer you to the book by Morris Kline on the history of mathematics. There is also an edited edition of this article available (Editor is Hermann who was already mentioned by Deane Yang). Whitney used the word "tensor" in connection with a group theoretic construction in 1938 (he obviuosly was influenced by what he knew about tensors; see end of the paper where he speaks about parallel transport which gives an impression on how people thought about thing like that in the 1930's). Later this was generalized to modules by Cartan and Dieudonné (see Weibel's history on homological algebra). Meanwhile Einstein and his friend Grossmann used this "absolute differential calculus" (as it is also called) to give a mathematical footing to general relativity. Later Einstein had a lot of correspondence with Levi-Civita as well as Cartan on topics like parallel transport which can be found in their scientific correspondence. 1917: Levi-Civita, T. Nozione di parallelismo in una varietà qualunque e conseguente specificazione geometrica della curvatura riemanniana. (Italian)
Palermo Rend. 42, 172-205 (1917). Here Levi-Civita gives, following an indication of his teacher Ricci, a geometric interpretation of the covariant differentiation by means of a associated parallel transport. I only know of the italian version of this article but there is book of Levi-Civita (from the 1930s?) on tensor analysis where he gives a more or less plausible derivation of this concept; though I found it hard to follow in all details. Felix Klein by the way was not very happy with the derivation given by Levi-Civita and in some book (currently I can not recall the title but it is about geometry and also available in English) he derived the whole thing from a physical experiment with some peculiar machine which can be used to detect curvature. I think the books title must be something like "Higher Geometry". The idea seems to be due to Radon (1918); but seemingly published nowhere else. Anyhow the review by Blaschke (one of the leading person in differential geometry at this time) for the Zentralblatt is very illuminating (almost a prophecy). between 1917 and 1920 (third German edition is published in 1919) Weyl published "Raum, Zeit, Materie" ("Space-time-matter"; available here: http://www.archive.org/details/spacetimematter00weyluoft ). There he uses the Christoffel symbols and their transformation behaviour to define "connection". Around the same time as Levi-Civita, Schouten and Hesse made similar observations. This can be found in an easy to find article by acclaimed German mathematics historian Karin Reich. Unfortunately, this article is in German. Nevertheless you can see pictures of models of surfaces with parallel transport drawn on them, that Schouten produced. Later Weyl in connection with his physical studies introduced the concept of a Weyl connection (whose most peculiar property is that it does not preserve length); this was already mentioned in the comments. More can be found in Chapter I of this brilliant book (again German would be required): http://books.google.de/books?id=oZLiqDQGnjgC&printsec=frontcover&dq=Scholz+and+Weyl&hl=de&ei=IcVfTeSULILXsgaRz422CA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwAA#v=onepage&q&f=false Around 1923 Élie Cartan starts to study connections (his so-called projective connections) More on this can be found in the splendid article "Vector bundles and connections in physics and mathematics: some historical remarks" by Varadarajan. Somewhere behind 1940 Norman Steenrod introduces the concept of a fibre bundle and
around 1950 Charles Ehresmann introduces the concept of a connection on a fibre resp. principal fibre bundle. His article is very readable and a valuable historical source.
|
{
"source": [
"https://mathoverflow.net/questions/55913",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12998/"
]
}
|
55,931 |
Let $\mbox{Rings}$ be the category of commutative rings with $1$ . Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ such that $$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$
|
The category $\mbox{CRing}$ of commutative rings is rigid , i.e. every equivalence $\mbox{CRing} \to \mbox{CRing}$ is isomorphic to the identity, Proof: Let $F : \mbox{CRing} \to \mbox{CRing}$ be an equivalence. The main part is to prove that $A := F(\mathbb{Z}[x])$ is isomorphic to $\mathbb{Z}[x]$ . Charles' answer shows that $A$ is a retract of a polynomial ring $\mathbb{Z}[x_1,...,x_n]$ , in particular an integral domain. Since $\mathbb{Z}$ is the initial ring, $F$ preserves it and since $\mathbb{Z} \subseteq \mathbb{Z}[x]$ is a split monomorphism, the same is true for $\mathbb{Z} \to A$ . Thus we have $\mathbb{Z} \subseteq A \subseteq \mathbb{Z}[x_1,...,x_n]$ . Now assume that we already knew that $A$ is a UFD with $\text{tr.deg}(Q(A)/\mathbb{Q}) = 1$ . Then [2], Theorem 4.1 would imply that $A \cong \mathbb{Z}[x]$ . Note that the only nontrivial input in the lemmas preceding this theorem is Lüroth's theorem applied to $A \otimes \mathbb{Q}$ , which is not needed here because be show below directly $Q(A) \cong \mathbb{Q}(x)$ . So let's check the two properties of $A$ . Since $A$ is a retract of $\mathbb{Z}[x_1,...,x_n]$ , which is UFD, we conclude from [1], Prop. 1.8 that also $A$ is UFD. Now it remains to prove that $Q(A) \cong \mathbb{Q}(x)$ , which Kevin Ventullo has already sketched in the comments. First note that $F$ preserves nontrivial rings since $0$ is the terminal ring. A monomorphism is the same as an injective ring homomorphism (since the forgetul functor from rings to sets is representable), thus $F$ preserves injective ring homomorphisms. Now fields are precisely the nontrivial rings with only one proper ideal, i.e. such that every homomorphism to a nontrivial ring is injective. Thus $F$ preserves fields (and also field extensions). The universal property of the quotient field and that we already know that $F$ preserves $\mathbb{Z}$ , integral domains, fields and injections implies $F(\mathbb{Q}) \cong \mathbb{Q}$ and $Q(A) = F(\mathbb{Q}(x))$ . Now the extension $\mathbb{Q}(x)$ of $\mathbb{Q}$ is characterized by the property it is transcendental and every transcendental extension factors over it. Thus it remains to characterize algebraic field extensions $L/K$ . To do this, we call $L/K$ locally finite if for every extension $E/K$ there are only finitely many $K$ -homomorphisms $L \to E$ . It is clear that every finite extension is locally finite. Besides, every locally finite extension is algebraic: If not, choose a transcendence basis $B$ and extend infinitely many homomorphisms from $K(B)$ to it's algebraic closure to get a contradiction. It follows that algebraic extensions are precisely the filtered colimits of locally finite extensions. So we have proved that $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ . Now let $R$ be a commutative ring. The usual coring structure on $\mathbb{Z}[x]$ induces a ring structure on $\text{Hom}(\mathbb{Z}[x],R)$ which is naturally isomorphic to $R$ . Every coring structure on $\mathbb{Z}[x]$ is isomorphic to the usual one ([3], Prop. 3.1). Thus, we get, naturally in $R$ , $R \cong \text{Hom}(\mathbb{Z}[x],R) \cong \text{Hom}(F(\mathbb{Z}[x]),F(R)) \cong \text{Hom}(\mathbb{Z}[x],F(R)) \cong F(R)$ . Thus $F$ is isomorphic to the identity. Actually [3] gives another, similar proof that $\mbox{CRing}$ is rigid, and more generally it classifies the automorphisms of the category of $R$ -algebras, where $R$ is an integral domain. I should have checked the literature in the first place. [1] Douglas L. Costa, Retracts of Polynomial Rings , J. Algebra 44 (1977), pp. 492 - 502 [2] S. Abhyankar, P. Eakin, W. Heinzer, On the uniqueness of the coefficient ring in a polynomial ring , J. Algebra 23 (1970), pp. 310 - 342 [3] W. E. Clark, G. M. Bergman, The Automorphism Class Group of the Category of Rings , J. Algebra 24 (1973), pp. 80 - 99
|
{
"source": [
"https://mathoverflow.net/questions/55931",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12259/"
]
}
|
55,933 |
I am afraid this post may show my naivety. At a recent conference, someone told me that there are some arguments in computability theory that don't relativize. Unfortunately, this person (who I think may be an MO regular) couldn't give me any examples off-hand. My naive understanding is that one can take any proof, fix an oracle and replace "Turing machine" with "Turing machine using that oracle". (Similarly, replace "computable" with "computable w.r.t. this oracle" and so on.) But I am probably missing something. So is it possible to relativize any proof? Is it not possible? Or is it technically possible to relativize any proof, but the relativization isn't what you may expect? If it is not always possible, can someone provide me a good example or reference? Also, any discussion on what makes a proof relativizable or not would be of help. Actually, while writing this, I may have come up with an example. It's known that $P=NP$ and $P \neq NP$ for different oracles. But this is a relativization of a theorem (conjecture), not it's proof. So this brings up related questions. Is relativizing the statement of a theorem different from relativizing the proof? What if the theorem/proof has nothing to do with the actual model of computation (e.g. the proof doesn't refer to ideas such as time-complexity, only computability)? Thanks!
|
Early in the history of recursion theory, the realization that all known proofs in the subject could be relativized in the manner you indicate led Hartley Rogers to make what is called the homogeneity conjecture. Let $\mathcal{D}$ be the structure of the Turing degrees with the partial order of Turing reducibility $\leq_T$. Let $\mathcal{D}(\geq \mathbf{x})$ be the structure of the Turing degrees that are $\geq_T \mathbf{x}$ also with the partial order $\leq_T$. The homogeneity conjecture says that for any Turing degree $\mathbf{x}$, $\mathcal{D}$ is isomorphic to $\mathcal{D}(\geq \mathbf{x})$. Richard Shore refuted the homogeneity conjecture in an elegant 1979 paper -- it's only one page long (though it relies on earlier work coding models of arithmetic in $\mathcal{D}$). A couple years later, Harrington and Shore showed that you can do even better. Not only are there Turing degrees $\mathbf{x}$ for which $\mathcal{D}$ and $\mathcal{D}(\geq \mathbf{x})$ aren't isomorphic as structures, there are Turing degrees $\mathbf{x}$ so that $\mathcal{D}$ and $\mathcal{D}(\geq \mathbf{x})$ aren't elementarily equivalent. So this means that there is a first order sentence $\varphi$ in the language only containing $\leq_T$ which is true about the Turing degrees, but which is false if you relativize the sentence to the Turing degrees $\geq_T \mathbf{x}$ for some $\mathbf{x}$ (and of course, working in the Turing degrees $\geq_T \mathbf{x}$ is equivalent to giving all Turing machines access to $\mathbf{x}$ as an oracle). Hence, the proof that $\varphi$ is true can't be relativized to $\mathbf{x}$. A somewhat misleading and oversimplified explanation of why this occurs is that you can code models of arithmetic into $\mathcal{D}$ (or $\mathcal{D}(\geq \mathbf{x})$) which can then interpret unrelativized concepts like being arithmetically definable or hyperarithmetically definable. A particularly spectacular form of this phenomenon would occur if Slaman and Woodin's biinterpertability conjecture is true. The conjecture says that the following relation (on $\overrightarrow{\mathbf{p}}$ and $\mathbf{d}$) is definable in $\mathcal{D}$: "$\overrightarrow{\mathbf{p}}$ codes a standard model of first order arithmetic and a real $X$ such that $X$ is of degree $\mathbf{d}$". All that being said, just about every proof in recursion theory that I know of relativizes. I'd be very interested in a proof which doesn't relativize and doesn't factor through the coding machinery I've mentioned above. By the way (and somewhat ironically), the Baker-Gill-Solovay theorem that you mentioned does itself relativize. The relativized version says that for any oracle $X$, there are oracles $A$ and $B$ so that $X$ is poly-time reducible to both $A$ and $B$, and $P^A = NP^A$ and $P^B \neq NP^B$. (We've relativized to $X$ here. The unrelativized result simply says that there are oracles $A$ and $B$ so that $P^A = NP^A$ and $P^B \neq NP^B$). Of course, the real point is that a proof that $P \neq NP$ can't use a technique that relativizes.
|
{
"source": [
"https://mathoverflow.net/questions/55933",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12978/"
]
}
|
55,988 |
Many authors (e.g Landsman, Gleason) have stated that in quantum mechanics, the observables of a system can be taken to be the self-adjoint elements of an appropriate C*-algebra. However, many observables in quantum mechanics - such as position, momentum, energy - are in general unbounded operators. Is there any way to reconcile these two apparently contradictory statements? I have looked at the notion of affiliation for a C*-algebra in the sense of Woronowicz. However, I can't see how you would extend states to elements affiliated with a C*-algebra and so it doesn't seem to solve the problem.
|
In addition to what has already been said I would like to add some more comments. I completely understand your suspicion that the passage from unbounded operators to bounded ones is at least tricky. For the canonical commutation relations of position and momentum operators this can be solved in a reasonable and also physically acceptable way by passing to the Weyl algebra (OK; there are zillions of Weyl algebras around in math, but I'm refering here to the $C^*$-algebra generated by the exponentials of $Q$'s and $P$'s subject to the heuristic commutation relations arising from $[Q, P] = \mathrm{i} \hbar$). However, there are other situations in physics where this is much more complicated: when one tries to quantize a classical mechanical system which has a more complicated phase space than just $\mathbb{R}^{2n}$ then one can not hope to get some easy commutation relations which allow for a Weyl algebra like construction. To be more specific, for general symplectic of Poisson manifolds the beast quantization scheme one can get in this generality is probably formal deformation quantization. Here the classical observable algebra (a Poisson algebra) is deformed into a noncommutative algebra in a $\hbar$-dependent way such that the new product, the so-called star product, depends on $\hbar$ in such a way that for $\hbar = 0$ one recovers the classical mutliplication and in first order of $\hbar$ one gets the Poisson bracket in the commutator. Now two difficulties arise: the most severe one is that in this generality one only can hope for formal power series in $\hbar$. Thus one has even changed the underlying ring of scalars from $\mathbb{C}$ to $\mathbb{C}[[\hbar]]$. So there is of course no notion of a $C^\ast$-algebra whatsoever on this ring. Nevertheless, there is a good notion of states in the sense of positive functionals already at this stage. Second, even if one succeeds to find a convergent subalgebra one does usually not end up with a $C^\ast$-algebra on the nose. Worse: in most of the explicit exampes one knows (and there are not really many of them...) the commutation relations one obtains are very complicated. In particular, it is not clear at all how one can affiliate a $C^\ast$-algebra to them.
Moreover, it is not clear which of the previous states survive this condition of convergence and yield reasonable representations by the GNS construction. It takes quite some effort to first represent the observables by typically very unbounded operators in a reasonable way and then show that they give rise to some self-adjoint operators still obying the relevant commutation relations. This is far from being obvious.
To get a flavour for the difficulties it is quite illustrative to take a look at the book of Klimyk and Schmüdgen on Quantum Groups adn their Representations. Note that in these examples one still has a lot of structure around which helps to understand the analysis. However, physics usually requires still much more general situations. Most important here are systems with gauge degrees of freedoms where one has to pass to a reduced phase space. Even if one starts with a geometrically nice phase space the reduced one can be horribly complicated. This problem is present in any contemporary QFT really relevant to physics :( For other quantization schemes things are similar, even though I'm not quite the expert to say something more substantial :) So one may ask the question why one should actually insist on $C^\ast$-algebras and this strong analytic background. There is indeed a physical reason and this is that quantum physics predicts not only expectation values of observables in given states (here the notion of a ${}^\ast$-algebra and a positive functional is sufficient) but also the possible outcomes of a measurement: they are given by particular numbers called the physical spectrum of the observable. To get a good description with predictive power the (as far as I know) only way to achieve this is to say that the physical spectrum is givebn by the mathematical spectrum of a self-adjoint operator in a Hilbert space. If one is here at this point, then the passage from a (unbounded) self-adjoint operator to a $C^\ast$-algebra is comparably easy: one has the spectral projections and takes e.g. the von Neumann algebra generated by them... So the point I would like to make is that it is very desirable from a physical point of view to have the strong analytic framework for observables as either self-adjoint operators or Hermitian elements in a $C^\ast$-algebra. But the quantum theory of many nontrivial systems requires a long long and nontrivial way before one ends up in this nice heavenly situation.
|
{
"source": [
"https://mathoverflow.net/questions/55988",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13087/"
]
}
|
56,011 |
It was Faltings who first proved in 1983 the Mordell conjecture, that a curve of genus 2 or more over a number field has only finitely many rational points. I am interested to know why Mordell and others believed this statement in the first place. What intuition is there that the statement must hold? Without reference to any proof, why should the conjecture 'morally' be true? Supposing one had to give a colloquium (to a general mathematically literate audience) on this, how could one convince them without going into details of heights or étale cohomology? Answers I'm not looking for will be of the form "You fool, because it's been proved already", or even "Read Faltings' proof".
|
If the curve $X$ (over the number field $k$) has no $k$-points at all, then Mordell's conjecture is true for $X$.
Otherwise, if $O$ is a given $k$-point on $X$, we can get a map (the Albanese map)
$X \to Jac(X)$ via $P \mapsto P - O$, embedding $X$ as a $k$-subvariety of $Jac(X)$. The Mordell--Weil theorem shows that $Jac(X)(k)$ is a finitely generated abelian group.
Since $g \geq 2$ by assumption, the curve $X$ is of positive codimension in $Jac(X)$,
and so it is not unreasonable to imagine that $X$ intersects $Jac(X)(k)$ in only finitely
many points. (George Lowther's answer, posted while I was writing this, describes the same
heuristic.) In fact, my understanding is that Weil proved the general form of the Mordell--Weil theorem
(in his thesis, I think) precisely to try to implement this strategy and so prove Mordell's
conjecture. Unfortunately, no-one has been able to implement this strategy quite as cleanly as the intuition above suggests, although it inspired an important approach, due to Chabauty, which established the Mordell conjecture in some non-trivial cases; see this paper by McCallum and Poonen for an explanation of Chabauty's method, and Coleman's strengthening of it. More recently, Minhyong Kim has developed an anabelian strengthening of Chabauty's method; see e.g. here and here . The second paper (joint with Coates) gives a new proof of Mordell for Fermat curve's, among other examples.
|
{
"source": [
"https://mathoverflow.net/questions/56011",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5744/"
]
}
|
56,062 |
in a recent MO question, link , discussing the current foundations of mathematics, the author linked a video lecture by Prof. Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's proof of the consistency of PA. In discussions arising from the question, some people commented that imagining an infinite descending chain in $\epsilon_{0}$ is "crazy". I would like to understand better this ordinal, since I actually don't know exactly how to depict it in my mind. I have clear in my mind the order associated with the finite ordinals. I use in my mind a notation of the following kind: $1 = I$ $2= II$ $3= III$ $4= IIII$ $\omega = (III\dots)$ $\omega+1= (III\dots)I$ $\omega +2 = (III\dots)II$ $\omega + \omega= \omega \cdot 2= (III\dots)(III\dots)$ In general I understand $\alpha + \beta$ as the juxtaposition of the two representations. $\omega\cdot 3 = (III\dots)(III\dots)(III\dots)$ $\omega\cdot \omega = \omega^{2} = \big( (III\dots)(III\dots)(III\dots)\dots\big)$ In general I understand $\alpha \cdot \beta$, by replacing each $I$ symbol in $\beta$ with the representation of $\alpha$. So $\omega^{3}=\omega^{2}\cdot \omega = \big( \omega^{2} \omega^{2} \omega^{2} \dots \big)$ This allows me to visualize every ordinal of the form $\omega^{n}\cdot m + k$, with $n,m,k$ naturals (i.e finite ordinals). So far I have absolutely no doubt that there are no infinite descending chain in ordinals of the form $\omega^{n}\cdot m + k$. However I start having problem with the ordinal $\omega^{\omega}= \bigsqcup_{n<\omega}\omega^{n}$. Do you have any idea on how to visualize $\omega^{\omega}$ is a way consistent with the representation used above (which i actually found here ) ? Anyway, looking at wikipedia , I still manage to visualize $\omega^{\omega}$ as the set of infinite strings of natural number, having only finitely many digits different from $0$. Still I have no doubt that there are no infinite descending chain in $\omega^{\omega}$. Perhaps i might be able to understand $\omega^{\omega^{\omega}}$, namely the set of infinite strings labeled with elements of $\omega^{\omega}$, having only finitely many elements different from $0$.
Or (i guess) equivalently a $\omega\times\omega$ square labeled with naturals, where only finitely many columns are different from $0^{\omega}$, and all of these non constant-$0$ columns, contains only finitely many digits different from $0$. However I do not know how to visualize $\epsilon_{0}$. I mean I know that the elements of $\epsilon_{0}$ can be represented by finite-branching finite trees labeled with natural numbers, but that doesn't give me a strong intuition about the fact that no infinite chain exists, so I guess its not a great picture (or at least I do not understand it properly, yet). Questions A) Could you suggest a way to visualize $\omega^{\omega^{\omega}}$? It should be in such a way to convince me about the fact that there are no infinite down-chain. B) Could you suggest a way to visualize $\epsilon_{0}$, again arguing that it should be very clear that there are no infinite down-chain. C) Could you please state your opinion about Prof. Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's?
This shouldn't be a duplicate of the previous, wider thread link , I'm only interested in this little bit of Voevodsky's talk. Thank you in advance, bye matteo
|
The standard way to visualize $\epsilon_0$ is by the Hydra game . Here the elements of $\epsilon_0$ are visualized as isomorphism classes of rooted finite trees. The inequality can be described by the "cutting off heads" rule: The tree $T_1$ is greater than $T_2$ is there is a series of head cuttings which reduces $T_1$ to $T_2$ . Writing out the inequality relationship between trees directly is a pain, see my blogpost . If you turn those nested sets into trees in the obvious way, you get the Hydra game. I am told that most people do not find it intuitive that the Hydra game ends. I find that, once I've played a few rounds (try this applet ) I find it "obvious", although writing down an actual proof is still painful. As far as an actual proof, you should directly show the following: Let $X$ be a totally ordered set. Let $\omega^X$ be the set of functions $X \to \omega$ which are $0$ for almost all $x \in X$ , ordered as follows: Let $f$ and $g$ be distinct elements of $\omega^X$ and let $x$ be the greatest element of $x$ for which $f(x)\neq g(x)$ . Then $f<g$ if and only if $f(x) < g(x)$ . Then $\omega^X$ is well ordered.
So every tower of $\omega$ 's is well ordered and, $\epsilon_0$ , being the union of all such towers, is also well-ordered. By the way, you don't ask this, but you might be curious what happens when you try to write out this proof within PA. Recall that PA can't directly talk about subsets of $\omega$ . The statement that $\omega$ is well-ordered is encoded as an axiom schema. Let $\phi(x, y_1, y_2, \ldots, y_N)$ be any statement with variables $x$ and $y_i$ running through $\omega$ . Then PA has the following axiom: $$\forall y_1, y_2, \ldots, y_n \in \omega: \left( \exists x \in \omega : \phi(x, y_{\bullet}) \implies \exists x' \in \omega : \left( \phi(x', y_{\bullet}) \wedge \forall x \in \omega \left( \phi(x, y_{\bullet}) \implies x \geq x' \right) \right) \right).$$ Please read this axiom until you understand that, in English, it says "For all $y$ 's, if there is some $x$ obeying $\phi$ , then there is a least $x$ obeying $\phi$ ." Let's call this axiom $W(\phi, \omega)$ . We'll use similar notation with $\omega$ replaced by other sets. Here is a challenging and important exercise: Let $X$ be an ordered set. Let $\phi$ be a statement about $\omega^X$ , which may have other variables $y_i$ in it. Construct a specific statement $\sigma(\phi)$ about $X$ , with other variables $z_i$ running through $X$ , such that $$ W(\sigma(\phi), X) \implies W(\phi, \omega^X) \quad (*).$$ For every specific $\phi$ , the statement $(*)$ can be proved in PA. Since $W(\psi, \omega)$ is a axiom of PA for every $\psi$ , we can prove $W(\phi, \omega^{\omega^{\ldots^{\omega}}})$ in PA for any $\phi$ and any specific height of tower. But, in order to show that $\epsilon_0$ is well-ordered, we need to show that $W(\phi, \omega^{\omega^{\ldots^{\omega}}})$ simultaneously for every height of tower. Tracing through the arguments here, you would need to know $W(\phi, \omega)$ , $W(\sigma(\phi), \omega)$ , $W(\sigma(\sigma(\phi)), \omega)$ , $W(\sigma^3(\phi), \omega)$ and so forth. As a human mathematician, that probably doesn't bother you at all. But, in the formal system PA, any proof can only use finitely many axioms. So there is no way to write a proof which uses all of the axioms $W(\sigma^k(\phi), X)$ , for all $k$ . Of course, this doesn't show that some more clever argument couldn't prove that $\epsilon_0$ is well-ordered while working with PA; you need the Kirby-Paris theorem for that. (More precisely Kirby-Paris plus Godel shows that, if PA proves $\epsilon_0$ is well-ordered, then PA is inconsistent.) But I find that seeing this obstacle, the need to use infinitely many versions of the well-ordering axiom, clarifies my understanding of what is gong on.
|
{
"source": [
"https://mathoverflow.net/questions/56062",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11618/"
]
}
|
56,082 |
To quote one source among many, "the general reference for vanishing cycles is [SGA 7] XIII and XV". Is there a more direct way to learn the main principles of this theory (i.e. without the language of derived categories), in particular as it applies to the study of certain integral models of curves?
|
For the purposes of intuition, let me write an "answer" which is probably close to the way Picard or Lefschetz would have thought about this. Suppose that you have a family of complex
nonsingular plane cubics $X_t$ degenerating to a nodal cubic $X_0$. It is possible to understand the change in topology rather explicitly. You can set up a basis
of (real) curves $\alpha_t,\beta_t\in H_1(X_t,\mathbb{Z})$, so that
$\beta_t\to \beta_0\in H_1(X_0)$ and $\alpha_t\to 0$
as $t\to 0$. So that $\alpha_t$ literally is a vanishing cycle. There is more.
As you transport these cycles around a loop in the $t$-plane, you end up with
a new basis $T(\alpha_t),T(\beta_t)$. This is related to the old basis by the
Picard-Lefschetz formula
$$T(\alpha_t)=\alpha_t$$
$$T(\beta_t) = \beta_t \pm (\alpha_t\cdot\beta_t)\alpha_t$$
You visualize this by cutting along $\alpha_t$ and giving a twist before regluing. You can jazz this up in various ways of course, and this is the modern theory of vanishing cycles. In modern notation you'll see a nearby cycle functor $R\Psi$, which corresponds roughly to $H^*(X_t) = H_*(X_t)^*$ and a vanishing cycle functor $R\Phi$ which measures
the difference between $H^*(X_t)$ and $H^*(X_0)$.
|
{
"source": [
"https://mathoverflow.net/questions/56082",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6121/"
]
}
|
56,107 |
Kummer proved that there are no non-trivial solutions to the Fermat equation FLT(n): $x^n + y^n = z^n$ with $n > 2$ natural and $x,y,z$ elements of a regular cyclotomic ring of integers $K$. I am looking for non-trivial solutions to the Fermat equation FLT(p) in the cyclotomic integer ring $\mathbb{Z}[\zeta_{p}]$ for irregular primes p or any information about how the solutions must be (as a step toward constructing them). George Lowther pointed out in an earlier discussion that by Kolyvagin's criterion any solution in $\mathbb{Z}[\zeta_{37}]$ must be in the second case.
|
This answer is a bit late; sorry for that. Kummer's proof of the nonsolvability of $x^p + y^p = z^p$ for regular primes $p$
used “ideal numbers” (in present-day language: ideals) and was intact, at least
basically. Hilbert in his Zahlbericht gave a modified proof. Both proofs cover not
only rational integers but also numbers in $\mathbb{Z}[\zeta_p]$. On the other
hand, Kummer’s second result concerning irregular primes that satisfy certain
additional conditions covers just the rational integers (although Hilbert, in the very
last section of Zahlbericht, erroneously says that Kummer had proven this result for
$\mathbb{Z}[\zeta_p]$ as well). Thus one cannot exclude the possibility that there
is indeed a solution $(x,y,z)$ for $p=37$. And because of "Kolyvagin's criterion"
about $(2^{37}-2)/37$, this solution must belong to the second case,
that is, at least one of these three numbers $x,y,z$ in $\mathbb{Z}[\zeta_{37}]$
must have a common factor with $37$ (as mentioned by George Lowther). By the way, this criterion was also proven by Taro Morishima in 1935 (Japan. J.
Math. 11, 241-252, Satz 1; but warning: Satz 2 or at least its proof is incorrect since
it is based on some incorrect result of Vandiver). I don’t know how to find such a solution $(x,y,z)$.
|
{
"source": [
"https://mathoverflow.net/questions/56107",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13121/"
]
}
|
56,188 |
Let $G$ be a compact group and let $R(G)$ be the representation ring of $G$. Additively, $R(G)$ is generated by the irreducible representations of $G$. Usually one only deals with those representations which are nonnegative integer combinations of the irreducible representations. However, often one has formulas which apply to an arbitrary element of $R(G)$, and so includes virtual representations which (for this question at least) are the elements of $R(G)$ whose decomposition into irreducible components includes negative coefficients. Question: Is there any 'natural' interpretation of virtual representations? In particular, aside from the obvious interpretation as the elements of the formal completion of the semiring of ordinary representations to a ring, is there a natural way to view these objects? Especially helpful would be pictures/ideas others use to assign meaning to virtual representations (if any). Motivation: Often in decomposing various formulas involving characters, virtual representations arise in one way or another. For example, in Lie groups, the notion of a highest weight representation $\rho_\omega$ can be extended to arbitrary weights $t$ of $G$ via: $\rho_{w(t)} = (-1)^w\rho_t$ In particular, if $t$ is not a dominant weight of $G$ then there is a unique $w$ in the Weyl Group of $G$ such that $w(t)$ is a dominant weight so that $\rho_t$ is defined for arbitrary weights $t$; note that the Weyl Dimension Formula agrees with this extension. If the length of $w$ is odd, then $\rho_t$ will have negative dimension from the dimension formula, hence is virtual. Another simple example is that when considering the action of the Adams operation $\psi^k$ on $R(G)$, one has that $\psi^k(\rho)$ is in general a virtual representation. It happens that from time to time I come across other instances of virtual representations appearing in equations I am considering, and I always work with them ignoring whether they have a physical interpretation or not, but at the same time it would be more satisfying if I could understand the equations as manifestations of some deeper structure. Edit: Per Qiaochu's comment, yes, virtual representations can be fit into the framework of super-representations. If it is the case that virtual representations are often viewed as super-representations, then perhaps someone could elaborate on why super-representations are so natural and how one works around the dimension mismatches between virtual representations and super-representations, i.e. $dim(\rho_1\ominus\rho_2) = dim(\rho_1)-dim(\rho_2)$ but the dimension of the corresponding super-representation is $dim(\rho_1)+dim(\rho_2)$.
|
Your question is really about virtual vector spaces: what is a virtual vector space? Once you know what a virtual vector space is, then there is only a small step to the answer of your question. There are a few possible answers: 1• A virtual vector space of a pair of vector spaces. Equivalently, it's a $\mathbb Z/2$-graded vector space. You should think of the pair $(V,W)$ as being the formal difference $V-W$. This approach has the downside that it makes it unclear what an isomorphism between virtual vector spaces should be. 2• A vector space of dimension $n$ is the same thing as a point (sic!) of the topological space $BU(n)$. A virtual vector space is then a point of the space
$BU:=\mathrm{colim}_{n\to \infty} BU(n)$. This approach is not geometric at all, but works very well for talking about virtual vector bundles on a space $X$: these are continuous maps $X\to BU$. 3• Fix an infinite dimensional vector space $U$ and a polarization $U=U_-\oplus U_+$ (both $U_-$ and $U_+$ are infinite dimensional).
A virtual vector space is a (necessarily infinite dimensional) subspace $V\subset U$ such that $V\cap U_-$ is of finite codimension inside $V+U_-$. 4• Fix an infinite dimensional Hilbert space $H$. A virtual vector space is a Fredholm operator $F:H\to H$ (i.e., an operator with finite dimensional kernel and cokernel). This is related to definition 1 by assigning to the Fredholm operator $F$ the pair $(\ker(F),\mathrm{coker}(F))$. 5• A virtual vector space is an object of the bounded derived category of vector spaces: i.e., it's a chain complex. All these definitions (with the exception of 2, for which it's more involved) can be easily adapted to the context of $G$-representation, you just need to replace "infinite dimensional vector space" with "$G$-rep that contains each irrep infinitely often".
|
{
"source": [
"https://mathoverflow.net/questions/56188",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12301/"
]
}
|
56,255 |
Let $A$ be a commutative ring with a unit element. Let $M$ and $N$ be $A$-modules. Let $M^v$ and $N^v$ be the dual modules. In general, do we have $M^v \otimes N^v \cong (M\otimes N)^v$? It is definitely true when M and N are free. I believe (though haven't worked out the details) that it is true when M and N are projective. Both Atiyah & Macdonald and Lang don't say anything on the matter. I came up with this question while studying $Pic(A)$: defined as the group of isomorphism classes of projective modules of rank 1. If I and J are projective $A$-modules of rank 1, then the fact that $Pic(A)$ is a group immediately implies $I^v \otimes J^v \cong (I\otimes J)^v$, as both are the inverse of $I\otimes J$.
|
In general, we have a map $\mu:M^\vee\otimes N^\vee\to (M\otimes N)^\vee$ given by $\mu(\phi\otimes\psi)(\sum_i m_i\otimes n_i)=\sum_i\phi(m_i)\psi(n_i)$; this is presumably what mephisto is referring to, and it is an isomorphism in the case that he mentions. If the ring $A$ is a principal ideal domain and $M$ and $N$ are both finitely generated then the map is again an isomorphism, for rather uninteresting reasons, because the functor $M\mapsto M^\vee$ kills all torsion modules. Now let $M$ be the free module over $A$ with basis $\{e_n\}_{n\in\mathbb{N}}$ and take $N=M$ and define $\xi:M\otimes M\to A$ by $\xi(e_i\otimes e_i)=1$ and $\xi(e_i\otimes e_j)=0$ for $j\neq i$. I claim that this is not in the image of $\mu$. Indeed, if $\zeta=\mu(\sum_{i=1}^r\phi_i\otimes\psi_i)$ then the matrix $(\zeta(e_i\otimes e_j))_{i,j=0}^k$ has rank at most $r$ for all $k$, and it is clear that $\xi$ does not have this property. For another interesting example, take $A=k[x,y]/(xy)$ and $M=A/x$ and $N=A/y$. Multiplication by $y$ gives a map $\phi:M\to A$, so $\phi\in M^\vee$. Clearly $x\phi=0$ so multiplication by $\phi$ gives a well-defined map $M=A/x\to M^\vee$, which is an isomorphism. Similarly we have an isomorphism $N\to N^\vee$, so $M^\vee\otimes N^\vee\simeq M\otimes N=A/(x,y)=k$. On the other hand, $(M\otimes N)^\vee=\text{Hom}_A(A/(x,y),A)=0$. Thus, we have a case where $\mu:M^\vee\otimes N^\vee\to(M\otimes N)^\vee$ is not injective.
|
{
"source": [
"https://mathoverflow.net/questions/56255",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13158/"
]
}
|
56,435 |
The Freudenthal suspension theorem states in particular that the map $$
\pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1})
$$ is an isomorphism for $n\geq k+2$ . My question is: What is the intuition behind the proof of the Freudenthal suspension theorem?
|
There are two proofs I particularly like: A Morse-theoretic proof, probably due to Bott, can be found in Milnors book. Idea: consider the space of all paths on $S^n$ from the north to the south pole, which is homotopy equivalent to $\Omega s^n$. There is the energy function on this space. One does Morse theory: critical points correspond to geodesics (they are not non-degenerate, but Bott proved that a good deal of Morse theory works nevertheless). The set of absolute minima is the space of minimal geodesics connecting the poles. This is homeomorphic to $S^{n-1}$. All other critical points have index at least (rouhgly) $2n$. Therefore, the inclusion of the set of absolute minima into the whole space is $2n$-connected. A spectral sequence proof (see Kirk, Davis, Lectures on Algebraic topology). Consider the homology Leray-Serre spectral sequence of the path-loop fibration $\Omega S^n \to PS^n \to S^n$. The total space $P S^n$ is contractible. Look at the shape of the spectral sequence; you'll see that for $k \leq 2n$ (again, only a rough estimate), all differentials out of the $E_{k,0}$-slot are zero, except of the last one, which gives an isomorphism $H_k (S^n)=E_{k,0}^{2} \to E_{0,k-1}^{2} = H_{k-1} (\Omega S^n)$. It is credible (but nontrivial to prove, this uses the transgression theorem) that this isomorphism is the same as the composition $H_k (S^n) \cong H_{k-1} (S^{n-1}) \to H_{k-1} (\Omega S^n)$ of the suspension isomorphism and the natural map $S^{n-1} \to \Omega S^n$.
Thus the natural map is a homology isomorphism in a range of degrees, and by the Hurewicz theorem, this holds for homotopy groups as well.
|
{
"source": [
"https://mathoverflow.net/questions/56435",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4676/"
]
}
|
56,524 |
A metric space $(X,d)$ is isometrically homogeneous if its isometry group acts transitively on points, i.e., for every $x,y \in X$ there is an isometry $\varphi:X\to X$ with $\varphi(x) = y$. I'd like to know an example of a compact isometrically homogeneous metric space which is not a manifold (a space with finitely many points counts as a 0-dimensional manifold). Googling a bit I've discovered enough recent literature on this general subject to be sure there must be classical examples known to experts, but I haven't managed to find them written down. For example, Theorem 1.2 of this paper implies: A compact isometrically homogeneous metric space is a finite-dimensional manifold if and only if it is locally contractible. So equivalently, I'd like an example of a compact isometrically homogeneous metric space which is not locally contractible. Added: Pete and Neil both gave very nice answers. I'm accepting Neil's since, as Pete points out, it essentially contains Pete's answer as a special case.
|
Take $X=\prod_{n=0}^\infty S^1$ with $d(x,y)=\sum_n|x_n-y_n|/2^n$. Then the metric topology is the same as the product topology, which is compact by Tychonov. There is an obvious group structure by pointwise multiplication, and multiplication by any fixed element is an isometry, so the space is isometrically homogeneous. It is path-connected but not locally contractible. More generally, I guess you can take any sequence of compact isometrically homogeneous spaces $X_n$, rescale the metric so that $d(x,y)\leq 2^{-n}$ for all $x,y\in X_n$, and then take $X=\prod_nX_n$ with $d(x,y)=\sum_nd(x_n,y_n)$.
|
{
"source": [
"https://mathoverflow.net/questions/56524",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1044/"
]
}
|
56,547 |
All of us have probably been exposed to questions such as: "What are the applications of group theory..." .
This is not the subject of this MO question. Here is a little newspaper article that I found inspiring: Madam, – In response to Marc Morgan’s question, “Does mathematics have any practical value?” (November 8th), I wish to respond as follows. Apart from its direct applications to electrical circuits and machinery, electronics (including circuit design and computer hardware), computer software (including cryptography for internet transaction security, business software, anti-virus software and games), telephones, mobile phones, fax machines, radio and television broadcasting systems, antenna design, computer game consoles, hand-held devices such as iPods, architecture and construction, automobile design and fabrication, space travel, GPS systems, radar, X-ray machines, medical scanners, particle research, meteorology, satellites, all of physics and much of chemistry, the answer is probably “No”. – Yours, etc, PAUL DUNNE, The Irish Times - Wednesday, November 10, 2010 The above article article seems to provide an ideal source of solutions to a perennial problem: How to tell something interesting about math to non-mathematicians, without losing your audience? However, I am embarrassed to admit that I have no idea what kind of math gets used for antenna designs, computer game consoles, GPS systems, etc. I would like to have a list applications of math, from the point of view of the applications . To make sure that each answer is sufficiently structured and developed, I shall impose some restrictions on their format. Each answer should contain the following three parts, roughly of the same size: • Start by the description of a practical problem that any layman can understand. • Then I would appreciate to have an explanation of why it is difficult to solve without mathematical tools. • Finally, there should be a little explanation of the kind of math that gets used in the solution. ♦♦ My ultimate goal is to have a nice collection of examples of applications of mathematics, for the purpose of casual discussions with non-mathematicians. ♦♦ As usual with community-wiki questions: one answer per post.
|
Sending a man to the Moon (and back). Hilbert once remarked half-jokingly that catching a fly on the Moon would be the most important technological achievement. "Why? "Because the auxiliary technical problems which would have to be solved for such a result to be achieved imply the solution of almost all the material difficulties of mankind." (Quoted from Hilbert-Courant by Constance Reid, Springer, 1986, p. 92). The task obviously required solving plenty of scientific and technological problems. But the key breakthrough that made it all possible was Richard Arenstorf's discovery of a stable 8-shaped orbit between the Earth and the Moon. This involved the development of a numerical algorithm for solving the restricted three-body problem which is just a special non-linear second order ODE (see also my answer to the previous MO question ). Another orbit, also mapped by Arenstorf, was later used in the dramatic rescue of the Apollo 13 crew.
|
{
"source": [
"https://mathoverflow.net/questions/56547",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5690/"
]
}
|
56,563 |
(I apologize that this is a vague question). I seems to me somehow that homotopy groups behave well with respect to (Serre)-fibrations. For example you get a long exact sequence of homotopy groups from it. On the other hand cofibrations and homotopy groups seems to be no good friends at all (e.g. $S^1\to D^2\to S^2$). But then again, the situation in homology seems to be the other way round. They behave well with respect to cofibrations (you get a long exact sequence) and fibrations are harder to investigate (Serre spectral sequence etc.). My question is: What is the intuition behind this difference? (in particular with respect to the fact that homology is just homotopy of another space.)
|
I agree that the long exact sequence in homotopy groups of a fibration follows from the fact that fibrations are defined using a mapping property in which the fibration is the target. One way to understand why homology behaves well with respect to cofibrations is to spell out your remark that
"homology is just homotopy of another space". This is true, but not obvious. There are a number of constructions of ordinary homology which take the following form. One finds a functor $F$ from (pointed) spaces to (pointed) spaces which takes cofibrations to quasifibrations. (A quasifibration is something for which you have a long exact sequence of homotopy groups, for example a Serre fibration). And then $H_* (X) \cong \pi_* F(X).$ If $X\to Y \to Z$ is a cofibration (maybe a cofibration of CW complexes), then $\dots \to \pi_* F(X) \to \pi_* F(Y) \to \pi_* F(Z) \to \ldots$ is the long exact sequence in homology associated to the cofibration. Here are several contexts in which one can describe such a functor $F$ . First, a formal approach. Let $\mathbf{S}$ denote the category of spectra: it is connected to the category $\mathbf{T}$ of spaces by adjoint functors $\Sigma^\infty: \mathbf{T} \to \mathbf{S}$ and $\Omega^\infty: \mathbf{S} \to \mathbf{T}.$ There a spectrum called the "Eilenberg-MacLane" spectrum, denoted $H\mathbb{Z}$ : its job is to represent singular cohomology, and one can take $F(X) = \Omega^\infty ((\Sigma^\infty X) \wedge H\mathbb{Z})$ . Why does $F$ have the cofibration-to-quasifibration property? Well, the way that this is set up, $\Sigma^\infty$ preserves cofibrations of CW complexes, and $\Omega^\infty$ preserves fibrations of fibrant objects, and in the category of spectra every cofibration is equivalent to a fibration. To be more explicit about $H\mathbb{Z}$ , you can define $F(X) = \lim (\ldots \Omega^k(X \wedge K(\mathbb{Z},k)) \to \Omega^{k+1} (X\wedge K(\mathbb{Z},k+1))a \ldots),$ where the limit is a colimit and the maps defining the system arise from the maps $K(\mathbb{Z},k) \simeq \Omega K(\mathbb{Z},k+1)$ . Second, the Dold-Thom theorem says that one can take $F(X) = Sp^\infty (X).$ Here $Sp^n(X) = X^n/\Sigma_n$ , and $Sp^\infty(X) = \lim \ldots Sp^n(X) \to Sp^{n+1}(X) \ldots$ , again the limit is a colimit. Third, if you're willing to allow $X$ to be a simplicial set, then one can take then one can take $F(X) = \mathbb{Z}X$ , the simplicial set whose $n$ simplices are the free abelian group on the $n$ -simplices of $X$ . (This approach is due to Dan Kan; see the proceedings of the Hurewicz conference) All of this is to focus attention on functors which take cofibrations to quasifibrations. In fact all $-1$ -connected generalized homology theories (at least the ones associated to cohomology theories: are there homology theories which are not? I don't know) are of the form $E_* X = \pi_* G(X)$ , where $G$ is a functor which takes cofibrations to quasifibrations. Indeed one takes $G(X) = \Omega^\infty (\Sigma^\infty X \wedge R)$ , where $R$ is the spectrum representing the cohomology theory. This approach goes back to G. W. Whitehead. One of the more compact discussions of such a functor, which I like, is in an article by G. Segal in Springer LNM 575; he gives a construction of connective real $K$ -homology there. Really he's showing how to generalize the work of Dold and Thom: Segal's argument applies just as well to $Sp^\infty(X)$ . I apologize that throughout I have done a poor job of saying how to handle basepoints.
|
{
"source": [
"https://mathoverflow.net/questions/56563",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2625/"
]
}
|
56,569 |
The following random construction is simple enough that I am guessing it must have been studied. Fix $d \ge 3$, and let $n > d$. For each of the $n$ vertices, pick exactly $d$ other vertices to connect it to, uniformly over all ${n-1 \choose d}$ possible choices, and making this choice independently over all $n$ vertices. Some edges may get put in twice, and that's fine, but we will consider the final graph to not have multiple edges. Has this model been studied, and if so does it have a name? Motivation: Erdos-Renyi random graphs $G(n,p)$ must have average vertex degree greater than, before they become connected and have nice expansion properties, etc. But this is because they still have isolated vertices. Here we don't have the problem of isolated vertices. (I think I can show that these graphs are a.a.s. connected.) I am aware of the large literature on random $d$-regular graphs, but I thought this might be an alternative model if one does not care about regularity. I am especially wondering about spectral properties of these graphs.
|
I agree that the long exact sequence in homotopy groups of a fibration follows from the fact that fibrations are defined using a mapping property in which the fibration is the target. One way to understand why homology behaves well with respect to cofibrations is to spell out your remark that
"homology is just homotopy of another space". This is true, but not obvious. There are a number of constructions of ordinary homology which take the following form. One finds a functor $F$ from (pointed) spaces to (pointed) spaces which takes cofibrations to quasifibrations. (A quasifibration is something for which you have a long exact sequence of homotopy groups, for example a Serre fibration). And then $H_* (X) \cong \pi_* F(X).$ If $X\to Y \to Z$ is a cofibration (maybe a cofibration of CW complexes), then $\dots \to \pi_* F(X) \to \pi_* F(Y) \to \pi_* F(Z) \to \ldots$ is the long exact sequence in homology associated to the cofibration. Here are several contexts in which one can describe such a functor $F$ . First, a formal approach. Let $\mathbf{S}$ denote the category of spectra: it is connected to the category $\mathbf{T}$ of spaces by adjoint functors $\Sigma^\infty: \mathbf{T} \to \mathbf{S}$ and $\Omega^\infty: \mathbf{S} \to \mathbf{T}.$ There a spectrum called the "Eilenberg-MacLane" spectrum, denoted $H\mathbb{Z}$ : its job is to represent singular cohomology, and one can take $F(X) = \Omega^\infty ((\Sigma^\infty X) \wedge H\mathbb{Z})$ . Why does $F$ have the cofibration-to-quasifibration property? Well, the way that this is set up, $\Sigma^\infty$ preserves cofibrations of CW complexes, and $\Omega^\infty$ preserves fibrations of fibrant objects, and in the category of spectra every cofibration is equivalent to a fibration. To be more explicit about $H\mathbb{Z}$ , you can define $F(X) = \lim (\ldots \Omega^k(X \wedge K(\mathbb{Z},k)) \to \Omega^{k+1} (X\wedge K(\mathbb{Z},k+1))a \ldots),$ where the limit is a colimit and the maps defining the system arise from the maps $K(\mathbb{Z},k) \simeq \Omega K(\mathbb{Z},k+1)$ . Second, the Dold-Thom theorem says that one can take $F(X) = Sp^\infty (X).$ Here $Sp^n(X) = X^n/\Sigma_n$ , and $Sp^\infty(X) = \lim \ldots Sp^n(X) \to Sp^{n+1}(X) \ldots$ , again the limit is a colimit. Third, if you're willing to allow $X$ to be a simplicial set, then one can take then one can take $F(X) = \mathbb{Z}X$ , the simplicial set whose $n$ simplices are the free abelian group on the $n$ -simplices of $X$ . (This approach is due to Dan Kan; see the proceedings of the Hurewicz conference) All of this is to focus attention on functors which take cofibrations to quasifibrations. In fact all $-1$ -connected generalized homology theories (at least the ones associated to cohomology theories: are there homology theories which are not? I don't know) are of the form $E_* X = \pi_* G(X)$ , where $G$ is a functor which takes cofibrations to quasifibrations. Indeed one takes $G(X) = \Omega^\infty (\Sigma^\infty X \wedge R)$ , where $R$ is the spectrum representing the cohomology theory. This approach goes back to G. W. Whitehead. One of the more compact discussions of such a functor, which I like, is in an article by G. Segal in Springer LNM 575; he gives a construction of connective real $K$ -homology there. Really he's showing how to generalize the work of Dold and Thom: Segal's argument applies just as well to $Sp^\infty(X)$ . I apologize that throughout I have done a poor job of saying how to handle basepoints.
|
{
"source": [
"https://mathoverflow.net/questions/56569",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4558/"
]
}
|
56,571 |
The statement of the ordinary non-categorical version of geometric Langlands conjecture, which was proven for GL(n) in around 2002 by Frenkel, Gaitsgory and Vilonen, is quite well-known and is easy to find in the literature. Recently, by talking to some students of Dennis Gaitsgory and postdocs working in this area, I understand that a stronger categorical version of the geometric Langlands conjectures has been in circulation for at least the past few years, and with recent advances Dennis is perhaps even close to proving it (at least in type A). My question is: what is the precise statement of the categorical version of geometric Langlands? I understand on the left you have something related to the category of $D$-modules on $Bun_G(X)$. After you take the moduli stack of $G^{\vee}$-local systems, on the right you have some category in between the categories of coherent sheaves on this stack, and the category of quasi-coherent sheaves on this stack. I also hear that the categorical version uses ideas from work of Lurie ( $(\infty,1)$-categories, $DG$-categories, etc). Am I correct that on the left we have the full category of D-modules on $Bun_G(X)$? (Where-as in the simpler non-categorical version we simply request a Hecke eigensheaf for each irreducible local system).
|
For context for Tom's answer,
let me state the naive version of the conjecture, which has been around since around 1997 I think (due to Beilinson-Drinfeld). It calls for an equivalence of (dg) categories
$$D(Bun_G(X))\simeq QC(Loc_{G^\vee}(X))$$
between (quasi)coherent $D$-modules on the stack of $G$-bundles on a curve $X$, and (quasi)coherent sheaves on the (derived) stack of flat $G^\vee$ connections on the curve.
Moreover (and this is where most of the content lies) this equivalence should be an equivalence as module categories for the spherical Hecke category $Rep(G^\vee)$ acting on both sides for every choice of point $x\in X$. The action on the right is given by simple multiplication operators (tensor product with a tautological vector bundle on $Loc_{G^\vee}(X)$ attached to a given representation of $G^\vee$ and point $x$. On the right hand side the action is by convolution (Hecke) functors, associated to modifications of $G$-bundles at $x$ (relative positions at $x$ of $G$-bundles are labeled by the affine Grassmannian, and in order to formulate this statement we use the geometric Satake theorem of Lusztig, Drinfeld, Ginzburg and Mirkovic-Vilonen).
The equivalence can be further fixed uniquely by using "Whittaker normalization",
a geometric analog of the identification of L-functions in the classical Langlands story. In this form the conjecture is a theorem for $GL_1$, using the extension of the Fourier-Mukai transform for D-modules. There are no other groups for which it's known (yet - though perhaps Dennis already has $SL_2$), and few curves (one can do $P^1$ for example). Also (as Scott points out) this is only the unramified case, and there are natural conjectures to make with at least tame ramification (a "parabolic" version of the above).
There's also a close variant of this conjecture which comes naturally from S-duality for N=4 super-Yang-Mills, thanks to Kapustin-Witten. [Edit: in fact the physics suggests a far more refined version of the whole geometric Langlands program.]
It is a theorem of Beilinson-Drinfeld if we restrict on the left hand side to D-modules generated by D itself, and on the right to coherent sheaves living on opers. It is also well known that this conjecture as stated is too naive, due to bad "functional analysis" of the categories involved (precisely analogous to the analytic issues appearing eg in the Arthur-Selberg trace formula). On the D-module side, the stack $Bun_G$ is not of finite type, and one might want to make more precise "growth conditions" on D-modules along the Harder-Narasimhan strata. On the coherent side, $Loc$ is a derived stack and singular, and one ought to modify the sheaves allowed at the singular points -- in particular at the most singular point, the trivial local system. [Edit:One approach to correcting this involves the "Arthur $SL_2$" -- roughly speaking looking at local systems with an additional flat $SL_2$-action that controls the reducibility..this comes up really beautifully from the physics in work of Gaiotto-Witten, one of the first points where the physics is clearly "smarter" than the math.] These two issues are very neatly paired by the duality. If one restricts to irreducible local systems
and "cuspidal" D-modules, these issues are completely avoided (though the conjecture even on this locus remains open except for $GL_n$, where I believe it can be deduced from work of Gaitsgory following his joint work with Frenkel-Vilonen, see his ICM). In any case, Dennis has now given a precise formulation of a conjecture, which was at some point at least on his website and follows the outline Tom explained. It is very clear that homotopical algebra a la Lurie is crucial to any attempt to prove this, and Dennis and Jacob have made (AFAIK) great progress on this. The basic idea of the approach is the same as that carried out by Beilinson-Drinfeld and suggested by (old) conformal field theory (in particular Feigin-Frenkel) and (new) topological field theory (Kapustin-Witten and Lurie) -- i.e. a local-to-global argument, deducing the result from a local equivalence which comes from representation theory of loop algebras. The necessary machinery for "categorical harmonic analysis" is now available, and I'm looking forward to hearing a solution before very long.. Edit: In response to Kevin's comment I wanted to make some very informal remarks about ramification. First of all one needs to keep in mind the distinction between reciprocity (which is what the above conjecture captures) and functoriality. In the geometric setting the former is strictly stronger than the latter, while in the arithmetic setting most of the emphasis is on the latter. It is in fact quite easy to formulate a functoriality conjecture in the geometric setting with arbitrary ramification. Namely, fix some ramification and look at the category of D-modules on the stack of G-bundles with corresponding level structure. Then this is a module category over coherent sheaves on the stack of $G^\vee$ connection
with poles prescribed by the ramification (eg we can take full level structure and allow arbitrary poles). Then one can conjecture that given a map of L-groups $G^\vee\to H^\vee$ the corresponding automorphic categories are given simply by tensoring the module categories from $QC(Loc_{G^\vee}(X))$ to $QC(Loc_{H^\vee}(X))$ (everything here must be taken on the derived level to make sense). It is not hard to see this follows from any form of reciprocity you can formulate. And there is also a geometric version of the Arthur-Selberg trace formula in the ramified setting (under development). Second, one can make a reciprocity conjecture with full ramification, though you have todecide to what extent you believe it. In the "completed"/"analytic" form of geometric Langlands that comes out of physics such a conjecture in fact appears in a paper of Witten on Wild Ramification. Roughly speaking in the above reciprocity rather than just looking
at the module category structure for D-modules over QC of local systems, you can ask for them to be equivalent... not stated anywhere since maybe I'm too naive and this is known to be too far from the truth, but I think more likely people haven't thought about it very much. Third, the local story: of course in the p-adic case Kevin discusses one restricts to $GL_n$. There are two things to point out: first of all in the geometric setting one is interested in all groups, and $GL_n$ is not much simpler as far as our understanding of the local story goes. Second, while everything is much harder and deeper in the p-adic setting than the geometric setting, it's worth pointing out that a formulation of a local geometric Langlands conjecture is a much more subtle proposition, involving the representation theory of loop groups on derived categories which is only beginning to be within the reach of modern technology (even at the level of formulation of the objects!) That being said, there are rough forms of local geometric Langlands conjectures developed
by Frenkel, Gaitsgory and Lurie. It is best understood in the so-called "quantum geometric Langlands program", a deformation of the above picture involving the representation theory of quantum groups, where Gaitsgory-Lurie give a precise general local conjecture and make progress on its resolution. The usual case above is a bit degenerate and one needs to be more careful. In any case the rough form of the local conjecture is an equivalence of 2-categories (again everything has to be taken in the appropriate derived sense) between "smooth" LG-actions on categories and quasicoherent sheaves of categories over the stack of connections on the punctured disc.. --THAT being said we don't have a proof of this for $GL_n$ so again the number theorists win! just wanted to give some sense that there is a reasonable understanding of full ramification.
|
{
"source": [
"https://mathoverflow.net/questions/56571",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2623/"
]
}
|
56,579 |
This question was inspired by this one . For every $n>m>0$ consider the polynomial $p_{m,n}=x^n-x^m-1$. For which $m,n$ is $p_{m,n}$ irreducible over $\mathbb Q$? In particular, if $m$ is odd, is it always irreducible?
|
MR0124313 (23 #A1627)
Ljunggren, Wilhelm
On the irreducibility of certain trinomials and quadrinomials.
Math. Scand. 8 1960 65–70.
12.30 The author considers the irreducibility over the field of rational numbers of the polynomials $f(x)=x^n+ε_1x^m+ε_2x^p+ε_3$, where $ε_1,ε_2,ε_3$ take the values $\pm1$. He proves that if $f(x)$ has no zeros which are roots of unity, then $f(x)$ is irreducible; if $f(x)$ has exactly $q$ such zeros, then $f(x)$ can be factored into two factors with rational coefficients, one of which is of degree $q$ with all these roots of unity as zeros, while the other is irreducible (and possibly merely a constant). He also determines all possible cases where roots of unity can be zeros of $f(x)$. As a corollary he is able to give a complete treatment of the trinomial $g(x)=x^n+ε_1x^m+ε_2$, where $ε_1,ε_2$ take the values $\pm1$ . The irreducibility of this trinomial was studied by E. S. Selmer, who gave a partial solution [Math. Scand. 4 (1956), 287--302; MR0085223 (19,7f); see also #A1628]. The methods used are direct and elementary.
Reviewed by H. W. Brinkmann
|
{
"source": [
"https://mathoverflow.net/questions/56579",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
56,677 |
"Everyone knows what a curve is, until he has studied enough mathematics to become confused through the countless number of possible exceptions." Felix Klein What notions are used but not clearly defined in modern mathematics? To clarify further what is the purpose of the question following is another quote by M. Emerton: "It is worth drawing out the idea that even in contemporary mathematics there are notions which (so far) escape rigorous definition, but which nevertheless have substantial mathematical content, and allow people to make computations and draw conclusions that are otherwise out of reach." The question is about examples for such notions. The question was asked by Kakaz
|
One of the most important contemporary mathematical concepts without a rigorous definition is
quantum field theory (and related concepts, such as Feynman path integrals). Note: As noted in the comments below, there is a branch of pure mathematics --- constructive field theory --- devoted to making rigorous sense of this problem via analytic methods. I should add that there is also a lot of research devoted to understanding various aspects of field theory via (higher) categorical points of view. But (as far as I understand), there remain important and interesting computations that physicists can make using quantum field theoretic methods which can't yet be put on a rigorous mathematical basis.
|
{
"source": [
"https://mathoverflow.net/questions/56677",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3811/"
]
}
|
56,930 |
What does the classical proof of the proposition "there exists irrational numbers a, b such that $a^b$ is rational" want to reveal? I know it has something to do with the difference between classical and constructive mathematics but am not quite clear about it. Materials I found online does not give quite clear explanations either. Could someone give a better explanation?
|
Presumably, the proof you have in mind is to use $a=b=\sqrt2$ if $\sqrt2^{\sqrt2}$ is rational, and otherwise use $a=\sqrt2^{\sqrt2}$ and $b=\sqrt 2$. The non-constructivity here is that, unless you know some deeper number theory than just irrationality of $\sqrt 2$, you won't know which of the two cases in the proof actually occurs, so you won't be able to give $a$ explicitly, say by writing a decimal approximation.
|
{
"source": [
"https://mathoverflow.net/questions/56930",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13012/"
]
}
|
56,932 |
This is just a general curiosity question: In the standard textbook treatments of characteristic classes, and in particular the treatment of universal Pontrjagin classes, it's standard to consider $H^\ast(BSO,Z[1/2])$ (or $H^\ast(BSO(n),Z[1/2])$) in order to kill the 2-torsion. But I'm curious about that 2-torsion, since it should still give us some extra characteristic class-type information about real oriented bundles. If nothing else, it would give a class that's characteristic in the sense that it behaves the right way with respect to pullbacks (though I imagine it might be too much for these to be stable or have any kind of product formulas). So I suppose my questions are: What's known about such classes? Are they useful for anything? Do these interact in any interesting way with the Stiefel-Whitney classes when everything is reduced mod 2? Why are they usually ignored (or obliterated by coefficient change)?
|
The basic fact is that the 2-torsion all has order exactly 2, so it injects into the mod 2 cohomology, forming a subalgebra of the polynomial algebra on the Stiefel-Whitney classes. This subalgebra can also be described as the image of the mod 2 Bockstein homomorphism, so it is computable although the answer is not easy to state. The conclusion of all this is that there is no new information in the torsion in the integral cohomology, beyond what one gets from Stiefel-Whitney classes. These facts are stated as an exercise in Chapter 15 of Milnor and Stasheff's book. A proof can also be found in my notes on Vector Bundles and K-Theory, Theorem 3.16, available on my webpage.
|
{
"source": [
"https://mathoverflow.net/questions/56932",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6646/"
]
}
|
56,938 |
Terms like "in the natural way" or "the natural X" are used frequently in mathematical writing. While it is certainly clear most of the time what is meant, on occasion, I have been confounded. The word "natural" seems to be one of the most ambiguous terms used in formal mathematics. I have never seen anyone actually define it. People just use it and expect others to understand it. What exactly is meant by "natural" in mathematical writing?
|
Actually, there is an exact meaning, but it is not always used in that sense. For two functors $\mathsf F,\mathsf G:\mathscr A\to \mathscr B$ a natural transformation is a morphism of functors $\eta:\mathsf F\to\mathsf G$ that is compatible with the functors in the obvious (sic!) way. For instance if $\mathsf F={\rm id}$ is the identity and $\mathsf G=(\underline{\quad} )^{*}$ is the dual of finite-dimensional vector spaces, then even though $\mathsf F(V)\simeq \mathsf G(V)$, there is no natural transformation between $\mathsf F$ and $\mathsf G$ that gives this isomorphism. On the other hand $\mathsf F$ is naturally isomorphic to $\mathsf D:=\mathsf G\circ\mathsf G$ via the natural transformation induced by the usual map to the double dual. Of course, often people say "there is a natural choice of" whatever. That usually means that the "choice" actually does not involve a "choice". In other words, two different people would be expected to make the same choice. There is however a danger involved that some authors overlook. There are situations when there are more than one natural choices to make. In other words, just because two choices are both natural, one should not assume that they are necessarily the same. For instance, often one can end up with $(-1)$-times the other choice (say for a map) and then whether they are equal or their sum is zero makes a big difference. For an example, consider choosing a generator of the infinite cyclic group, a.k.a., $(\mathbb Z,+)$. One might be led to believe that "the" natural choice of a generator for $(\mathbb Z,+)$ is $1\in \mathbb Z$, but as long as this is only a group there is no way to distinguish $1\in \mathbb Z$ and $-1\in \mathbb Z$. They both generate the group and they are each other's inverses. In other words, there are two natural choices of a generator of the group $(\mathbb Z,+)$. Of course, once it is a ring, then $1\in \mathbb Z$ is the unity, while $-1\in \mathbb Z$ is not, and there is only one choice for the unity, but actually then the question of this choice being natural is moot since there is only one choice at all, so it's kind of silly to say it is natural. Addendum (to answer the question raised in the comment below by unknown ) A two dimensional vector space does not have a "natural" inner product. A two dimensional vector space with a chosen basis does: If $V$ is a vector space (over the field $k$) with basis $\mathbf v_1,\mathbf v_2\in V$ then one can define a "natural" inner product by $<\mathbf a,\mathbf b>:=a_1b_1+a_2b_2$ where $\mathbf a=a_1\mathbf v_1+a_2\mathbf v_2$ and $\mathbf b=b_1\mathbf v_1+b_2\mathbf v_2$. But this is only natural after the basis has been chosen. Basically the problem is that the definition of the inner product depends on the choice of the basis, so this definition is only natural if there is a natural choice of a basis (assuming there is no other structure present that could give a natural inner product; for more on this see the example of the wedge product below). For instance if $V$ is given with a basis as above, then there is a natural choice of a basis (called the dual basis ) for $V^*$ the dual space of $V$: Let $\phi_1: V\to k$ be defined by $\phi_1(\mathbf v_1)=1$, $\phi_1(\mathbf v_2)=0$ and extended by linearity and similarly $\phi_2: V\to k$ be defined by $\phi_2(\mathbf v_1)=0$, $\phi_2(\mathbf v_2)=1$ and extended by linearity. So, if you already have a chosen basis on $V$ you can find a natural choice of a basis on $V^*$ and with that you can find a natural choice of an inner product, but this will not be a natural choice if you consider the vector space without the given basis. Otherwise, I agree, it is hard to tell what people mean by "natural". As said by many people in various answers, the essence is whether you can do the construction without making a sort of a random choice when choosing a different element would be equally good. In this example, to give an inner product you need to give a basis and unless you have some extra structure, choosing any given basis is equal to choosing any other, so the choice is non-natural. On the other hand if there is some extra structure on the vector space then there may be a natural choice for an inner product, or more generally for a bilinear form. For instance if your vector space is a space of differential forms on a manifold, then there is no natural choice of basis, but there is a natural choice of an alternating bilinear form: the wedge product. This is actually pretty good, because this means that it is possible to define this on a manifold locally : picking a chart is a non-natural choice and defining the wedge product of two differential forms locally seems like it depends on a lot of choices, but it ends up being independent of these in the sense that choosing a different chart you get the same wedge product it just looks different because it is in a different basis. Addendum 2 (I realized that this might be an interesting comment while writing this answer to another MO question). Here is an example of the importance of naturality: Suppose $M$ is a manifold and $U\subseteq M$ and open set. Then there is a natural homomorphism from the ring of regular functions on $M$ to the ring of regular functions on $U$. (If you like adjust "regular" for your favourite category; continuous, smooth, holomorphic, etc.). It can happen that this homomorphism is an isomorphism and then it has nice consequences. However, it is often important that in order to get the nice consequence one needs that the homomorphism induced by the embedding is an isomorphism and not that there exists some isomorphism. In other words, one could say that the natural homomorphism (=the one induced by the embedding $U\subseteq M$) is an isomorphism. For an explicit example where this matters see the above mentioned answer .
|
{
"source": [
"https://mathoverflow.net/questions/56938",
"https://mathoverflow.net",
"https://mathoverflow.net/users/6137/"
]
}
|
57,025 |
I'm soon giving an introductory talk on de Rham cohomology to a wide postgraduate audience. I'm hoping to get to arrive at the idea of de Rham cohomology for a smooth manifold, building up from vector fields and one-forms on Euclidean space. However, once I've got there I'm not too sure how to convince everyone that it was worth the journey. What down-to-earth uses could one cite to prove the worth of the construct?
|
The motivation that most appeals to me is very simple and can come up in a freshman vector calculus course. We say that a vector field $F$ in $\mathbb{R}^3$ is conservative if $F = \nabla f$ for some scalar-valued function $f$. This has natural applications in physics (e.g. electric fields). It's easy to see this happens iff line integrals of $F$ are path independent, iff line integrals around closed loops vanish, etc. On a simply connected domain, $F$ is conservative iff $\nabla \times F = 0$ (use the freshman version of Stokes' theorem). On a non-simply connected domain, this may fail (e.g. $\mathbb{R}^3$ minus a line). The extent to which it fails is of course the de Rham cohomology of the domain. So this suggests that the de Rham cohomology is a good way to detect the "shape" of a domain, and one goes on from there. Maybe this is too basic to be interesting, but I like it myself.
|
{
"source": [
"https://mathoverflow.net/questions/57025",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12653/"
]
}
|
57,072 |
In an interview (at http://www.alainconnes.org/docs/Inteng.pdf ) Connes remarks that I had been working on non-standard analysis, but after a while I had found a catch in the theory.... The point is that as soon as you have a non-standard number, you get a non-measurable set. And in Choquet’s circle, having well studied the Polish school, we knew that every set you can name is measurable; so it seemed utterly doomed to failure to try to use non-standard analysis to do physics. What does he mean; what is he referring to?
|
...as soon as you have a non-standard number, you get a non-measurable set. Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$ , by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$ , the nonstandard analogue of $X$ . In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$ , which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$ , but $U$ is invariant by the operation of flipping any finite number of bits, and so must have measure $0$ or $1$ by Kolmogorov's zero-one law . (See also this article by Blackwell and Diaconis proving the same fact.) ...every set you can name is measurable. Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy , and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
|
{
"source": [
"https://mathoverflow.net/questions/57072",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10946/"
]
}
|
57,079 |
Consider $(M^{n},g)$ to be a Riemannian manifold and suppose that $X$ is a smooth non-trivial Killing vector field on $M$. Away from the zeros of $X$ we have a natural distribution $D$ of $(n-1)$-planes defined so that $D_p$ is orthogonal to $X_p$. If the distribution $D$ is (completely) integrable then it is straightforward to verify that the one form $\omega$ defined by
$$\omega(\cdot )=\frac{1}{g(X,X)} g(X, \cdot).$$
is closed (away from $\lbrace X=0\rbrace$). Moreover, the converse also holds. Examples in $\mathbb{R}^n$ with the euclidean metric include the the translations along the $x_i$-axis, $T_i$ and rotations around the $x_i$-axis, $R_i$. The Killing fields $T_i+R_i$ are non-examples. My question is whether this concept already has a name and where it might appear in the literature.
|
...as soon as you have a non-standard number, you get a non-measurable set. Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$ , by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$ , the nonstandard analogue of $X$ . In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$ , which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$ , but $U$ is invariant by the operation of flipping any finite number of bits, and so must have measure $0$ or $1$ by Kolmogorov's zero-one law . (See also this article by Blackwell and Diaconis proving the same fact.) ...every set you can name is measurable. Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy , and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
|
{
"source": [
"https://mathoverflow.net/questions/57079",
"https://mathoverflow.net",
"https://mathoverflow.net/users/26801/"
]
}
|
57,129 |
Obvious necessary condition is that the center must be a cyclic group. Is it sufficient (doubt here)? If not, is there any nice characterization in terms of group structure, without appealing to representations?
|
A finite abelian group has a faithful irreducible representation if and only if it is cyclic. The case of finite groups was solved by Gaschütz in W. Gaschütz, Endliche Gruppen mit treuen absolut-irreduziblen Darstellungen. Math. Nach. 12 (1954) From Mathematical Reviews: "In the present formulation the author calls the direct product $$ S= M_1\times M_2\times\cdots\times M_t $$ of the minimal normal subgroups $M_i$ of $G$ the `base' of $ G$ , and writes $S=A\times H$ , where $A$ is abelian and $H$ contains no normal abelian subgroup. The condition is as follows: a finite group $G$ has a faithful irreducible representation in an algebraically closed field of characteristic zero if and only if the base $S$ (or alternatively, $A$ ) of $G$ is generated by a single class of conjugates in $G$ . The proof is based on an elegant application of the exclusion principle." Maybe you also want to look at Bekka, Bachir; de la Harpe, Pierre, Irreducibly represented groups. Comment. Math. Helv. 83 (2008), no. 4, 847–868 where the case of infinite groups and unitary representations on Hilbert spaces was studied. One of the main results of the paper is the following: Theorem: A countable group $G$ is irreducibly represented, if one of the following conditions hold: $G$ is torsion-free $G$ is icc; this means the all non-trivial conjugacy classes are infinite, $G$ has a faithful primitive action on an infinite set.
|
{
"source": [
"https://mathoverflow.net/questions/57129",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4312/"
]
}
|
57,337 |
What are people's views on this? To be specific: suppose a PhD student has produced a piece of original mathematical research. Suppose that student's supervisor suggested the problem, and gave a few helpful comments, but otherwise did not contribute to the work. Should that supervisor still be named as a co-author, or would an acknowledgment suffice? I am interested in two aspects of this. Firstly the moral/etiquette aspect: do you consider it bad form for a student not to name their supervisor? Or does it depend on that supervisor's input? And secondly, the practical, career-advancing aspect: which is better, for a student to have a well-known name on his or her paper (and hence more chance of it being noticed/published), or to have a sole-authored piece of work under their belt to hopefully increase their chances of being offered a good post-doc position? [To clarify: original question asked by MrB ]
|
In my opinion, as a rule the supervisor should not be a co-author in the main paper taken from a student's thesis, even if he has contributed substantially to it, and even more so in the circumstances you suggest. The student needs to publish much more than the advisor does. If the advisor him/herself is a junior person and has given a lot of help and a very good idea to the student, then I suppose that an exception might be reasonable. Also, a thesis project might spawn more than one paper, of course, in which case it's fine if the advisor is a co-author in some of them (always assuming that he has done much more than suggesting the initial idea). Of course, it may happen that the student is weak, is given a good project, and needs to be guided step by step, so that at the end the advisor has contributed much more to the thesis than the student. Then a joint publication is in order. Such a student will most likely not pursue an academic career, so it does not really matter. [Edit] Let me try to clarify my thought, and perhaps be less radical. What I am going to say applies to pure mathematics; I am very much aware that in other fields things may be completely different. A good thesis project is one that is both interesting and feasible. Devising such a project in pure mathematics is hard; most beginning students, even very bright one, need guidance, particularly in countries, like Italy, where the PhD program is 3 years. A student has a lot to learn before getting to a level to understand and appreciate a research project; it is clear that a student in a short program does not have a lot a time for trying and failing (which is, of course, very educational, but also time-consuming). Now, some students come up with their own problems and solve them, but in my experience they are exceptions. I consider it part of my job as an advisor to suggest a problem, or an area of investigation that can be profitably mined from the student. After that, I follow the student, teaching her (let's say she's a woman, purely to avoid the "him or her") whatever I can, trying to dissuade to pursue lines of work that seem barren, uninteresting or risky to me, and also giving ideas. Sometimes she will get stuck; and then I'll think about the problem, to see if there is a difficulty that seems unsurmountable, or if there is an approach that she can try. After some time of this, if she is good she will take off on her own, and understand the problem better than I do; then I will consider that I have done my job. When she writes the paper, I will not be a co-author, even if I have obviously contributed a lot to the project. Of course, different students require very different levels of involvement; but in my experience, it is not necessary true that the best student are the ones needing less help. Also, a lot depends on the problem. Now, some people tend to give students substantial parts of their research agenda; in this case the advisor is directly interested in making progress, gets more involved, and is more likely to be a co-author. This is another case in which joint authorship is perfectly reasonable. I would not want to conclude anything about a student from the fact that have published the main paper from their thesis with their advisor.
|
{
"source": [
"https://mathoverflow.net/questions/57337",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2189/"
]
}
|
57,437 |
Question: Let $\mathcal{A}$ be an abelian category and $D^?(\mathcal{A})$ be its derived category, where ? could be empty, +, - or b (for boundedness). Is it possible to recover the homological dimension of $\mathcal{A}$ from the derived category? Here I'm using the term homological dimension in the sense of Gelfand-Manin, i.e. if for all $X,Y\in\mathcal{A}$, $\text{Ext}\_{\mathcal{A}}^i(X,Y):=\text{Hom}_{D(\mathcal{A})}(X[0],Y[i])=0$, then the homological dimension is said to be less than $i$. The homological dimension is the maximal $n$ such that there exists a non-vanishing $\text{Ext}^n(X,Y)$. Note that in the derived category one could have all kinds of non-vanishing $\text{Ext}^n(X,Y)$, as $X,Y $ can be complexes shifted arbitrarily. Is it still possible to recover this information via other method?
|
Let $V$ be a finite-dimensional vector space, $\mathcal{A}$ be the abelian category of finitely generated graded modules over the symmetric algebra $S(V)$, and $\mathcal{B}$ be the abelian category of finitely generated graded modules over the exterior algebra $\Lambda(V^*)$. Then the bounded derived categories $\mathcal{D}^b(\mathcal{A})$ and $\mathcal{D}^b(\mathcal{B})$ are naturally equivalent. This is called the Bernstein-Gelfand-Gelfand duality, a particular case of Koszul duality. On the other hand, the homological dimension of $\mathcal{A}$ is equal to $\dim V$, while the homological dimension of $\mathcal{B}$ is infinite. To obtain a similar example with unbounded derived categories, let $\mathcal{A}^+$ be the abelian category of (infinitely generated) nonnegatively graded $S(V)$-modules and $\mathcal{B}^+$ be the abelian category of nonnegatively graded $\Lambda(V^\ast)$-modules. Here it is presumed that $S(V)$ is graded so that $V$ is placed in the degree $1$, while $\Lambda(V^\ast)$ is graded so that $V^*$ is placed in the degree $-1$. Then the unbounded derived categories $\mathcal{D}(\mathcal{A}^+)$ and $\mathcal{D}(\mathcal{B}^+)$ are equivalent. UPDATE. I was asked in the comments to provide an example with both homological dimensions being finite. This can be done by yet another modification of the above examples. Pick an integer $n>\dim V$. Let $\mathcal{A}_n$ be the abelian category of finitely generated graded $S(V)$-modules concentrated in the gradings $0\leq i\leq n$. Similarly, let $\mathcal{B}_n$ be the abelian category of finitely generated graded $\Lambda(V^\ast)$-modules concentrated in the gradings $0\leq i\leq n$. Then the homological dimension of $\mathcal{A}_n$ is equal to $\dim V$, the homological dimension of $\mathcal{B}_n$ is equal to $n$, and $\mathcal{D}^b(\mathcal{A}_n)\simeq\mathcal{D}^b(\mathcal{B}_n)$. A similar example with unbounded derived categories can be obtained by removing the finitely generatedness assumption. All of the above counterexamples presume that $\dim V>0$. The only positive result in the direction of the original question that I can think of at the moment is that if the derived categories of $\mathcal{A}$ and $\mathcal{B}$ are equivalent, and $\mathcal{A}$ has homological dimension $0$, then so does $\mathcal{B}$. Indeed, $\mathcal{A}$ is a semisimple abelian category if and only if $\mathcal{D}^b(\mathcal{A})$ and $\mathcal{D}(\mathcal{A})$ are.
|
{
"source": [
"https://mathoverflow.net/questions/57437",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1657/"
]
}
|
57,465 |
The question is the extent to which we can unify addition
and multiplication, realizing them as terms in a single
underlying binary operation. I have a number of questions. Is there a binary operation $n\star m$ on the integers
$\mathbb{Z}$ such that both addition $+$ and multiplication
$\cdot$ can be expressed as specific composition
expressions using only $\star$? A more relaxed version of
the question would allow constants into the expressions;
for example, perhaps we can arrange that $a+b=0\star(a\star
b)$ and $ab=1\star(a\star b)$. One obvious idea is to try somehow to use a pairing
function ,
so that $a\star b$ codes up both $a$ and $b$ into one
number, which then can appear as one argument to $\star$,
whose other argument will signal whether or not addition or
multiplication is desired. But there is the difficulty of
making these two tasks not interfere with one another. Note that if we allow a trinary operation, then we can
easily do it simply by defining $\star(0,a,b)=a+b$ and
$\star(1,a,b)=ab$ and extending this arbitrarily. Can we
get rid of the need for parameters here? Can we prove that there is no associative such binary
operation $\star$ on the integers, from which both $+$ and
$\cdot$ are expressible by terms? Does every ring have such a binary operation $\star$
from which both addition $+$ and multiplication $\cdot$ are
expressible as $\star$-terms? Does it matter if the ring is
finite or infinite? Can every countable family $F$ of finitary operations on
an infinite set $Z$ be unified by a single binary operation
$\star$ on $Z$, so that every function in $F$ is the same
function as that induced by some $\star$-term? This question arose from what I found interesting about a recent question on
math.SE ,
asked by someone who was interested in the phenomenon that
logical and and or are expressible from nand and also
from nor .
|
The answer to Q4 is yes. Let $X$ be any infinite set. Wlog $X= Z\times\mathbb N$, where
there is a bijection $i:X\to Z\times \lbrace0\rbrace $. For $x=(z,n)$ write $x+1$ for $(z,n+1)$. [Typo corrected.] You are given countably many finitary functions $g_1, g_2, \ldots$. We may assume there is a pairing function $x*y$ among them, so we may as well assume that all of them are binary. (Due to Sierpinski, I think. E.g., $g(x,y,z) = h(x*(y*z)) $ for some unary $h$.) Now there is a binary function $f$ satisfying the following for all $x,y\in X$: $f(x,x) = x+1$. $f(x, x+1) = i(x)$. $f(i(x)+k,i(y)) = g_k(x,y)$ for $k=1,2,\ldots$. Clearly $f$ generates the functions $x+1$, $i(x)$, and $g_k$ for all $k$.
|
{
"source": [
"https://mathoverflow.net/questions/57465",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
57,515 |
I was trying to understand completely the post of Terrence Tao on Ax-Grothendieck theorem . This is very cute. Using finite fields you prove that every injective polynomial map $\mathbb C^n\to \mathbb C^n$ is bijective. It seems to me that the only point in the proof presented in the post that is not explained completely is the following lemma: Take any finitely generated ring over $\mathbb Z$ and quotient it by a maximal ideal. Then the quotient is a finite field. Is there some comprehensible reference for the proof of this lemma? In slightly different wording, the question is the following: assuming Nullstellensatz, can one really give a complete proof of Ax-Grothendick theorem in two pages, so that it can be completely explained in one (2 hours) lecture of an undergraduate course on algebraic geometry?
|
To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \mathrm{Spec}(R)\to\mathrm{Spec}(\mathbb Z)$ is of finite type, a theorem of Chevalley says that the image of any constructible subset is constructible. So the image of any closed point by $f$ is a point which is a constructible subset. This can not be the generic point of $\mathrm{Spec}(\mathbb Z)$, so it must be a closed point. Note that this does not hold in general. For example, over the ring of $p$-adic integers, the ideal $(pX-1)\mathbb{Z}_p[X]$ is maximal, but its preimage in $\mathbb{Z}_p$ is $0$ and it not maximal. [ EDIT ] Another proof using Noether's normalization lemma: Noether's normalization lemma over a ring A : if a maximal ideal $\mathfrak m$ of $R$ is such that $\mathfrak m\cap \mathbb Z=0$, then $R/\mathfrak m$ is finite type over (and contains) $\mathbb Z$. So there exits $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z_f[X_1,\dots, X_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z_f[X_1,\dots, X_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers.
|
{
"source": [
"https://mathoverflow.net/questions/57515",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13441/"
]
}
|
57,520 |
It seems to me, that a typical science often has simple and important examples whose formulation can be understood (or at least there are some outcomes that can be understood). So if we consider mirror symmetry as science, what are some examples there, that can be understood? I would like to explain a bit this question. If we consider the article
"Meet homological mirror symmetry" http://arxiv.org/abs/0801.2014 it turns out, that in order to understand something we need to know huge amount of material, including $A_{\infty}$ algebras, Floer cohomology, ect. Here, on the contrary, is an example, that "can be understood" (for my taste): According to Arnold, the first instance of symplectic geometry was "last geometric theorem of Poincare". This is the following statement:
Let $F: C\to C$ be any area-preserving self map of a cylinder $A$ to itself, that rotates the boundaries of $A$ in opposite directions. Then the map has at least two fixed points.
(this was proven by Birkhoff http://en.wikipedia.org/wiki/George_David_Birkhoff ) So, I would like to ask if there are some phenomena related to mirror symmetry that can be formulated in simple words. Added. I would like to thank everyone for the given answers! I decided to give a bit of bounty for this question, to encourage people share phenomena related to mirror symmetry that can be simply formulated (or at least look exciting). Since there are lot of people in this area I am sure there must be more examples.
|
Here is my biased view of a simple example: the two-torus.
Everything I know about homological mirror symmetry
stems from this example. Because the example is one-dimensional, a symplectic form
is just an area form, and Lagrangians are simply
curves, and the holomorphic maps which are part of the Fukaya
category are simply topological disks. (By uniformization of
Riemann surfaces, there is one holomorphic map for each topological
disk satisfying the appropriate boundary conditions.)
Even better, you can go to the universal cover, which is $R^2,$
and just draw Lagrangians as straight lines with rational
slope. The holomorphic disks which determine
compositions in the category are simply triangles. On the mirror side, we're talking about a complex two-torus, or
elliptic curve. A typical object would be a line bundle on
the elliptic curve, such as the theta line bundle, whose
sections are theta functions, once we lift them up to the
complex plane. The two-torus is circle-fibered over a base circle,
and the elliptic curve is circle-fibered by the dual circle
(i.e., $U(1)$ local systems on the original circle).
This is called T-duality, and it explains how to construct the
mirror equivalence going from Lagrangians to line
bundles, or vice versa. For example, the Lagrangian $\{y=0\}$
represents a family of trivial $U(1)$ local systems, corresponding to
the trivial holomorphic line bundle whose sections are just holomorphic
functions. The Lagrangian $\{y=nx\}$ corresponds to a line bundle
of degree $n$. After making these definitions, one checks
that compositions match up.
|
{
"source": [
"https://mathoverflow.net/questions/57520",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13441/"
]
}
|
57,543 |
This question is related to this previous question where I asked about ordinary Fourier coefficients. ##Special case: is Möbius nearly orthogonal to Morse ! Harold Calvin Marston Morse (24 March 1892 – 22 June 1977),
August Ferdinand Möbius (November 17, 1790 – September 26, 1868) Consider the sequence of values of the Möbius functions on nonnegative integers. (Starting with 0 for 0.) 0, 1, −1, −1, 0, −1, 1, −1, 0, 0, 1, −1, 0, −1, 1, 1, 0, −1, 0, −1, 0, 1, 1, −1, 0, 0, ... And the Morse sequence 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1 Are these two sequences nearly orthogonal? Remark: This case of the general problem follows from the solution of Mauduit and Rivat of a 1968 conjecture of Gelfond. They show that primes are equally likely to have odd or even digit sum in base 2. See Ben Green's remark below.) The Problems Start with the Möbius function $\mu (m)$ . (Thus $\mu(m)=0$ unless all prime factors of $m$ appear once and $\mu (m)=(-1)^r$ if $m$ has $r$ distinct prime factors.) Now, for a n-digit positive number $m$ regard the Mobius function as a Boolean function $\mu(x_1,x_2,\dots,x_n)$ where $x_1,x_2,\dots,x_n$ are the binary digits of $m$ . For example $\mu (0,1,0,1)=\mu(2+8)=\mu(10)=1$ .
We write $\Omega_n$ as the set of 0-1 vectors $x=(x_1,x_2,x_m)$ of length $n$ . We also write $[n]=\{1,2,\dots,n\}$ , and $N=2^n$ . Next consider for some natural number $n$ the Walsh-Fourier transform $$\hat \mu (S)= \frac{1}{2^n} \sum _{x\in \Omega_n} \mu(x_1,x_2,\dots,x_n)(-1)^{\sum\{x_i:i\in S\}}.$$ So $\sum_{S \subset [n]}|\hat \mu (S)|^2$ is roughly $6/\pi ^2$ ; and the prime number theorem asserts that $\hat \mu(\emptyset)=o(1)$ ; In fact the known strong form of the prime number theorem asserts that $$|\hat \mu (\emptyset )| \lt n^{-A} =(\log N)^{-A},$$ for every $A>0$ . (Note that $|\hat \mu (\emptyset)=\sum_{k=0}^{N-1}\mu(k)$ .) My questions are: Is it true that the individual coefficients tend to 0? Is it known even that $|\hat \mu (S)| \le n^{-A}$ for every $A>0$ . Solved in the positive by Jean Bourgain (April 12, 2011) : Moebius-Walsh correlation bounds and an estimate of Mauduit and Rivat ; (Dec, 2011) For even stronger results see Bourgain's paper On the Fourier-Walsh Spectrum on the Moebius Function . Is it the case that $$(*) \sum \{ \hat \mu ^2(S)~:~|S|<(\log n)^A \} =o(1), $$ for every $A>0$ . (This does not seem to follow from bounds we can expect unconditionally on individual coefficients.) Solved in the positive by Ben Green (March 12, 2011) : On (not) computing the Mobius function using bounded depth circuits . (See Green's answer below.) The Riemann Hypothesis asserts that $$|\hat \mu (\emptyset )| < N^{-1/2+\epsilon}.$$ Does it follows from the GRH that for some $c>0$ , $$| \hat \mu (S)| < N^{-c},$$ for every $S$ ? An upper bound of $(\log N)^{-{\log \log N}^A}$ suffices to get the desired application. ##The motivation The motivation for these questions from a certain computational complexity extension of the prime number theorem. It asserts that every function on the positive integers that can be represented by bounded depth Boolean circuit in terms of the binary expansion has diminishing correlation with the Mobius function. This conjecture that we can refer to as the $AC^0$ - prime number conjecture is discussed here, on my blog and here, on Dick Lipton's blog . The conjecture follows from formula (***) by a result of Linial Mansour and Nisan on Walsh-Fourier coefficients of $AC^0$ functions. Question 3 suggests that perhaps we can deduce the $AC^0$ prime number conjecture from the GRH which would be of interest. Of course, it will be best to prove it unconditionally. ( Ben Green proved it unconditionally ). For polynomial size formulas , namely for functions expressible by depth-2 polynomial size circuits we may need even less. A result of Mansour shows that the inequality $|\hat \mu (S)| \le n^{-(\log \log n)^A}$ for every $A>0$ , would suffice! Moreover, a conjecture of Mansour (which also follows from a more general conjecture called the influence/entropy conjecture, see this blog post for a description of both conjectures) implies that it will be enough to prove that $$|\hat \mu (S)| \le n^{-A}$$ for every $A>0$ , to deduce the PNT for formulas.) Some background Let me mention that the question follows to a large extent a line of research associating $AC^0$ formulas with number theoretic questions. See the papers by Anna Bernasconi and Igor Shparlinski and the paper by Eric Allender Mike Saks Igor Shparlinski, and the paper Complexity of some arithmetic problems for binary polynomials by Eric Allender, Anna Bernasconi, Carsten Damm, Joachim von zur Gathen, Michael Saks, and Igor Shparlinski. Related MO question: Odd-bit primes ratio
|
An update to my earlier answer. I've written a proof of this "AC0 prime number conjecture" as a short paper, which can be found here. https://arxiv.org/abs/1103.4991 I thought a bit about establishing a nontrivial bound on the Fourier-Walsh coefficients $\hat{\mu}(S)$ for all sets $S$ . My paper does this when $|S| < cn^{1/2}/\log n$ (here $S \subseteq \{1,\dots,n\}$ ). On the GRH it works for $|S| = O(n/\log n)$ . I remarked before that the extreme case $S = \{1,\dots,n\}$ follows from work of Mauduit and Rivat. I still believe that there is hope of proving such a bound in general, but this does seem to be pretty tough. At the very least one has to combine the work of Mauduit and Rivat with the material in my note above, and neither of these (especially the former) is that easy.
|
{
"source": [
"https://mathoverflow.net/questions/57543",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1532/"
]
}
|
57,548 |
The second stage of elliptic curve factorization has the drawback of large memory usage. Let $n=pq$, $E(\mathbb{Z}/n\mathbb{Z})$ is elliptic curve and $P$ point on $E(\mathbb{Z}/n\mathbb{Z})$. On $E(\mathbb{Z}/p\mathbb{Z})$ the order of $P$ is small $r$. Set $Q=kP$ for pseudo-random $k$. On $E(\mathbb{Z}/p\mathbb{Z})$ one can solve the discrete logarithm $Q=xP$ in time $O(\sqrt{r})$ and constant memory using Pollard's rho algorithm (more precisely one can find $aP=bQ$ if $r$ is unknown) basically by doing random walks and exploiting the birthday paradox. The question is: Working on $E(\mathbb{Z}/n\mathbb{Z})$ ($p,q$ are unknown) can one solve $Q=xP$ (or $aP=bQ$) on $E(\mathbb{Z}/p\mathbb{Z})$ using the rho algorithm in $O(\sqrt{r})$ and constant memory: note that $r$ can be significantly smaller than the order of $P$ on $E(\mathbb{Z}/n\mathbb{Z})$. The only significant choice appears to be the random walk. I failed to do this yet discrete logarithms on $E(\mathbb{Z}/n\mathbb{Z})$ appear to work as expected. Update : I suppose part of the problem with constant memory is that it may happen $Q_i \ne Q_j \mod n$ while $Q_i = Q_j \mod p$ If $E(\mathbb{Z}/n\mathbb{Z})$ were a ring, one could simply iterate (A) $Q_{i+1}=Q_i^2+c$. Would it be possible instead on an elliptic curve to work in some ring where (A) would be trivially stage 2 and stage 1 would be $Q_1 = k P \ k \in \mathbb{N}$?
|
An update to my earlier answer. I've written a proof of this "AC0 prime number conjecture" as a short paper, which can be found here. https://arxiv.org/abs/1103.4991 I thought a bit about establishing a nontrivial bound on the Fourier-Walsh coefficients $\hat{\mu}(S)$ for all sets $S$ . My paper does this when $|S| < cn^{1/2}/\log n$ (here $S \subseteq \{1,\dots,n\}$ ). On the GRH it works for $|S| = O(n/\log n)$ . I remarked before that the extreme case $S = \{1,\dots,n\}$ follows from work of Mauduit and Rivat. I still believe that there is hope of proving such a bound in general, but this does seem to be pretty tough. At the very least one has to combine the work of Mauduit and Rivat with the material in my note above, and neither of these (especially the former) is that easy.
|
{
"source": [
"https://mathoverflow.net/questions/57548",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11847/"
]
}
|
57,597 |
Every countable order type, such as the countable ordinals, $\mathbb Z$, etc. can be embedded in $\mathbb Q$, so it is universal for countable order types. Is there a universal space for all linear orders of cardinality continuum? Or more generally, a universal space for all linear orders of any given cardinality?
|
You are looking for the concept of saturated model . A model $M$ is $\kappa$-saturated if any type consisting of fewer than $\kappa$ many assertions that is consistent with the elementary diagram of $M$ is realized in $M$. One can easily build saturated models simply by realizing more and more types. Any linear order that is $\kappa$-saturated will be universal for all linear orders of size $\kappa$, simply by a length $\kappa$ analogue of the forth part of Cantor's back-and-forth construction, which is how he proved the countable universal case you mentioned. That is, the way that you prove that $\mathbb{Q}$ is universal for countable linear orders, is that if we have a countable linear order $\langle L,\leq\rangle$, then we enumerate the elements of $L$ as $a_0,a_1,a_2,\ldots$ and having assigned an order-preserving partial isomorphism on an initial segment of this order $a_i\mapsto q_i$, for $i\lt n$, we look at how $a_n$ sits with respect to the $a_i$ for $i\lt n$ and choose an image $q_n$ that sits the same way with respect to $q_i$ for $i\lt n$. That is to say, we pick a $q_n$ that realizes the same type over the earlier $q_i$ that $a_n$ realizes over the earlier $a_i$. In other words, the point is that $\mathbb{Q}$ is $\omega$-saturated, which amounts to the trivial fact that if we have finitely many rational numbers, then any type describing how an unknown point $x$ might relate to those finitely many points of course just says that $x$ is between two of them or on top or on the bottom and hence is realized in $\mathbb{Q}$, since this is an endless dense linear order. This idea generalizes to transfinite constructions, provided that we can be guarranteed that the type will be realized. If the target model is $\kappa$-saturated, then any linear order of cardinality $\kappa$ can be enumerated $a_\alpha$ for $\alpha\lt\kappa$, and at stage $\beta$ we want to pick an image $q_\beta$ for $a_\beta$ that realizes the same type over the earlier $q_\alpha$ that $a_\beta$ realizes over the earlier $a_\alpha$. If the target is $\kappa$-saturated, then this type will be realized in $M$, and so we will have our desired target $q_\beta$ and the construction will proceed. Apart from the abstract saturated concept, however, one can construct universal orders directly in a variety of ways. First, of course, there are only $2^\kappa$ many different order types for an order of size $\kappa$, and by simply placing them one after the other, one gets a linear order that is clearly universal. But the size of that order may be wastefully high. If CH holds, then one can make a linear order of size $\aleph_1$ that is saturated for countable types. Given any other linear order of size $\aleph_1$, we may enumerate it in an $\omega_1$-sequence, and build an embedding by the "forth" construction as above, since the image of the next point will simply be any point in our saturated order that realizes the corresponding type. If CH fails, however, then we cannot necessarily get a saturated model of size $\aleph_1$, but we can still build a $c$-saturated model of larger size, and this is enough to carry out the "forth" construction. The Surreals. Lastly, let me say that the construction of the surreal numbers can be viewed as a construction focused relentlessly on building a universal linear order. That is, one way to construct the surreal numbers is to proceed in a transfinite recursion, which at each stage $\alpha$ we have the linear order constructed so far, and the next linear order simply fills all cuts in this order. So one starts with nothing, and then fills the empty cut, to get one point. Then, one fills the lower cut and the upper cut, which in effect adds a point below and a point above. Now, one adds points at the bottom, top and in between all the current points, and so on. At stage $\omega$, one can therefore constructed a countable dense linear order. At the next stage, all the Dedekind cuts are filled in, and also a lower point and an upper point are added. At the next stage, we start to get infinitesimals, by filling the cut between $0$ and all the positive numbers, and the same above and below any point. At each step of the construction, by filling a cut one is realizing the type that describes what it means to be the thing that fills that cut. It follows that the class of surreal numbers is universal for all set-sized linear orders. The restriction of the surreals to the numbers born before any stage $\beta$ will be universal for linear orders of size $\beta$, since each next point in the original order fills some cut in that order, and so one can map it to the correspondingly newly-born surreal number filling the corresponding cut in the surreals.
|
{
"source": [
"https://mathoverflow.net/questions/57597",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4903/"
]
}
|
57,627 |
Not sure if this is a "good" question for this forum or if it'll get panned, but here goes anyway... Consider this problem. I've been trying to find a formula to expand the "regular iteration" of "exp". Regular iteration is a special kind of complex function that is a solution of the equation $$f(z+1) = \exp(f(z))$$ (or more generally for functions other than $\exp$. It is called "regular" because as a solution it is characterized by the fact the the functional iterates $F^t(z) = f(t + f^{-1}(z))$, with $F$ being the function that is $\exp$ in this case, are "regular", or analytic, at a chosen fixed point of $F$, for all non-integer $t$. There are regular iterations for every fixed point.) This regular iteration in particular is an entire function. To get it, we take a fixed point $L$ of $\exp$ and expand a solution in powers of $L^z$. The result is to obtain a Fourier series $$f(z) = \sum_{n=0}^{\infty} a_n L^{nz}$$ where $$a_0 = L$$
$$a_1 = 1$$
$$a_n = \frac{B_n(1! a_1, 2! a_2, ..., (n-1)! a_{n-1}, 0)}{n!(L^{n-1} - 1)}$$ with $B_n$ being the nth "complete" Bell polynomial. This recursive formula yields the following expansions: $$a_2 = \frac{1}{2L - 2}$$
$$a_3 = \frac{L + 2}{6L^3 - 6L^2 - 6L + 6}$$
$$a_4 = \frac{L^3 + 5L^2 + 6L + 6}{24L^6 - 24L^5 - 24L^4 + 24L^2 + 24L - 24}$$
$$a_5 = \frac{L^6 + 9L^5 + 24L^4 + 40L^3 + 46L^2 + 36L + 24}{120L^{10} - 120L^9 - 120L^8 + 240L^5 - 120L^2 - 120L + 120}$$
... It appears that, by pattern recognition and factoring the denominators, $$a_n = \frac{\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} mag_{n,j} L^j}{\prod_{j=2}^{n} j(L^{j-1} - 1)}$$ where $\mathrm{mag}_{n,j}$ is a sequence of "magic" numbers (integers) that looks like this (with the leftmost column being $j = 0$): [update]: remark: for readability I transposed the original table. But I did not adapt the use of "columns" and "rows" in the surrounding text, so that must be translated in mind (Gottfried Helms) n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 ...
--------------------------------------------------------------
1 1 2 6 24 120 720 5040 40320 362880 ...
1 6 36 240 1800 15120 141120 1451520
5 46 390 3480 33600 352800 4021920
1 40 480 5250 58800 695520 8769600
24 514 7028 91014 1204056 16664760
9 416 8056 124250 1855728 28264320
1 301 8252 155994 2640832 44216040
160 7426 177220 3473156 64324680
64 5979 186810 4277156 88189476
14 4208 181076 4942428 114342744
1 2542 163149 5395818 141184014
1295 134665 5561296 166279080
504 102745 5433412 187614312
139 71070 5021790 202901634
20 44605 4391304 210825718
1 24550 3625896 210403826
11712 2820686 201934358
4543 2056845 186191430
1344 1398299 164980407
265 879339 140216446
27 504762 114231817
1 260613 88934355
117748 66047166
45178 46576620
13845 31071602
3156 19460271
461 11365652
35 6112650
1 2987358
1298181
488878
153094
37692
6705
749
44
1 But what is the simplest (or at least "reasonably" simple) non-recursive formula for these numbers, or perhaps the numerators in general? Like a sum formula, or something like that. Is there some kind of "combinatorical"-like formula here (sums/products, perhaps nested, of factorials and powers and stuff like that, binomial coefficients, special numbers, etc.)? I notice that the first column is factorials... (how can one prove that?) And regardless of the formula for the "mag", can one prove from the recurrence formula that the $a_n$ have the form given, and if so, how? Especially, how can one prove the numerator has degree $\frac{(n-1)(n-2)}{2}$? Perhaps that might provide insight into how to find the formula for the "mag". The ultimate goal here is to try and obtain a series expansion for the "tetration" function $^z e$, more specifically, Kneser's tetrational function, described in Kneser's papers on solutions of $f(f(x)) = \exp(x)$ and related equations (paper is in German, I only saw the translations.). Though this may not be the best way to go, since after constructing this regular iteration function, we then need a special mapping derived from a Riemann mapping to "distort" it so it becomes real-valued at the real axis, and I don't know if there's any good way to construct Riemann mappings even as "non-closed" infinite expansions. But I'm still curious to see if at least a formula for this function is possible. EDIT: Oh, and for all its worth, apparently $$\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} = \frac{n!(n-1)!}{2^{n-1}}$$ if that helps any (don't see how it would, and this is not proven, I just got it by looking up the sums on the integer sequences dictionary site.). Perhaps maybe some hints as to why it has that value could help in finding the formula, though... Justification for thinking a formula exists Why do I think this even exists, when there's no guarantee that this kind of really non-trivial recurrence relation should even have a non-recursive solution in the first place? Well, for one, the fact that so much of it could be put in simple form as given, and also I did manage to come up with an explicit formula from a very roundabout way but this formula is excessively complicated and based on very general techniques. It is difficult to describe that formula here, but the outline of the process to construct it is this, for all its worth: A general recurrence of the form $$A_1 = r_{1, 1}$$
$$A_n = \sum_{m=1}^{n-1} r_{n,m} A_m$$ has a non-recursive solution formula. This I found myself, but it is hideous and involves binary bit operations. This kind of recurrence is very general, and it also includes the recurrence for the Bernoulli numbers and other kinds of recurrences. The "regular Schroder function" of $F(z) = e^{uz} - 1$, i.e. the function satisfying $\mathrm{RSF}(F(z)) = K \mathrm{RSF}(z)$ (sometimes called the Schroder functional equation, hence the name) which is "regular" in that it can be turned into the regular iteration of $F$ (as we do next), can be given as a Taylor series $$\mathrm{RSF}(z) = \sum_{n=1}^{\infty} A_n z^n$$ where $A_n$ is given by the recurrence-solving formula with $r_{1,1} = 1$ and $r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} S(n, m)$ (here, $S(n, m)$ is a Stirling number of the 2nd kind). This is hideous, involving lots of "binary bit manipulation" stuff such as counting 1 bits and positions of 1 bits, which have not-so-nice formulas (the latter involves a set indicator function, at least in the formulation I found myself...). Not sure at all how this could be simplified. The formulas just don't seem to lend themselves to simplification, at least not any that I know of. Invert the regular Schroder function using the Lagrange inversion theorem. This can be expanded in an explicit "non-recursive" form, but it needs so-called "potential polynomials" and other complexity. Plug the huge $A_n$ formula into this. Horrific! Now $U(z) = \mathrm{RSF}^{-1}(u^z)$ is a "regular iteration" of $e^{uz} - 1$, giveable as a Fourier series, or Taylor series in $u^z$. Apply the topological conjugation to conjugate it to iteration of $e^{vz}$ by taking $v = ue^{-u}$ thus $u = -W(-v)$ (Lambert's W-function). Take $H(z) = e^{-u} z - 1$ then find $H^{-1} o U o H$. This gives a regular iteration of $e^{vz}$, thus set $v = 1$ ($u = -W(-1) = \mathrm{fixed\ point\ of\ exponential}$). Though, there may be a constant displacement of some kind offsetting this regular from the one given by the $a_n$-formula.
EDIT: Oops!!!! That should be $H^{-1}(U(U^{-1}(H(U(0))) + z))$, but wait, that's just a constant-shift of $H^{-1} o U$, so just take $H^{-1} o U$ as the regular iteration of $e^{vz}$, probably displaced (in $z$) from the one we're trying to solve for by a constant, but should be structurally identical (and you can try and compute $U^{-1}(H(U(0)))$. Perhaps that is the shift required, but I don't know.). (EDIT: Apparently the step-numbering above isn't working right for some reason.) So by this, I think an explicit formula exists (though that constant-shift at the end may be a little problem, but not much, since it is immaterial to the structure of the function). I'm just interested in something simpler than this, preferably something to "fill out" the "mag" formula I gave... EDIT: Now I'm pratically sure explicit non-recursive solution is possible. Using some numerical tests, I figured the constant shift should be (for $v = 1$, i.e. $u = L$) simply -1, that is, take $H^{-1}(U(z - 1))$ and the coefficients of the Fourier expansion will be equal to $a_n$ in explicit non-recursive form (but atrocious , hence my question, to find something more elegant . This at least evidences that an explicit non-recursive solution is possible , addressing any skeptics' concerns that it isn't and so an elegant one wouldn't exist either. And it is a good bet that if an atrocious formula exists derived from very general principles (note Step 1 above), there may be a more elegant one derived from more specific principles.). So, almost a proof. It could probably be turned into one with a little more work, though that would be much too long to post here.
|
Let $\beta_n$ denote the flag $h$-vector (as defined in EC1, Section
3.13) of the partition lattice $\Pi_n$ (EC1, Example 3.10.4). Then
$$ \mathrm{mag}_{n,{n-1\choose 2}-j} = \sum_S \beta_n(S), $$
where $S$ ranges over all subsets of $\lbrace 1,2,\dots,n-2\rbrace$ whose
elements sum to $j$. An explicit formula for $\beta_n(S)$ is given by
$$ \beta_n(S) = \sum_{T\subseteq S} (-1)^{|S-T|}
\alpha_n(T), $$
where if the elements of $T$ are $t_1<\cdots < t_k$, then
$$ \alpha_n(T) = S(n,n-t_1)S(n-t_1,n-t_2)
S(n-t_2,n-t_3)\cdots S(n-t_{k-1},n-t_k). $$
Here $S(m,j)$ denotes a Stirling number of the second kind. Addendum. A combinatorial description of the mag numbers is
somewhat complicated. Consider all ways to start with the $n$ sets
$\lbrace 1 \rbrace,\dots, \lbrace n \rbrace$. At each step we take two
of our sets and replace them by their union. After $n-1$ steps we will
have the single set $\lbrace 1,2,\dots,n \rbrace$. An example for
$n=6$ is (writing a set like $\lbrace 2,3,5\rbrace$ as 235)
1-2-3-4-5-6, 1-2-36-4-5, 14-36-2-5, 14-356-2, 14356-2, 123456. At the
$i$th step suppose we take the union of two sets $S$ and $T$. Let
$a_i$ be the least integer $j$ such that $j$ belongs to one of the
sets $S$ or $T$, and some number smaller than $j$ belongs to the other
set. For the example above we get $(a_1,\dots,a_5)=(6,4,5,3,2)$. If
$\nu$ denotes this merging process, then let $f(\nu) = \sum i$,
summed over all $i$ for which $a_i>a_{i+1}$. For the above example,
$f(\nu) = 1+3+4=8$. (The number $f(\nu)$ is called the major index of the sequence $(a_1,\dots,a_{n-1})$.) Then
$\mathrm{mag}_{n,{n-1\choose 2}-j}$ is the number
of merging processes $\nu$ for which $f(\nu)=j$. This might look
completely contrived to the uninitiated, but it is very natural within
the theory of flag $h$-vectors.
|
{
"source": [
"https://mathoverflow.net/questions/57627",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11576/"
]
}
|
57,657 |
A short version of my question is: Is there a $p$-adic theory of integration? Now let me expand a little further. In introductory texts such as Koblitz' book $p$-adic numbers,.. a bunch of $p$-adic analysis is developed. However, since all applications are towards number theory, the exposition stops at some point. In particular, there is no theory of integration developed for $p$-adic numbers.
By this I do not mean putting a measure on $\mathbb{C}_p$ and integrating real or complex valued functions, but instead putting a "$p$-adic measure"(whatever this may be) on it and integrating $\mathbb{C}_p$-valued functions on it. To rephrase my question: Is there are an integration theroy for $\mathbb{C}_p$-valued functions on $\mathbb{C}_p$. In particular I would like to know if an analogue of Cauchy's theorem holds. Where can I read more about such a theory?
|
There is an important difference, relevant to the original question, between the two kinds of $p$-adic integrals mentioned by Kevin in his comments. Because I see frequent confusion on this issue, I thought I'd comment. The 'usual' $p$-adic integrals as you might see in, say, Tate's thesis on L-functions or the adelic theory of automorphic forms, are volume integrals, with respect to a measure, typically on some group. This kind of volume integral can also be easily defined on arbitary varieties, and you will see plenty in Weil's book on Tamagawa numbers, or in papers on motivic integration. Coleman integration, on the other hand, is a $p$-adic analogue of line integrals , and comes up most naturally in discussing the holonomy of vector bundles with connection on a variety over a $p$-adic field (often interpreted as isocrytals). These, therefore, should be the right quantities to relate to a Cauchy formula. However, unfortunately (and fortunately), it doesn't work. The reason is that Coleman integration is a line integral along a canonical path between two points on a variety over the $p$-adics. So there is a canonical holonomy in the theory, at least if you just want to compute it for a bundle with unipotent connection, that is, one that has a strictly upper-triangular connection form. This is where a mysterious 'crystalline' structure on the space of paths is used, whereby there is a unique path invariant under the action of the Frobenius. The notion of a path, by the way, uses the Tannakian formalism in this context. For a very quick overview of this approach, you can look at section 2 of this paper: http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf Breuil's paper linked from Chandan's answer should provide a more systematic overview. Anyways, because of the canonical paths in Coleman's theory, there can be no holonomy around a loop, and hence, no Cauchy formula. I was told quite a few years ago by Berkovich that he has a theory of line integrals on Berkovich spaces that are path dependent in interesting ways, but I've never looked into it. Added: I realize I didn't mention above the connection between holonomy and usual integration of a one-form $A$. You get this by considering the connection
$$d+\begin{bmatrix}0& A; \\
0& 0\end{bmatrix}$$ on the trivial bundle of rank two. One view of Coleman integration is that the holonomy $H_a^b$ from $a$ to $b$ is defined first. And then, the naive integral is defined by the fomula
$$H_a^b=\begin{bmatrix}1& \int_a^bA ;\\
0& 1\end{bmatrix}$$
|
{
"source": [
"https://mathoverflow.net/questions/57657",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3757/"
]
}
|
57,667 |
Is it known which Kahler manifolds are also Einstein manifolds? For example complex projective spaces are Einstein. Are the Grassmannians Einsein? Are all flag manifolds Einstein?
|
This question can be interpreted in two different ways. 1) Which Kahler manifolds admit a Kahler metric that is at the same time Einstein? 2) Which Kahler manifolds admit an Einstein metric? If you want 1) , then you need to start with a manifold whose canonical bundle is either a) ample (like hypersurfaces of degree $\ge n+2$ in $\mathbb CP^n$), or b) trivial (Calabi-Yau), c) is dual to an ample line bundle - Fano case. In a) and b) there is always a Kahler-Einstein metric by a theorem of Aubin and Yau. In the case c) we get a very subtle question, which is expected to be governed by Yau-Tian-Donaldson conjecture. But all homogenious varieties are Kahler-Einstein. If you want 2) , then the amount of Einstein metrics clearly becomes much larger. For example, $\mathbb CP^2$ blown up in one or two point do not admit a Kahler-Einstein metric, but they do admit an Einstein metric. For a reference to this statement you can check the article of Lebrun http://arxiv.org/abs/1009.1270 . In general the question weather a given Kahler surface admits an Einstein metric is quite subtle. But at least there exists an obstruction. We can blow up any surface in sufficient number of points so that the obtained manifold violates Hitchin-Thorpe inequality http://en.wikipedia.org/wiki/Hitchin%E2%80%93Thorpe_inequality , hence not Einstein. Finally, it was speculated (for example by Gromov here: http://www.ihes.fr/~gromov/topics/SpacesandQuestions.pdf ), that starting from real dimension 5 each manifold admits an Einstein metric. Added reference. "Every compact, simply connected, homogeneous Kahler manifold admits a unique (up to homothety) invariant Kahler-Einstein metric structure"- this result can be found in Y. Matsushima. Remakrs on Kahler-Einstein manifolds, Nagoya Math J. 46. (I found this reference in the book Besse, Einstein manifolds, 8.95).
|
{
"source": [
"https://mathoverflow.net/questions/57667",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2612/"
]
}
|
57,741 |
I recently attended a lecture where the speaker mentioned that what he was talking about was connected to the algebraic version of the $P$ vs. $NP$ problem. Could someone explain what that means in a simple way or point me to a source suitable for a non-expert mathematician? Thanks.
|
I suspect that the question under consideration is whether or not $VP=VNP$; this is the problem directly studied by geometric complexity theorists, as I understand their work. This project is described in some detail here (this paper is by Burgisser, Landsberg, Manivel, and Weyman, describing work of Mulmuley and related people)--it is aimed at algebraic geometers; so you will likely be comfortable with it. The description of the complexity problem under consideration is in section 9. The algebraic $VP$ vs. $VNP$ conjeture is due to Valiant, in this paper and his paper "Reducibility by algebraic projections" which I can't find online at the moment, unfortunately; these are references [63] and [64] in the paper I link to above. Valiant is, as I recall, a very clear writer, so hopefully you will find these papers readable as well. Essentially, Valiant argues that some algebraic properties of the permanent and related varieties should have complexity-theoretic implications; a reasonable heuristic for this might be the many combinatorial interpretations of the permanent. Unfortunately, as far as I know there are few implications between these algebraic versions of P vs. NP and the problem itself; there are some results assuming GRH. See e.g. this paper by Burgisser . Hopefully, this is the algebraic analogue of P vs. NP you were looking for.
|
{
"source": [
"https://mathoverflow.net/questions/57741",
"https://mathoverflow.net",
"https://mathoverflow.net/users/10076/"
]
}
|
57,766 |
On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can extend your Jordan arc to a Jordan curve. That is, as long as you can extend an embedding of an interval to an embedding of a circle. However, I have to admit I don't readily see how to do this, and there are examples which show this is a subtle question. E.g. you can take an arc which spirals infinitely into a point. So my question is how one can show an embedding of an arc into $\mathbb R^2$ extends to an embedding of a circle into $\mathbb R^2$, or, failing that, if someone knows some other proof that all planar Jordan arcs are standard. Edit: I want to highlight Bill Thurston's elegant answer (in a comment) to the question of whether you can extend an arc to a circle, even though I accepted his other answer using Caratheodory's theorem. Namely, assume your arc runs from $0$ to $\infty$ in $\mathbb C\cup\{\infty\}$. Then take the pre-image under the double branched cover $z\mapsto z^2$. The original arc can be identified with one of its two preimages, while the other preimage fills it in to a circle. Then one can apply the Schönflies theorem.
|
One way to show all arcs are tame is to apply the Riemann mapping theorem to the complement of the arc on $S^2$. Caratheodory proved that whenever the complement of a simply-connected domain is locally connected, then the Riemann mapping extends to a continuous map of the disk to the plane. One half of the disk parametrizes the arc. Another (related) method is to take the stereographic projection of the arc to a sphere
in $\mathbb E^3$, and form its convex hull. If the interior of the ball is interpreted as the projective model for hyperbolic 3-space, then the boundary of the convex hull intersect the interior of the ball is a developable surface, i.e. its path metric (that is, distance between points is the minimum arc length of a path connecting them) is a complete metric locally isometric to the path-metric of the hyperbolic plane. Since it's simply-connected, it's globally isometric to the hyperbolic plane. For reasons parallel to Caratheodory's the map extends continuously to a map of the closed disk (2-dimensional projective disk model of
the hyperbolic plane, together with its circle at infinity) to the closed ball.
|
{
"source": [
"https://mathoverflow.net/questions/57766",
"https://mathoverflow.net",
"https://mathoverflow.net/users/9417/"
]
}
|
57,820 |
Since Euclid's axiomatization of space, we have developed a sophisticated mathematical model of space. Given a category of structures (measures), local space is modeled the spectrum of measurements that these structures can possibly make. Global space is construed as patches connected via transport, which identifies measurements across patches. I'm troubled that I have not come across any axiomatization of time. Assuming that mathematics is a priori science, the great varieties of theories of space in physics can be attributed to our sophisticated mathematical model of space. There is relativity, string theory, quantum theory and M theory. Perhaps the reader may object that these theories are theories of space-time, rather than theories of space. However, I wish to note that in these theories, time is essentially treated in the same manner as space. In classical physics, time is but another dimension of space. In relativity, time is distinguished from time by the (3,1) signature, but this is just a metric. Riemannian geometry is still considered a theory of space rather than a theory of time. I'm wondering, then, whether you have encountered is a mathematical axiomatization of time, that treats time in a way not that is not inherently spatial? Assuming once more that mathematics is a priori science, perhaps such an axiomatization can lead to breakthroughs in physics and finance. Finally, there is a physical theory that I think comes close to a model of time. Namely, entropy. Just as space is dual to measures, we can think of time as dual to entropy. Given as entropy can be defined using combinatorics and probability, this could be viewed as a mathematical theory. EDIT: Steve mentioned that perhaps one can view entropy as a theory of time via the Thermal Time Hypothesis. Other than entropy, are there any other axiomatizations of time? ANOTHER EDIT: In the answers given below, most of the models of time are archimedean. I'm wondering, this models can be tweaked to allow a cyclic conceptualization of time. Many ancient cultures, eg from India, consider time to be cyclic rather than archimedean. Should I ask this as a separate question? I think of this cyclic/archimdean dichotomy as something like Euclidean/non-Euclidean geometry.
|
In probability, time is usually handled as a nested sequence of $\sigma$-algebras (say $B_t$, with $B_t \subset B_s$ if $t\leq s$), and to find the reality (call the reality $f$, and it includes the state at all times past and future) at time $t$, one takes the conditional expectation $f_t := E[f | B_t ]$. The sequence $(f_t)$ is then a martingale (a uniformly integrable martingale, more precisely), and this construction is the essence of what the big deal is about martingales. Brownian motion is a martingale that you've probably heard of, but this also handles simpler situations. For example, consider the experiment: toss a coin repeatedly, and keep track of how many heads you've thrown, minus how many tails. We can capture this experiment in the following way: For $0\leq x <1$, let $f_n(x)$ be the number of 1's minus the number of 0's among the first $n$ digits of the binary expansion of $x$, and let $B_n$ be the $\sigma$-algebra (in this case, a boolean algebra) generated by the intervals $[i/2^n,(i+1)/2^n)$. Then $f_n$ is $B_n$-measurable, and $E[f_t | B_s]=f_s$ for any natural numbers $s < t$, and the sequence $(f_n)_{n=1}^\infty$ is a martingale (albeit different from the type mentioned above). If you want to play any fair game on the "coin tosses" as they come up (allowing use of knowledge of all previous-in-time tosses), then your fortune at time $t$ is still a martingale. In other words, the passage of time is captured as un-conditional-expectating a function. For a practical introduction to martingales, I recommend Williams' "Probability with martingales." It is a marvel of writing, and in my humble opinion should be taken as a model for how to write a monograph.
|
{
"source": [
"https://mathoverflow.net/questions/57820",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
57,952 |
Fix $p>0$ a rational prime, and $K$ an algebraic number field with Galois group $G_K:=Gal(\bar{\mathbb{Q}}/K) $. The Fontaine-Mazur conjecture predicts that if $\rho:G_K\rightarrow GL(V)$ is a finite dimensional $\mathbb{Q}_p$-representation, then it comes from a motive over $K$ (like a subquotient of $H^i_c(X\times_K\bar{\mathbb{Q}}, \mathbb{Q}_p)$) exactly when it is unramified over almost every finite place, and potentially semi-stable over those finite prime dividing $p$. My question is the contrary: how many examples do we have for $p$-adic Galois representations having infinite images but for which the conditions of Fontaine-Mazur fails? Maybe it should be difficult to construct them when the dimension of the representation is large? Could one get continuous $p$-adic representations that ramifies at infinitely many places?
|
Here are two things that can occur: 1) If V is the representation attached to an overconvergent modular form f, then V will be unramified at almost every prime but will not be de Rham at p (unless f is classical). 2) Ramakrishna has written an article "Infinitely ramified Galois representations". Here's part of the introduction: "In this paper we show how to construct [...] representations [...] that are ramified at an infinite number of primes." Under GRH, these repns are crystalline at p. So both conditions in FM can fail independently.
|
{
"source": [
"https://mathoverflow.net/questions/57952",
"https://mathoverflow.net",
"https://mathoverflow.net/users/9246/"
]
}
|
58,004 |
I saw the functional equation and its proof for the Riemann zeta function many times, but usually the books start with, e.g. tricky change of variable of Gamma function or other seemingly unmotivated things (at least for me!). My question is, how does one really motivate the functional equation of the zeta function? Can one see there is some hidden symmetry before finding/proving it? For example, I think $\Gamma(s+1)=s\Gamma(s)$ for $s>1$ "motivates" the analytic continuation of the Gamma function.
|
You do not try to motivate it! Even Riemann didn't see a nice argument right away. Riemann's first proof of the functional equation used a contour integral and led him to a yucky functional equation expressing $\zeta(1-s)$ in terms of $\zeta(s)$ multiplied by things like $\Gamma(s)$ and $\cos(\pi{s}/2)$. Only after finding this functional equation did Riemann observe that the functional equation he found could be expressed more elegantly as $Z(1-s) = Z(s)$, where $Z(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$. Then he gave a proof of this more symmetric functional equation for $Z(s)$ using a transformation formula for the theta-function $\theta(t) = \sum_{n \in {\mathbf Z}} e^{-\pi{n^2}t},$ which is $$\theta(1/t) = \sqrt{t}\theta(t).$$ In a sense that transformation formula for $\theta(t)$ is equivalent to the functional equation for $Z(s)$. The transformation formula for $\theta(t)$ is itself a consequence of the Poisson summation formula and also reflects the fact that $\theta(t)$ is a modular form of weight 1/2. Instead of trying to motivate the technique of analytically continuing the Riemann zeta-function I think it's more important to emphasize what concepts are needed to prove it: Poisson summation and a connection with modular forms (for $\theta(t)$). These ideas are needed for the analytic continuation of most other Dirichlet series with Euler product which generalize $\zeta(s)$, so an awareness that the method of Riemann continues to work in other cases by suitably jazzing up Poisson summation and a link to modular forms will leave the reader/student with an appreciation for what goes into the proof. This proof is not intuitive and I think it's a good illustration of von Neumann's comment that in mathematics we don't understand things, but rather we just get used to them.
|
{
"source": [
"https://mathoverflow.net/questions/58004",
"https://mathoverflow.net",
"https://mathoverflow.net/users/11286/"
]
}
|
58,040 |
In the study of special functions there are three levels of objects, classical, basic and elliptic. These correspond to classical hypergeometric functions , basic (q-) hypergeometric functions , and elliptic hypergeometric functions . In combinatorics these notions are related to enumeration, q-enumeration and "elliptic enumeration" (see this article of Schlosser). Now, I always related the passing to q-analogs by analogy to the "way" one passes from groups to quantum groups. And indeed, q-analogs and quantum groups are not entirely unrelated concepts. But this makes me ask the question in the title, whether someone has considered quantum groups at the "elliptic level", and if so what are they?
|
That's a wonderful question, but I think there's a fundamental confusion here about two possible roles of the rational/trigonometric/elliptic trichotomy -- the one asked in the question and the one that leads to elliptic quantum groups -- which are in some sense "Fourier dual". (Everything I understand about this I learned talking to Tom Nevins.) For example the R/T/E trichotomy in R-matrices corresponds in quantum group world to the trichotomy Yangians/quantum affine algebras/elliptic quantum groups, not to group/quantum group/elliptic quantum group, or in nonquantum world to the trichotomy "Lie algebra/Lie group/elliptic group", with no quantums around. Not that there's an independent notion of an "elliptic group", but you can define a lot about it in terms of moduli of bundles on an elliptic curve. What the question asks is to fill in the trichotomy group/quantum group/"elliptically quantum" group.. Perhaps the easiest setting to see these two roles is in the study of many-body systems, eg the Calogero-Moser systems (or equivalently of meromorphic solutions of the KP and Toda hierarchies). These are completely integrable hamiltonian systems describing the motion of particles in the line. Initially it looks like they come in three flavors - rational, trigonometric and elliptic - labeled by whether the dependence of the potential on positions is rational, periodic or doubly periodic. This trichotomy is nicely explained by the trichotomy in R-matrices or in one-dimensional algebraic groups over $C$ or in irreducible genus one curves (Weierstrass cubics). The phase space can be described in terms of a cotangent bundle to a configuration space of points (on $C$, $C^\times$ or an elliptic curve), and the corresponding quantum systems can be described in terms of differential operators on the corresponding configuration spaces.
In representation theory this trichotomy appears in studying three versions of the loop algebra - current algebras from $C$, $C^\times$ or $E$ (the latter needs to be interpreted more sophisticatedly).One can (as I mentioned) invent something you call an "elliptic group" by studying G-bundles on an elliptic curve, in such a way that if your elliptic curve acquires a node you get the usual group, and if it acquires a cusp you get the Lie algebra.. but again this is not the question. I claim this R/T/E trichotomy is naturally identified with the one in the above answers, but "linearly independent" (in fact Fourier dual) to the one in the question. In integrable systems world this is expressed as follows: there's a deformation of the Calogero-Moser particle systems (called Ruijsenaars or Ruijsenaars-Schneider or Macdonald) in which we change the rational dependence on MOMENTUM to trigonometric dependence on momentum. This corresponds to changing from the linear Poisson structure on the cotangent bundle to configuration space to a quadratic one on the multiplicative cotangent bundle to config space (replace all $C$'s by $C^\times$'s). When we quantize this quadratic Poisson bracket we get DIFFERENCE rather than differential operators, and a relation to quantum groups rather than groups as in the CM case. Integrable systems people (eg Nekrasov, Gorsky and collaborators) like to express this by a 3 x 3 square : we can havce R/T/E dependence on position as above, or R/T/E dependence on momenta.. except no one has a good definition (AFAIK) of what elliptic dependence on momenta MEANS - except via Fourier transform, exchanging position and momenta - so you can define elliptic momenta/rational positions eg by switching the order.. In any case the row "rational momenta" relates to (loop) groups, "trig momenta" relates to quantum groups. So the question is what is the elliptic version?
There are various hints from string theory (see works of Nekrasov eg--- the rational row relates to SUSY 4dimensional gauge theory via the Seiberg-Witten solution, the trigonometric row to 5dimensional gauge theory, and the elliptic row should come from the mysterious "5-brane theory" or "6-dimensional (0,2) CFT")... But maybe the most concrete answer I can give is motivated by Nakajima's work (see eg his ICM). You can realize representations of (loop) groups (of simple groups of type ADE) in equivariant cohomology of quiver varieties.
If you replace equivariant cohomology by equivariant K-theory, you find the representation theory of quantum (affine) algebras. So this gives a natural place to look for the elliptic analogue --- try to understand the equivariant elliptic cohomology of quiver varieties! such ideas were put forth by Grojnowski, Ginzburg-Kapranov-Vasserot, and others but a good theory of equivariant elliptic cohomology was only developed fairly recently by Lurie, so one can contemplate such questions anew.
|
{
"source": [
"https://mathoverflow.net/questions/58040",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2384/"
]
}
|
58,061 |
This is a very naive question, and I'm hoping that it will be matched by a correspondingly elementary answer. It is well known that not every topological 4-manifold admits a smooth structure. So what's wrong with the following very sketchy proof that, actually, a topological 4-manifold does admit a smooth structure (apart from the sketchiness)? Step 1: Embed the manifold into $\mathbb{R}^9$, which, as I understand it, can be done . Step 2: "Iron out the kinks" in the embedded manifold. Step 3: Once the embedded manifold looks nice enough, give it an obvious smooth structure coming from $\mathbb{R}^9$. The second step looks the dodgiest to me, because my intuition comes from cases that are presumably much too special, such as a 2-dimensional manifold sitting in $\mathbb{R}^3$. Take, for instance, the surface of a cube. We can easily smooth off the corners and edges and obtain a smooth manifold. There are many smoothing methods around (such as convolving with nice objects). So why can't we find one that works in general? When I try to think how I would actually go about it, then I do of course run into difficulties. For instance, in the cube case I could take all points outside the cube of some fixed small distance from the cube. That would give me a smoother version. But if I try a trick like that when the codimension is not 1, then I get a set of the wrong dimension. That suggests that I have to make a clever choice of direction, and I don't see an obvious way of doing that. I have similar questions about other wacky (as they seem to me) facts about manifolds, such as the existence of topological manifolds that cannot be given piecewise linear triangulations. I'm not looking for an insight into why such results are true. All I want to understand is why they are not obviously false. Can anyone say anything that might be helpful?
|
Let's try doing this with a compact, closed, connected $1$-manifold $M$. Certainly I can choose a topological embedding $f:M\to\mathbb{R}^2$. However, the image might be very fractal, like a Koch snowflake for example. If I try to take a short piece of the image and straighten it out by projecting in some direction, the fractalness will ensure that I lose injectivity. If I consider all points at a fixed distance $\epsilon$ from $f(M)$ and try to retract back to $f(M)$, the same phenomenon will mean that the retraction is not well-defined. If I instead try to smooth it out by convolving, then it seems that I need a convolution kernel $u:M\times M\to\mathbb{R}$ that is supported near the diagonal, together with a measure on $M$, so I can define $g(x)=\int u(x,y)f(y)dx$. If $M$ already has a smooth structure and $u$ and the measure are smooth, then $g$ will be smooth and close to $f$. It isn't obviously injective but perhaps that could be arranged. However, I can't see anything like this that you could do if $M$ did not already have a smooth structure. In this two-dimensional case the Riemann mapping theorem gives a nearly unique conformal isomorphism $h$ from the open unit disc to the bounded component of $\mathbb{R}^2\setminus f(M)$. You could try to define $h_1:S^1\to f(M)$ by $h_1(z)=\lim_{t\to 1}h(tz)$. If I recall correctly, there are people in Finland who have thought about this a lot and decided that $h_1$ can be arbitrarily bad.
|
{
"source": [
"https://mathoverflow.net/questions/58061",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1459/"
]
}
|
58,096 |
Hello, Given a finitely presentable group $G$, I'm interested in the cup-product from $H^1$ to $H^2$ with real coefficients. I want to know if this is explicitly computable (with a computer) with a presentation of the group.
More precisely, I want a program that takes the generators and relations as entries and returns the dimension of the $H^1$ and a finite generating set of linear relations between the cup-products of every couple of elements in a basis of $H^1$. (I am not really interested in all the $H^2$)
Does this seem possible ? I precise that I am not really familiar with group cohomology and I ask this question because it is certainly known if such a problem cannot be resolved with an efficient algorithm. The problem comes in the study of Kähler groups where this cup-product plays an important role. Thank you.
|
EDIT: I was explaining this to a grad student today, and I realized that I didn't give any references. The result I describe below was first stated by Sullivan in Sullivan, Dennis
On the intersection ring of compact three manifolds.
Topology 14 (1975), no. 3, 275-277. He claims it is true for a 3-manifold, but all he says about the proof is that it is "a certain amount of soul searching classical algebraic topology.". In fact, the result is true for any connected CW-complex (including an Eilenberg-MacLane space, as in the group cohomology question I was answering). This whole picture was later subsumed into Sullivan's theory of 1-minimal models and rational homotopy theory in Sullivan, Dennis
Infinitesimal computations in topology.
Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 269-331 (1978). An accessible reference for this is Griffiths, Phillip A.; Morgan, John W.
Rational homotopy theory and differential forms.
Progress in Mathematics, 16. Birkhäuser, Boston, Mass., 1981. xi+242 pp. ISBN: 3-7643-3041-4 The stuff on fundamental groups is in Chapter 13. I don't know where the proof I gave first appeared (I came up with it myself, but I doubt I was the first). An alternate and very pretty geometric proof is in De Michelis, Stefano,
A remark on cup products in $H^1(X)$ ,
Rend. Accad. Naz. Sci. XL Mem. Mat. (5) 14 (1990), no. 1, 323-325. This is very computable. Let $G^{(k)}$ be the lower central series of $G$ , ie $G^{(1)}=G$ and $G^{(k+1)} = [G^{(k)},G]$ . There are algorithmic ways to compute the quotients $G^{(k)}/G^{(k+1)}$ (eg using the Fox free differential calculus -- see Fox's series of papers on the free differential calculus for the details). The direct sum $$\oplus_{k=1}^{\infty} G^{(k)} / G^{(k+1)}$$ has the structure of a graded Lie algebra with the Lie bracket induced by conjugation (this is explained in many places -- I recommend the last chapter of Magnus-Karass-Solitar's book on combinatorial group theory or Serre's book "Lie Algebras and Lie Groups"). This Lie algebra is generated by the degree 1 piece, namely $G^{(1)} / G^{(2)} \cong G^{ab}$ . The degree 2 piece is a quotient of $\wedge^2 G^{ab}$ by some subgroup $R$ . I claim that understanding $R$ is exactly what you need to know to understand the kernel of the cup product map. Namely, we have a surjection $$\wedge^2 G^{ab} \rightarrow \wedge^2 G^{ab} / R$$ and thus a dual injection $$(\wedge^2 G^{ab} / R)^{\ast} \hookrightarrow \wedge^2 (G^{ab})^{\ast}.$$ The image of this injection is exactly the kernel of the cup product map. Let me sketch a proof. To simplify things, let's assume that everything in sight is torsion-free (it will simplify our statements). Set $H = H_1(G)$ and $H^{\ast} = H^1(G) = Hom(H,\mathbb{Z})$ . The above will allow you to compute the kernel of the cup product map $\wedge^2 H^{\ast} \rightarrow H^2(G)$ as follows. Consider the short exact sequence $$1 \longrightarrow G^{(2)} \longrightarrow G \longrightarrow H \longrightarrow 1.$$ There is an associated 5-term exact sequence in group cohomology which takes the form $$0 \longrightarrow H^1(H) \longrightarrow H^1(G) \longrightarrow (H^1(G^{(2)}))^H \longrightarrow H^2(H) \longrightarrow H^2(G).$$ Now, the map $H^1(H) \rightarrow H^1(G)$ is an isomorphism. Also, $H^2(H) = \wedge^2 H^{\ast}$ , and the map $H^2(H) \rightarrow H^2(G)$ is easily seen to be the cup product map. What we deduce is that we have an exact sequence $$0 \longrightarrow (H^1(G^{(2)}))^H \longrightarrow \wedge^2 H^{\ast} \longrightarrow H^2(G).$$ In other words, the kernel of the cup product map is the subgroup $(H^1(G^{(2)}))^H$ of $\wedge^2 H^{\ast}$ . Let us now interpret this subspace. It is easiest to dualize. The dual of the above inclusion is the surjection $$H_2(H) \rightarrow (H_1(G^{(2)}))_H.$$ Now, $H_1(G^{(2)})$ is just $G^{(2)} / [G^{(2)},G^{(2)}]$ , and we are killing off the action of $H$ , which is the same as killing off the conjugation action of $G$ . In other words, we have an isomorphism $$(H_1(G^{(2)}))_H \cong G^{(2)} / [G,G^{(2)}] = G^{(2)} / G^{(3)}.$$ The desired claim is an immediate consequence.
|
{
"source": [
"https://mathoverflow.net/questions/58096",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12517/"
]
}
|
58,131 |
Are the Milnor's seven dimensional exotic spheres parallelizable?
|
A much more general result is true. Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma $ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$. It says that exotic spheres cannot be distinguished by looking at the tangent bundle.
This result is one of the hidden gems of the golden age of topology and the proof invokes the whole plethora of topology of the 1950s. The argument can be recollected from the old literature, but I do not know a coherent reference. To start with, there are several invariants of the tangent bundles that do not depend on the smooth structure.
Let $\Sigma$ be a homotopy sphere. Then: the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (Gauss-Bonnet, relatively easy). $T \Sigma^n \oplus \mathbb{R}$ is trivial. This is a deep result by Kervaire and Milnor (not in the Annals paper, but a small note published before). $T \Sigma \oplus \mathbb{R}$ is given by an element of $\pi_{n-1} (O)$, which is known, by Bott periodicity, to be either $Z$, $0$ or $Z/2$. The $Z$ groups are detected by the Pontrjagin class, which has to vanish by Hirzebruchs signature formula because the sphere evidently has signature $0$. In the $Z/2$ case, the argument is more delicate. Essentially, the normal spherical fibration of a manifold does not depend on the smooth structure. Since the normal fibration of the standard sphere is trivial, so is that of $T \Sigma$. The process of associating to a vector bundle its spherical fibration is the J-homomorphism which is injective in these dimensions by Adams' $J(X)$ paper. Now look at the homotopy sequence of the fibration $O(n)\to O(n+1) \to S^n$, i.e. the piece $$
\mathbb{Z}=\pi_n (S^n) \to \pi_{n-1} (O(n)) \to \pi_{n-1} (O(n+1)) = \pi_{n-1} (O).
$$ It is known that $TS^n$ (for the standard smooth structure) is the image of the generator of $\pi_n (S^n)$ (not hard, see Steenrods book).
By the above deep result, $T \Sigma$ lies in the kernel of
$\pi_{n-1}(O(n)) \to \pi_{n-1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n) $. The image of $Z \to \pi_{n-1} (O(n)$ can be computed.
It is $Z$ if $k$ is even (using the Euler class),
it is $0$ if $n=1,3,7$ (follows directly from Adams' result on the parallelizability of the standard spheres) and it is $Z/2$ in the
remaining cases.
You can find the (not so hard, but clever) argument for the last assertion in Levine's lectures on homotopy spheres (which can be viewed as the sequel to the Kervaire-Milnor paper). How to proceed? If $n=1,3,7$, it follows that $T \Sigma$ is trivial, as $TS^n$. If $n$ is even, then the kernel of
$\pi_{n-1}(O(n))\to \pi_{n-1}(S^{n+1})$ is detected by the Euler class,
and by Gauss-Bonnet, this characterizes $T \Sigma$. It remains the case of odd $n$ apart from the "Adams dimensions". One has to argue that in these dimensions, $T \Sigma$ is nontrivial.
In the introduction to his Hopf invariant paper, Adams attributes to Dold the result ''$T \Sigma$ parallel implies that $\Sigma$ (and hence $S^n$) is an H-space''.
But he (Adams) proved that is not the case $n=1,3,7$. Adams does not give a reference for Dolds result, but in his answer to this question (and the subsequent comments), John Klein sketches a proof that looks like a 1950s argument. EDIT: I supervised a Master's thesis (written by Julia Heller), who worked out the details of this argument.
Here is the argument for the fact that if a homotopy sphere $\Sigma^n$ is parallelizable, then $n=0,1,3,7$. Consider the diagonal $\Sigma \subset \Sigma \times \Sigma$. Its normal bundle $N$ is isomorphic to $T \Sigma$, hence trivial. We have the Pontrjagin-Thom collapse
$c:\Sigma \times \Sigma \to Th (N) $. Any trivialization $N \cong \Sigma \times \mathbb{R}^n$ induces a map $a: Th (N) \to S^n$. Let
$$
\mu: \Sigma \times \Sigma \to Th (N) \to S^n
$$
be the composition of the maps just explained. By counting intersections, it can be seen that the restriction of $\mu$ to the submanifolds $\{x\} \times \Sigma$ and $\Sigma \times \{x\}$ has degree $\pm 1$, hence is a homotopy equivalence. By composing with $(h,k):S^n \times S^n \to \Sigma \times \Sigma$ for suitably chosen homotopy equivalences, we get a map
$$
\mu' : S^n \times S^n \to S^n
$$
which restricts to degree 1 maps $S^n \times \{x\} \to S^n$ and $\{x\} \times S^n \to S^n$. Since maps of degree 1 are homotopic, it follows that $\mu'$ gives $S^n $ the structure of an $H$-space. By Adams' theorem, $n=0,1,3,7$. Note: it is not important for Adams' theorem that the $H$-space structure is homotopy associative, but homotopy unitality is essential.
|
{
"source": [
"https://mathoverflow.net/questions/58131",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13431/"
]
}
|
58,188 |
The Diophantine equation
$$x^3+y^3+z^3=3$$
has four easy integer solutions: $(1,1,1)$ and the three permutations of $(4,4,-5)$. Elsenhans and Jahnel wrote in 2007 that these were all the solutions known at that time. Are any other solutions known? By a conjecture of Tyszka, it would follow that if this equation had finitely many roots, then each component of a solution tuple would be at most $2^{2^{12}/3} \lt 2^{1365.34}$ in absolute value. (To see this, it is enough to express the equation using a Diophantine system in 13 variables in the form considered by Tyszka.) This leaves a large gap, since Elsenhans and Jahnel only considered solutions with components up to $10^{14} \approx 2^{46.5}$ in absolute value. It is also not obvious whether Tyszka's conjecture is true. OEIS sequence A173515 refers to equations of the form $x^3+y^3=z^3-n$, for $n$ a positive integer, as "Fermat near-misses". Infinite families of solutions are known for $n=\pm 1$, including one constructed by Ramanujan from generating functions (see Rowland's survey). Andreas-Stephan Elsenhans and Jörg Jahnel, New sums of three cubes , Math. Comp. 78 (2009), 1227–1230. DOI: 10.1090/S0025-5718-08-02168-6 . ( preprint ) Apoloniusz Tyszka, A conjecture on integer arithmetic , Newsletter of the European Mathematical Society (75), March 2010, 56–57. ( issue ) Eric S. Rowland, Known Families of Integer Solutions of $x^3+y^3+z^3=n$ , 2005. ( manuscript )
|
Just noticed this question. I agree with L.H.Gallardo that the problem is old (see e.g. Problem D5 in UPINT = Unsolved Problems in Number Theory by R.K.Guy), but not that it is hopeless: the usual heuristics suggest that the number of solutions with $\max(|x|,|y|,|z|) \leq H$ should be asymptotic to a multiple of $\log H$, so further solutions should eventually emerge (though it may indeed be hopeless to prove anything close to the $\log H$ heuristic). See also my article Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63 = math.NT/0005139 on the arXiv. Among other things it gives an algorithm for finding all solutions of $|x^3 + y^3 + z^3| \ll H$ with $\max(|x|,|y|,|z|) \leq H$ that should run (and in practice does run) in time $\widetilde{O}(H)$; since we expect the number of solutions to be asymptotically proportional to $H$, this means we find the solutions in little more time than it takes to write them down. D.J.Bernstein has implemented the algorithm efficiently, and reports on the results of his and others' extensive computations at http://cr.yp.to/threecubes.html . EDIT: for the specific problem $x^3+y^3+z^3=3$, Cassels showed that any solution must satisfy $x\equiv y\equiv z \bmod 9$ in this brief article: A Note on the Diophantine Equation $x^3+y^3+z^3=3$, Math. of Computation 44 #169 (Jan.1985), 265-266. This uses cubic reciprocity, and is stronger than what one can obtain from congruence conditions. See also Heath-Brown's paper "The Density of Zeros of Forms for which Weak Approximation Fails" ( Math. of Computation 59 #200 (Oct.1992), 613-623), where he gives corresponding conditions for the homogeneous equation $x^3 + y^3 + z^3 = 3w^3$ and also $x^3 + y^3 + z^3 = 2w^3$, and reports that In a letter to the author, Professor Colliot-Thélène has shown that the above congruence restrictions are exactly those implied by the Brauer-Manin obstruction. Moreover, for the general equation $x^3 + y^3 + z^3 = kw^3$, with a noncube integer $k$, there is always a nontrivial obstruction, eliminating two-thirds of the adèlic points.
|
{
"source": [
"https://mathoverflow.net/questions/58188",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7252/"
]
}
|
58,193 |
Where can I find a calculus textbook that emphasizes differentials?
Is there such a book that I could realistically require my calculus students to use? I want a textbook that supports me when I tell my students something like: $\Delta((x^2+1)^5)\approx5(x^2+1)^4\Delta(x^2+1)\approx5(x^2+1)^4(2x\Delta x)$ $d((x^2+1)^5)=5(x^2+1)^4d(x^2+1)=5(x^2+1)^4(2x\ dx)$ Or: $\Sigma_{k=1}^n 3x_k^2\Delta x_k\approx\Sigma_{k=1}^n\Delta(x_k^3)=x_n^3-x_1^3$ $\int_{x=0}^{x=4}3x^2\ dx=\int_{x=0}^{x=4}d(x^3)=4^3-0^3=64$ Perhaps I could write this book someday, but it'd be a lot easier for me
if my students and I could just buy and/or download a book that takes
this approach without neglecting to provide a cornucopia of exercises, examples, and applications similar to what's available in today's most popular calculus textbooks.
|
There is a marvelous old book (19th Century if I recall correctly) where I learned Calculus the first time, called "Calculus Made Easy" by Silvanus P. Thompson, and subtitled "What one fool can do another can". He explains that dx means a "little bit of x" and shows a square with sides x and x + dx and you can see why you can "ignore dx^2". Of course it isn't rigorous in any sense, but it uses differentials to get all the essential ideas of both differential and integral Calculus across quickly and smoothly. Needless to say, once I had absorbed all these essential ideas I went on to read more rigorous books where limits were introduced and used to make precise what I already understood well from this intuitive introduction. If I recall correctly Calculus Made Easy was republished some years back (Dover?) and was quite popular. I would suggest that you recommend it to your students, with appropriate caveats. (Added later) I checked online and indeed there is a recent reprinting (available from Amazon and the other usual places). Moreover it has three new chapters written by the late great Martin Gardner aimed at the modern reader. I'm going to buy myself a copy!
|
{
"source": [
"https://mathoverflow.net/questions/58193",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12106/"
]
}
|
58,276 |
The motivation for asking this question is a passage (3.2) in an article by Greg Hjorth where he said that "...it is also an attractive feature of the theory of Borel cardinalities and of the theory of $L(\mathbb{R})$ cardinality that these are largely immunized against independence." My question is: What other parts of Set Theory are so immunised? Assuming something extra is okay(say Determinacy), but whatever it is, should settle 'most' of the questions. The wording above is not too clear, but I'm not sure how I can make a stronger statement(suggestions about this would be great). I guess that if you ask high-level enough questions(definability hierarchy wise), independence will come in sooner or later. However, I'm not really asking this from the point of view of absoluteness, but from the point of view of what large class of questions can be settled by what small set of tools. A bonus question is: Why is the part about Borel cardinalities true? I guess this might have some simple absoluteness explanation, but the 'largely' tells me there is more to it than I think. Thanks in advance. P.S. I'm not sure if I've tagged this appropriately, perhaps a 'soft-question' or 'big-list' one would be okay (although I wonder how big the list would be!). I hope people will retag it if they find it suitable.
|
It is a theorem of Woodin that if there is a proper class of Woodin cardinals, then the theory of $L(\mathbb{R})$ can not be changed by forcing. Since forcing and large cardinals are essentially our only means for establishing independence results, this can be interpreted as saying that the theory of $L(\mathbb{R})$ is immune to independence phenomena (except for that which G\" odel's theorem imposes). Here $L(\mathbb{R})$ is the smallest model of ZF which contains all of the reals. $L(\mathbb{R})$ does satisfy the Axiom of Dependent Choice under this assumption, as well as the Axiom of Determinacy. Most of not all theorems in real and complex analysis, measure theory (in the setting of standard Borel space), manifolds, geometry, and number theory can be regarded as statements about what is true in $L(\mathbb{R})$. It is the ideal model in which to study descriptive set theory. Uncountable sets and cardinals, however, often behave strangely in this model (largely because of the influence of the Axiom of Determinacy). For instance, in $L(\mathbb{R})$ and under the above assumptions, $\omega_1$ and $\omega_2$ are measurable cardinals, $\omega_n$ is a singular cardinal for each $n > 2$, there is no uncountable well orderable set of reals, and there are no non-principal ultrafilters on $\omega$. Ironically, Woodin's theorem was the culmination of several deep results concerning iterated forcing, the combinatorics of $\omega_1$, the study of large cardinals, and the fine structure of inner models generalizing $L$.
|
{
"source": [
"https://mathoverflow.net/questions/58276",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3462/"
]
}
|
58,325 |
I remember reading somewhere that the complex Fourier coefficients were introduced by an engineer sometime around 1900, but I can't find anymore this information. Does anyone know the name of this person and where I can find a reference to it? EDIT: I state the question more clearly: "Who was it that first wrote a Fourier series not as a sum of sines and cosines but as a sums of complex exponentials, with the relative formula for the coefficients?". I may be totally wrong about this all, since I don't remember well and that's why I'm asking. Also don't take the 1900 thing seriously, I may be off by 50+ years.
|
In the book Fourier Series and Wavelets , J.P. Kahane, P.G. Lemarié-Rieusset, Gordon and Breach Publishers, 1995, pp. 1 (available as a Publications Mathématiques d'Orsay here ), the authors state that "The subject matter of Fourier series consists essentially of two formulas : (1) $$f(x) = \sum c_n e^{inx}, $$ (2) $$c_n = \int f(x) e ^{-inx} \frac{dx}{2 \pi}.$$ The first involves a series and the second an integral." In the last paragraph of page 2, they add: "It is time to say that formulas (1) and (2) were never written by Fourier. Complex exponentials were not used in Fourier Series until well into the twentieth century ". Unfortunately, no references are given to this statement. -- UPDATE 1: 1935 G.H.Hardy, J.E.Littlewood, "Notes on the theory of series (XIX): A problem concerning majorants of fourier series", Quartely Journal of Mathematics, Vol os-6, Issue 1, pp. 304-315, 1935, equation (1.1.1) explicit complex Fourier series equation. I have no access to the full paper so I cannot search for references. A. Zygmund, Trigonometrical Series, 1935: §§1.13 (p. 2) and 1.43 (p. 6) give formulas (1) and (2) above. -- UPDATE 2: 1892 J. de Séguier, "Sur la série de Fourier", Nouvelles annales de mathématiques 3e série, tome 11, p. 299-301, 1892 The paper begins with a very beautiful equation: "Considérons la série $$S = \sum_{n =-\infty}^{+\infty} \frac{e^{\frac{2n i \pi z}{\omega}}}{\omega} \int_{u_0}^{u_0 + \omega} f(u)
e^{-\frac{2n i \pi u}{\omega}}du.\text{"} $$ As we can see, the integral part of this equation is the complex Fourier coefficients. Therefore, $S$ represents the complex Fourier series. One interesting conclusion: By the equation published in this paper, complex exponentials were used in Fourier Series BEFORE twentieth century -- UPDATE 3: 1875 M.M. Briot et Bouquet, Théorie des Fonctions Elliptiques, Deuxième Édition, Gauthien-Villars, Paris, 1875 Under the title "Série de Fourier" (page 161), at page 162 we can see equations (2) and (3) that are expressions of Fourier series with complex exponentials. The link for page 162: http://gallica.bnf.fr/ark:/12148/bpt6k99571w/f172.image
|
{
"source": [
"https://mathoverflow.net/questions/58325",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13108/"
]
}
|
58,329 |
Consider the polynomial $(1+x)(1+x^2)\dots (1+x^n)=1+x+\dots+x^{n(n+1)/2}$, which enumerates subj. How to prove that it's coefficients increase up to $x^{n(n+1)/4}$ (and hence decrease after this)? Or maybe this is false? This problem was proposed long ago on some Russian high school competition, but nobody managed to solve, including jury.
|
In fact, you cannot prove unimodality of the coefficients of $P_n(x)=(1+x)\cdots (1+x^n)$ using the result about $B_m/G$ mentioned by Qiaochu. The $P_n(x)$ result is implicit in work of Dynkin on the principal sl(2) subalgebra of a complex semisimple Lie algebra. Hughes was the first to realize that a special case of Dynkin's result implied the unimodality of $P_n(x)$. Proctor removed all the Lie algebra theory from the proof, yielding a proof involving only elementary linear algebra. See http://math.mit.edu/~rstan/pubs/pubfiles/72.pdf , esp. equation (23). See also http://math.mit.edu/~rstan/algcomb/algcomb.pdf , beginning on page 90. It is a long-standing open problem to find a combinatorial proof of this unimodality result. I would be extremely impressed if a high school student could prove unimodality by any method.
|
{
"source": [
"https://mathoverflow.net/questions/58329",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4312/"
]
}
|
58,339 |
These terms have become common in Lie theory and related algebraic geometry and combinatorics, as seen in many questions posted on MO, but it's unclear to me where they first came into use. Probably the concrete notion of (complete) flag of subspaces $0 \subset V_1 \subset \dots \subset V_n =V$ with $\dim V_k = k$ in an $n$ -dimensional vector space $V$ occurs very early in the literature (Grassmann?), though I'm not sure whether a specific national flag (drapeau, Flagge, ... ) is supposed to come to mind. Those wanting to visualize the simplest case may find the picture here useful (or not). At some point in the development of Lie groups and their homogeneous spaces, the notion of flag manifold got attached to the set of all complete flags in $\mathbb{C}^n$ or $\mathbb{R}^n$ : this is realized as the quotient of the corresponding general linear group $G$ by the isotropy subgroup $B$ of a standard flag (say the group of upper triangular nonsingular matrices). In the 1950s such connected maximal solvable subgroups became known as Borel subgroups, while the notion of flag manifold came to mean the quotient $G/B$ for an arbitrary connected reductive Lie group $G$ and a Borel subgroup $B$ . In the setting of real compact Lie groups, the analogue takes the form $G/T$ for a maximal torus $T$ : this is apparently the earliest version of the flag manifold. The work of Borel and Chevalley led to parallel developments for reductive algebraic groups over fields of arbitrary characteristic, with algebraic geometry replacing differential geometry and the term flag variety becoming common. While the people I've mentioned certainly deserve most of the credit for recognizing the essential role of flag manifolds or varieties in the study of geometry and representation theory associated to reductive groups, I'm still left with some uncertainty: What are the earliest sources in the literature for these terms? ADDED: The original reason for using the word "flag" in this context is a minor though interesting part of my question; but as the comments here indicate there is some variation in the folklore. In old literature on traditional projective geometry a "flag" is sometimes defined as a pair consisting of a point and a line through it, but that doesn't help much with the etymology. Is there more than folklore? Following the answer by Charles, I've looked further at the thesis of Ehresmann and the May 1951 Bourbaki talk by Borel (which like other expository talks is not included in his collected papers). I see more clearly how the notion of flag variety or flag manifold evolved from the older and still somewhat mysterious use of the term flag ( Flagge, Fahne , ...) in projective geometry, for instance to refer to an incident point-line pair. Near the end of Borel's talk he notes: Cela permet en particulier de montrer que les "varietes de drapeaux" complexes considerees par EHRESMANN dans sa these sont sans torsion ... . His use of quotation marks suggests to me that this might actually be the first time that label was applied to what later became known as the flag variety $G/B$ of a reductive Lie or algebraic group. (I wish I had thought to ask him at the right time.)
|
Armand Borel's Bourbaki Seminar 121 Groupes algébriques is from 1955, and uses "drapeau" (page 7). (It's online at archive.numdam.org .) This may not be the earliest occurrence, but there is a good reason for attention to the full flag variety in this context (the theory of Borel subgroups). The concept traces back some way, to Ehresmann's thesis in the 1930s; Kolchin's work (the Lie-Kolchin theorem) uses the non-intrinsic language of upper triangular form. Hodge & Pedoe talks about Schubert spaces in general, which would be natural in the enumerative geometry tradition, for which full flags is just one of the cases. Edit: A further data point is Chern's paper On the Characteristic Classes of Complex Sphere Bundles and Algebraic Varieties (1953), which relies on Ehresmann's work to some extent. The word "flag" is absent (though used by Chern discussing it in his Selected Papers ).
|
{
"source": [
"https://mathoverflow.net/questions/58339",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4231/"
]
}
|
58,495 |
I have recently run into this Wikipedia article on mereology . I was surprised I had never heard of it before and indeed it seems to be seldom mentioned in the mathematical literature. Unlike set theory, which is founded on the idea of set membership, mereology is built upon what I consider conceptually more elementary, namely the relation between parts and the whole. Personally, I have always found a little bit unsatisfactory (philosophically speaking) the fact that set theory postulates the existence of an empty set. But of course there is the technical aspect and current axiomatizations of set theory seem to be quite good regarding what it allows us to prove. Now it seems there have been some attempts to relate mereology and set theory, and according to the article, some authors have recently tried to deduce ZFC axioms as theorems in certain axiomatizations of it. Yet, apparently only a couple of well trained mathematicians (one of them Tarski) have discussed mereology, since most people have shown indifference towards the whole subject. So my questions are: how is it that mereology had no success as a possible foundation for mathematics? Are axiomatizations based on mereology not suitable for most developments or simply not worth the while? If so, which would be the technical reason behind?
|
I have long found this question interesting, and in some recent joint work with Makoto Kikuchi, now available, we consider various aspects of the question of whether a set-theoretic version of mereology can form a foundation of mathematics. In particular, for our main thesis we argue that the particular understanding of mereology by means of the inclusion relation $\subseteq$ cannot, by itself, form a foundation of mathematics. Joel David Hamkins and Makoto Kikuchi , Set-theoretic mereology , Logic and Logical Philosophy, special issue “Mereology and beyond, part II”, vol. 25, iss. 3, pp. 285-308, 2016. arxiv.org/abs/1601.06593 , ( blog post ). Abstract. We consider a set-theoretic version of mereology based on the inclusion relation $\newcommand\of{\subseteq}\of$ and analyze how well it might serve as a foundation of mathematics. After establishing the non-definability of $\in$ from $\of$ , we identify the natural axioms for $\of$ -based mereology, which constitute a finitely axiomatizable, complete, decidable theory. Ultimately, for these reasons, we conclude that this form of set-theoretic mereology cannot by itself serve as a foundation of mathematics. Meanwhile, augmented forms of set-theoretic mereology, such as that obtained by adding the singleton operator, are foundationally robust. Please follow through to the arxiv for a pdf version of the article. Update. Here is a link to a follow-up article: Joel David Hamkins and Makoto Kikuchi , The inclusion relations of the countable models of set theory are all isomorphic , manuscript under review. arxiv.org/abs/1704.04480 , ( blog post ). Abstract. The structures $\langle M,\newcommand\of{\subseteq}\of^M\rangle$ arising as the inclusion relation of a countable model of sufficient set theory $\langle M,\in^M\rangle$ , whether well-founded or not, are all isomorphic. These structures $\langle M,\of^M\rangle$ are exactly the countable saturated models of the theory of set-theoretic mereology: an unbounded atomic relatively complemented distributive lattice. A very weak set theory suffices, even finite set theory, provided that one excludes the $\omega$ -standard models with no infinite sets and the $\omega$ -standard models of set theory with an amorphous set. Analogous results hold also for class theories such as Gödel-Bernays set theory and Kelley-Morse set theory. And see the related question, Do all countable models of ZF with an amorphous set have the same inclusion relation up to isomorphism? That question remains an open question in the paper.
|
{
"source": [
"https://mathoverflow.net/questions/58495",
"https://mathoverflow.net",
"https://mathoverflow.net/users/12976/"
]
}
|
58,497 |
My (limited) understanding is that simplicial methods tend to be used whenever you want some kind of nontrivial homotopy theory -- for instance, to get a nice model structure, you use simplicial sets and not just plain sets; to make $\mathbb{A}^1$-homotopy work, you work with simplicial (pre?)sheaves and not just plain sheaves or schemes; to construct the cotangent complex (which if I understand correctly is a homotopical construction, hopefully a Quillen derived functor on the category of simplicial algebras), you use simplicial commutative rings. But why does "simplicial" make everything work so well? For instance, a simplicial set is a contravariant functor $\Delta \to \mathbf{Sets}$ for $\Delta$ the simplex category: what is so wonderful about $\Delta$ that allows a model structure (and one, moreover, Quillen equivalent to topological spaces) appear?
|
I don't think I have a compelling answer to this question, but maybe some bits and pieces that will be helpful. One point is that all of the examples that you bring up are related to the first: simplicial sets can be used as a model for the homotopy theory of spaces. Pretty much any homotopy theory can be "described" in terms of the homotopy theory of spaces, just like any category can be "described" in terms of the category of sets (via the Yoneda embedding, for example). So if you've decided that "space" means simplicial set, then it's pretty natural to start thinking about presheaves of simplicial sets when you want to think about the homotopy theory of (pre)sheaves of spaces, as in motivic homotopy theory. But that just brings us to the question "why use simplicial sets as a model for the homotopy theory of spaces"? It's certainly not the only model, and some alternatives have been listed in the other responses. Another alternative is more classical: the category of topological spaces can be used as a model for the homotopy theory of spaces. So, you might ask, why not develop the theory of the cotangent complex using topological commutative rings instead of simplicial commutative rings? There's no reason one couldn't do this; it's just less convenient than the alternative. There are several things that make simplicial sets very convenient to work with. 1) The category of simplicial sets is very simple: it is described by presheaves on a category with not too many objects and not too many morphisms, so the data of a simplicial set is reasonably concrete and combinatorial. The category of topological spaces (say) is more complicated in comparison, due in part to pathologies in point-set topology which aren't really relevant to the study of homotopy theory. 2) The category of simplices is (op)-sifted. This is related to the concrete observation that the formation of geometric realizations of simplicial sets (or simplicial spaces) commutes with finite products. More generally it guarantees a nice connection between the homotopy theory of simplicial sets and the homotopy theory of bisimplicial sets, which is frequently very useful. 3) The Dold-Kan correspondence tells you that studying simplicial objects in an abelian category is equivalent to studying chain complexes in that abelian category (satisfying certain boundedness conditions). So if you're already convinced that chain complexes are a good way to do homological algebra, it's a short leap to deciding that simplicial objects
are a good way to do homological algebra in nonabelian settings. This also tells you that when you "abelianize" a simplicial construction, you're going to get a chain complex
(as in the story of the cotangent complex: Kahler differentials applied to a simplicial commutative ring yields a chain complex of abelian groups). 4) Simplicial objects arise very naturally in many situations. For example, if
U is a comonad on a category C (arising, say, from a pair of adjoint functors), then applying iterates of U to an object of C gives a simplicial object of C. This sort of thing comes up often when you want to study resolutions. For example, let C be the category of abelian groups, and let U be the comonad U(G) = free group generated by the elements of G
(associated to the adjunction {Groups} <-> {Sets} given by the forgetful functor,free functor). Then the simplicial object I just mentioned is the canonical resolution of any group by free groups. Since "resolutions" play an important role in homotopy theory, it's convenient to work with a model that plays nicely with the combinatorics of the category of simplices. (For example, if we apply the above procedure to a simplicial group, we would get a resolution which was a bisimplicial free group. We can then obtain a simplicial free group by passing to the diagonal (which is a reasonable thing to do by virtue of (2) )). 5) Simplicial sets are related to category theory: the nerve construction gives a fully faithful embedding from the category of small categories to the category of simplicial sets.
Suppose you're interested in higher category theory, and you adopt the position that
"space" = "higher-groupoid" = "higher category in which all morphisms are invertible". If you decide that you're going to model this notion of "space" via Kan complexes, then working with arbitrary simplicial sets gives you a setting where categories (via their nerves)
and higher groupoids (as Kan complexes) both sit naturally. This observation is the starting point for the theory of quasi-categories. All these arguments really say is that simplicial objects are nice/convenient things to work with. They don't really prove that there couldn't be something nicer/more convenient. For this I'd just offer a sociological argument. The definition of a simplicial set is pretty simple (see (1)), and if there was a simpler definition that worked as well, I suspect that we would be using it already.
|
{
"source": [
"https://mathoverflow.net/questions/58497",
"https://mathoverflow.net",
"https://mathoverflow.net/users/344/"
]
}
|
58,577 |
I've been doing some light(?) reading on motives and the standard conjectures in an attempt to put various things that I tangentially know in perspective. The question is this: the Weil conjectures assert that $Z=\frac{P_1(t)...P_{2r-1}(t)}{P_0(t)...P_{2r}(t)}$ where the $P_i$'s are certain polynomials. (the assertion is of course stronger, but the rest of it is besides the point) What is the deep reason that these $P_i$'s alternate between numerator and denominator? In Milne's notes about motives ( http://www.jmilne.org/math/xnotes/MOT.pdf ), he asserts the following: let $hX$ be the motive corresponding to $X$, and let $h^0X$,...,$h^{2r}X$ be the conjectured decomposition into pure motives (conjecture C in Milne). He then says: define for a pure motive of weight $k$ the zeta function as the characteristic polynomial of the Frobenius if $k$ is odd, and its inverse if $k$ is even. Then extend the definition to motives by: the zeta function of a direct product of motives goes to the product of the zeta functions of the individual motives. Then indeed:
$Z(X,s)=Z(hX,s)=Z(h^0X,s)...Z(h^{2r}X,s)$, where $Z(h^kX,s)=P_k(t)$ for $k$ odd and $\frac{1}{P_k(t)}$ for $k$ even. So one can reduce this question to: why are we defining the zeta function of a motive of weight $k$ to be the characteristic polynomial of the Frobenius or its inverse depending on the parity?
|
The zeta function of a variety $X$ over a finite field is a priori defined to be a point counting function, i.e. it is the following product over the closed points of $X$ (thought of as a scheme):
$$\zeta_X(s) = \prod_{x}(1 - | \kappa(x)|^{-s})^{-1},$$
where $\kappa(x)$ is the residue field of $x$ and $|\kappa(x)|$ denotes its order. (This is motivated by analogy with the Riemann zeta function, which is what we get if we apply the same definition with $X$ replaced by Spec $\mathbb Z$.) Now this will be a Dirichlet series involving only powers of $p^{-s}$ (if $p$ is the char. of the finite field), and so replacing $p^{-s}$ by $T$, we obtain a power series in $T$, whose
log can be reinterpreted in the usual way as a generating function counting the number of points of $X$ with values in the various extensions of $\mathbb F_p$. Now one can count these points by the Lefschetz fixed point formula (applied to the $\ell$-adic cohomology), and this gives the alternating product of char. polys. of Frobenius that you write down in your question. Of course, one could write down their product, rather than their alternating product, but the resulting power series would not have any particular interpretation; in particular, it wouldn't be related to counting points of $X$ in the same way that the zeta function is. Milne's definition of the $\zeta$-function directly in terms of $\ell$-adic cohomology is to some extent putting the cart before the horse; as Stopple notes, it is a reasonable definition only because of the back story about counting points and so on. Nevertheless, if you want to take the definition in terms of cohomology as the basic one, then you can ask yourself: how should you define such a quantity if you want it to behave well under chopping up varieties (which is what motives essentially are --- pieces of varieties cut out by correspondences). The basic quantity that is defined in terms of cohomology and which is additive with respect to cutting up spaces is the Euler characteristic. And for this additivity to hold, it is crucial that involve an alternating sum, with the sign being dictated by the cohomogical degree. The reason is that the behaviour of cohomology under chopping up and/or gluing is given by the excision and Mayer--Vietoris long exact sequences, and it is the alternating sum of the dimensions which is additive in exact sequences. Viewed cohomologically, the zeta function is like an enhanced, multiplicative version of the Euler characteristic, and like the Euler characteristic, for it to be multiplicative with respect to cutting up varieties, we must form it via an alternating product. In conclusion: I think that the "deep reason" that you are looking for is the yoga of Euler characteristics.
|
{
"source": [
"https://mathoverflow.net/questions/58577",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5309/"
]
}
|
58,633 |
(Previously posted on math.SE with no answers.) Let $G$ be a compact Lie group and $V$ a faithful (complex, continuous, finite-dimensional) representation of it. Is it true that every (complex, continuous, finite-dimensional) irreducible representation of $G$ occurs in $V^{\otimes n} (V^{\ast})^{\otimes m}$ for some $n, m$? The proof I know for finite groups doesn't seem to easily generalize. I want to apply Stone-Weierstrass, but can't figure out if the characters I get will always separate points (in the space of conjugacy classes). Ben Webster in his answer to this MO question seems to be suggesting that this follows from the first part of Peter-Weyl, but I don't see how this works. Certainly the corresponding algebra of characters separates points whenever the eigenvalues of an element $g \in G$ acting on $V$ determine its conjugacy class (since one can get the eigenvalues from an examination of the exterior powers). Does this always happen?
|
You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$ , with characters $\chi_X$ and $\chi_Y$ , then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \chi_Y$ , where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$ . The proof of this is exactly as in the finite case. Now, let $W=1 \oplus V \oplus \overline{V}$ . Your goal is to show that, for any nonzero representation $Y$ , $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else. I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$ . The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$ , of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$ , we have $f=1+2 \sum \cos \theta_i$ . In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have $$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$ where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$ .
And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$ . So the contribution to our integral near the identity can be approximated by $$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$ where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$ . You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$ , I'll leave that to you as well. For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial. Now, look at the contribution from the rest of $G$ . For any point in the unitary group $U(d)$ , other than the identity, we have $|f| < 2n+1$ . (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$ . So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar .
|
{
"source": [
"https://mathoverflow.net/questions/58633",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
58,696 |
I don't mean to be rude asking this question, I know that the theory of Lie groups and Lie algebras is a very deep one, very aesthetic and that has broad applications in various areas of mathematics and physics. I visited a course on Lie groups, and an elementary one on Lie algebras. But I don't fully understand how those theories are being applied. I actually don't even understand the importance of Lie groups in differential geometry. I know, among others, of the following facts: If $G$ and $H$ are two Lie groups, with $G$ simply connected, and $\mathfrak{g,h}$ are their respective Lie algebras, then there is a one to one correspondance between Lie algebra homomorphisms $\mathfrak{g}\rightarrow\mathfrak{h}$ and group homomorphisms $G\rightarrow H$ . The same remains true if we replace $H$ with any manifold $M$ : any Lie algebra homomorphism from $\mathfrak{g}$ to the Lie algebra $\Gamma(TM)$ of smooth vector fields on $M$ gives rise to a local action of $G$ on $M$ . Under some conditions like (I think) compactness, the cohomology of $\mathfrak{g}$ is isomorphic to the real cohomology of the group $G$ . I know that calculating the cohomology of $\mathfrak{g}$ is tractable in some cases. There is a whole lot to be said of the representation theory of Lie algebras. Compact connected centerless Lie groups $\leftrightarrow$ complex semisimple Lie algebras How do people use Lie groups and Lie algebras? What questions do they ask for which Lie groups or algebras will be of any help? And if a geometer reads this, how, if at all, do you use Lie theory? How is the representation theory of Lie algebras useful in differential geometry?
|
Here is a brief answer: Lie groups provide a way to express the concept of a continuous family of symmetries for geometric objects. Most, if not all, of differential geometry centers around this. By differentiating the Lie group action, you get a Lie algebra action, which is a linearization of the group action. As a linear object, a Lie algebra is often a lot easier to work with than working directly with the corresponding Lie group. Whenever you do different kinds of differential geometry (Riemannian, Kahler, symplectic, etc.), there is always a Lie group and algebra lurking around either explicitly or implicitly. It is possible to learn each particular specific geometry and work with the specific Lie group and algebra without learning anything about the general theory. However, it can be extremely useful to know the general theory and find common techniques that apply to different types of geometric structures. Moreover, the general theory of Lie groups and algebras leads to a rich assortment of important explicit examples of geometric objects. I consider Lie groups and algebras to be near or at the center of the mathematical universe and among the most important and useful mathematical objects I know. As far as I can tell, they play central roles in most other fields of mathematics and not just differential geometry. ADDED: I have to say that I understand why this question needed to be asked. I don't think we introduce Lie groups and algebras properly to our students. They are missing from most if not all of the basic courses. Except for the orthogonal and possibly the unitary group, they are not mentioned much in differential geometry courses. They are too often introduced to students in a separate Lie group and algebra course, where everything is discussed too abstractly and too isolated from other subjects for my taste.
|
{
"source": [
"https://mathoverflow.net/questions/58696",
"https://mathoverflow.net",
"https://mathoverflow.net/users/13700/"
]
}
|
58,721 |
There are many optimization problems in which the variables are symmetric in the objective and the constraints; i.e., you can swap any two variables, and the problem remains the same. Let's call such problems symmetric optimization problems. The optimal solution for a symmetric optimization problem - like many of the ones that show up in calculus texts - frequently has all variables equal. To take some simple examples, The rectangle with fixed area that minimizes perimeter is a square. (Minimize $2x+2y$ subject to $xy = A$ and $x,y \geq 0$.) The rectangle with fixed perimeter that maximizes area is a square. (Maximize $xy$ subject to $2x + 2y = P$ and $x,y \geq 0$.) The difference between the arithmetic mean and the geometric mean of a set of numbers is minimized (and equals $0$) when all the numbers are equal. There are also more complicated symmetric optimization problems for which the variables are equal at optimality, such as the one in this recent math.SE question . However, it is not true that every symmetric optimization problem has all variables equal at optimality. For example, the problem of minimizing $x +y$ subject to $x^2 + y^2 \geq 1$ and $x, y \geq 0$ has $(0,1)$ and $(1,0)$ as the optimal solutions. Does anyone know of general conditions on a symmetric optimization problem that guarantee the optimal solution has all variables equal? The existence of such conditions might be very nice. Unless the conditions themselves are ugly, they ought to vastly simplify solving a large class of symmetric optimization problems. (Maybe convexity plays a role? My last example has a nonconvex feasible region.)
|
The Monthly article " Do symmetric problems have symmetric solutions? " by William Waterhouse discusses this issue. For global optimality one really needs strong global constraints on the objective function, such as convexity (as in Igor's answer), and there's nothing one can say otherwise. However, local results hold in considerably greater generality. Waterhouse calls this the Purkiss Principle: in a symmetric optimization problem, symmetric points tend to be either local maxima or local minima. Specifically, suppose we are optimizing a smooth function $f$ on a manifold $M$. The symmetry can be expressed by saying some finite group $G$ acts on $M$ and preserves $f$. If $x$ is a point in $M$ that is fixed by $G$, we would like to understand the behavior of $f$ near $x$. To analyze it, we can study the action of $G$ on the tangent space $T_x M$. The key hypothesis is that $T_x M$ should be a nontrivial irreducible representation of $G$. In that case, $x$ is automatically a critical point for $f$, and if it is a nondegenerate critical point, then it must be a local maximum or minimum (i.e., not a saddle point). In fact, even more is true: the Hessian matrix will have only one eigenvalue, which is either zero (degenerate), positive (local minimum), or negative (local maximum). This is one of those results where finding the right statement is the real issue, and once the statement has been found it takes just a few lines to prove it. In his article, Waterhouse builds up to this formulation in several steps, and he shows how it encompasses various more concrete cases and applications. He also gives a wonderful historical overview. If the representation on the tangent space is reducible, then the Purkiss Principle can fail. If we break the representation into a direct sum of irreducible subrepresentations, then the Hessian matrix for $f$ will have an eigenvalue for each irreducible (with multiplicity equal to the dimension of the irreducible), and there's no reason why they should all have the same sign. However, this decomposition is nevertheless very useful for doing local calculations, because even if $M$ is high-dimensional, $T_x M$ may decompose into just a few irreducibles. So the upshot is this: if you are applying the second derivative test to a symmetric optimization problem, then representation theory will tell you what the symmetry implies, and irreducibility leads to the simplest possible answer.
|
{
"source": [
"https://mathoverflow.net/questions/58721",
"https://mathoverflow.net",
"https://mathoverflow.net/users/9716/"
]
}
|
58,870 |
I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook Differential Topology by
Guillemin and Pollack, supplemented by Milnor's book . My question is: What are good topics to cover that are not in assigned textbooks?
|
I nominate Ehresmann's theorem according to which a proper submersion between manifolds is automatically a locally trivial bundle. It is incredibly useful, in deformation theory for example, but is sadly neglected in introductory courses and books on manifolds. It is completely elementary: witness these lecture notes by Peter Petersen, where it is proved in a few lines on page 9, the prerequisites being about two pages long. Bjørn Ian Dundas and our friend Andrew Stacey also have online documents proving this theorem.
|
{
"source": [
"https://mathoverflow.net/questions/58870",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5337/"
]
}
|
58,988 |
Are there non-compact complex manifolds that
a) Don't embed in C^n (holomorphically)
and
b) Cannot be covered by a finite number of coordinate open sets?
If b) can be satisfied, then I think so can a) be by taking a product with a compact complex manifold. If one takes a Riemann surface of infinite genus, one does not have a "good" finite open cover, but I allow non-contractible open covers as well. Apologies in advance for this elementary question.
|
Fornaess and Stout proved that EVERY complex manifold (connected and second countable) can be covered by finitely many open subsets biholomorphic to a polydisc (Lemma II.1 in MR0470251). They even have an explicit bound on the size of the cover in terms of the dimension of the manifold. Further results of a similar flavour are contained in their papers MR0435441 and MR0662439.
|
{
"source": [
"https://mathoverflow.net/questions/58988",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3709/"
]
}
|
59,071 |
I'm a graduate student who's been learning about schemes this year from the usual sources (e.g. Hartshorne, Eisenbud-Harris, Ravi Vakil's notes). I'm looking for some examples of elementary self-contained problems that scheme theory answers - ideally something that I could explain to a fellow grad student in another field when they ask "What can you do with schemes?" Let me give an example of what I'm looking for: In finite group theory, a well known theorem of Burnside's is that a group of order $p^a q^b$ is solvable. It turns out an easy way to prove this theorem is by using fairly basic character theory (a later proof using only 'elementary' group theory is now known, but is much more intricate). Then, if another graduate student asks me "What can you do with character theory?", I can give them this example, even if they don't know what a character is. Moreover, the statement of Burnside's theorem doesn't depend on character theory, and so this is also an example of character theory proving something external (e.g. character theory isn't just proving theorems about character theory). I'm very interested in learning about similar examples from scheme theory. What are some elementary problems (ideally not depending on schemes) that have nice proofs using schemes? Please note that I'm not asking for large-scale justification of scheme theoretic algebraic geometry (e.g. studying the Weil conjectures, etc). The goal is to be able to give some concrete notion of what you can do with schemes to, say, a beginning graduate student or someone not studying algebraic geometry.
|
A smooth projective variety over $\mathbb{Q}$ has only finitely many places of bad reduction. Shimura had a horribly complicated proof of this in the language of Weil's foundations in a paper from the 50's. With schemes, it's completely obvious, as smoothness is an open condition. Even stating this without schemes is painful. The whole field of arithmetic geometry is an example of what you want. Actually, this is my only serious complaint about Hartshorne. He doesn't do any Number Theory and $\operatorname{Spec}\mathbb{Z}$ is where schemes really shine.
|
{
"source": [
"https://mathoverflow.net/questions/59071",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4891/"
]
}
|
59,086 |
Suppose you start from partial differential equations and functional analysis (on $\mathbb R^n$ and on real manifolds). Which prominent example problems lead you to work with pseudo-differential operators? I would appreciate any good examples, as well as some historical outlines on the topic's development. (Shubin's classical book spends a few lines on history and motivation in the preface, but no "natural" examples. I am not aware of any historical outlines in the literature.)
|
I don't know the history at all, but I have to imagine that the language was invented to provide a context for talking about solution operators for differential equations. Consider, for example, the PDE $D f = f_0$ where $D$ is a nice differential operator. Taking Fourier transforms, this says that $P(\xi)\hat{f} = \hat{f}_0$, where $P$ is the principal symbol of $D$ (a polynomial). Everyone in the world just wants to write $\hat{f} = \frac{1}{P(\xi)}\hat{f}_0$ and take inverse Fourier transforms. In other words, solving the PDE is the same thing as finding an operator $S$ whose Fourier multiplier is $\frac{1}{P(\xi)}$. This most likely fails to be a polynomial, so $S$ is evidently not a differential operator. As far as I can tell, many of those big fat books on pseudodifferential operator theory are all about how to invert as many operators as possible in this sense while salvaging as much regularity as you can. It gets extremely subtle, but I think the motivation is fairly close to the surface. Aside from that, you might also be led to invent pseudodifferential operators if you cared deeply about the spectral theory of differential operators. The spectral theorem for an operator $T$ is more or less equivalent to the existence of a "functional calculus", i.e. a sensible way to form operators $f(T)$ out of various classes of functions $f$ on the spectrum of $T$. For differential operators (especially on non-compact domains where there need not be a nice eigenspace decomposition), the functional calculus is often obtained via the Fourier transform, and the pseudodifferential calculus manifests itself.
|
{
"source": [
"https://mathoverflow.net/questions/59086",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2082/"
]
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.