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https://mathoverflow.net/questions/24280 | 4 | $(A, \mathfrak{m})$ a Noetherian local ring, $a\in\mathfrak{m}$ a *zero divisor*. Then is it true that $\mbox{dim}\ A/(a) = \mbox{dim}\ A$ ?
| https://mathoverflow.net/users/5292 | Converse of Principal Ideal Theorem | No. Let A be the ring k[x,y,z]/(xz,yz) localized at (x,y,z). Then the dimension of A is 2, but the dimension of A/(x) is one. Note that xz = 0 in A.
The geometric picture is this: In 3-space, take the union of the horizontal z=0 plane with the vertical line x = y = 0, and look at the local ring at the origin. Then restricting to x=0, we get the union of the y-axis and the z-axis.
| 15 | https://mathoverflow.net/users/5094 | 24285 | 15,942 |
https://mathoverflow.net/questions/24265 | 62 | Let $s\_n = \sum\_{i=1}^{n-1} i!$ and let $g\_n = \gcd (s\_n, n!)$. Then it is easy to see that $g\_n$ divides $g\_{n+1}$. The first few values of $g\_n$, starting at $n=2$ are $1, 3, 3, 3, 9, 9, 9, 9, 9, 99$, where $g\_{11}=99$. Then $g\_n=99$ for $11\leq n\leq 100,000$.
Note that if $n$ divides $s\_n$, then $n$ divides $g\_m$ for all $m\geq n$. If $n$ does not divide $s\_n$, then $n$ does not divide $s\_m$ for any $m\geq n$.
If $p$ is a prime dividing $g\_n$ but not dividing $g\_{n-1}$ then $p=n$, for if $p<n$ then $p$ divides $(n-1)!$ and therefore $p$ divides $s\_n-(n-1)!=s\_{n-1}$, whence $p$ divides $g\_{n-1}$.
So to show that $g\_n\rightarrow \infty$ it suffices to show that there are infinitely many primes $p$ such that $1!+2!+\cdots +(p-1)! \equiv 0$ (mod $p$).
| https://mathoverflow.net/users/1243 | What is the limit of gcd(1! + 2! + ... + (n-1)! , n!) ? | This is so close to the Kurepa conjecture which asserts that $\gcd\left(\sum\_{k=0}^{n-1}k!,n!\right)=2$ for all $n\geq 2$, which was settled in 2004 by D. Barsky and B. Benzaghou "Nombres de Bell et somme de factorielles". So what they proved is that $K(p)=1!+\cdots+(p-1)!\neq -1\pmod{p}$ for any odd prime $p$. This goes against Kevin Buzzard's heuristic that $K(p)$ is random mod $p$. Let me mention two ways you can restate the fact $p|K(p)$:
a) It is equivalent to $K(\infty)=\sum\_{k=1}^{\infty}k!$ not being a unit in $\mathbb Z\_p$.
b) It is equivalent to $\mathcal B\_{p-1}=2\pmod{p}$ where $\mathcal{B} \_n$ is the $n$th Bell number. (It is easy to show that $\mathcal B \_{p}=2\pmod{p}$)
I forgot to mention that the conjecture that $p>11$ doesn't divide $K(p)$ is in question B44 of R. Guy's "Unsolved Problems in Number theory".
| 25 | https://mathoverflow.net/users/2384 | 24286 | 15,943 |
https://mathoverflow.net/questions/24020 | 5 | I am reading the book 'surface in 4-space' about the unknotting conjecture (Page 97): a 2-knot (2-sphere in 4-sphere) is trival if and only if the fundamental group of the exterior is infinite cyclic.
It said that in TOP category, Freedman proved the statement is true. I don't know why it is also true for general surface. in top category?
| https://mathoverflow.net/users/5980 | Is there a knotted torus in 4-sphere whose complement's fundamental group is infinite cyclic ? | I have been avoiding addressing this since almost all that I know about the question is in that book. I don't recall exactly, but I think that Kawauchi showed that a torus with the fundamental group of the complement being Z is topologically unknotted. Recent work of Hillman <http://arxiv.org/pdf/1003.5408>
addresses some problems of 2-knot groups, but he deals with the spherical case.
| 1 | https://mathoverflow.net/users/36108 | 24295 | 15,948 |
https://mathoverflow.net/questions/24318 | 30 | If there are not, then would it be easier to say that 2 objects are identical as ordered fields as opposed to being isomorphic as ordered fields? Or is the word isomorphism used to emphasise the fact that the objects are different as sets?
| https://mathoverflow.net/users/4692 | Are there situations when regarding isomorphic objects as identical leads to mistakes? | Inside of the complex numbers there are lots of examples of distinct fields which are isomorphic. For instance, there are three subfields of the form ${\mathbf Q}(\alpha)$ where $\alpha^3 = 2$: take for $\alpha$ any of the three complex cube roots of 2 and you get a different subfield. What are the consequences of treating them as literally equal? You can't make any sense of Galois theory if you do that! Similarly, all $p$-Sylow subgroups of a finite group are isomorphic (since conjugate subgroups are isomorphic groups), but it would kind of destroy a lot of the content of the Sylow theorems by
trying to say the $p$-Sylow subgroups are identical.
More generally, anytime you have isomorphic but unequal objects inside a larger object, it can lead to confusion if not outright incomprehensibility if you try to regard them all as identical. (There was a paper by Chevalley about unit groups in number fields where he made a genuine error by an abuse of the "square root" notation and I think one might be able to express the mistake in the form of an isomorphism being confused with an equality, but I'd have to look at the paper again to be sure about this.)
The word isomorphism does not emphasize that two objects are different; any group or vector space admits an isomorphism with itself using the identity map. The word emphasizes that in a structural way the two objects look like each other even though they are not literally the same. Never say two objects are identical if they are not actually identical. Having said that, I must admit that in mathematics one meets phrases like "since $X$ and $Y$ are isomorphic we can identify $X$ with $Y$" and then $X$ is replaced with $Y$. The usefulness of doing this depends on the application you have in mind. Note, however, that replacing $X$ with $Y$ is not saying that $X$ and $Y$ *are* the same thing.
This question sounds like it is being asked by someone who hasn't had a lot of experience with isomorphisms and is trying to get a feel for what it means. In a year or two, after seeing more appearances of the concept and its uses, you'll get a better feel for it, but for now do *not* think the word isomorphism is a synonym for identical.
| 42 | https://mathoverflow.net/users/3272 | 24320 | 15,959 |
https://mathoverflow.net/questions/10039 | 11 | **Originally asked by [Ali Dino Jumani](https://mathoverflow.net/users/2866/ali-dino-jumani); EXTENSIVELY EDITED by David Speyer**. The previous version was a bit confused, but Steven Sivek and Graham, in the comments, figured out what was going on.
---
G. C. Shephard, in his paper "[Twenty Problems on Convex Polyhedra: Part I](http://www.jstor.org/pss/3612678)", associates to a three dimensional polyhedron the sequence $(p\_3, p\_4, p\_5, p\_6, \ldots)$, with $p\_k$ being the number of facets that are $k$-gons. He poses the problem of characterizing all sequences of integers which arise in this way.
Are there any developments and references on this problem?
the original question as appeared in Shephard's paper "Given any finite sequence (f3, f4, ...., fm) of non-negative integers, find a necessary and sufficient condition for it to be assocoated with some convex polyhedron". Following Grunbaum's Convex Polytope 2e notation, section 13.3 under Eberhard's Therorem heading, f3 is triangle renamed as p3 and f4 as p4 is square and fm as pk is n-gon. A convex polyhedron containing these faces satisfies the Eberhard's criterion.
Revisions and corrections are highly appreciated, Thanks.
| https://mathoverflow.net/users/2866 | Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:) | Eberhard Theorem
----------------
Consider a simple 3-polytope P, (so every vertex has 3 neighbors). If $p\_k$ is the number of faces of P which are k-gonal, Euler's theorem implies that
$$(\*) ~~\sum\_{k \ge 3} (6-k)p\_k=12.$$
Note that 6-gonal faces do not contribute to the LHS.
One way to think about it is that polygonal faces with 7 and more sides contribute "negative curvature", small faces namely triangles quadrangles and pentagons, contribute positive "curvature" and hexagons are "flat".
Eberhard's theorem asserts that if you have a sequence of numbers $p\_k, k \ne 6$ such that $\sum\_{k \ge 3} (6-k)p\_k=12$ then you can find a simple 3-polytope with $p\_k$ k-gonal faces. (But you have no control on $p\_6$).
Extensions of Eberhard theorem
------------------------------
There are various results extending Eberhard's theorem in various directions. Chapter 13 in Grunbaum's book "[Convex Polytopes](http://books.google.co.il/books?id=ISHO86XJ1CsC&printsec=frontcover&dq=grunbaum%27s+convex+polytopes&source=bl&ots=S7T41O6hvF&sig=wRkbvA0TtPVeRuiRFHo6FjVQv1I&hl=iw&ei=QkrqS7LzCaOiOLaTsPMK&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAgQ6AEwAA#v=onepage&q=grunbaum%27s%2520convex%2520polytopes&f=false)" and especially the supplemantary material at the end of the chapter in the new 2nd edition is a good source. Another general source is [my chapter](http://www.ma.huji.ac.il/%7Ekalai/ch19.pdf) from the "Handbook of Discrete and Computational geometry" on garphs and skeleta of polytops.
A relatively recent paper on the subject is by Stanislav Jendrol "[On the face vectors of trivalent convex polyhedra](http://pdf.dml.cz/bitstream/handle/10338.dmlcz/136327/MathSlov_33-1983-2_7.pdf)". Another paper by Jendrol which deals with general 3-polytopes from the same year is "On face vectors and vertex vectors of convex polyhedra" Discrete Math 118 (1993)119-144. There are analogs of Eberhard theorem for 4-regular planar graphs, for toroidal graphs and in other directions.
A far as I know, there is no good answer known for the question posed by Shephard of characterizing all sequences $(p\_3,p\_4,\dots)$, and no such characterization is known even for the simple case.
High dimensions
---------------
In higher dimensions and even in four dimensions there are various different ways to extend these problems, these problems become very difficult and very little is known.
### 2-dimensional faces
You can ask again about the numbers of k-gonal 2-dimensional faces. While the formula above implies that in dimension 3 and more $p\_3+p\_4+p\_5>0$ it is known that in dimensions 5 and more $p\_3+p\_4>0$.
### Types of facets
Perhaps an even more natural extension is to consider the type of facets a given d-simensional polytope have. You can ask for a simple 4 polytope (a 4-polytope whose graph is 4-regular) what are the number of facets $p\_Q$ isomorphic to a given 3-polytope Q. This gives you a vector indexed by combinatorial types of simple 3-polytopes, but I am not aware of any Eberhard type theorem and I do not know even which 3-polytopes should be considered as the analogs of the hexagons in the above formula. Dually stated and extented to triangulations of 3-spheres the question is to associate to a triangulated 3-simensional sphere (or just simplicial 3-polytope) the list of links of vertices (with multiplicities) it has. A related MO question is [this one](https://mathoverflow.net/questions/10503/simplicial-and-cubical-decompositions-of-low-valence).
### Type of facets according to their number of their facets
Rather than classifying the facets according to their full combinatorial type we can classify them according to their own number of facets. Here, for dimension greater than 4 there is no analog for (\*). It is possible that under wide circumstences the numbers of facets with k facets can be prescribed, but I am not aware of results in this direction.
### Type of facets according to their f-vectors
Precscribing the entire vector of face numbers of the individual facets, may well be the most interesting extension from 3 to higher dimensions.
There are some reasons to jump from dimension 3 directly to dimension 5. The nature of the problem is different in even and odd dimensions. Let me try to elaborate on that.
Suppose you have a simple 5-polytope P and denote by $p\_{a,b}$ the number of facets F so that $f\_3(F)=a$ and $f\_2(F)=b$. (For a polytope or some other cell complex X, $f\_i(X)$ denotes the number of $i$-dimensional faces of $X$.) Dually, we can consider a triangulation $K$ of the 4-simensional sphere $S^4$ and let $p\_{a,b}$ be the number of vertices whose **link** has $a$ vertices and $b$ edges.
Now the Dehn-Sommervill relations imply that
$$(\*\*) \sum p\_{a,b} (b-6a+30) = 60 .$$
So now we can consider 4 dimensional facets as "small" if b-6a+30 is negative, as "large" if $b-6a+30$ is positive, and as neutral if $b=6a-30$. An analog of Eberhard theorem will say that you can prescribe the types of small and large facets and realize the polytope (or the triangulated sphere) by allowing to add many "neutral" facets.
### Facets and non facets
A related question that was studied in several papers is: for which d-polytopes P there is a (d+1) polytopes Q such that all faces of Q are isomorphic to P. Such polytopes P are called "facets". An intriguing open problem is if the icosahedron is a facet.
### Cubical complexes
Simple polytopes are dual to simplicial polytopes, and there is some special interest in these problems for cubical complexes (or their duals). Indeed there is an analog of Eberhard's theorem for 3-polytopes with all vertex degrees 4. There should be an analog for (\*\*) and this may be related to interesting questions about curvature of cubical complexes.
### Stacked polytopes
A stack polytope is a polytope obtained by gluing simplices along facets. (They are related to Appolonian circle packing.) Stacked polytopes are simplicial. I do not know if there is an analog of Eberhard's theorem when we consider only simple 3-polytope which are dual to stacked 3-polytopes. This is also interesting for other dimensions. All facets of stacked 5 polytopes are stacked and their number of edges is four times the number of vertices minus 10. Therefore, for a dual-to-stacked 5 polytope formula (\*\*) reduces to $$\sum p\_{a,b}(10-a)=30.$$ (Here, $b=4a-10$ so we can omit the subscript $b$.) So in this case, the distinction is between facets with $a<10$, those with $a>10$ and those with $a=10$. Again we can look for some analogs for Eberhard's theorem, which in this case, might be easier.
| 20 | https://mathoverflow.net/users/1532 | 24340 | 15,972 |
https://mathoverflow.net/questions/24345 | 3 | Let $G$ be a reductive algebraic group and $\varrho$ a representation of $G$ in $GL(n)$. Is it true that $\varrho$ is completely reducible? Moreover, how are related the representations of the Lie algebra $\mathfrak{g}$ of $G$ with the one of $G$?Finally, the centre of the identity component of $G$ consists of semisimple transformations, is it true also for $\mathfrak{g}$?
| https://mathoverflow.net/users/4821 | Representations of reductive Lie group | You need to be over a field of zero characteristic and your representation needs to be rational, i.e. matrix entries need to be algebraic functions on $G$. Then it is completely reducible, see any book on algebraic groups, e.g., Jantzen or Humphreys.
You can always differentiate, so a differential of a map $G\rightarrow GL(V)$ is a representation of ${\mathfrak g}$. In the opposite direction, a certain care is required. To integrate a vector field, you need exponential function, which is not, in general, algebraic. However, for a semisimple group in characteristic zero, you have enough nilpotent elements $X\in{\mathfrak g}$, so that the polynomials $e^{\rho (X)}$ define a representation of the group.
Finally, the answer is no. Take ${\mathfrak g}$ to be one-dimensional Lie algebra acting on $K^2$ by the nilpotent nonzero transformation.
| 1 | https://mathoverflow.net/users/5301 | 24349 | 15,977 |
https://mathoverflow.net/questions/24321 | 1 | Let $\Lambda$ be an $n$ dimensional sublattice of the integer lattice $\mathbb{Z}^n$. The quotient $\mathbb{Z}^n/\Lambda$ has order $\sqrt{\det{\Lambda}}$.
What is the best/standard way to compute a set of coset representatives for this quotient?
Edit: I initially forgot to take the square root of $\det{\Lambda}$, which is likely the reason for KCronrad's initial comment.
| https://mathoverflow.net/users/5378 | Computing a set of coset representatives for $\mathbb{Z}^n / \Lambda$ | As KConrad suggested (why only in the comments?), Smith's normal formal is your best bet. Its running time is insensitive to $m=|{\mathbb Z}^n/\Lambda |$ (unless you need to use arbitrary long entries in your matrix) and behaves as $n^3$.
You may also try coset enumeration, whose running time is usually unbounded but may be bounded in this case by something like $(mn)^2$.
| 3 | https://mathoverflow.net/users/5301 | 24351 | 15,978 |
https://mathoverflow.net/questions/24361 | 6 | Let $X$ be a topological space and let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over $X$.
Of course, if one has a morphism $f : \mathcal{F} \to \mathcal{G}$ such that for all $x\in X$, $f\_x : \mathcal{F}\_x \to \mathcal{G}\_x$ is an isomorphism, then it is known that $f$ itself is an isomorphism.
My question is the following: if we don't have such a morphism $f$, but if we know that for all $x\in X$, $\mathcal{F}\_x$ and $\mathcal{G}\_x$ are isomorphic, is it true that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic ?
| https://mathoverflow.net/users/2330 | Are two sheaves that are locally isomorphic globally isomorphic ? | Definitely not. If $X$ is a ringed space, then a $\mathcal{O}\_X$-module $F$ is called locally free of rank $1$, if $X$ is covered by open subsets $U\_i$ such that $F|\_{U\_i}$ is free of rank $1$ over $\mathcal{O}\_{U\_i}$. The correspond to line bundles on $X$. Line bundles form a group, called the Picard group of $X$ and denoted by $\text{Pic}(X)$. This group does not have to vanish: If $X$ is a CW complex and we take the sheaf of continuous functions, then $\text{Pic}(X)$ is isomorphic to $H^1(X,\mathbb{Z}/2)$. For $X=S^1$, the moebius strip is the nontrivial element here. The corresponding example in algebraic geometry is $\text{Pic}(\mathbb{P}^n\_k)=\mathbb{Z}$, the generator given by the Serre twist $\mathcal{O}(1)$.
Of course, there are more easy counterexamples, but I wanted to indicate that there is a rich theory coming from the observation that two locally isomorphic sheaves are not isomorphic.
| 15 | https://mathoverflow.net/users/2841 | 24364 | 15,987 |
https://mathoverflow.net/questions/24358 | 2 | I need a reference which states which of the "normal properties of vector spaces" carry over to free $\mathbb{Z}$-modules.
Especially I am interested in things like: If you have a linear map between two free $\mathbb{Z}$-modules and you choose a basis for its kernel, can you choose a basis of a complementary space so that both together form a basis of the whole space (and the map, viewed only on this complementary space, is an isomorphism on its image)?
Probably this is an easy question for algebra guys.
| https://mathoverflow.net/users/3816 | free Z-modules: Bases etc. | What carries over?
As Peter pointed out, a submodule of a free $\mathbb{Z}$-module though
free need not have a complement. Indeed each submodule of a free
$\mathbb{Z}$-module is free, but a quotient module need not be, for instance
$\mathbb{Z}/2\mathbb{Z}$. Also a $\mathbb{Z}$-module is free if and oly if
it is projective; this entails that a kernel of a map of free modules
does have a complement.
The set $\mathrm{Hom}(F,G)$ for free $\mathbb{Z}$-modules need not be free.
If $F$ is free of countably infinite rank and $G=\mathbb{Z}$, then
$\mathrm{Hom}(F,G)\cong\prod\_{j=1}^\infty\mathbb{Z}$ which remarkably
is not free over $\mathbb{Z}$. But $F\otimes G$ is free for free $F$ and $G$.
| 3 | https://mathoverflow.net/users/4213 | 24371 | 15,994 |
https://mathoverflow.net/questions/24307 | 17 | Henry Segerman and I recently considered the following question:
Given a fixed area $A < \pi$ and two fixed points in the upper half-plane model for hyperbolic $2$-space, what is the locus of points which give rise to a hyperbolic triangle of the given area?
We found it a fun exercise in hyperbolic geometry to show that the answer is a Euclidean straight line, or an arc of a Euclidean circle. As this requires only elementary properties of hyperbolic geometry, we strongly suspect it should be known, but have thus far been unable to find a reference for it. Does anyone know whether it's known, and if so, where one can find it?
| https://mathoverflow.net/users/6040 | Locus of equal area hyperbolic triangles | Nice problem! After googling "hyperbolic triangles of equal area on a fixed base" I found the paper "Extremal properties of the principal Dirichlet eigenvalue for regular polygons in the hyperbolic plane" by Karp and Peyerimhoff. Their Theorem 7 looks like the result you ask for. In a footnote KP refer to a 1965 book of Fejes Toth, "Regulare Figuren", as containing the same result. There appears to be an English version "Regular Figures" published in 1964.
This question seems very interesting in spherical geometry, as well.
| 8 | https://mathoverflow.net/users/1650 | 24380 | 15,999 |
https://mathoverflow.net/questions/24350 | 12 | As I understand it, mathematics is concerned with correct deductions using postulates and rules of inference. From what I have seen, statements are called true if they are correct deductions and false if they are incorrect deductions. If this is the case, then there is no need for the words true and false. I have read something along the lines that Godel's incompleteness theorems prove that there are true statements which are unprovable, but if you cannot prove a statement, how can you be certain that it is true? And if a statement is unprovable, what does it mean to say that it is true?
| https://mathoverflow.net/users/4692 | What does it mean for a mathematical statement to be true? | Tarski defined what it means to say that a first-order statement is true in a structure $M\models \varphi$ by a simple induction on formulas. This is a completely mathematical definition of truth.
Goedel defined what it means to say that a statement $\varphi$ is *provable* from a theory $T$, namely, there should be a finite sequence of statements constituting a *proof*, meaning that each statement is either an axiom or follows from earlier statements by certain logical rules. (There are numerous equivalent proof systems, useful for various purposes.)
The Completeness Theorem of first order logic, proved by Goedel, asserts that a statement $\varphi$ is true in all models of a theory $T$ if and only if there is a proof of $\varphi$ from $T$. Thus, for example, any statement in the language of group theory is true in all groups if and only if there is a proof of that statement from the basic group axioms.
The Incompleteness Theorem, also proved by Goedel, asserts that any consistent theory $T$ extending some a very weak theory of arithmetic admits statements $\varphi$ that are not provable from $T$, but which are true in the intended model of the natural numbers. That is, we prove in a stronger theory that is able to speak of this intended model that $\varphi$ is true there, and we also prove that $\varphi$ is not provable in $T$. This is the sense in which there are true-but-unprovable statements.
The situation can be confusing if you think of *provable* as a notion by itself, without thinking much about varying the collection of axioms. After all, as the background theory becomes stronger, we can of course prove more and more. The true-but-unprovable statement is really unprovable-in-$T$, but provable in a stronger theory.
Actually, although ZFC proves that every arithmetic statement is either true or false in the standard model of the natural numbers, nevertheless there are certain statements for which ZFC does not prove which of these situations occurs.
Much or almost all of mathematics can be viewed with the set-theoretical axioms ZFC as the background theory, and so for most of mathematics, the naive view equating *true* with *provable in ZFC* will not get you into trouble. But the independence phenomenon will eventually arrive, making such a view ultimately unsustainable. The fact is that there are numerous mathematical questions that cannot be settled on the basis of ZFC, such as the Continuum Hypothesis and many other examples. We have of course many strengthenings of ZFC to stronger theories, involving large cardinals and other set-theoretic principles, and these stronger theories settle many of those independent questions. Some set theorists have a view that these various stronger theories are approaching some kind of undescribable limit theory, and that it is that limit theory that is the true theory of sets. Others have a view that set-theoretic truth is inherently unsettled, and that we really have a multiverse of different concepts of set. On that view, the situation is that we seem to have no *standard* model of sets, in the way that we seem to have a standard model of arithmetic.
| 24 | https://mathoverflow.net/users/1946 | 24385 | 16,003 |
https://mathoverflow.net/questions/24406 | 0 | Suppose $A\_0$ is an abelian variety over $\mathbb{C}$, $E$ is a CM field ,denote $A= A\_0\otimes\_Q E$, is there an isomorphism $ H\_1(A\_0\otimes\_Q E,Q)=H\_1(A\_0,Q)\otimes\_Q E$?how it comes?
| https://mathoverflow.net/users/3945 | homology of abelian variety ? | If $A$ over $\mathbb{C}$ is an abelian variety of dimension $g$, then for all $i \in \mathbb{N}$, $H\_i(A,\mathbb{Q}) \cong H^i(A,\mathbb{Q}) \cong \mathbb{Q}^{ {2g \choose i}}$.
Taking $i = 1$, the conclusion you are asking about is true if and only if $\operatorname{dim} A\_0 \otimes\_\mathbb{Q} E = [E:\mathbb{Q}] \operatorname{dim} A\_0$.
Which it probably is, if $A\_0 \otimes\_{\mathbb{Q}} E$ means the Serre tensor construction. I can't quite remember how this goes at the moment (and I'll wait for you to confirm your notation before trying).
If you tell/remind us exactly what $A\_0 \otimes\_{\mathbb{Q}} E$ means, we could probably give you a natural isomorphism between these two homology groups.
**Addendum**: I found a nice online treatment of Serre's tensor construction [here](http://math.stanford.edu/~conrad/vigregroup/vigre04/serre_constr.pdf).
| 3 | https://mathoverflow.net/users/1149 | 24409 | 16,017 |
https://mathoverflow.net/questions/24411 | 11 | Ok the question is pretty dumb: suppose you have a torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ and a vector $\bar{v}=(v\_1,\ldots,v\_n)\in\mathbb{R}^n$.
Consider the torus $T\_{\bar{v}}$ given by the closure of the one parameter group in $T^n$ generated by $\bar{v}$:
$T\_{\bar{v}}=\overline{ \{t\cdot\bar{v}\mod\mathbb{Z}^n|\phantom{a}t\in\mathbb{R}\}}$
My questions are:
1. what is the dimension of $T\_{\bar{v}}$?
2. How can i find a basis of vectors spanning the tangent space of $T\_{\bar{v}}$ at the origin?
My guess for question 1. is $\dim T\_{\bar{v}}=\dim\_{\mathbb{Q}}\langle v\_1,\ldots, v\_n\rangle$, but i don't know what the answer to question 2. can be.
Thanks!
| https://mathoverflow.net/users/3465 | sub-tori of a torus, generated by 1-dimensional subgroup | The key to solving both problems is the use of the following two facts: 1) Any
closed subgroup of $T^n$ is the intersection of the kernels of characters of
$T^n$, i.e., continuous group homomorphisms $T^n \rightarrow S^1$. 2) Any
continuous homomorphism $T^n \rightarrow S^1$ is of the form $(\overline
x)\mapsto e^{2\pi i x\cdot m}$ for a unique $m\in\mathbb Z^n$. Hence, the
closure of $T\_{\overline \nu}$ is the intersection of the kernels of the
characters corresponding to $m$ for which $\nu\cdot m\in\mathbb Z$. Picking a
basis $m\_1,\dots,m\_k$ of the group of such $m$ gives a surjection $T^n
\rightarrow T^k$ for which $T\_{\overline\nu}$ is the kernel ($T\_\nu$ is not
necessarily a torus as it might not be connected but it doesn't change
anything). By the above this map is just given by an $n\times k$ integer matrix
specified by $m\_1,\dots,m\_k$. The tangent map at the origin is then obtained by
regarding this matrix as a real matrix and thus the tangent space of
$T\_{\overline \nu}$ is the null space of this matrix.
In particular this gives that the dimension of $T\_{\overline \nu}$ is equal to
$\dim\_{\mathbb Q}\langle1,\nu\_1,\nu\_2,\ldots,\nu\_n\rangle -1$. This is off by one
from your guess if $1$ is in the span of of the $\nu\_i$ but equal to it if it isn't.
[[Added]]
I misread the question and the above is for the closed subgroup generated by $\overline\nu$ while the question was about the closure of the $1$-parameter subgroup generated by it. To answer the question everything works the same only the condition is that $r\nu\cdot m\in\mathbb Z$ for all real $r$ which gives $\nu\cdot m=0$ and indeed the dimension is $\dim\_{\mathbb Q}\langle\nu\_1,\nu\_2,\ldots,\nu\_n\rangle $.
| 5 | https://mathoverflow.net/users/4008 | 24415 | 16,022 |
https://mathoverflow.net/questions/24419 | 6 | Let $L$ be the poset (ordered by set inclusion) that is the power set of some set $X$.
A *state* is a function $s:L \rightarrow [0,1]$ satisfying
i) for {$p\_1,p\_2,...$}, $p\_i \in L$ a
pairwise orthogonal (i.e. $p\_i \leq p\_j'$
where $a'$ is the complement of $a$)
countable sequence, $\bigvee\_i p\_i$
exists, and $s(\bigvee\_i p\_i) =
\sum\_i s(p\_i)$.
ii) s(X) = 1.
also consider functions $f:X \rightarrow [0,1]$.
Then for $X$ **countably infinite**, every state $s$ is in one to one correspondance to a function $f$ by the following argument (skip the next three paragraphs if you are OK with this):
---
As $L$ is atomistic, we have for arbitrary $p \in L$, $p = \bigvee\_i a\_i$ ($= \bigcup\_i a\_i$ as the join of the poset corresponds to the union of subsets) where $a\_i \in L$ are atoms (which are pairwise orthogonal).
The atoms $a\_i$ are in one to one correspondence to the elements of $X$ such that, given $f:X \rightarrow [0,1]$, we can define $f$ on the set of atoms via $f(a) = f(x\_a)$ where $x\_a$ is the element of $X$ associated to the atom $a$.
Then for a given state $s$ and arbitrary $p \in L$, $s(p) = s(\bigvee\_i a\_i) = \sum\_i s(a\_i)$. So $s$ is determined by its values on the atoms and we can associate a state to a function $f$ by setting for all atoms $a$, $s\_f(a) = f(a)$. This is bijective. (End of argument.)
---
Questions:
1) Now for $X = R^n$ (or uncountable) is there a similar correspondence?
2) Or do I need to adapt condition i) ?
What I fail to show in the uncountable case, is whether condition i) is strong enough to ensure that any state on the power set of such an $X$ is uniquely determined by its values on the atoms.
I hope this question is worthy of a response, it is my first one and I hesitated for the last 4 days.
| https://mathoverflow.net/users/5887 | Correspondence between functions on a set and "states" on its power set | Your correspondence is equivalent to the existence of a [real-valued measurable cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal), a large cardinal concept equiconsistent with the existence of a measurable cardinal.
First, note that if $\kappa$ is a measurable cardinal, then there is a 2-valued measure $\mu$ on $P(\kappa)$ which is not only countably-additive but $\kappa$-additive, in the sense the measure of the union of fewer than $\kappa$ many disjoint sets is the sum of the measures. For this measure, every set gets measure either 0 or 1, and so there are no disjoint sets of positive measure, and also every singleton gets measure 0. So it is an instance of a violation of your correspondence.
More generally, if $\kappa$ is a [real-valued measurable cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal), then there is a real-valued $\kappa$-additive measure $\mu$ on $P(\kappa)$ giving measure 0 to singletons. In particular, such a measure would be countably additive, and it would not correspond to function in your sense.
Conversely, suppose that there were a countably additive real-valued measure $\mu$ on $P(X)$ for some set $X$. If this measure does not correspond to a function, let's subtract from it the sum measure on singletons, to arrive without loss of generality at a measure that gives measure zero to singletons, but positive measure to the whole space. In this case, let $\kappa$ be the additivity of $\mu$, the largest cardinal such that the $\mu$ measure of any less-than-$\kappa$ sized disjoint union is equal to the sum of the measures individually. In this case, there is a set $Y\subset X$ of positive measure and a $\kappa$ partition of $Y=\cup\_{\alpha\lt\kappa} Y\_\alpha$ such that each $Y\_\alpha$ has measure $0$. We may now define a $\kappa$-additive measure on $P(\kappa)$ by $\mu\_0(I)=\Sigma\_{\alpha\in I}\mu(Y\_\alpha)$. Thus, $\kappa$ is a real-valued measurable cardinal.
So your question is equivalent to the existence of a real-valued measurable cardinal. Such a hypothesis is equiconsistent with the existence of a measurable cardinal.
The particular case when the set $X$ has size continuum $c$ corresponds to the situation where $c$ is a real-valued measurable cardinal. This implies a strong failure of the Continuum Hypothesis, since in this case $c$ would be weakly inaccessible. It is equivalent to the existence of a countably-additive extension of Lebesgue measure measuring all sets. (Such an extension cannot be translation invariant by Vitali.)
| 3 | https://mathoverflow.net/users/1946 | 24420 | 16,025 |
https://mathoverflow.net/questions/24424 | 14 | I could not find any references about this fact. I apologize if this is completely trivial, but is the product of two Baire spaces, or for that matter of finitely many of them a Baire space? Now is a countable product of Baire spaces a Baire space?
What about an uncountable product of Baire space? This fact seems to be treated in an article I can't access.
It seems to work for the Sorgenfrey line: $S$ is a Baire space and $SxS$ is a Baire space since if you consider the diagonal $A$={($-x$,$x$): $x\in S$ and $x\in$ℚ} then this is a closed discrete subspaces which is the union of countably many closed nowhere dense sets but its interior is empty.
Is that true?
Thx
| https://mathoverflow.net/users/3859 | Products of Baire spaces | See:
Cohen, Paul E.
Products of Baire spaces.
Proc. Amer. Math. Soc. 55 (1976), no. 1, 119--124.
$ $
>
> MathSciNet review by Douglas Censer: A topological space is said to be Baire if any countable intersection of its dense open sets is dense. Assuming the continuum hypothesis (CH), J. C. Oxtoby [Fund. Math. 49 (1960/61), 157--166; MR0140638 (25 #4055); errata, MR 26, p. 1543] constructed a Baire space whose square is not Baire. The author shows that the assumption of CH is unnecessary here. The spaces considered in this paper are partially ordered $(P,\leq)$ with topology generated by the initial segments. The construction of the desired Baire space with non-Baire product is based on the study of $P$ as a set of forcing conditions, as outlined by R. M. Solovay [Ann. of Math. 92 (1970), 1--56; MR0265151 MR0265151 (42 #64)].
>
>
>
Note that googling "products of Baire spaces" returns Cohen's article as the second hit. (The first hit is this question!)
For a positive result, see
Zsilinszky, László
Products of Baire spaces revisited.
Fund. Math. 183 (2004), no. 2, 115--121.
>
> Excerpted from the MathSciNet review by Paul Bankston: Without extra assumptions, the product of two Baire spaces need not be Baire [see, e.g., J. C. Oxtoby, Fund. Math. 49 (1960/1961), 157--166; MR0140638 (25 #4055); P. E. Cohen, Proc. Amer. Math. Soc. 55 (1976), no. 1, 119--124; MR0401480 (53 #5307)]. This brings us to the notion of a $\pi$-base; i.e., a collection of open sets such that every nonempty open set in the space contains a member of the collection. A $\pi$-base each of whose members contains only countably many members of the $\pi$-base is called countable-in-itself.
>
>
> Oxtoby's theorem, from his 1961 paper, states that any Tikhonov (resp., finite) product of Baire spaces with countable (resp., countable-in-itself) $\pi$-bases is a Baire space. The main result of the present paper is a significant strengthening of this; in particular it implies that arbitrary Tikhonov products of Baire spaces with countable-in-itself $\pi$-bases are Baire spaces.
>
>
> A space $X$ is called universally Kuratowski-Ulam (uK-U for short, first considered in [C. Kuratowski and S. Ulam, Fund. Math. 19 (1932), 247--251; Zbl 0005.18301]) if whenever $Y$ is a space and $E$ is a meager subset of $X\times Y$, the set $ Y \setminus \{y\in Y\:\ \{x\in X\:\ (x,y)\in E\}\ \text{is meager in}\ X\} $ is meager in $Y$. The author now defines a space $X$ to be almost locally uK-U if the set $\{x\in X\:\ x \text{has an open uK-U neighborhood} \}$ is dense in $X$.
>
>
> After showing that the property of being almost locally uK-U is a proper generalization of having a countable-in-itself $\pi$-base, the author proves his main theorem: only a Tikhonov (or countable box) product of Baire spaces that are almost locally uK-U is a Baire space.
>
>
> The proof for the box product case is a variant of that for the Tikhonov case, and partially answers a question raised in [W. G. Fleissner, in General topology and its relations to modern analysis and algebra, IV (Proc. Fourth Prague Topological Sympos., Prague, 1976), Part B, 125--126, Soc. Czechoslovak Mathematicians and Physicists, Prague, 1977; MR0464181 (57 #4116)].
>
>
>
| 12 | https://mathoverflow.net/users/1149 | 24427 | 16,030 |
https://mathoverflow.net/questions/24426 | 13 | Mac Lane defines a subcategory as a subset of objects and a subset of morphisms that form a category. But the first rule of category theory is that you do not talk about equality of objects. Up to equivalence, the definition becomes a faithful functor. This is a useful concept, but I don't think it fits the name. I don't want groups to be a subcategory of sets!
This is not a question about aesthetics, but about usage. I don't think people tend to use Mac Lane's definition. Maybe they're just wrong, but I'd like to know if there is another definition which fits the usage better. [All the time](https://mathoverflow.net/questions/23515/cofinal-inclusions-of-waldhausen-categories) I see people say things like "we may assume that our subcategory contains every object isomorphic to an object of the subcategory." I guess we can expand to an equivalent subcategory to achieve this, but we probably have to choose *how* the objects are isomorphic (though it may be easier if to change the ambient category). This is a much more natural thing to do (and safer) if the subcategory is full, or at least contains all the automorphisms of its objects. This leads me to suspect that people are assuming or thinking of some stronger definition than faithful.
Do people tend to mean the official definition? or do they also require full? containing all the automorphisms? Are there other useful intermediate notions?
| https://mathoverflow.net/users/4639 | What do people mean by "subcategory"? | Do you want the notion of "subcategory" to be invariant under categorical equivalence? If so, then "pseudomonic" functors are the right thing: faithful, and full on isomorphisms.
But I don't think one would want this any more than the notion of "inclusion" of topological spaces to be invariant under homotopy equivalence (which would make it meaningless).
| 6 | https://mathoverflow.net/users/4183 | 24438 | 16,035 |
https://mathoverflow.net/questions/24447 | 6 | Is there a way to dissect any tetrahedron into a finite number of orthoschemes?
I know that for a tetrahedron which only has acute angles, one can take the center of the inscribed circle and project the center on all the faces and edges and connect it with the vertices to get the orthoschemes. This however does not work when the tetrahedron is allowed to have obtuse angles since the projection of the center of the inscribed circle on the plane containing a face for instance may fall outside of the tetrahedron.
| https://mathoverflow.net/users/1612 | Dissecting a tetrahedron into orthoschemes | Yes, this is known (12 is always enough). Interestingly, in higher dimension it is open whether every simplex in $\Bbb R^d$ can be dissected into finitely many orthoschemes (also called path-simplices). This is called Hadwiger's conjecture. See [this survey](http://dare.uva.nl/document/162823) for results and refs to proofs of the conjecture for $d \le 5$.
P.S. In 1993, Tschirpke showed that 12,598,800 orthoschemes suffices in $\Bbb R^5$.
| 8 | https://mathoverflow.net/users/4040 | 24449 | 16,043 |
https://mathoverflow.net/questions/24448 | 2 | There is a proof of [Mittag-Leffler's theorem](http://en.wikipedia.org/wiki/Mittag-Leffler_theorem) with an explicit construction of a holomorphic function with the prescribed poles with prescribed order and residues, for a countable discrete set of points. I do not remember the reference; but my memory from my graduate course is that one defines a series sum and make certain adjustments. I was never quite good in this type of processes; so I am facing problem with the following exercise, which is nagging me for a long time. I thought of using Math Overflow with the hope that somebody can help me out.
Now I want to prove that every open set in the complex plane is now a [domain of holomorphy](http://en.wikipedia.org/wiki/Mittag-Leffler_theorem). We take the boundary $\partial \Omega$ of the open set $\Omega$, and we take a countable dense sequence of points $z\_i$ in $\partial \Omega$. If we are able to construct a series sum with poles at $z\_i$, but so that it converges absolutely and uniformly on every compact set in the interior of $\Omega$, then we are done.
I would be most grateful if somebody can show me how to do the above.
| https://mathoverflow.net/users/6031 | Domains of holomorphy in the complex plane | Let $\zeta\_k$ be a countable dense sequence of points in the boundary and consider $f(z) = \sum \frac{1}{2^k} \frac{1}{z-\zeta\_k}$. The sum is plainly uniformly convergent on any subset of finite distance from the boundary, in particular on any compact subset of the interior.
| 8 | https://mathoverflow.net/users/373 | 24454 | 16,046 |
https://mathoverflow.net/questions/24373 | 6 | Background
----------
Recall that a functor $G\colon A\to X$ is called
[monadic](http://ncatlab.org/nlab/show/monadic+functor) if it has a
left adjoint $F$ for which the Eilenberg--Moore comparison functor $K\colon A\to
X^{\mathbb{T}}$ is an equivalence of categories, where $\mathbb{T}$ is
the monad in $X$ defined by the adjunction $\langle
F,G,\ldots\rangle\colon X\rightharpoonup A$, and $X^{\mathbb{T}}$ is
the category of $\mathbb{T}$-algebras in $X$.
This means that a monadic functor is the forgetful functor
$G^{\mathbb{T}}\colon X^{\mathbb{T}}\to X$ up to composition with an
equivalence of categories (the comparison functor $K$). Now, it can be
verified that $G^{\mathbb{T}}$ creates
limits (Ex. 6.2.2 of Mac Lane). If the comparison functor is an
isomorphism, then it is straightforward to verify that $G$ creates
limits. In fact, I think that even if it is only assumed that $K$ is an
equivalence for which the object function is surjective, then $G$
creates limits.
However, in Proposition 4.4.1 on [p. 178](http://books.google.com/books?id=SGwwDerbEowC&lpg=PP1&dq=moerdijk%20sheaves&pg=PA178#v=onepage&q&f=false) of Mac Lane--Moerdijk, it is
stated that *any* monadic functor creates limits. The proof starts
with the following words (with minor omissions):
>
> Let $G$ be monadic. Then by definition, $G$ is the
> forgetful functor $G^\mathbb{T}$ up to an equivalence of
> categories. It thus suffices to show that such a forgetful functor
> $G^\mathbb{T}$ creates limits.
>
>
>
I simply do not understand this statement: In general, the composition
of an equivalence and a functor that creates limits need not create
limits. For example, the identity $\mathbf{Set}\to\mathbf{Set}$
creates limits, and for any skeleton $X$ of $\mathbf{Set}$ the inclusion
$X\subseteq \mathbf{Set}$ is an equivalence. Let $X$ be some skeleton
of $\mathbf{Set}$ (for which I am happy to assume any necessary axiom
of choice), and take a one-element set $1$ that is not in $X$. Then
$1$ is a limit of the functor obtained by composing the unique functor
from the empty category to $X$ with
$X\subset\mathbf{Set}\stackrel{\operatorname{Id}}{\to}\mathbf{Set}$,
but $1$ has no lifting in $X$.
So it seems that there are 4 possibilities:
1. The above Proposition 4.4.1, as stated, is wrong. There is a
counter example where a monadic functor (for which the comparison
functor is not an isomorphism) does not create limits.
2. The proof in ML-M covers just some of the cases, and for the other
cases it is not known if the assertion is true (namely, for a monadic
functor $G$ for which the comparison functor is not an isomorphism, it
is not known whether in general $G$ creates limits).
3. The proposition is correct because the comparison functor has some
additional special property (e.g., its object function must be
surjective whenever it is an equivalence).
4. (Most likely) I am wrong, and the quoted argument from Mac Lane--Moerdijk is
correct.
I would like to note that in Theorem 3.4.2 on p. 105 of Barr-Wells, it
is only claimed that monadic functors reflect limits.
Question
--------
Which one of the above 4 possibilities is true? In essence, my
question is: If $G$ is monadic and the comparison functor is an equivalence
that is not an isomorphism, does $G$ create limits?
| https://mathoverflow.net/users/2734 | If G is monadic and the comparison functor is an equivalence that is not an isomorphism, does G create limits? | Mac Lane-Moerdijk slipped up; they really should have said "reflects limits". Now it's true that the forgetful functor from the literal category of algebras to the underlying category creates limits (according to the definition of "creates" in CWM), but this notion doesn't transfer across equivalences, as you have observed.
An example where the distinction is important is the statement that for a topos E, the power object functor P: E^{op} --> E is monadic. It wouldn't make much sense to say that this creates limits in Mac Lane's sense of the term.
| 2 | https://mathoverflow.net/users/2926 | 24466 | 16,053 |
https://mathoverflow.net/questions/24453 | 8 | This is a foundational doubt I have. How does singular homology H\_n capture the number of n-dimensional holes in a space?
We disregard the case of $H\_0$ as it has the very satisfactory explanation that it is the direction sum of $\mathbb Z$ over the path-connected components of the space.
Now, handwaving aside, we consider the most important example of this "detecting hole" phenomenon, viz,, the fact that for $i \geq 1$ $H\_i(S^n) = \mathbb(Z)$ if and only if $i = n$. For this we use Mayer-Vietoris and a decomposition of $S\_n$ into a union of two open sets which are the complements of the north pole and south pole. And the intersection deformation retracts to $S^{n -1}$ and from the long exact sequence we get the isomorphisms $H\_i \cong H\_{i -1}$.
Now, by the above computation, it seems that the "hole detection" is achieved via Mayer-Vietoris and going up from the dimension below, using the long exact sequence. Mayer-Vietoris on the other hand depends on the snake lemma, which is very un-geometric and difficult to visualize.
So I would be most grateful for a more intuitive explanation of this hole capturing phenomenon. I can see that it is very natural that boundaries should be cancelled out as the solid simplices can be contracted to the central point. I can also "feel" that a hollow $n$-simplex, there should be a nontrivial $n$-chain which is not a boundary of an $n+1$-chain. But I am still left with a feeling of partial understanding. I hope this fundamental vagueness of understanding of mine can be cleared here.
| https://mathoverflow.net/users/6031 | How does singular homology H_n capture the number of n-dimensional "holes" in a space? | The "hole detection" is rather in the very definition of homology. Consider, for example, $H\_2$: it is morally the set of closed surfaces in you space modulo those that bound a $3$-dimensional body, and if a surface is not the boundary of any $3$-dimensional body then surely there must be a hole entrapped in it, no?
(Morally because when you want to actually implement this, you get a slightly different thing... Although I'd be thrilled to be informed that, in the case of a manifold at least, say, one can somehow construct a free abelian group on the set of maps $\Sigma\to M$ from $k$-manifolds $\Sigma$ to $M$ which, when one mods out the subgroup of those maps that extend to a manifold-with-boundary $N$ such that $\partial N=\Sigma$, gets you $H\_k(M)$ or something close)
| 6 | https://mathoverflow.net/users/1409 | 24468 | 16,055 |
https://mathoverflow.net/questions/24430 | 7 | The real question is both more serious and somewhat longer than the title.
For the definition of the rational Cherednik algebra attached to a complex reflection group $W$, see for instance 5.1.1 of Rouquier's paper [here](http://arxiv.org/abs/math/0509252). It is a flat family of algebras depending on some parameters $h\_{H,j}$ indexed by pairs consisting of a $W$-orbit of reflecting hyperplanes $H$ and an integer $0 \leq j \leq e\_H-1$, where $e\_H$ is the order of the pointwise stabilizer of $H$.
Various features of the structure of the Cherednik algebra turn out to be governed by systems of linear equations with *rational* coefficients in the parameters. For instance, in [this paper](http://arxiv.org/abs/math/0108185) Dunkl and Opdam show that the polynomial representation is irreducible exactly if the parameters avoid a certain locally finite system of rational hyperplanes, and in [this paper](http://arxiv.org/abs/1002.4607) a couple of guys show that it is Morita equivalent to its spherical subalgebra off a certain set of rational hyperplanes. In [this](http://arxiv.org/abs/0911.3208) paper, Etingof shows (for real reflection groups) that the set of parameters for which the irreducible head of the polynomial representation is finite dimensional is a set described by linear conditions with rational coefficients on the parameters.
Every time someone discovers the set of parameters for which the rational Cherednik algebra satisfies some reasonable properties, it winds up being described linearly in the parameters with rational coefficients. Why?
(I'm asking for a conceptual reason---in each case I mentioned I know the proof. For instance, I'd love to know an a priori proof that the set of parameters where the RCA is not Morita equiv. to its spherical subalgebra is a finite union of rational hyperplanes, without necessarily giving the union explicitly)
| https://mathoverflow.net/users/nan | Why are rational Cherednik algebras so... rational? | The phenomenon seems the same as for affine Kac-Moody algebras, where rational level is where everything special happens, or quantum groups, where roots of unity are the exceptional locus (and of course these examples are related). The parameter for Cherednik algebras is an additive/Lie algebra type parameter, like the KM level, as opposed to the exponentiated version as in the quantum groups case. If you mod out by the action of translation functors, you're asking why finite order points of the parameter are special. I don't know that I can give a completely uniform answer, but it certainly seems reasonable.
For example in type A, modules for Cherednik algebras are realized using twisted D-modules on some stack. The parameter for twisting is an additive one, but the abstract category of twisted D-modules depends only on this parameter mod integral translations (hence translation functors). At integral points (or rather "the" integral points) many special things happen -- there are geometric obstructions to the existence of objects with various supports, and these obstructions vanish integrally -- which is why say category O for Lie algebras becomes much bigger integrally. More generally twisted D-modules can be described as sheaves on a gerbe, which depends only on the twist mod integers. If (and certainly only if) the parameter is rational - ie the twist is a torsion element - then you might expect to represent your gerbe by an Azumaya algebra, or equivalently to have finite rank twisted sheaves. I would imagine this general type of phenomenon is behind the results you mention, though I haven't thought about the specifics. But in any case this is a geometric phenomenon about categories of twisted D-modules in general, and as we know "basically all interesting categories of representations are some categories of twisted D-modules" so this is quite general.
| 6 | https://mathoverflow.net/users/582 | 24470 | 16,057 |
https://mathoverflow.net/questions/24490 | 5 | Hi!
I've always read that on a complex manifold (obviously not kahler), with a given
hermitian metric on tangent bundle, the chern connection and the levi civita connection
on the underlying real bundle could be different.
Please can someone give me an explicit example of this fact?
Thank you in advance
| https://mathoverflow.net/users/4971 | chern connection vs levi-civita connection | You need a non-Kahler complex manifold. Then the Chern connection will have nontrivial torsion. And the torsion corresponds to the non-closed Kahler form of the metric.
| 16 | https://mathoverflow.net/users/2555 | 24491 | 16,070 |
https://mathoverflow.net/questions/24493 | 6 | Let $G$ be a compact connected topological group and let $H$ be a subgroup of $G$. Suppose that $H$ is measurable with respect to the normalised Haar measure $\mu$ on $G$. Do we necessarily have $\mu(H)=0$ or $\mu(H)=1$?
Maybe this is well--known, I ask it just out of curiosity. The question is related to [this](https://mathoverflow.net/questions/22934/a-question-of-erds-on-equidistribution) one: If you provide a measurable subgroup $H$ of $\mathbb R/\mathbb Z$ of measure not 0 or 1, then the characteristic function of $H$ violates the conjecture stated there.
| https://mathoverflow.net/users/5952 | Measurable subgroups. | Don't we still have this: if $A$ is measurable of positive measure, then $A A^{-1}$ contains a neighborhood of the identity...? So: a measurable subgroup of positive measure itself contains a neighborhood of the identity, and thus by connectedness is all of $G$.
| 10 | https://mathoverflow.net/users/454 | 24495 | 16,071 |
https://mathoverflow.net/questions/24492 | 3 | Hi,
I'm actually physics student and I've not been able to understand how the following integration has been performed:
>
> [...]
>
>
> The fields e and A have support on these discrete structures. The $su(2)$-valued 1-form
> field $e$ is represented by the assignment of an $e \in su(2)$ to each 1-cell in $\Delta$. The connection
> field A is represented by the assignment of group elements $g\_e \in SU(2)$ to each edge in $\mathcal{J}\_\Delta$
>
>
> The partition function is defined by
>
>
> ${\cal Z}(\Delta)=\int \prod\_{f \in {\cal J}\_{\Delta}} de\_f\prod\_{e \in {\cal J}\_{\Delta}} dg\_e e^{i {\rm Tr}\left[e\_f U\_f\right]}$
>
>
> where $de\_f$ is the regular Lebesgue measure on $R^3$,
> $dg\_e$ is the Haar measure on $SU(2)$, and $U\_f$
> denotes the holonomy around faces, i.e., $U\_f=g^1\_e\dots g^{N}\_e$ for $N$ being the number of edges bounding the corresponding
> face. Since $U\_f \in SU(2)$ we can write it as $U\_f=u^0\_f\ {{1}} + F\_f$
> where $u^0\_f\in C$ and $F\_f \in su(2)$. $F\_f$ is interpreted as
> the discrete curvature around the face $f$. Clearly ${\rm Tr}[e\_f U\_f]={\rm Tr}[e\_f F\_f]$.
> An arbitrary orientation is assigned to faces when computing
> $U\_f$. We use the fact that faces in ${\cal J}\_{\Delta}$ are
> in one-to-one correspondence with $1$-cells in $\Delta$ and label
> $e\_f$ with a face subindex.
>
>
> Integrating over $e\_f$, we obtain
>
>
> ${\cal Z}(\Delta)=\int \ \prod\_{e \in {\cal J}\_{\Delta}} dg\_e \prod\_{f \in {\cal J}\_{\Delta}}{\delta}(g^1\_e\dots g^{N}\_e),$
>
>
> where $\delta$ corresponds to the delta distribution defined on
> ${\cal L}^2(SU(2))$.
>
>
> [...]
>
>
>
The details are not important, what I need to understand is how the integration over the $e\_f$'s has been performed
thanks
| https://mathoverflow.net/users/2597 | Delta distribution as an integral... | Whoever performed that integration is using the following fact from Fourier analysis:
The "delta function supported at the position 0" is the Fourier transform of the constant function 1.
$\delta(x) = \frac{1}{2\pi}\int\_{\mathbb{R}} 1 e^{ikx}dk$
You can prove this by approximating the delta-function with a sequence of Gaussian bump functions. Fourier transforming Gaussians inverts the variance, so as the Gaussians approach a "delta function spike", their Fourier transforms approach a constant.
| 7 | https://mathoverflow.net/users/35508 | 24498 | 16,073 |
https://mathoverflow.net/questions/24432 | 8 | Let $X$ be a Banach space, $X' = \mathcal{L}(X, \mathbb{K})$ its dual space. Denote by $\mathcal{B}(X)$ the $\sigma$-algebra of Borel sets and denote by $\sigma(X')$ the $\sigma$-algebra which is generated by all sets of the form $u^{-1}(C)$ for $u \in X'$ and $C \in \mathcal{B}(\mathbb{K})$.
For $X$ separable we have that
$\mathcal{B}(X) = \sigma(X')$ (\*)
see e.g. "Gaussian measures in Banach spaces" by Hui-Hsiung Kuo, p. 74 - 75.
Now the author of this book does not bother to discuss the case of $X$ non-separable.
In [1] is a halfway believable counterexample for $X = \ell^2(\mathbb{R})$.
I'm specifically interested in the case $X = \ell^{\infty}$.
Does (\*) hold in this case and why or why not?
Thanks.
[1] <http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst;task=show_msg;msg=1533.0001.0001.0001>
| https://mathoverflow.net/users/2258 | Borel(X) = \sigma(X') for X non-separable | If $I$ is uncountable, then in space $l^2(I)$ no countable set of functionals separates points. Consequently, for any set $A$ in the sigma-algebra generated by these functionals [the Baire sets for the weak topology, see reference below], if $0 \in A$, then an entire subspace is contained in $A$. So all elements of this sigma-algebra are unbounded. Thus this sigma-algebra is not all of the norm-Borel sets.
My papers on measurability in Banach space:
Indiana Univ. Math. J. 26 (1977) 663--677
Indiana Univ. Math. J. 28 (1979) 559--579
**edit**
For gaussian measures in Banach space, you *really* want the example
of Fremlin and Talagrand, "A Gaussian measure on $l^{\infty}$". Ann. Probab. 8 (1980), no. 6, 1192--1193. This gaussian measure on $l^\infty$ with the cylindrical sigma-algebra has total mass 1, yet every ball of radius 1 has measure 0.
| 8 | https://mathoverflow.net/users/454 | 24499 | 16,074 |
https://mathoverflow.net/questions/24481 | 11 | To put this in context: I am in the process of developing a package for Macaulay 2 (a commutative algebra software,) called "Permutations", which will add permutations as a type of combinatorial object that M2 will handle, hopefully integrating nicely with the (still in development) Posets package, the (still in development) Graphs package, and various current M2 functions.
One of the first things that I'd wanted to put in was functions which compute the poset of the Bruhat order on $S\_n$ and compute when one permutation covers another in the Bruhat order. This was all coded and "working" - but has been producing a poset which is decidedly NOT the desired poset (among other things, the graph of the Hasse diagram isn't regular.) I'd like to ask whether the (probably somewhat naive) algorithm I was using to check covering relations seems reasonable (so the problem is just in the coding of it, not in the theory behind it) or not.
To see if $P\leq R$ in the Bruhat order:
Given a pair of permutations P and R, compute their lengths (the number of simple transpositions in their decomposition, or [as implemented right now] the sum of entries in their inversion vectors.) If length(P)=length(R)+1, then we compute
$(P^{-1})\*R$.
If R covers P in the Bruhat order, then length$((P^{-1})\*R)=1.$
Am I missing some subtlety of the Bruhat order? I thought one permutation covered enough exactly when they differed by a single, simple transposition. This seemed to capture that - but is giving me an incorrect poset.
Coding error or theory error? I'd love to hear it.
| https://mathoverflow.net/users/5002 | Computing Bruhat Order Covering Relations | My M2 permutation code is here:
<http://www.math.cornell.edu/~allenk/permutation.m2>
It's got a bunch of specialized stuff about Rothe diagrams and Fulton's essential set, but at the end it's got a BruhatLeq, for strong Bruhat order (so, not what you're computing).
If you want covering relations in strong Bruhat or right weak Bruhat, find the first place F and the last place L that your permutations differ. Weak Bruhat Covering: L=F+1. Strong Bruhat covering: each value w(F+1)...w(L-1) is not in the interval (w(F),w(L)).
If you want all relations in weak Bruhat order, w >= v if l(w) = l(v) + l(v^-1 w).
For all relations in strong Bruhat order, best to compute the corresponding rank matrices, and compare those entrywise. (That's what I do in the above code.)
| 8 | https://mathoverflow.net/users/391 | 24505 | 16,076 |
https://mathoverflow.net/questions/24487 | 4 | Let $\Lambda$ denote Connes's cyclic category. It is an extension of the simplex category $\Delta$ (of nonempty finite linearly ordered sets) obtained by adding an automorphism of order $n+1$ to the object $\textbf{n}$.
**Question:** Suppose $X: \Lambda^{op} \to Top$ is a cyclic space. What is a description of the homotopy colimit of this functor?
Just to put this in a bit of context, if $Y: \Delta^{op} \to Top$ is a simplicial space then it has a geometric realisation $|Y|$. One can also take the homotopy colimit of $Y$, and under some reasonable hypotheses there will be an equivalence $\mathrm{hocolim}\_{\Delta^{op}} Y \simeq |Y|$.
There is an inclusion $\Delta \to \Lambda$, so a cyclic space $X$ can be considered as a simplicial space and one can thus make a geometric realisation $|X|$. This space is supposed to have a circle action. I suppose my question should be:
How is the hocolim of $X$ over $\Delta^{op}$ related to the hocolim of $X$ over $\Lambda^{op}$.
| https://mathoverflow.net/users/4910 | Homotopy colimits of cyclic spaces | The homotopy theory of cyclic spaces is equivalent to that of spaces over $BS^1$ (Dwyer-Hopkins-Kan). The colimit over the simplicial category is as you say a space $X$ with $S^1$ action, and the colimit over the cyclic category is the quotient (Borel construction) $X/S^1$ as a space over $BS^1$.
| 8 | https://mathoverflow.net/users/582 | 24507 | 16,077 |
https://mathoverflow.net/questions/24503 | 16 | This is quite possibly a stupid question, but it is pretty far from what I normally do, so I wouldn't even know where to look it up.
If $X$ is a projective variety over an algebraically closed field of arbitrary characteristic and $Y\subset X$ a smooth divisor. Under which conditions can I contract $Y$ to a point, i.e. under which conditions is there a projective (smooth!?) variety $V$, and a morphism $f:X\rightarrow V$, such that $f$ is an isomorphism away from $Y$, and $Y$ is mapped to a point.
What can one say if $Y$ is a strict normal crossings divisor?
Hints and references are very appreciated!
| https://mathoverflow.net/users/259 | Contracting divisors to a point | For a smooth $Y$, a necessary condition for contractibility is that the conormal line bundle $N\_{Y,X}^\\*$ is ample. It is also sufficient for contracting to an algebraic space. The reference is *Algebraization of formal moduli. II. Existence of modifications.* by M. Artin.
$Y$ can be contracted to a point on an algebraic (projective) variety if in addition $Y=\mathbb P^{n-1}$, $n=\dim X$. You can prove this easily by hands. Start with an ample divisor $H$ and then prove that an appropriate linear combination $|aH+bY|$ is base point free and is zero exactly on $Y$. You will find the argument in Matsuki's book on Mori's program for example.
So if $X$ is a surface and $Y=\mathbb P^1$ with $Y^2<0$ then it is contractible to a projective surface. For a reducible divisor $Y=\sum Y\_i$ a necessary condition (which is also sufficient in the category of algebraic spaces) is that the matrix $(Y\_i.Y\_j)$ is negative definite. The strongest elementary sufficient condition for contractibility to a variety is that $\sum Y\_i$ is a *rational configuration* of curves. This is contained in *On isolated rational singularities of surfaces* by M. Artin.
This paper also contains an example of an elliptic curve $Y$ with $Y^2=-1$ which is not contractible to an algebraic surface. The surface $X$ is the blowup of $\mathbb P^2$ at 10 sufficiently general points lying on a smooth cubic, $Y$ is the strict preimage of that cubic.
Finally, for an irreducible divisor $Y$ the resulting space $V$ is smooth iff $Y=\mathbb P^{n-1}$ and $N\_{Y,X}=\mathcal O(-1)$. Indeed, $X\to V$ has to factor through the blowup of $V$ at a point by the universal property of the blowup. But then $X$ has to coincide with this blowup by Zariski main theorem. And on the blowup at a point the exceptional divisor is $\mathbb P^{n-1}$ with the normal bundle $\mathcal O(-1)$.
| 28 | https://mathoverflow.net/users/1784 | 24516 | 16,082 |
https://mathoverflow.net/questions/24521 | 11 | Fix a finite set of primes $S$ and an additional prime $p$. Let $K$ be the maximal extension of $\mathbb{Q}$ that is unramified outside $S$ and $\infty$ and totally split at $p$. Is the extension $K$ finite?
My intuitive guess would be no, but the simple constructions (based on class field theory) I tried so far do not prove this, at least for the ground field $\mathbb{Q}$. In contrast, for imaginary quadratic fields, there are extensions with Galois group $\mathbb{Z}\_{\ell}^2$ ramified only above $\ell$, and any place not above $\ell$ splits completely in an infinite subextension, as $\mathbb{Z}\_{\ell}^2$ has no procyclic subgroups of finite index.
| https://mathoverflow.net/users/6074 | Maximal extension almost everywhere unramified and totally split at one place | Nope.
I'm lacking a reference in front of me at the moment (see NSW's Cohomology of Number Fields, or Gras's Class Field Theory -- I'll update with a precise reference later), but there are remarkably clean formulas for the generator and relation ranks for the Galois group of the maximal $\ell$-extension of $\mathbb{Q}$ unramified outside $S$ and completely split at $T$, for finite sets of primes $S$ and $T$. Throwing out some silly cases, these depend only on $|S|$ and $|T|$ (and, in your problem, maybe even just $|S|-|T|$). In your case, where $|T|=1$, it's just a matter of making $S$ big enough (again, a reference will say how big, but right now, I think $|S|=4$ does the trick.)
**Edit** to add in in a precise reference (though the above book references certainly contain the results as well): Christian Maire's "Finitude de tours et p-tours T-ramifiees moderees, S-decomposees".
| 11 | https://mathoverflow.net/users/35575 | 24523 | 16,086 |
https://mathoverflow.net/questions/24528 | 12 | Let *G* be a semisimple algebraic group.
Following work of Matsumoto [1], Brylinski and Deligne [2] constructed a central extension of the functor
*G* : *Rings* → *Groups* by the second algebraic *K*-theory functor.
Plugging in ℂ((t)) into those functors, we get the well known central extension $\widetilde{G\big(\mathbb C((t))}\big)$ of
the loop group *G*(ℂ((t))) by the multiplicative group ℂ\*=*K*2(ℂ((t))).
It is interesting to note that the above group comes from an algebraic
group defined over the subfield ℂ of ℂ((t)). Namely, $\widetilde{G\big(\mathbb C((t))}\big)$ = $\widetilde{LG}(\mathbb C)$.
Doing all this with ℚp instead of ℂ((t)),
we get a central extension $\widetilde{G(\mathbb Q\_p)}$ of *G*(ℚp) by the group *K*2(ℚp) = **F**p\*.
Now, here's an idea: maybe that central extension is defined over... the subfield
**F**1 of ℚp?...
**My questions:**
• Has this been considered before?
• If yes, among all the exitsing notion of "*defined over* **F**1",
which one(s) make this possible?
• If no: is my heuristic argument is convincing?
---
**References:**
[1] Matsumoto,
"Sur les sous-groupes arithmétiques des groupes semi-simples déployés".
[2] Brylinski, Deligne, "Central extensions of reductive groups by $K\_2$".
| https://mathoverflow.net/users/5690 | Are centrally extended p-adic groups defined over F_1? | First a small thing. I am pretty sure we don't have $K\_2(\mathbb C((t)))=\mathbb
C^\*$, we have a surjective residue homomorphism $K\_2(\mathbb C((t)))\rightarrow
\mathbb C^\*$ but, I believe, with a non-trivial kernel. In any case, we can look
at the induced central extension and then the rest of what you say is
OK. Similarly, we have a surjective map $K\_2(\mathbb Q\_p)$.
Disrergarding this, there is a much simpler analogy between the two cases which on the
one hand, I think, makes the analogy that you want less likely and on the other
hand can be proven... To begin with it is not quite true that even $G(\mathbb
C((t)))$ is defined over $\mathbb C$ at least not as a group scheme. What
happens is that $G(\mathbb C[[t]])$ is a group scheme, it is the inverse limit
of the $G(\mathbb C[t]/(t^n))$ and these have a natural structure of algebraic
group over $\mathbb C$ (through the Greenberg functor). then $G(\mathbb C[[t]])$
as the inverse limit of algebraic groups is a group scheme (it is not of finite
type hence convention forces us to call it a group scheme rather than algebraic
group). Now, if we try to pass to $G(\mathbb
C((t)))$ we get into trouble. It is an infinite union of schemes (bound the
valuations of the entries of the elements of $G(\mathbb
C((t)))$ in some faithful linear representation of $G$) but an infinite union of
schemes does in general not have a scheme structure. There are ways of extending
the scheme notion to cover this case and what we get is what is called an
ind-group scheme over $\mathbb C$. Also the loop group type extension of $G(\mathbb
C((t)))$ by $\mathbb C^\*$ has such an extension (as does every Kac-Moody type
group).
The situation for $G(\mathbb Q\_p)$ is almost identical; $G(\mathbb Z/p^n)$ are
the $\mathbb Z/p$-points of a $\mathbb Z/p$-algebraic group, $G(\mathbb Z\_p)$
are the $\mathbb Z/p$-points of a group scheme over $\mathbb Z/p$ and $G(\mathbb
Q\_p)$ are the $\mathbb Z/p$-points of an ind-group scheme over $\mathbb Z/p$. I
think that the same thing is true for the central extension. The upshot is that
there is a close analogy to the $\mathbb C$ case but in that analogy $\mathbb C$
is replaced by $\mathbb F\_p$ not by $\mathbb F\_1$.
Note that in the Connes-Consani version of $\mathbb F\_1$ $G$ is defined over $\mathbb F\_{1^2}$ so perhaps that is the place to look for a version of the Brylinski-Deligne result.
**Addendum**: Just to add even more concreteness to George's answer about the explicit form of Greenberg's functor for $\mathbb G\_m$. We have that $W\_n(B)$ is just $B^n$ with a funny multiplication and addition. They are however given by polynomials (which are independent of $B$). The units in this ring are the tuples of the form $B^\ast\times B^{n-1}$ and multiplication is given by polynomials. This means that the algebraic group associated to this is just $\mathbb G\_m\times\mathbb A^{n-1}$ as scheme but with a funny product structure. In particular its $\mathbb F\_p$-points are just $(\mathbb Z/p^n)^\ast$.
| 10 | https://mathoverflow.net/users/4008 | 24535 | 16,093 |
https://mathoverflow.net/questions/24506 | 15 | In model theory, two structures $\mathfrak{A}, \mathfrak{B}$ of identical signature $\Sigma$ are said to be *elementarily equivalent* ($\mathfrak{A} \equiv \mathfrak{B}$) if they satisfy exactly the same first-order sentences w.r.t. $\Sigma$. An astounding theorem giving an algebraic characterisation of this notion is the so-called Keisler-Shelah isomorphism theorem, proved originally by Keisler (assuming GCH) and then by Shelah (avoiding GCH), which we state in its modern strengthening (saying that only a single ultrafilter is needed):
$\mathfrak{A} \equiv \mathfrak{B} \ \iff \ \exists \mathcal{U} \text{ s.t. } (\Pi\_{i\in\mathcal{I}} \ \mathfrak{A})/\mathcal{U} \cong (\Pi\_{i\in\mathcal{I}} \ \mathfrak{B})/\mathcal{U},$
where $\mathcal{U}$ is a non-principal ultrafilter on, say, $\mathcal{I} = \mathbb{N}$. That is, two structures are elementarily equivalent iff they have isomorphic ultrapowers.
My question is the following (admittedly rather vague): Does anyone know of constructions in which an ultrafilter is **chosen** by an appeal to this characterisation and then used for other means? An example of what I have in mind would be something like this (using the fact that any two real closed fields are elementarily equivalent w.r.t. the language of ordered rings): In order to perform some construction $C$ I ``choose'' a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$ by specifying it as a witness to the following isomorphism induced by Keisler-Shelah:
$\mathbb{R}^\mathbb{N}/\mathcal{U} \cong \mathbb{R}\_{alg}^\mathbb{N}/\mathcal{U},$
where $\mathbb{R}\_{alg}$ is the field of real algebraic numbers. So the construction $C$ should be dependent upon the fact that $\mathcal{U}$ is a non-principal ultrafilter bearing witness to the Keisler-Shelah isomorphism between some ultrapower of the reals and the algebraic reals, resp.
Also, a follow-up question: Let's say I'd like to ``solve'' the above isomorphism for $\mathcal{U}$. Are there interesting things in general known about the solution space, e.g., the set of all non-principal ultrafilters bearing witness to the Keisler-Shelah isomorphism for two fixed elementarily equivalent structures such as $\mathbb{R}$ and $\mathbb{R}\_{alg}$? What machinery is useful in investigating this?
| https://mathoverflow.net/users/4915 | Ultrafilters arising from Keisler-Shelah ultrapower characterisation of elementary equivalence | Under the Continuum Hypothesis, your solution space is *all* nonprincipal ultrafilters. This is because under CH, the ultrapower $M^N/U$ of a mathematical structure $M$ of size at most continuum does not actually depend on the (nonprincipal) ultrafilter $U$. One can see this by using the fact that the ultrapower will be saturated, and so one can run a back-and-forth argument to achieve the isomorphism. In particular, it follows under CH that any $U$ will witness your desired isomorphism for $R^N/U\cong (R\_{alg})^N/U$.
(See Corollary 6.1.2 in Chang-Keisler's book Model Theory.)
A similar fact holds for larger cardinals and larger structures under GCH, but here, one needs an additional assumption on the ultrafilter. Namely, Theorem 6.1.9 in Chang-Keisler asserts that if $2^\alpha=\alpha^+$ and $A$ and $B$ are two structures of size at most $\alpha^+$, then they are elementarily equivalent if and only if $\Pi\_DA\cong\Pi\_D B$ for any $\alpha^+$-good incomplete ultrafilter $D$ on $\alpha$. The proof uses the same saturation idea, and this establishes the Keisler-Shelah theorem in the case that GCH holds.
Chang-Keisler states (page 393-394) that it is open whether the assertion of Theorem 6.1.9 stated above holds under $\neg CH$.
| 10 | https://mathoverflow.net/users/1946 | 24538 | 16,095 |
https://mathoverflow.net/questions/24540 | 3 | e.g. is $\widehat{\mathbf{SET}}$ locally small?
| https://mathoverflow.net/users/6082 | is the presheaf category of a locally small category locally small? | No, not in general. Todd gave a useful general answer and here is a concrete counter-example.
Let $\mathbf{C}$ be a category whose objects are sets and where the only morphisms are the identities. Consider the functors $F, G : \mathbf{C} \to \mathbf{Set}$ given by $F(X) = 1 = \lbrace 0 \rbrace$ and $G(X) = 2 = \lbrace 0,1\rbrace$. Because $\mathbf{C}$ is discrete, every family of maps $(\eta\_X : F(X) \to G(X))\_{X \in \mathbf{C}}$ is natural. Therefore, natural transformations $F \to G$ are in bijective correspondence with classes:
* a natural transformation $\eta : F \to G$ corresponds to the class $\lbrace X \in \mathbf{Set} \mid \eta\_X(0) = 1\rbrace$.
* a class $C$ corresponds to the natural transformation $\eta : F \to G$ given by
>
> $\eta\_X(0) = 1$ if $X \in C$ and $\eta\_X(0) = 0$ if $X \not\in C$.
>
>
>
Thus the presheaf category $\widehat{\mathbf{C}}$ is not locally small (note that $\mathbf{C}^{op} = \mathbf{C}$).
I cannot think at the moment of concrete presheaves $F$ and $G$ on $\mathbf{Set}$ for which the natural transformations $F \to G$ form a proper class. Someone please help.
| 3 | https://mathoverflow.net/users/1176 | 24551 | 16,102 |
https://mathoverflow.net/questions/24550 | 10 | Let $H$ and $K$ be affine group schemes over a field $k$ of characteristic zero. Let $\varphi:H\to Aut(K)$ be a group action. Then we can form the semi-direct product $G = K\ltimes H$.
**Problem**: Describe the tannakian category $Rep\_k(G)$ in terms of $Rep\_k(K)$, $Rep\_k(H)$ and $\varphi$.
If $\varphi$ is the trivial action, $G$ is the direct product $K\times H$. In this case, $Rep\_k(G) = Rep\_k(K) \boxtimes Rep\_k(G)$ is just the Deligne tensor product of the tannakian categories.
**Question**: What happens in the opposite scenario where the action $\varphi$ is faithful? Can one characterize this situation in terms of Ext groups of simple objects in $Rep\_k(K)$ and $Rep\_k(H)$?
I'm mostly insterested in the case where $K$ and $H$ are extensions of $\mathbb{G}\_m$ by a pro-unipotent group but I have no idea where to start.
| https://mathoverflow.net/users/1985 | Tannakian description of a semi-direct product | The algebraic stack BG is the quotient of BK by the action of H, induced by its action on K. So we can describe coherent sheaves on it (aka reps of G) via descent from BK - i.e. reps of G are H-equivariant sheaves on BK, or H-equivariant objects in Rep K. Now this is not yet the answer you want, since it involves Rep K and H, rather than Rep H.
One indirect (and maybe slightly imprecise) answer is to use the Morita equivalence between Rep H-module categories and
categories with H action: the Rep H-module category Rep G corresponds under this equivalence to Rep K as a category with H action. One direction takes a category with H
action to its H-equivariant objects, which carry a natural Rep H action. The other
direction takes a Rep H category to the category of its eigenobjects.
Here an eigenobject is an object $M\in C$ with a functorial identification
$$V\ast M \simeq \underline{V}\otimes M$$ of the module action of $V\in Rep K$ with the naive tensor product by the underlying vector space of $V$. (see e.g. [this paper](http://arxiv.org/abs/math/0010270)) -- again this is just playing with monadic formalism.. So I would then characterize Rep G as the Rep H-module (via induction of representations) whose eigenobject category is Rep K as an H-category..
This Morita equivalence for finite groups appears in papers of Mueger and Ostrik cited in [here](http://arxiv.org/abs/0805.0157). For affine group schemes I proved a derived version of this result with Nadler and Francis, but it's not available sadly. I haven't thought this through in the usual Tannakian setting, so maybe I'm missing something obvious, but it should be a fairly straightforward (though 2-categorical!) application of Barr-Beck I would think: we're simply claiming that categories over BH are described via descent from a point, and the descent data is an action of H.
| 8 | https://mathoverflow.net/users/582 | 24553 | 16,103 |
https://mathoverflow.net/questions/24524 | 6 | Consider two (distinct) octahedral diagrams i.e. diagrams mentioned in the octahedron axioms of triangulated categories (with four 'commutative triangular faces' and four 'distinguished triangular faces'). Is it true than one can extend to a morphism of such diagrams:
1. any morphism of one of the 'commutative faces' of the octahedron
2 any morphism of the pair of morphisms whose target is the upper vertex of the octahedron (i.e. a morphism of commutative triangles not lying on the faces of the octahedrons)?
Is there any text where I could look for various facts of this sort?
P.S. It seems that the answer is 'no' in general. Having a morphism of 'commutative faces', one can extend it to a morphism of three neighbouring 'triangulated faces'. Thus one obtains morphisms of each of six vertices. Yet (all possible) compositions of edges of the 'first' commutative triangles and the neigbouring distinguished faces do not yield all edges of the octahedron; two of the edges (in the 'lower hat') are missing.
Yet it would be very interesting to know which additional conditions are needed in order for the morphism of the octahedrons desired to exist. I would be deeply grateful for any comments!!
| https://mathoverflow.net/users/2191 | Can one extend a morphism of commutative triangles to a morphism of octahedral diagrams? | It is true in a Heller triangulated category aka $\infty$-triangulated category (although strictly speaking one only needs a 3-triangulation for octahedra) that any morphism between the bases of octahedra (by which I mean the 3 objects and two composable morphisms from which the octahedron is built) extends to a morphism between the octahedra (it is part of the axiomatics).
It turns out that the triangulated categories which turn up in "nature" all satisfy these stronger axioms. For instance the homotopy category of a stable model category is $\infty$-triangulated (a more general statement holds which is Theorem 2 in Maltsiniotis' preprint).
References are M. Künzer. Heller triangulated categories and G. Maltsiniotis. Catégories Triangulées Supérieures
| 8 | https://mathoverflow.net/users/310 | 24558 | 16,107 |
https://mathoverflow.net/questions/24556 | 9 | It is shown in Lang's Algebra (and many other books I assume) that:
if A if a principal entire ring, then A is a factorial ring.
The proof uses Zorn's Lemma. Is this theorem equivalent to the axiom of choice?
| https://mathoverflow.net/users/4002 | Factorial Rings and The Axiom of Choice | Lang uses Zorn's lemma only in the step that nonzero nonunits in a PID admit irreducible factorizations (not the uniqueness of irreducible factorizations, once we know such factorizations exist). The way he uses Zorn's lemma, I think, is excessive. What follows is how I work out the existence of irreducible factorizations when I teach the abstract algebra class.
Claim: In a PID which is not a field, any nonzero nonunit is a product of irreducibles.
We will use the following lemma (which is not Zorn's lemma).
Lemma: If $R$ is an integral domain and $a \in R$ is
a nonzero nonunit which
does not admit a factorization into irreducibles then
there is a strict inclusion of principal ideals
$(a) \subset (b)$ where $b$ is some other nonzero nonunit which
does not admit a factorization into irreducibles.
Proof of lemma: By hypothesis $a$ is not irreducible, so (since it is
neither 0 nor a unit either) there is some factorization
$a = bc$ where $b$ and $c$ are nonunits (and obviously are
not 0 either). If both $b$ and $c$ admitted
irreducible factorizations then so does $a$, so
at least one of $b$ or $c$ has no irreducible factorization.
Without loss of generality it is $b$ which has no
irreducible factorization. Since $c$ is not a unit, the inclusion
$(a) \subset (b)$ is strict. QED lemma.
Now we can prove the claim.
Proof of claim: Suppose there is an element $a$ in the PID which
is not 0 or a unit and has no irreducible factorization.
Then by the lemma there is a strict inclusion
$$
(a) \subset (a\_1)
$$
where $a\_1$ has no irreducible factorization.
Then using $a\_1$ in the role of $a$ (and the lemma again)
there is a strict inclusion
$$
(a\_1) \subset (a\_2)
$$
where $a\_2$ has no irreducible factorization.
This argument (repeatedly applying the lemma to
the generator of the next larger principal ideal)
leads to an infinite increasing chain of principal ideals
$$
(a) \subset (a\_1) \subset (a\_2) \subset (a\_3) \subset \cdots
$$
where all inclusions are strict. (At this step I suppose you may say we need the Axiom of Choice to get an infinite ascending chain, but it's only *countably* many choices, so really not the full thrust of Zorn's lemma and in any case it feels like a less pedantic use of Zorn's lemma than the way Lang does this.) Such a chain of ideals is impossible in a PID.
Indeed, suppose a PID contains an infinite strictly increasing chain of
ideals:
$$
I\_0 \subset I\_1 \subset I\_2 \subset I\_3 \subset \cdots
$$
and set
$$
I = \bigcup\_{n \geq 0} I\_n.
$$
This union $I$ is an ideal. The reason is that
the $I\_n$'s are strictly increasing, so any *finite* set
of elements from $I$ lies in a common $I\_n$.
Therefore $I$ is closed under addition and arbitrary multiplications from
the ring since each $I\_n$ has these properties. Because we are in a PID, $I$ is principal:
$I = (r)$ for some $r$ in the ring. But because
$I$ is the union of the $I\_n$'s, $r$ is in some $I\_N$.
Then $(r) \subset I\_N$ since $I\_N$ is an ideal, so
$$
I = (r) \subset I\_N \subset I,
$$
which means
$$
I\_N = I.
$$
But this is impossible because the inclusion $I\_{N+1} \subset I$
becomes $I\_{N+1} \subset I\_N$ and we were assuming
$I\_N$ was a proper subset of $I\_{N+1}$.
Because of this contradiction, nonzero nonunits in a PID without
an irreducible factorization do not exist. QED
Lang's argument only uses the axiom of choice in a countable way, as above, but the way he pulls it in makes the application of Zorn's lemma feel a lot more fussy.
| 19 | https://mathoverflow.net/users/3272 | 24559 | 16,108 |
https://mathoverflow.net/questions/24399 | 5 | Let $k$ be a field, $K/k$ a separable quadratic extension,
and $D/K$ a central division algebra of dimension $r^2$ over $K$
with an involution $\sigma$ of second kind
(i.e. $\sigma$ acts non-trivially on $K$ and trivially on $k$).
Does there exist a field extension $F/k$ such that $L:=K\otimes\_k F$
is a field, and $D\otimes\_K L$ splits (i.e. is isomorphic to the matrix algebra $M\_r(L)$ over $L$)?
Motivation: Let $h\in D$ be a Hermitian element ($h^\sigma =h$), and let
$G$ be the $k$-group with
$G(k)=${$g\in D^\times\ | \ ghg^\sigma=h$}.
I want to find a field extension $F/k$ such that $G\times\_k F$
is a unitary group over a field $L$ (and not over a division algebra over $L$).
| https://mathoverflow.net/users/4149 | Splitting of a division algebra with an involution of second kind | I answer my own question. The answer is *yes*.
Since there are no non-trivial division algebras over finite fields,
we may assume that $k$ and $K$ are infinite.
Let $H=${$h\in D\ |\ h^\sigma=h$} denote the $k$-space of Hermitian elements of $D$.
Consider the embedding $D\hookrightarrow M\_r(\bar K)$ induced
by an isomorphism $D\otimes\_K \bar K\simeq M\_r(\bar K)$.
An element x of $D$ is called semisimple regular,
if its image in $D\otimes\_K \bar K\simeq M\_r(\bar K)$
is a semisimple matrix that has $r$ different eigenvalues.
A standard argument using an isomorphism
$D\otimes\_k \bar K\simeq M\_r(\bar K)\times M\_r(\bar K)$
shows that there is a dense Zariski open subset
$H\_{reg}$ consisting of semisimple regular elements in $H$.
Clearly $H\_{reg}$ contains $k$-points.
Let $h\in H\_{reg}$ be a semisimple regular Hermitian element.
Let $L$ be the centralizer of $h$ in $D$.
Since $h$ is Hermitian ($\sigma$-invariant), the $k$-algebra $L$ is $\sigma$-invariant.
Since $h$ is semisimple and regular, the algebra $L$
is a commutative étale $K$-subalgebra of $D$
of dimension $r$ over $K$ (we calculate in $D\otimes\_K K\_s$).
Clearly $L$ is a field, $[L:K]=r$.
Since $L\subset D$ and $[L:K]=r$, the field $L$ is a splitting field for $D$,
see e.g. Scharlau, Quadratic and Hermitian Forms, Ch. 8, Thm. 5.4.
Since $L\supset K$, we see that $\sigma$ acts non-trivially on $L$.
Let $F$ denote the subfield of fixed points of $\sigma$ in $L$,
then $[L:F]=2$ and $[F:k]=r$.
Clearly $F\cap K=k$ and $FK=L$, hence $L=K\otimes\_k F$.
The extension $F/k$ is separable.
Another version of the proof vas proposed by Uzi Vishne.
| 5 | https://mathoverflow.net/users/4149 | 24561 | 16,110 |
https://mathoverflow.net/questions/24513 | 3 | I need to determine the minimal polynomial for a quotient in (1).
(1) B = C / A
C is known as a root of a 36th degree polynomial and A is known as a root of a 24th degree polynomial.
However I have not been able to succeed in recovering the coefficients nor the degree of the polynomial for B.
Any suggestions? I have tried to use GP-Pari's algdep(number,power) command, but so far with little success, even though I know the decimal value of B to 10,018 digits.
Thanks for your help.
Randall
P.S. This is a repost after a suggestion
After working with the resultant method, I was able to successfully recover a 144th degree polynomial whose highest power term has the expected square coefficient. This polynomial was one of 3 polynomials factored from a 864th degree polynomial originally obtained.
I guessed 72nd degree, but it would have taken too long using GP-Pari's algdep(number,144) to recover the polynomial.
Thanks for your suggestions, I now have a valuable tool to help me work with algebraic vectors in R3.
| https://mathoverflow.net/users/6046 | Question on determining the minimal polynomial for an algebraic quotient | Let $F$ be the polynomial for $A$, let $G$ be the polynomial for $C$. Consider the resultant of $x^{24}F(y/x)$ and
$G(y)$. This will be a polynomial whose roots are all the numbers of the form $\gamma/\alpha$, where $\gamma$ (resp., $\alpha$) runs through the roots of $G$ (resp., $F$). The resultant is the determinant of a $60\times60$ matrix.
| 4 | https://mathoverflow.net/users/3684 | 24568 | 16,117 |
https://mathoverflow.net/questions/24569 | 8 | I am teaching a combinatorics class in which I introduced the notion of a "mass formula". My terminology is inspired by the [Smith–Minkowski–Siegel mass formula](http://en.wikipedia.org/wiki/Smith%25E2%2580%2593Minkowski%25E2%2580%2593Siegel_mass_formula) for the total mass of positive-definite quadratic forms of a given size and genus. That famous mass formula is much too fancy of an example for my class. All that I really do is define the concept of the "mass" of a combinatorial object to be $1/|G|$ if $G$ is its automorphism group, and then argue that it can be easier to find the total mass of a collection of objects than to count them straight (using Polya counting theory). For example, the total mass of unlabeled trees of order $n$ is $n^{n-2}/n!$, because there are $n^{n-2}$ labeled trees.
So I have two questions for which a quick answer (i.e. sooner than two weeks) would be most convenient:
1. Is "mass formula" a standard name for this concept? Is there a standard name?
2. Can someone suggest a free on-line reference, comparable to a Wikipedia page or a little longer? The class textbook doesn't have a discussion.
| https://mathoverflow.net/users/1450 | Seeking reference for the enumerative "mass formula" concept | I do call such things "mass formulas", but then again I am a number theorist, and one of my colleagues is a quadratic form theorist who specializes in such things. So this is mostly an expression of my specific mathematical culture.
I do not think that it is a standard term, at least not the only standard term. For instance, from another MO answer I noticed that some categorists call this the [groupoid cardinality](http://ncatlab.org/nlab/show/groupoid%20cardinality). This term in fact seems quite sensible to me, because the concept seems closely related to taking a quotient by the action of a group with nontrivial stabilizers and regarding the quotient set as a groupoid rather than a mere set.
As you say, combinatorially minded people speak of "Polya theory" or "counting with symmetry". Many algebraic geometers, upon seeing this phenomenon, would use the word "stacky". I wouldn't be surprised if there were other terms as well.
Overall I think this has the effect that a lot of people are partially rediscovering what is essentially the same concept. I would very much like to see a reasonably authoritative treatment of this subject appealing to mathematicians from different fields. Of course, I also look forward to seeing (better!) answers to this question.
| 6 | https://mathoverflow.net/users/1149 | 24571 | 16,118 |
https://mathoverflow.net/questions/24573 | 37 | If $k$ is a characteristic $p$ field containing a subfield with $p^2$ elements (e.g., an algebraic closure of $\mathbb{F}\_p$), then the number of isomorphism classes of supersingular elliptic curves over $k$ has a formula involving $\lfloor p/12 \rfloor$ and the residue class of $p$ mod 12, described in Chapter V of Silverman's *The Arithmetic of Elliptic Curves*. If we weight these curves by the reciprocals of the orders of their automorphism groups, we obtain the substantially simpler Eichler-Deuring mass formula: $\frac{p-1}{24}$. For example, when $p=2$, the unique supersingular curve $y^2+y=x^3$ has endomorphisms given by the [Hurwitz integers](http://en.wikipedia.org/wiki/Hurwitz_quaternion) (a maximal order in the quaternions), and its automorphism group is therefore isomorphic to the [binary tetrahedral group](http://en.wikipedia.org/wiki/Binary_tetrahedral_group), which has order 24.
Silverman gives the mass formula as an exercise, and it's pretty easy to derive from the formula in the text. The proof of the complicated formula uses the Legendre form (hence only works away from 2), and the appearance of the $p/12$ boils down to the following two facts:
1. Supersingular values of $\lambda$ are precisely the roots of the Hasse polynomial, which is separable of degree $\frac{p-1}2$.
2. The $\lambda$-line is a 6-fold cover of the $j$-line away from $j=0$ and $j=1728$ (so the roots away from these values give an overcount by a factor of 6).
**Question:** Is there a proof of the Eichler-Deuring formula in the literature that avoids most of the case analysis, e.g., by using a normal form of representable level?
I suppose any nontrivial level structure will probably require some special treatment for the prime(s) dividing that level. Even so, it would be neat to see any suitably holistic enumeration, in particular, one that doesn't need to single out special $j$-invariants.
(This question has been troubling me for a while, but [Greg's question](https://mathoverflow.net/questions/24569/seeking-reference-for-the-enumerative-mass-formula-concept) inspired me to actually write it down.)
| https://mathoverflow.net/users/121 | Is there a nice proof of the fact that there are (p-1)/24 supersingular elliptic curves in characteristic p? | One argument (maybe not of the kind you want) is to use the fact that the wt. 2 Eisenstein series on $\Gamma\_0(p)$ has constant term (p-1)/24.
More precisely: if $\{E\_i\}$ are the s.s. curves, then for each $i,j$,
the Hom space $L\_{i,j} := Hom(E\_i,E\_j)$ is a lattice with
a quadratic form (the degree of an isogeny), and we can form the corresponding
theta series $$\Theta\_{i,j} := \sum\_{n = 0}^{\infty} r\_n(L\_{i,j})q^n,$$
where as usual $r\_n(L\_{i,j})$ denotes the number of elements of degree $n$.
These are wt. 2 forms on $\Gamma\_0(p)$.
There is a pairing on the $\mathbb Q$-span $X$ of the $E\_i$ given by $\langle E\_i,E\_j\rangle
= $ # $Iso(E\_i,E\_j),$ i.e. $$\langle E\_i,E\_j\rangle = 0 \text{ if } i \neq j\text{ and equals # }Aut(E\_i) \text{ if }i = j,$$
and another formula for $\Theta\_{i,j}$ is
$$\Theta\_{i,j} := 1 + \sum\_{n = 1}^{\infty} \langle T\_n E\_i, E\_j\rangle q^n,$$
where $T\_n$ is the $n$th Hecke correspondence.
Now write $x := \sum\_{j} \frac{1}{\text{#}Aut(E\_j)} E\_j \in X$. It's easy to see
that for any fixed $i$, the value of the pairing $\langle T\_n E\_i,x\rangle$
is equal to $\sum\_{d |n , (p,d) = 1} d$. (This is just the number of $n$-isogenies
with source $E\_i,$ where the target is counted up to isomorphism.)
Now
$$\sum\_{j}
\frac{1}{\text{#}Aut(E\_j)} \Theta\_{i,j} =
\bigg{(}\sum\_{j} \frac{1}{\text{#}Aut(E\_j)}\bigg{)} + \sum\_{n =1}^{\infty} \langle T\_n E\_i, x\rangle
q^n
= \bigg{(}\sum\_{j}\frac{1}{\text{#}Aut(E\_j)}\bigg{)} + \sum\_{n = 1}^{\infty} \bigg{(}\sum\_{d | n, (p,d) = 1} d\bigg{)}q^n.$$
Now the LHS is modular of wt. 2 on $\Gamma\_0(p)$, thus so is the RHS. Since we know
all its Fourier coefficients besides the constant term, and they coincide with those of the Eisenstein series, it must *be* the Eisenstein series.
Thus we know its constant term as well, and that gives the mass formula.
(One can replace the geometric aspects of this argument, involving s.s. curves and Hecke
correspondences, with pure group theory/automorphic forms: namely the set $\{E\_i\}$ is
precisely the idele class set of the multiplicative
group $D^{\times}$, where $D$ is the quat. alg. over $\mathbb Q$ ramified at $p$ and $\infty$. This formula, writing the Eisenstein series as a sum of theta series, is then
a special case of the Seigel--Weil formula, I believe, which in general, when you pass to constant
terms, gives mass formulas of the type you asked about.)
| 31 | https://mathoverflow.net/users/2874 | 24582 | 16,125 |
https://mathoverflow.net/questions/24585 | 16 | There seems to be a few papers around with Erdős written as Erdös. For example:
MR0987571 (90h:11090) Alladi, K.; Erdös, P.; Vaaler, J. D. Multiplicative functions and small divisors. II. J. Number Theory 31 (1989), no. 2, 183--190. (Reviewer: Friedrich Roesler) 11N37
>
> Would it be incorrect to cite such papers using Erdős instead?
>
>
>
| https://mathoverflow.net/users/2264 | If Erdős is published as Erdös in a paper, which do I cite? | We cite papers to show our respect to the authors and to help our readers find stuff. For the second purpose, I suspect most people would just type in names without diacritical marks, and most search facilities would find what you're looking for based on the letters alone, so it doesn't really matter. But for the first purpose, I think you should spell the name the way its owner would want it spelled, regardless of what some journal may have done.
| 33 | https://mathoverflow.net/users/3684 | 24587 | 16,128 |
https://mathoverflow.net/questions/24497 | 5 | The following seems to be a question related to standard calculus, but I am not quite sure
where to look for an answer.
Suppose $f,g:\mathbb{N} \to \mathbb{C}$ are such that the have the same asymptotical behaviour, i.e. $f(n)/g(n) \to 1$ as
$n \to \infty$. Of course, suppose that one of the sums $\sum\_{n=0}^\infty f(n)$ and $\sum\_{n=0}^\infty g(n)$ converges absolutely,
then so does the other. This can be proven by a standard estimate. However this standard estimate fails
if we do not have absolute convergence. I do not see how to prove convergence of one of sums implies the convergence of the other.
I feel that it may be actually false.
So the first question is:
>
> $1$. Is it true that one series converges iff the other does?
>
>
>
If this is not the case, however, in the problem I am studying, I want to prove convergence for both series. For my application in mind, you may assume that $f(n)/g(n)$ is always in $\mathbb{R}$. So the second question is
>
> $2$. Under which additional conditions (which do not! imply absolute convergence) can
> we deduce both series have the same behaviour. Are there books treating such topics?
>
>
>
| https://mathoverflow.net/users/3757 | Non-absolute convergence of series with asymtotically equal coefficients | Following Theo Johnson--Freyd's suggestion, I am making my above comment an answer:
Just subtracting one series from the other, it seems that you need $\sum\_{n=0}^{\infty} (f(n)−g(n))$ to converge. Writing $$f(n)/g(n)=1+\delta\_n,$$ $$\text{so that } \qquad \qquad\sum\_{n=0}^{\infty} (f(n)−g(n))=\sum\_{n=0}^{\infty} \delta\_n g(n),$$ you see need control over the signs of the $\delta\_n$, or (as Theo notes in his comment, on their
rate of growth).
E.g. if they are all of the same sign, you are okay, while if the sign of $\delta\_n$ is always the same as, or always opposite to, that of $\delta\_n$, then you could be in bad shape. (This is what goes wrong in Xandi Tuni's example.)
As Theo notes in his comment, you are also okay if $\sum\_{n = 0}^{\infty} \delta\_n$
converges absolutely. Whether this
applies in your case will depend on how closely $f$ and $g$ approximate one another.
(This is illustrated by the example in Julian Aguirre's answer.)
| 2 | https://mathoverflow.net/users/2874 | 24595 | 16,134 |
https://mathoverflow.net/questions/24593 | 1 | I'm reading the book 'A course in Modern Mathematical Physics' by
'Szekeres' and encountered a problem in interpreting the proof of the
following corollary of Schur's lemma.
The corollary and the proof in the book is as follows (I mark my problem
area's in the proof with (<1>) and (<2>) :
Corollary 4.2. Let $T : G \rightarrow {GL} (V)$ be a irreducible
representation of a finite (<1>) group G on a complex vector space $V$ and $A
: V \rightarrow V$ an operator such that $\forall g \in G$ we have $T (g)
\circ A = A \circ T (g)$ then $A = \alpha {id}\_V$ for some complex scalar
$\alpha$.
Proof: As $\alpha {id}\_V$ commutes with $T (g)$ for every $g$ and every complex $\alpha$
we have $\forall g \in G$ that $T (g) \circ (A - \alpha
{id}\_V) = (A - \alpha {id}\_V) \circ T (g)$ and from Shur's lemma it
follows then that either $A - \alpha {id}\_V = 0$ or $A - \alpha
{id}\_V$ is a isomorphism (and thus invertible). Now if $V$ is complex
then $A$ must have a eigenvalue $\alpha$ (<2>) so that $A - \alpha
{id}\_V$ is not invertible and thus we must have $A = \alpha {id}\_V$
The problems I have with this proof are as follows :
1. For (<2>) to be valid (that $A$ has a eigenvalue) must we not have that $V$
is finite dimensional. The corollary tells nothing about finite dimensionality
of $V$.
2. For what do we need (<1>) ($G$ is finite). In the same book at the
formulation of Shur's lemma $G$ does not need to be finite, so I think the
fact that $G$ is finite should be used somewhere in the proof of the corollary
but I fail to see where (unless from $G$ finite it follows that $V$ is finite
which would solve my previous problem but I fail to see how this could be
true). This is my biggest problem, the previous problem is in my opinion a typo in the text of the book, but here I have the impression that I'm missing something in the proof, in my opinion none of the steps in the proof do not need finiteness of G so can we omit the requirement that G must be finite in the corollary.
Thanks a lot in advance
Marc Mertens
| https://mathoverflow.net/users/5338 | Problem with the proof of a corollary of Schur's lemma | What Emerton says is of course correct: an irreducible representation of a finite group is necessarily finite-dimensional. Indeed, the same holds for any continuous irreducible representation of a compact group on a Hilbert space.
It seems to me that you can get away with milder assumptions: $G$ can be any group so long as $V$ is a Banach space and $A$ is a bounded linear operator. Then in the proof you take $\alpha$ to be the Banach space analogue of an eigenvalue for $A$, i.e., an element of the [spectrum](http://en.wikipedia.org/wiki/Spectrum_%28functional_analysis%29) of $A$.
You did of course look ahead and try to see what form of Schur's Lemma is actually used? (I mentioned a generalization because of the vague impression that operators discussed in physics are usually on infinite-dimensional spaces.) There are certainly multiple related results all going under that name. If you feel you need a different form than is proved in the text, let us know. I'm sure someone here (e.g. Emerton) can help you out.
| 3 | https://mathoverflow.net/users/1149 | 24599 | 16,138 |
https://mathoverflow.net/questions/24594 | 15 | Are there general surveys or introductions to the homotopy groups of spheres? I'm interested especially in connections to low-dimensional geometry and topology.
| https://mathoverflow.net/users/nan | Survey articles on homotopy groups of spheres | While my Algebraic Topology book and my unfinished book on spectral sequences (referred to in other answers to this question) contain some information about homotopy groups of spheres, they don't really qualify as a general survey or introduction. One source that fits this bill more closely is Chapter 1 of Doug Ravenel's "green book" Complex Cobordism and Stable Homotopy Groups of Spheres, from 1986. This introductory chapter starts at a reasonably accessible level, with increasing prerequisites in the later sections of the chapter. More recent surveys ought to exist, although at the moment I can't recall any. With the recent solution of the Kervaire invariant problem by Hill-Hopkins-Ravenel, this would be a good time for an updated survey.
Connections between homotopy groups of spheres and low-dimensional geometry and topology have traditionally been somewhat limited, with the Hopf bundle being the thing that comes most immediately to mind. A fairly recent connection is Soren Galatius' theorem that the homology groups of $Aut(F\_n)$, the automorphism group of a free group, are isomorphic in a stable range of dimensions to the homology groups of "loop-infinity S-infinity", the space whose homotopy groups are the stable homotopy groups of spheres.
| 32 | https://mathoverflow.net/users/23571 | 24624 | 16,153 |
https://mathoverflow.net/questions/24570 | 11 | Given a smooth vector bundle $E$ with *non-compact* base, let
$\Gamma(E)$ be the space of $C^\infty$ sections equipped with *compact-open* $C^\infty$-topology.
1. I have heard that $\Gamma(E)$ is not locally-contractible. Why not?
2. Is $\Gamma(E)$ contractible? Visibly any section can be joined to the zero section by "straight line", doesn't this prove that $\Gamma(E)$ is contractible?
3. Is it true that every convex subset of $\Gamma(E)$ is contractible? The argument of 2 seems to apply, but then it seems plausible that each section has an arbitrary small convex neighborhood, contradicting 1.
CLARIFICATION: One source of "rumor 1" is the book "The Convenient Setting of Global Analysis" freely available
[here](http://www.ams.org/publications/online-books/surv53-index). On page
429 one reads: "Unfortunately, for non-compact $M$, the space $C^\infty(M, N)$ is not locally contractible in the compact-open $C^\infty$-topology". Another source is the discussion in Hirsch's book in the beginning of Chapter 2, which says "It can be shown that
$C^\infty(M, N)$ has very nice features, e.g. it has a complete metric, and a countable base; if $M$ is *compact*, it is locally contractible and $C^r(M, \mathbb R^n)$ is a Banach space for $2\le r<\infty$".
Thus I assumeed that in general, if $M$ is non-compact, then the space $\Gamma(E)$ is not (or maybe just need not be?) locally contractible. Also I am uncertain whether $C^\infty(M, \mathbb R^n)$ or $\Gamma(E)$ is a topological vector space, is it really? There seems to be a sequence of semi-norms giving these spaces a structure of Frechet spaces, but then they must be locally convex, hence locally contractible. Obviously, I am missing something.
In response to comments I ask a more specific question.
**Question.** Let $T\_{r,s}(M)$ denote the space of $C^\infty$-smooth $(r,s)$-tensors on a connected non-compact $C^\infty$ manifold $M$. For $k$ with $2\le k\le \infty$, give $T\_{r,s}(M)$ the weak $C^k$-topology as in Hirsch's book (roughly for $k$ finite we require that given $\epsilon>0$ and compact subset $K$ all derivatives up to $k$ are $\epsilon$-close over $K$, and for $k=\infty$ we take the union of all $C^k$-topologies for all finite $k$ under the inclusions $C^\infty\to C^k$). Now I ask
*Is $T\_{r,s}(M)$ a Fréchet space with respect to the weak $C^k$-topology?*
In particular, I want to conclude that $T\_{r,s}(M)$ is locally contractible, and any convex subset of $T\_{r,s}(M)$ is contractible; I think Fréchet spaces must have this property.
| https://mathoverflow.net/users/1573 | What's wrong with compact-open topology on the space of maps? | (For the more specific question)
Yes for $k = \infty$ if $M$ can be exhausted by a countable number of compact sets, no otherwise. However, the failure is due to a lack of completeness (for $k \ne \infty$) or size issues (if $M$ can't be exhausted) rather than anything else and thus the local contractibility still holds. Indeed, contractibility holds simply by contracting the vector bundle itself down to the image of the zero section.
The reason is due to the fact that the topology can be described by a family of semi-norms, as Sergei indicates in his comment to his earlier answer, so you get a locally convex topological vector space. I recommend that you read about these spaces; Schaefer's book is a good place to start (as in Jarchow's but that doesn't seem to be available any more).
(In particular, be wary of saying "union of all $C^k$-topologies"; actually you are taking a projective limit here which means that the space has a $0$-neighbourhood base which is a union of the $0$-neigbourhood bases from each of the $C^k$ topologies, but that doesn't mean that the final topology is the union of all of the $C^k$ topologies. Simply take a set of point $x\_n$ that are "far apart" and put a set $U\_n$ about each one so that $U\_n$ is open in the $C^n$-topology but not in $C^{n-1}$. Then $\bigcup U\_n$ is open in the $C^\infty$ topology but not in any of the $C^n$-topolgies.)
| 5 | https://mathoverflow.net/users/45 | 24625 | 16,154 |
https://mathoverflow.net/questions/24608 | 8 | Hi. Are there nice/simple examples of cyclic extensions $L/K$ (that is, Galois extensions with cyclic Galois group) for which $L$ cannot be written as $K(a)$ with $a^n\in K$?
Thanks.
| https://mathoverflow.net/users/416 | Cyclic extensions | Dear Justin, your question is subtly ambiguous.
**First interpretation:** Is there a cyclic extension $L/K$ of degree $n$ that cannot be written $L=K(a)$ with $a\in L$ and $a^n\in K ?$.
**Answer** Yes. For example, let $p$ be a prime number and $n$ an integer.The extension
$\mathbb F\_p\subset \mathbb F\_{p^{p^n}}$ is cyclic and not of the required form since $a^{p^n} \in \mathbb F\_p$ implies $a\in \mathbb F\_p .\;$ So the simplest possible example for your question is $\mathbb F\_2 \subset \mathbb F\_4$ ! [Franz and David give excellent examples in characteristic zero]
**Second interpretation:** Is there a cyclic extension $L/K$ of degreee $n$ that cannot be written $L=K(a)$ with $a\in L$ and $a^N\in K$ *for some N that might be different from n ?*
**Partial answer** Some examples in the preceding interpretation disappear! For instance if you take
$\mathbb F\_2 \subset \mathbb F\_4$, you CAN write $\mathbb F\_4= \mathbb F\_2(a)$ with $a^\textbf{3}=1$ : just take for $a$ one of the two elements in $\mathbb F\_4 \setminus \mathbb F\_2$. Actually if $K\subset L$ are finite all examples disappar: all such extensions are cyclic and can be written $L=K(a)$ with $a^N\in K$. Indeed if $a\in L$ is a primitive element (which always exists: our extension is separable), we have $L=K(a)$ and if $q=card(L)$ we can be sure that $a^{q-1}=1 \in K$.
| 9 | https://mathoverflow.net/users/450 | 24628 | 16,156 |
https://mathoverflow.net/questions/24626 | 5 | The symmetric product of a variety $M$ is the quotient of $M^n/S\_n$ where $S\_n$ is the symmetric group permuting components of n-fold product $M^n$. IF $M$ is an affine plane $C^k$ over complex numbers, the coordinate ring of the symmetric product is the invariant polynomials in $R:=C[x^1\_1,...,x^1\_k, x^2\_1,...,x^2\_k,... ,x^n\_1,...,x^n\_k]$ under the action of $S\_n$ where $S\_n$ permutes the variables $x\_i^1,...,x\_i^n$ simultaneously for $i=1,...,k$. I want to know the invariant subring $R^{S\_n}$ in terms of generators and relations. Could anybody help me?
| https://mathoverflow.net/users/6093 | What is the coordinate ring of symmetric product of affine plane? | Those invariant polynomials are called *multisymmetric functions*. There are several papers on them; you could start with J. Dalbec, Multisymmetric functions, Beiträge Algebra Geom. 40(1) (1999), 27-51 <http://www.emis.de/journals/BAG/vol.40/no.1/b40h1dal.ps.gz>.
| 5 | https://mathoverflow.net/users/4790 | 24629 | 16,157 |
https://mathoverflow.net/questions/24548 | -5 | Hi
On page 98 "Stochastic differential equations" of Øksendal, 6th edition,
the author writes that $$\int\_{0}^{u}\Big(\int\_{0}^{t}\frac{\partial}{\partial t}f(s,t)dR\_{s}\Big)dt=\int\_{0}^{u}\Big(\int\_{s}^{u}\frac{\partial}{\partial t}f(s,t)dt\Big)dR\_{s}$$
could some one please tell me why? thanks so much for your time !
| https://mathoverflow.net/users/5136 | Another question on Øksendal's book | i am also having some problems with oksendal sometimes but here i can help u: Its basically Fubini theorem where one can change the order of integration: on the left the inner integrand is being integrated over 0<=s<=t and outer integral over 0<=t<=u. When u combine these two inequalities you get 0<=s<=t<=u. So on the right u can now have the inner integral wrt t as s<=t<=u and the outer over 0<=s<=u. hope that helps.
| 0 | https://mathoverflow.net/users/6114 | 24634 | 16,162 |
https://mathoverflow.net/questions/24609 | 5 | $\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? I suspect there is some simple proof that no such p, q can exist, but I haven't been able to find one.
Solving the Pell equation gives candidates for p^2=x and q=y, with x=y=1 as the first candidate solution and subsequent ones given by x'=3x+4y, y'=2x+3y; chances of a prime square seem vanishingly unlikely as x increases, but I don't have a proof.
Meta: how do you search for a question like this? I looked for a searching HOWTO here and on meta, and couldn't find one. That the search appears to strip '^' and '=' makes it all the harder.
| https://mathoverflow.net/users/6089 | Are there primes p, q such that p^4+1 = 2q^2 ? | This is not my solution, but I don't remember where I learned it.
Square both sides, subtract $4p^4$, and divide by 4 to obtain
$({p^4-1\over 2})^2=q^4-p^4$.
However, $z^2=x^4-y^4$ has no solutions in non-zero integers.
This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem.
The proof uses the representation of Pythagorean triples and infinite
descent.
So you must have $p=\pm 1$.
| 20 | https://mathoverflow.net/users/nan | 24636 | 16,163 |
https://mathoverflow.net/questions/24603 | 3 | It is well known and easy to prove that any Denjoy counterexample in the circle is approximated by homeomorphisms of the circle which have the same rotation number and are transitive (in particular, they are conjugated to the rotation).
My question is the following. Given a $C^1$ denjoy counterexample $f:S^1\to S^1$, does there exists a sequence $f\_n$ of $C^1$ diffeomorphisms of the circle such that:
1-the rotation number of $f\_n$ is the same as the one of $f$.
2-the diffeomorphisms $f\_n$ are conjugated to the rotation.
3-$f\_n \rightarrow f$ in the $C^1$ topology.
If the diffeomorphisms $f\_n$ are of class $C^2$, there is no need to ask for the second hypothesis by the Theorem of Denjoy.
I would guess that for diophantine rotation number, one could even get that the $f\_n$ are $C^1$ conjugated to the rotation. I believe that this must be known, but I could not find any reference.
| https://mathoverflow.net/users/5753 | Denjoy counterexamples $C^1$ close to conjugates of rotations. | Dear Raphael,
perhaps I misunderstood the question, but in order to get your sequence $f\_n$, one can $C^1$-approximate $f$ by $C^2$-diffeomorphisms $g\_n$ and then compose each $g\_n$ with an adequate small rotations $R\_{\epsilon\_n}$ to adjust the rotation number (so that $f\_n=R\_{\epsilon\_n}\circ g\_n$ do the job), right?
Best,
Matheus
| 5 | https://mathoverflow.net/users/1568 | 24637 | 16,164 |
https://mathoverflow.net/questions/24649 | 0 | As part of working with dot products of vectors in an algebraic field, having previously solved the algebraic quotient, I need to find the algebraic square root as shown in (1)
Let $r\_i$ be a root of polynomial P and $s\_j$ be a root of polynomial Q i.e., P($r\_i$)=0, Q($s\_j$)=0.
I seek to find a third polynomial R and its root $t\_k$, such that R($t\_k$)=0, so that
(1) $t\_k$ = $\sqrt{1 - r\_i^2 - s\_j^2}$
is satisfied. How can R be found, knowing $t\_k$?
| https://mathoverflow.net/users/6046 | Algebraic square root question | **Lemma 1:** If $P(x)$ is an integer polynomial with root $r$, then $P(x^2)$ is an integer polynomial with root $\sqrt{r}$.
**Lemma 2:** If $P(x)$ is an integer polynomial with root $r$, then $P(\sqrt{x})P(-\sqrt{x})$ is an integer polynomial with root $r^2$.
**Lemma 3:** If $P(x)$ is an integer polynomial with root $r$, then $P(x - t)$ is an integer polynomial with root $r + t$, for $t$ an integer.
**Lemma 4:** If $P, Q$ are integer polynomials with roots $r, s$, then an integer polynomial with root $r + s$ is given by the characteristic polynomial of $A \otimes I + I \otimes B$ where $A, B$ are the [companion matrices](http://en.wikipedia.org/wiki/Companion_matrix) of $P, Q$ and $\otimes$ denotes the [Kronecker product](http://en.wikipedia.org/wiki/Kronecker_product). Multiplication by $r + s$ defines a $\mathbb{Q}$-linear transformation on $\mathbb{Q}[r, s]$, which has $\mathbb{Q}$-basis $r^i s^j$ where $i, j$ range from $0$ to one less than the degrees of $P$ and $q$, and the matrix above is the matrix of this linear transformation in that basis.
So apply Lemma 2 twice, then Lemma 3, then Lemma 4, then Lemma 1.
| 4 | https://mathoverflow.net/users/290 | 24651 | 16,174 |
https://mathoverflow.net/questions/20355 | 22 | I am looking for a good book to study probability. My advisor suggested the "Probability" by Leo Breiman. I am reading it now, it seems rather a dense book, so I would like to ask you guys advice on which book you guys often start with for Probability.
| https://mathoverflow.net/users/5136 | Book for probability | I would definitely go for "Probability" by Jim Pitman. It is a very good book for learning Probability Theory, one of the best text books I have encountered in my studies.
| 9 | https://mathoverflow.net/users/1539 | 24664 | 16,181 |
https://mathoverflow.net/questions/24680 | 6 | The question is in the title exactly as I want to ask it, but let me provide some background and motivation.
Many of the properties of fields studied in the algebraic theory of quadratic forms are manifestly *elementary* properties in the sense of model theory: that is, if one field has this property, then any other field which has the same first-order theory in the language of fields has that same property. Examples:
being quadratically closed, being formally real, being real-closed, being Pythagorean (sum of two squares is always a square), for any fixed positive integer n, having I^n = 0 (follows from the Milnor conjectures!), the u-invariant, the level, the Pythagoras number...
These properties imply that at least for some fields $K$, if $L$ is any field elementarily equivalent ot $K$, then $W(L) \cong W(K)$: e.g. $K$ is quadratically closed, $K$ is real-closed, $K = \mathbb{C}((t))$. Is it always the case that $K \equiv L$ implies $W(K) \cong W(L)$? I am pretty sure the answer is no because for instance if $\operatorname{dim}\_{\mathbb{F}\_2} K^{\times}/K^{\times 2}$ is infinite, I think it is not an elementary invariant. And if you take a field with vanishing Brauer group, then $W(K)$ is, additively, an elementary $2$-group of dimension $\operatorname{dim}\_{\mathbb{F}\_2} K^{\times}/K^{\times 2} + 1$.
But are there known positive results in this direction?
| https://mathoverflow.net/users/1149 | If two fields are elementarily equivalent, what can we say about their Witt rings? | As you point out, one cannot hope that the Witt ring, up to isomorphism, be an elementary invariant of a field. The strongest statement which I might conjecture would be that if $K \preceq L$ is an elementary extension of fields, then $W(K) \to W(L)$ is an elementary extension of rings. If this statement were true, then the theory of the Witt ring would be an elementary invariant as any two elementarily equivalent fields have a common elementary extension.
It is true that if $K \preceq L$ is an elementary extension of fields, then map $W(K) \hookrightarrow W(L)$ is an inclusion [Why? Being zero in the Witt ring is defined by an existential condition.] One might try to prove that $W(K) \hookrightarrow W(L)$ is elementary by induction where the key step would be to show that if $W(L) \models (\exists x) \phi(x;a)$ where $a$ is a tuple from $W(K)$, $x$ is a single variable, and $\phi$ is a formula in the language of rings, then $W(K) \models (\exists x) \phi(x;a)$. The witness in $W(L)$ would be represented by a quadratic space of some finite dimension $n$. One would like to argue that the set defined by $\phi(x;a)$ in the space of $n$-dimensional quadratic forms is definable in the field language in $K$ in which case a witness could be found in $K$ via elementarity. This last part of the argument is delicate as it would require knowing bounds for checking equalities in the Witt ring.
The Witt ring construction is an example of an ind-definable set modulo an ind-definable equivalence relation. These are discussed in some detail in Hrushovski's paper on approximate groups ( arXiv:0909.2190 ). With Krajiceck, I considered similar issues (how does the Grothendieck ring of a first-order structure depend on its theory) in Combinatorics with definable sets: Euler characteristics and Grothendieck rings. Bull. Symbolic Logic 6 (2000), no. 3, 311--330.
| 8 | https://mathoverflow.net/users/5147 | 24685 | 16,191 |
https://mathoverflow.net/questions/24688 | 16 | Is there an efficient way to sample uniformly points from the unit n-sphere? Informally, by "uniformly" I mean the probability of picking a point from a region is proportional to the area of that region on the surface of the sphere. Formally, I guess I'm referring to the Haar measure.
I guess "efficient" means the algorithm should take poly(n) time. Of course, it's not clear what I mean by an algorithm since real numbers cannot be represented on a computer to arbitrary precision, so instead we can imagine a model where real numbers can be stored, and arithmetic can be performed on them in constant time. Also, we're given access to a random number generator which outputs a real in [0,1]. In such a model, it's easy to sample from the surface of the n-hypercube in O(n) time, for example.
If you prefer to stick with the standard model of computation, you can consider the approximate version of the problem where you have to sample from a discrete set of vectors that $\epsilon$-approximate the surface of the n-sphere.
| https://mathoverflow.net/users/1042 | Efficiently sampling points uniformly from the surface of an n-sphere | Generate $X\_1, X\_2, \ldots, X\_n$ independent, normally distributed random variables See [wikipedia](http://en.wikipedia.org/wiki/Normal_distribution#Generating_values_from_normal_distribution) for information on how to do this given some standard source of randomness, for example uniform(0,1) random variables.
Then let $Y\_i = X\_i/\sqrt{X\_1^2 + \cdots + X\_n^2}$ for $i = 1, \ldots, n$. Then $(Y\_1, \ldots, Y\_n)$ is uniformly distributed on the surface of the sphere. The time this takes is linear in $n$.
This works because the multivariate normal $(X\_1, \ldots, X\_n)$ with covariance matrix the identity (that is, $n$ independent unit normals) is rotationally symmetric around the origin.
| 24 | https://mathoverflow.net/users/143 | 24690 | 16,194 |
https://mathoverflow.net/questions/24659 | 6 | I'm a computer programmer who's caught on to denotational semantics. I mostly work with Ruby, JavaScript and C, but I know a little Haskell and ML. I've taken my first steps towards reasoning about what my software *means*, but my knowledge of domain theory is weak. DCPOs, chains, new notation – can you recommend a coherent introduction to this stuff?
| https://mathoverflow.net/users/6100 | Resources for learning domain theory? | The [book recommended by jef](https://mathoverflow.net/a/24677/49269) is the domain-theory bible. It may be a bit overwhelming for a beginner. For an easier and more compressed introduction I recommend that you have a look at Abramsky and Jung's chapter on domain theory from the Handbook of Logic in Computer Science. It is [available in PDF](https://www.cs.bham.ac.uk/~axj/pub/papers/handy1.pdf) from [Achim's publications](https://www.cs.bham.ac.uk/~axj/papers.html) lits.
| 4 | https://mathoverflow.net/users/1176 | 24702 | 16,203 |
https://mathoverflow.net/questions/24622 | 9 | My general question concerns what we can learn about an arbitrary, three-dimensional convex polytope (or convex hull of an arbitrary polytope) strictly from the surface areas of its two-dimensional projections on a plane as it 'tumbles' in 3-space (i.e. as it rotates along an arbitrary, shifting axis).
If it's helpful, please imagine the following physical set-up:
We take an arbitrary three-dimensional convex polytope, and fix the center of mass to a coordinate in 3-space, $(x\_0, y\_0, z\_0)$, located at some distance, $D$, above a flat surface. While this prohibits translation of the center of mass, the polytope is still allowed to tumble freely (i.e. it is allowed rotation around an arbitrary axis centered at the fixed coordinate).
There is no 'gravity' or other force to stabilize the tumbling polytope in a particular orientation. Over time it will continue to tumble randomly. (The 'physical' set-up is only meant for descriptive reasons.)
We shine a beam of coherent light on the tumbling polytope, larger than the polytope's dimensions, and continually record the area of the resulting shadow, or two-dimensional projection on the surface. To be clear, the area of the two-dimensional projection is the only information we are allowed to observe or record, and we are allowed to do so over an arbitrary length of time.
---
My question is - From observing the area of the tumbling polytope's shadow, or two-dimensional projection over time, what can we learn about it's geometry? To what extent can we characterize and/or reconstruct the polytope from its changing shadow (extracting the surface area for example - hat tip to Nurdin Takenov)?
Do we gain anything by watching the evolution of the convex polytopes shadow as it tumbles (part of the point for the physical example), as opposed to an unordered collection of two-dimensional projections?
Update - Nurdin Takenov (and Sergei Ivanov in later comments) nicely points out that we can use the average surface area of the two-dimensional projection to find the surface area of the tumbling convex polygon. Might we be able to find it's volume?
(Addendum - I would be really neat if somebody could point me to any algorithms in the literature... or available software.... that let's me calculate and characterize two-dimensional surface projections of convex polytopes!)
| https://mathoverflow.net/users/3248 | Characterizing a tumbling convex polytope from the surface areas of its two-dimensional projections | You cannot recover a convex polytope from its projection areas, even if you know the whole function (unit vector) $\mapsto$ (area of projection along this vector). There exist two different polytopes $P\_1$ and $P\_2$ such that for every unit vector $v\in\mathbb R^3$, the areas of the two projections along $v$ are equal.
Let me begin with a two-dimensional example. In this case, the projections are one-dimensional, and the "projection area" is width. Consider a regular triangle $T$ with side 1 and regular hexagon $H$ with side 1/2, positioned so that their sides are parallel. Their widths are the same in every direction. One can prove this without computation: the hexagon is the Minkowski symmetrization of the triangle and the symmetrization preserves widths.
Now go to dimension 3. Let $P\_1=T\times[0,1]$, the prism with height 1 based on the regular triangle with side 1. Let $P\_2=(\sqrt{2/3}H)\times[0,\sqrt{3/2}]$, the prism with height $\sqrt{3/2}$ based on the regular hexagon with side $1/\sqrt6$ (I hope that I get the constants right). I claim that they have the same projection area in every direction.
More generally, consider a prism of height $h$ based on a convex figure $F$ of area $A$ such that the base is parallel to the $xy$-plane. Its area of projection along a vector $v=(\cos\alpha\cos\theta,\sin\alpha\cos\theta,\sin\theta)$, is given by
$$
A\cdot \sin\theta+w(\alpha)\cdot h \cdot \cos\theta
$$
where $w(\alpha)$ is the width of $F$ in the horizontal direction that forms oriented angle $\alpha$ with the $x$-axis. The constants in the above example are chosen so that the two prisms have the same $A$ and the difference in $w(\alpha)$ is compensated by the difference in $h$.
| 10 | https://mathoverflow.net/users/4354 | 24707 | 16,206 |
https://mathoverflow.net/questions/24579 | 166 | I saw a while ago in a book by Clifford Pickover, that whether the [Flint Hills series](https://mathworld.wolfram.com/FlintHillsSeries.html) $\displaystyle \sum\_{n=1}^\infty\frac1{n^3\sin^2 n}$ converges is open.
I would think that the question of its convergence is really about the density in $\mathbb N$ of the sequence of numerators of the standard convergent approximations to $\pi$ (which, in itself, seems like an interesting question). Naively, the point is that if $n$ is "close" to a whole multiple of $\pi$, then $1/(n^3\sin^2n)$ is "close" to $\frac1{\pi^2 n}$.
[Numerically there is some evidence that only some of these values of $n$ affect the overall behavior of the series. For example, letting $S(k)=\sum\_{n=1}^{k}\frac1{n^3\sin^2n}$, one sees that $S(k)$ does not change much in the interval, say, $[50,354]$, with $S(354)<5$. However, $S(355)$ is close to $30$, and note that $355$ is very close to $113\pi$. On the other hand, $S(k)$ does not change much from that point until $k=100000$, where I stopped looking.]
I imagine there is a large body of work within which the question of the convergence of this series would fall naturally, and I would be interested in knowing something about it. Sadly, I'm terribly ignorant in these matters. Even knowing where to look for some information on approximations of $\pi$ by rationals, or an *ad hoc* approach just tailored to this specific series would be interesting as well.
| https://mathoverflow.net/users/6085 | Convergence of $\sum(n^3\sin^2n)^{-1}$ | As Robin Chapman mentions in his comment, the difficulty of investigating
the convergence of
$$
\sum\_{n=1}^\infty\frac1{n^3\sin^2n}
$$
is due to lack of knowledge about the behavior of $|n\sin n|$ as $n\to\infty$,
while the latter is related to rational approximations to $\pi$ as follows.
Neglecting the terms of the sum for which $n|\sin n|\ge n^\varepsilon$
($\varepsilon>0$ is arbitrary),
as they all contribute only to the `convergent part' of the sum, the question
is equivalent to the one for the series
$$
\sum\_{n:n|\sin n|< n^\varepsilon}\frac1{n^3\sin^2n}.
\qquad(1)
$$
For any such $n$, let $q=q(n)$ minimizes the distance $|\pi q-n|\le\pi/2$.
Then
$$
\sin|\pi q-n|=|\sin n|< \frac1{n^{1-\varepsilon}},
$$
so that $|\pi q-n|\le C\_1/n^{1-\varepsilon}$ for some absolute constant $C\_1$
(here we use that $\sin x\sim x$ as $x\to0$). Therefore,
$$
\biggl|\pi-\frac nq\biggr|<\frac{C\_1}{qn^{-\varepsilon}},
$$
equivalently
$$
\biggl|\pi-\frac nq\biggr|<\frac{C\_2}{n^{2-\varepsilon}}
\quad\text{or}\quad
\biggl|\pi-\frac nq\biggr|<\frac{C\_2'}{q^{2-\varepsilon}}
$$
(because $n/q\approx\pi$) for all $n$ participating in the sum (1).
It is now clear that the convergence of the sum (1) depends
on how often we have
$$
\biggl|\pi-\frac nq\biggr|<\frac{C\_2'}{q^{2-\varepsilon}}
$$
and how small is the quantity in these cases. (Note that
it follows from Dirichlet's theorem that an even stronger inequality,
$$
\biggl|\pi-\frac nq\biggr|<\frac1{q^2},
$$
happens for infinitely many pairs $n$ and $q$.)
The series (1) converges if and only if
$$
\sum\_{n:|\pi-n/q|< C\_2n^{-2+\varepsilon}}\frac1{n^5|\pi-n/q|^2}
$$
converges. We can replace the summation by summing over $q$
(again, for each term $\pi q\approx n$) and then sum the result
over all $q$, because the
terms corresponding to $|\pi-n/q|< C\_2n^{-2+\varepsilon}$ do not
influence on the convergence:
$$
\sum\_{q=1}^\infty\frac1{q^5|\pi-n/q|^2}
=\sum\_{q=1}^\infty\frac1{q^3(\pi q-n)^2}
\qquad(2)
$$
where $n=n(q)$ is now chosen to minimize $|\pi-n/q|$.
Summarizing, *the original series converges if and only if the series in* (2)
*converges.*
It is already an interesting question of what can be said about the
convergence of (2) if we replace $\pi$ by other constant $\alpha$,
for example by a "generic irrationality". The series
$$
\sum\_{q=1}^\infty\frac1{q^3(\alpha q-n)^2}
$$
for a real quadratic irrationality $\alpha$ converges because the best
approximations are $C\_3/q^2\le|\alpha-n/q|\le C\_4/q^2$, and they are achieved on
the convergents $n/q$ with $q$ increasing geometrically. A more delicate
question seems to be for $\alpha=e$, because one third of its convergents satisfies
$$
C\_3\frac{\log\log q}{q^2\log q}<\biggl|e-\frac pq\biggr|< C\_4\frac{\log\log q}{q^2\log q}
$$
(see, e.g., [C.S.Davis, *Bull. Austral. Math. Soc.* 20 (1979) 407--410]).
The number $e$, quadratic irrationalities, and even algebraic numbers
are `generic' in the sense that their irrationality exponent is known to be 2.
What about $\pi$?
The *irrationality exponent* $\mu=\mu(\alpha)$ of a real irrational number
$\alpha$ is defined as the infimum of exponents $\gamma$
such that the inequality $|\alpha-n/q|\le|q|^{-\gamma}$ has
only finitely many solutions in $(n,q)\in\Bbb Z^2$ with $q\ne0$.
(So, Dirichlet's theorem implies that $\mu(\alpha)\ge2$. At the same
time from metric number theory we know that it is 2 for almost all real irrationals.)
Assume that $\mu(\pi)>5/2$, then there are infinitely many
solutions to the inequality
$$
\biggl|\pi-\frac nq\biggr|<\frac{C\_5}{q^{5/2}},
$$
hence infinitely many terms in (2) are bounded below by $1/C\_5$, so that
the series diverges (and (1) does as well). Although the general
belief is that $\mu(\pi)=2$, the best known result of V.Salikhov (see
[this answer](https://mathoverflow.net/questions/23547/does-pi-contain-1000-consecutive-zeroes-in-base-10/23551#23551)
by Gerry and my comment)
only asserts that $\mu(\pi)<7.6064\dots$,.
I hope that this explains the problem of determining the behavior of the series in question.
| 132 | https://mathoverflow.net/users/4953 | 24712 | 16,209 |
https://mathoverflow.net/questions/24718 | 7 | By a graph I will understand an undirected graph without multiple edges or loops. By a morphism of graphs I will understand a map $f$ between the underlying sets of vertices, such that if $x$ and $y$ are adjacent, then $f(x)$ and $f(y)$ are either adjacent or equal.
Let $G$ be a finite graph. One can realise $G$ as a CW-complex $|G|$ and look at topological invariants, such as singular homology. But this captures only very little information about $G$, because except from $H\_0(|G|)$ and $H\_1(|G|)$ all homology groups are zero.
Consider the following alternative construction: Let us write $\Delta\_n$ for the complete graph on $n$ vertices, and let us re-baptise this graph by the name "standard $n$-simplex". There are obvious codegeneracy and coface maps between standard simplices, so that we obtain a cosimplicial object $\Delta\_\bullet$ in the category of graphs. Now, proceed as usual: Morphisms $\Delta\_\bullet \to G$ form a simplicial set, applying the free group construction yields then simplicial group, and the associated chain complex is the one whose homology $H\_i^{\mathrm{sing}}(G)$ I shall call "singular homology of $G$".
Obvious properties of $H\_i^{\mathrm{sing}}(G)$ are: It is a finitely generated commutative group ($G$ is finite), covariantly functorial in $G$. In particular, if we work with coefficients in a field, we obtain representations of the automorphism group of $G$. The homology of the point is $\mathbb Z$ in degree $0$ and trivial in higher degrees. We can define singular cohomology accordingly, and get then a natural pairing between homology and cohomology.
The list of all natural questions one must ask after making such a definition is long, so I will not ask everything.
(a) Is there a comparison map $H\_i(|G|) \to H\_i^{\mathrm{sing}}(G)$, maybe even on the level of chain complexes? Is there some more elaborate CW-complex $||G||$ one can naturally associate with $G$ such that $H\_i(||G||)$ gives back singular homology of $G$? In that case, one would ask for a natural map $|G| \to ||G||$.
(b) Given a graph $G$, is there a largest integer $i$ such that $H\_i^{\mathrm{sing}}(G)$ is nonzero? Assuming yes, is this integer less or equal the size of the largest complete subgraph of $G$.
(c) Is there a Künneth morphism in singular cohomology? --is there a natural ring strucure on cohomology?
(d) What is a homotopy between morphisms of graphs? Given an answer to that, do homotopically equivalent morphisms induce the same maps in homology?
(z) Can you give an example of a graph with nontrivial $H\_2^{\mathrm{sing}}(G)$?
| https://mathoverflow.net/users/5952 | Singular homology of a graph. | You talk about morphisms from $\Delta\_\bullet$ to a graph $G$.
I presume a morphism from $\Delta\_n$ to $G$ is just an embedding
of $\Delta\_n$ in $G$, that is an $(n+1)$-clique in $G$ with a
labelling of its vertices from $0$ to $n$. It seems to me that
defining (co)homology in this way will be the same as the standard
simplicial/singular (co)homology of the space $\widehat G$ obtained from $G$
by "filling in" each $(n+1)$-clique with an $(n+1)$-simplex.
As a example consider the graph $G$ consisting of the vertices
and edges of a regular icosahedron. Then its $3$-cliques
correspond to the faces of the icosahedron and it has no $4$-cliques.
Therefore $\widehat G$ will consist of the boundary of the
solid icosahedron in $\mathbb{R}^3$ and so your second homology group
will be nonzero.
I hasten to add that I have not checked any details, and admit
in advance that my thoughts here may be msiguided or just complete nonsense.
| 3 | https://mathoverflow.net/users/4213 | 24728 | 16,215 |
https://mathoverflow.net/questions/24700 | 1 | My question concerns the existence of a nice (deterministic?) method/algorithm for calculating the distribution of surface areas for two-dimensional projections of an arbitrary polytope (or convex approximation of a polytope). Less optimistically, a method of finding the minimum, maximum, and perhaps, mean surface area of the polytope's projections.
It is a relatively straightforward procedure to calculate a given two-dimensional surface projection along some orientational vector, and then calculate the approximate surface area of the projection (or its convex hull). But, beyond statistical sampling or methods related to simulated annealing, I'm having trouble imagining how to go about characterizing the full set of projections along all arbitrary vectors... and I haven't had any luck with a literature search (so far).
Note - This question is directly related to computations one might like to perform for - [Characterizing a tumbling convex polytope from the surface areas of its two-dimensional projections](https://mathoverflow.net/questions/24622/characterizing-a-tumbling-convex-polytope-from-the-surface-areas-of-its-two-dimen). I hope this follow-up post is appropriate...
| https://mathoverflow.net/users/3248 | Calculating the surface area distribution of two-dimensional projections for a polytope | If you take the arrangement of planes determined by the faces of your polytope, then the combinatorial structure of the projection is constant throughout all the viewpoints within one cell of the arrangement. An alternative viewpoint is to partition $S^2$ by these planes moved to the center of that sphere. Within each cell of this arrangement of great circles on $S^2$, the area of the projection changes in a regular, computable manner (as a function of coordinates on $S^2$). None of this would be easy to implement, but it is computable in roughly $O(n^2)$ time for a 3-polytope of $n$ vertices. (Note here I am using $n$ for the number of vertices, and assuming you are working in $R^3$, whereas in Robby McKilliam's posting, $n$ is the dimension).
For this arrangements viewpoint, see the paper by Michael McKenna and Raimund Seidel,
"Finding the optimal shadows of a convex polytope,"
<http://portal.acm.org/citation.cfm?id=323237> .
| 3 | https://mathoverflow.net/users/6094 | 24730 | 16,217 |
https://mathoverflow.net/questions/24726 | -3 | Given E[v|X=x]=g[x] and the pdf of X (f[x]), how to calculate E[v|x>=x0]? The pdf of V or the joint pdf of V,X are unknown. My guess is that this problem has no solution.
| https://mathoverflow.net/users/3899 | Conditional expectation | So if $p(v,x)$ is the unknown pdf,
$f(x) = \int p(v,x) \mathrm{d}v$
$g(x) = E[v|X=x] = \int v \ P[v|X=x] \mathrm{d}v = \int v \ \frac{p(v,x)}{f(x)} \mathrm{d}v$
and then
$E[v|X \ge x\_0] = \frac{1}{P[X \ge x\_0]} \int\_{x\_0}^\infty E[v|X = x] P[x] \mathrm{d}x = \frac{ \int\_{x\_0}^\infty f(x)g(x) \mathrm{d}x}{\int\_{x\_0}^\infty f(x)\mathrm{d}x}$
| 2 | https://mathoverflow.net/users/5789 | 24731 | 16,218 |
https://mathoverflow.net/questions/24713 | 10 | Being far from analysis, I recently learned about the [Invariant subspace problem](http://en.wikipedia.org/wiki/Invariant_subspace_problem) and came up with the following (perhaps simple or well-known) question.
Let $H$ be a separable complex Hilbert space and $T:H\to H$ a bounded operator. Assume that the spectrum of $H$ is $\{0\}$, i.e. $T-\lambda I$ has a bounded inverse for every $\lambda\in\mathbb C\setminus\{0\}$. In finite dimensions, this would imply that $T$ is nilpotent ($T^n=0$ for some $n$). I wonder if there is something similar in the infinite dimensional case. The precise formulation I have in mind is the following.
It is easy to see that there is a maximal subspace $X\subset H$ such that $T(X)=X$. This is a purely set-theoretic fact, a sort of explicit construction is the intersection of the images of iterations of $T$ up to and beyond infinity (via transfinite induction).
**Question:** can it happen that $X\ne\{0\}$?
Here are some observations that I made:
* $T$ cannot be onto. (I derived this from some random Wikipedia quotes so there are high chances of error; please correct me if I am wrong.) It follows that a nontrivial $X$ cannot be closed.
* If there is an example, then there is one where $X$ is dense. Just take the closure of $X$ for $H$.
* It is possible that $T(H)$ is dense. My example is the shift in $\ell^2(\mathbb Z)$ composed with a mutiplication by a positive function (sequence) that goes to zero at both ends. Again, please correct me if I am wrong.
| https://mathoverflow.net/users/4354 | Are operators with trivial spectrum nilpotent in a sense? | Perhaps I am missing something. Isn't the example you mention quasinilpotent and and maps the finitely non zero sequences onto themselves?
I am not an expert on invariant subspaces, but I think it is widely believed among experts that the answer to the question is the same for quasinilpotent operators as for general operators.
| 6 | https://mathoverflow.net/users/2554 | 24739 | 16,221 |
https://mathoverflow.net/questions/24733 | 17 | I was in Paris recently for a meeting about motives or motifs, and since I'm too jet lagged
for real work let me ask the following somewhat frivolous question. The word "motif" is
usually translated as "motive" in English. However, I wonder if this is really the best choice. "Motive" has, for me, a primarily psychological meaning, whereas "motif" -- which is
a perfectly good English word -- means pattern or theme. I guess my question is which word better captures the intended meaning?
Incidentally, it appears that this usage of "motive" goes back to Grothendieck himself, cf."Standard conjectures on algebraic cycles". So perhaps, one should allow him to have the last word and not question his motives, which have wonderful if unintended consequences.
| https://mathoverflow.net/users/4144 | Motives versus Motifs | Dear Donu, here are Grothendieck's own words:
"Contrary to what occurs in ordinary
topology, one finds oneself confronting
a disconcerting abundance of different
cohomological theories. One has the
distinct impression (but in a sense that
remains vague) that each of these theories
“amount to the same thing”, that
they “give the same results”. In order to
express this intuition, of the kinship of
these different cohomological theories,
I formulated the notion of “motive” associated
to an algebraic variety. By this
term, I want to suggest that it is the
“common motive” (or “common reason”)
behind this multitude of cohomological
invariants attached to an
algebraic variety, or indeed, behind all
cohomological invariants that are a
priori possible"
They can be found in his autobiographical "Récoltes et Semailles", where there is also an allusion to a musical meaning of "motif".
The translation is Barry Mazur's in his article "What is... a Motive?" which is, needless to say, a fascinating short survey (plus bibliography) .Here is the reference:
<http://www.ams.org/notices/200410/what-is.pdf>
| 16 | https://mathoverflow.net/users/450 | 24749 | 16,228 |
https://mathoverflow.net/questions/24754 | 29 | Let $(G, n)$ be a pair such that $n$ is a natural number, $G$ is a finite group which is abelian if $n \geq 1$. It is well-known that $\pi\_n(K(G,n)) = G$ and $\pi\_i (K(G,n)) = 0$ if $i \neq n$.
Also it is known that these spaces $K(G,n)$ play a very important role for cohomology. For any abelian group $G$, and any CW-complex $X$, the set $[X, K(G,n)]$ of homotopy classes of maps from $X$ to $K(G,n)$ is in natural bijection with the $n^{\mathrm{th}}$ singular cohomology group $H^n(X; G)$ with coefficients in $G$.
But what is known about the cohomology of the $K(G,n)$ themselves? It is interesting in the light of the above. Here I mean the singular cohomology with integral coefficients.
| https://mathoverflow.net/users/6031 | (Co)homology of the Eilenberg-MacLane spaces K(G,n) | Computing the integral cohomology of $K(\pi,n)$'s is feasible but a bit tricky. In fact the only reference I know is [exposé 11 of H. Cartan's seminar, year 7](http://archive.numdam.org/ARCHIVE/SHC/SHC_1954-1955__7_1/SHC_1954-1955__7_1_A11_0/SHC_1954-1955__7_1_A11_0.pdf). I'd be interested if there are other sources that cover that.
| 19 | https://mathoverflow.net/users/2349 | 24759 | 16,235 |
https://mathoverflow.net/questions/24758 | -1 | I think it is helpful to keep a hoard of helpful sites pertaining to unsolved problems and pay a regular/casual visit to them.
Please when you answer, add a comment on the content of the site for faster surfing.
Thanks in advance.
| https://mathoverflow.net/users/5627 | List of sites related to the Riemann Hypothesis and recent developments? | <http://aimpl.org/pl>
Quote
>
> The AIM Problem Lists are part of the Bibliographic Knowledge Network (BKN) project funded by a Cyber Enabled Discovery and Innovation (CDI) grant from the National Science Foundation. Major partners in the project are UC Berkeley, Harvard, Stanford, and AIM.
>
>
>
It currently contains four large categories:
* Braid Groups, Clusters, and Free Probability
* Low Eigenvalues of Laplace and Schroedinger Operators
* Equivalences to the Riemann Hypothesis
* The Riemann Hypothesis and related problems
| 2 | https://mathoverflow.net/users/3948 | 24761 | 16,237 |
https://mathoverflow.net/questions/24715 | 8 | The question I ask is in the title. This should be quite well-known, and in fact probably I am going to get the response that it is the definition. To convey my confusion, I have to convey my understanding of what is a differential form and what is the contangent bundle. To simplify things, we assume that our whole setup is immersed in the Euclidean space.
$1$. **Differential forms**.
We take the definitions from Rudin, Principles of Mathematical Analysis. This is for an open set in $\mathbb R^n$.
Suppose $E$ is an an open set in $\mathbb R^n$. A $k$-surface in $E$ is a differentiable mapping $\Phi$ from a compact subset $D \subset \mathbb R^k$ into $E$. $D$ is called the parameter domain of $\Phi$ consisting of points $\mathbf u = ( u\_{i\_1}, \cdots , u\_{i\_k} )$.
A differential form of order $k \geq 1$ in $E$ is a function $\omega$, symbolically represented by the sum
$$\omega = \sum a\_{i\_1, \cdots , i\_k}(\mathbf x) dx\_{i\_1}\wedge \cdots \wedge dx\_{i\_k}$$
where the indices $i\_1, \cdots , i\_k$ range independently from $1$ to $n$, and so that $\omega$ assigns to each $k$-surface $\Phi$ in $E$ a number$\omega(\Phi) = \int\_\Phi \omega$
, according to the rule
$$\int\_\Phi \omega = \int\_D \sum a\_{i\_1, \cdots , i\_k}(\Phi((\mathbf{u})) \frac{\partial ( x\_{i\_1}, \cdots , x\_{i\_k})}{\partial ( u\_{i\_1}, \cdots , u\_{i\_k})}d\mathbf u $$
where $D$ is the parameter domain of $\Phi$, and the functions $a\_{i\_1}, \cdots, a\_{i\_k}$ are assumed to be real and continuous in $D$.
So in the above definition the differential $k$-form is a certain integral for functions on compact $k$-surfaces. Thus a differential form can be treated as a measure for the $k$-surfaces, which can be integrated.
$2$. **Cotangent bundle**
We take this from wikipedia.
Let $M \times M$ be the Cartesian product of $M$ with itself. The diagonal mapping $\Delta$ sends a point $p$ in $M$ to the point $(p,p)$ of $M \times M$. The image of $\Delta$ is called the diagonal. Let $\mathcal{I}$ be the sheaf of germs of smooth functions on $M \times M$ which vanish on the diagonal. Then the quotient sheaf $\mathcal{I}/\mathcal{I}^2$ consists of equivalence classes of functions which vanish on the diagonal modulo higher order terms. The cotangent sheaf $\Omega$ is the pullback of this sheaf to $M$.
Now, Def 2: A differential form $k$-form $\omega$ is a section of $\wedge^k\ \Omega$.
**Question**.
We consider an open set in the Euclidean space and look at the two definitions. A priori, to my eyes, both appear to be different things. It needs to be proved that they are the same. Please help me out with a reference with the required proofs.
| https://mathoverflow.net/users/6031 | Why are order-k differential forms sections of the kth exterior power of the cotangent bundle? | You have to work a bit to get those two definitions to agree, but it is all standard lore in differential geometry. Both of the references in the comments - Spivak and Madsen & Tornehave - are good and should have what you need, the latter a bit more useful in my opinion. But I am writing this answer because in neither text (or virtually any other introductory differential geometry text) will you encounter explicitly your definition of the cotangent bundle, or for that matter words like "sheaf" and "germ". Such notions are useful in differential geometry, but it is not so crucial to incorporate them into the foundations the way it is in algebraic geometry.
A definition that you will see in books (in some form) proceeds as follows. Given a point $p$ in $M$, define the tangent space $T\_p M$ to be the vector space of point derivations of $C^\infty(M)$ at $p$. If $(x\_1, \ldots, x\_n): U \subseteq M \to \mathbb{R}^n$ is a local coordinate system near $p$ then the directional derivative derivations $\frac{\partial}{\partial x\_i}|\_{p}$ form a basis of $T\_p M$. Construct a vector bundle $TM$ whose fiber over a point $x$ in $M$ is $T\_x M$, and form its dual $T^\*M$. This is the cotangent bundle, and it takes only some basic techniques with sheaves to prove that it is the same as what you defined.
The equivalence of this definition with your first definition is just a bunch of coordinate calculations, complicated by the fact that Rudin defines surfaces extrinsically (in contrast to the intrinsic definitions that you and I produced). From the intrinsic point of view, a differential k-form is a smooth section of the bundle $\wedge^k T^\*M$. As above, a local coordinate system $(x\_1 \ldots x\_n)$ on an open neighborhood $U$ of $M$ yields a trivializing frame $\frac{\partial}{\partial x\_1}, \ldots, \frac{\partial}{\partial x\_n}$ for $TM$ over $U$, and the corresponding dual frame for $T^\*M$ over $U$ is denoted by $dx\_1 \ldots dx\_n$. Thus a trivializing frame for $\wedge^k T^\*M$ is given by $\{ dx\_{i\_{1}} \wedge \ldots \wedge dx\_{i\_{k}}: 1 \leq i\_{1} \leq \ldots \leq i\_{k} \leq n\}$, which more or less explains the local formula in your question. The details of your integral formula are explained by noting that Rudin is defining a $k$ form on $\mathbb{R}^n$ with its standard coordinate system and pulling it back to a $k$ form on the surface along the embedding of the surface in $\mathbb{R}^n$. Thus it is necessary to sort out how an intrinsic differential form behaves under coordinate change (from $\mathbb{R}^n$ coordinates to coordinates on the surface), and the whole point of the theory is that they change in a way which makes integration coordinate invariant.
One last comment. Rudin defines differential k-forms for k-dimensional surfaces in $\mathbb{R}^n$ which comes naturally equipped with a notion of integration. Of course most interesting manifolds (e.g. the sphere) are not the images of compact domains in $\mathbb{R}^k$ embedded in $\mathbb{R}^n$; this is the right LOCAL picture, but not the right global picture. So to properly define the integral of a k-form over a k-manifold, it is necessary to define the global integral in terms of patched-together local integrals via a partition of unity. Most books set up the theory this way and prove that the integral doesn't depend on the choice of partition of unity, but no books explain why, for example, the intrinsic integral of an n-form $f dx\_1 \wedge \ldots \wedge dx\_n$ over an open set in $\mathbb{R}^n$ (involving partitions of unity) agrees with standard Lebesgue integral of $f$ over that open set. I suspect that one would have to sort out this kind of issue in order to REALLY prove that your two notions of differential form agree. It's possible to work this out on your own, but Brian Conrad sorted it out in his "How to compute integrals" handout here:
<http://math.stanford.edu/~conrad/diffgeomPage/handouts.html>
You might find some of the other handouts helpful too.
| 10 | https://mathoverflow.net/users/4362 | 24767 | 16,239 |
https://mathoverflow.net/questions/24764 | 5 | A finite groups of $\mathrm{GL}\_n(\mathbb C)$ of exponent $m$ necessarily have order $C$ verifying $C\leqslant m^n$ and $n! m^n$ divides $C$, but this condition is not sufficient, for instance $\mathrm{GL}\_3(\mathbb C)$ has no subgroup of exponent $7$ and order $42$.
So can we determine fully the possible orders of the subgroups of $\mathrm{GL}\_n(\mathbb C)$ of exponent $m$?
~~~~~~~~~~~~OLD QUESTION - Which was obvious, sorry and thanks Pete ~~~~~~~~~~~~~~~~
After some computation and search it seems that finite subgroups of $\mathrm{GL}\_3(\mathbb C)$ of exponent $5$ have order $5$, $5^2$ or $5^3$.
Similarly, finite subgroups of $\mathrm{GL}\_3(\mathbb C)$ of exponent $7$ have order $7$, $7^2$ or $7^3$.
So, for $n >2$ and a prime $p >2$, is it true that finite subgroups of $\mathrm{GL}\_n(\mathbb C)$ of exponent $p$ necessarily have order of the form $p^k$? I apologize in advance if this is obvious and I missed it.
| https://mathoverflow.net/users/3958 | Finite subgroups of GL_n(C) | The paper:
Herzog, Marcel; Praeger, Cheryl E. "On the order of linear groups of fixed finite exponent."
J. Algebra 43 (1976), no. 1, 216–220.
[MR424960](http://www.ams.org/mathscinet-getitem?mr=424960)
[DOI:10.1016/0021-8693(76)90156-3](http://dx.doi.org/10.1016/0021-8693(76)90156-3)
contains the important bound, if G ≤ GL(n,F) where F has characteristic coprime to |G|, then |G| ≤ exp(G)n. Obviously these bounds can be obtained over large enough F (containing exp(G) roots of unity), as the diagonal subgroup generated by *e*th roots of unity has exponent e and order en.
The exponent of a finite group divides the order of the group: the exponent of a group is the product of the exponents of its Sylow subgroups, and a p-group always contains an element whose order is equal to the exponent of the group, so by Lagrange the exponent divides the order. Also, every prime dividing the order of the group divides the exponent of the group, by Cauchy's theorem.
In particular, the possible orders of finite groups of exponent p, p a prime, that are contained in GL(n,C) are exactly p1, p2, …, pn. The elementary abelian subgroups generated by diagonal matrices whose entries are *p*th roots of unity shows the existence, and Herzog and Praeger eliminate all other orders. Note that when p is large, these are all possible anyways, so that the theorem is probably only interesting for small p.
For instance, GL(3,C) contains no non-abelian group of exponent 5.
| 12 | https://mathoverflow.net/users/3710 | 24768 | 16,240 |
https://mathoverflow.net/questions/24773 | 38 | This is a question that I have seen asked passively in comments relating to the separation of category theory from set theory, but I haven't seen it addressed in full.
I know that it's possible to formulate category theory within set theory while still being albe to construct the useful things one would want from category theory. So as far as I understand, all normal mathematics that involves category theory can be done as long as a little caution is taken.
I also know that some people (categorical foundationalsists) would still like to formulate category theory without use of or reference to set theory. While I admit that I am curious about this for curiosity's sake, I'm not sure if there are any practical motivations for doing this. The only reason for wanting to separate category theory from set theory that I have read about is for the sake of `autonomy of category theory'.
So my question is twofold: What other reasons might categorical foundationalists have for separating category theory from set theory, and what practical purposes might it serve to do this?
| https://mathoverflow.net/users/3664 | Why do categorical foundationalists want to escape set theory? | I don't agree that this is what (most) categorists who are interested in foundations are doing.
It is true that Lawvere in the mid-60's (and perhaps to this day) wanted to develop a theory of categories independent of a theory of sets, but I don't think that represents the main thrust of modern-day categorical work on "foundations". Much more work has been directed toward developing a full-fledged categorical *theory of sets*, either as in Lawvere's Elementary Theory of a Category of Sets and extensions thereof, or understanding classical theories of sets such as ZF through a categorical lens, as in Algebraic Set Theory. There is also ongoing discussion of what strength of set theory is suitable for doing what category theorists would like to do. As one can see with even a casual perusal of such work, there is no antagonism toward set theory per se, or a desire to somehow get away from sets.
I think some confusion might stem from over-hasty identification of set theory with a "canonized" form of set theory, such as ZFC (or something in that family such as Gödel-Bernays set theory), based on a single binary predicate called "membership". In ordinary ZFC, a set is characterized by its membership tree, so that the elements of sets are sets themselves, possessing their own internal structure. This may be termed a "materialist" form of set theory (material because elements of sets are considered as having "substance"). If there is antagonism toward this type of set theory on the part of some category theorists, it's because it lends itself to a conception of "set" that is largely irrelevant to the actual practice of core mathematics, insofar as mathematicians don't care what elements are "made of".
The prevailing trends of mathematical practice today and throughout most of the twentieth century promote a more "structuralist" view: that what counts is not what the elements of a structure "are" particularly, but rather how they are interrelated in a structure, and where two structures are considered abstractly the same if they are isomorphic. This seems like a truism today, but it is precisely this view which drives a more categorically-minded view, which looks toward not what sets "are", but of how we use them, what abstract constructions we want to perform on them, and so on. Thus, concepts such as "power set" are in this view more relevantly captured by suitable universal properties which serve to characterize their structure up to specified isomorphism. A theory of sets which takes this point of view seriously and axiomatically may be termed a "structural set theory".
Thus the real contrast is between "material" and "structural" theories of sets, with category theorists tending to prefer structural set theory. An example of such is Lawvere's aforementioned Elementary Theory of the Category of Sets (ETCS). A different and more recent example is Mike Shulman's SEAR (Sets, Elements, and Relations), which you can read about at the nLab.
As for practical benefits of structuralist set theory: they are huge! It should be borne in mind that elementary topos theory was largely inspired by Lawvere's insight that Grothendieck toposes themselves model most of the axioms of the kind of structuralist set theory he was investigating in ETCS, and this has been revolutionary. This answer is already long enough, so I won't enter on a discussion of that here.
| 62 | https://mathoverflow.net/users/2926 | 24783 | 16,250 |
https://mathoverflow.net/questions/24403 | 8 | Let $\varphi:\mathcal S^{d-1}\longrightarrow \mathbb R\_{>0}$ be a strictly positive function describing the boundary $\varphi(\mathbf x)\mathbf x,\mathbf x\in\mathbb S^{d-1}$ of a $d-$dimensional closed convex set $C$ with barycenter at the origin.
We consider the decomposition $\varphi=\oplus\_{\lambda\in \mathbf{Spec}(\Delta)}\varphi\_\lambda$ into eigenvectors associated
to distinct eigenvalues of the Laplacian.
To what extend does the sequence $\parallel \varphi\_\lambda\parallel\_{\lambda\in\mathbf{Spec}(\Delta)}$ determine $C$?
(This is certainly the case for a sphere of radius $\rho$ since then $\varphi=\varphi\_0=\rho$.)
Are there examples of two non-isometric convex sets (or even two non-isometric polytopes) for which the two corresponding sequences of norms coincide?
Moreover, the convexity of $C$ gives probably constraints for the norms $\parallel \varphi\_\lambda\parallel$ since spherical harmonics associated to high eigenvalues wiggle a lot
and have thus to be involved with coefficients that are small with respect to
$\parallel \varphi\_0\parallel$. What are these constraints?
| https://mathoverflow.net/users/4556 | A variation on "Hearing the shape of a drum" for polytopes. | The short answer is that there are no particular constraints on the spectral decomposition of the function $\varphi$, as long as a basic convexity condition is satisfied.
>
> **Lemma.**.Assume that $\varphi\in C^2(\mathbb S^{d-1},\mathbb R)$ satisfies for all $x\in \mathbb S^{d-1}$, $i$, $j=1,\dots,d$, and some $r>0$ the inequalities
> $$|\varphi(x)|+\sqrt{2d^3}|D^i\varphi(x)|+2d^3|D^{ij}\varphi(x)|< r, \qquad\qquad (1)$$
> where
> $$D^i\varphi(x)=\left(\frac{\partial \varphi(u/|u|)}{\partial u\_i}\right)\_{u=x},\quad
> D^{ij}\varphi(x)=\left(\frac{\partial^2 \varphi(u/|u|)}{\partial u\_i\partial u\_j}\right)\_{u=x},\quad u\in \mathbb R^d,\ x\in \mathbb S^{d-1}.$$
> Then $\varphi\_r(x)= r+\varphi(x)$ is the support function of a compact convex set in $\mathbb R^d$.
>
>
>
Condition (1) can be translated into constraints on the harmonic expansion of the support function.
>
> **Theorem.** Let $P\_i=P\_i(x)$, $x\in \mathbb S^{d-1}$ denote a spherical harmonic of order i.
>
>
> * There exists a constant $c\_0$ such that for all $c\geq c\_0$
> $$c+P\_{n\_1}+\dots+P\_{n\_m} $$
> is the support function of a compact convex set in $\mathbb R^d$.
> * The subset of those compact convex sets whose support functions are finite sums of spherical harmonics is dense (with respect to the Hausdorff metric) in the set of all convex bodies in $\mathbb R^d$.
> * If $C\subset \mathbb R^d$ is a compact convex set whose principal radii of curvature exist and srtictly positive and whose support function
> $\varphi=\sum\limits\_{n=1}^{\infty} P\_n $
> belongs to the class $C^k(\mathbb S^{d-1},\mathbb R)$ with $k>\frac{d+4}{2}$, then there is
> an $n\_0$ such that the partial sum
> $$P\_{0}+P\_1\dots+P\_{n} $$
> is the support function of a convex body in $\mathbb R^d$ for any $n>n\_0$.
>
>
>
**Reference.** *Geometric Applications of Fourier Series and Spherical Harmonics* by H. Groemer.
| 4 | https://mathoverflow.net/users/5371 | 24786 | 16,253 |
https://mathoverflow.net/questions/24787 | 1 | If $f\_n=1\_{(n,n+1)}(x)$, where $1\_A(x)$ is the indicator function. Why is $f\_n
\rightarrow0$? Same is true for $f\_n=1\_{(n,\infty)}$.
i just dont get it.
i thought $f\_n$ was always 0 for all n so i think $f\_n\rightarrow1$ but its not
the case. i try to reason it by the integral which is 1 for all n
but then i dont go anyway.
| https://mathoverflow.net/users/6114 | strick inequality for Fatou theorem | This is just the definition of convergence for sequences of functions : $\forall x \in \mathbb R, \lim\_{n \to+\infty} f\_n(x) =0$, which is of course the case here, all sequences $(f\_n(x))$ being stationary.
| 4 | https://mathoverflow.net/users/5659 | 24788 | 16,254 |
https://mathoverflow.net/questions/24678 | 15 | Dear everyone,
(i) Who is the father of the adjective “syntomic” in algebraic geometry?
(ii) And why did he choose to introduce a new term for what we already know from EGA IV.19.3.6 and SGA 6.VIII.1.1 as “flat, locally of finite presentation, and local complete intersection”?
Thanks!
| https://mathoverflow.net/users/307 | Why “syntomic” if “flat, locally of finite presentation, and local complete intersection” is already available? | Mazur gives the following beautiful justification, which explains the “syn-” in “syntomic” as well.
>
> Dear Thanos,
>
>
> Thanks for your question. I'm thinking of “ local complete
> intersection” as being a way of cutting out a (sub-) space from an
> ambient surrounding space; the fact that it is flat over the parameter
> space means that each such "cutting" as you move along the parameter
> space, is---more or less---cut out similarly. I'm also thinking of the
> word "syntomic" as built from the verb temnein (i.e., to cut) and the
> prefix "syn" which I take in the sense of "same" or "together". So I
> think it fits.
>
>
> Best wishes,
>
>
> Barry
>
>
>
| 25 | https://mathoverflow.net/users/307 | 24790 | 16,255 |
https://mathoverflow.net/questions/24789 | 0 | Assume that you have a set S of having 2^2 elements first, let S={0,1,2,3}
Then the desired 2 partitions would be
1-{{0,1},{2,3}}
2-{{0,2},{1,3}}
3-{{0,3},{1,2}}
If S={0,1,2,3,4,5,6,7} having 2^3 elements then similarly the 2 partitions would be
1- {{0,1},{2,3},{4,5},{6,7}}
2- {{0,2},{1,3},{4,6},{5,7}}
3- {{0,3},{1,5},{2,6},{4,7}}
4- {{0,4},{1,6},{2,5},{3,7}}
5- {{0,5},{1,4},{2,7},{3,6}}
6- {{0,6},{1,7},{2,4},{3,5}}
7- {{0,7},{1,2},{3,4},{5,6}}
So it seems that for S having 2^n elements the initial table would like
1: {{0,1},{2,3},{4,5},{6,7},...,{(2^n)-2,(2^n)-1}}
2: {{0,2},... }
.
.
.
(2^n)-1: {{0,(2^n)-1},... }
I am just wondering whether there is an direct approach to generate the above table for a given n.
Thanks
| https://mathoverflow.net/users/6113 | Is there any direct approach to generate discrete 2 partitions of a set of having 2^n elements for a given n ? | I presume you want $2^n-1$ partitions of ${0,\ldots,2^n-1\}$
into parts of size two so that each possible pair occurs
exactly once in a partition. You can do this as follows.
Write the numbers in question base $2$. Then each can be represented
by $n$ binary digits going from $00\cdots0$ to $11\cdots 1$. Define
an operation $\oplus$ on these as follows: we obtain $a\oplus b$
by add corresponding digits in $a$ and $b$ modulo $2$.
For example $11\oplus 14=5$ as in base $2$, $11$ and $14$ are
$1011$ and $1110$, and $1+1$ is $0$ mod $2$, $0+1$ is $1$ mod $2$,
$1+1$ is $0$ mod $2$ and $1+0$ is $1$ mod $2$ so in binary $a\oplus b$
is $0101$ which is $5$.
The operation $\oplus$ is sometimes known as "nim addition" as it is used
in the analysis of the game nim.
To get the $j$-th partition where $1\le j\le 2^n-1$ we just pair off
$a$ and $a\oplus j$. So for $n=4$, the partition for $j=5$
is $\{ \{0,5 \}, \{1,4 \}, \{2,7 \}, \{3,6 \}, \{8,13 \},
\{9,12 \}, \{10,15 \}, \{11,14 \} \}$
as $0\oplus 5=5$, $1\oplus 5=4$, $2\oplus 5=7$ etc.
| 1 | https://mathoverflow.net/users/4213 | 24791 | 16,256 |
https://mathoverflow.net/questions/24737 | 0 | I would like to draw an octahedral diagram in my paper; I would prefer to present it as the 'upper hat' + the 'lower hat' (as it is common in the texts on triangulated categories). Could anyone tell me where I can find this diagram (certainly, the 'upper hat part' is sufficient) written down in latex. Maybe, someone could just share with me his own latex realization of this diagram (so that I could replace the original names of objects and morphisms by the ones I need)?
| https://mathoverflow.net/users/2191 | The (upper hat of) an octahedral diagram in (la)tex | Mikhail,
Here's an upper cap in xy-pic
\xymatrix{
X'\ar[rd]^{[1]}\ar[dd]^{[1]} & & Z\ar[ll] \\
& Y\ar[ru]\ar[ld] & \\
Z'\ar[rr]^{[1]} & & X\ar[lu]\ar[uu]
}
at least it's enough of one to get you started. Note that I didn't construct it by hand.
I have a script for building these kinds of diagrams visually:
<http://www.math.purdue.edu/~dvb/scripts/arraymaker>
If it's useful to you, help yourself.
| 4 | https://mathoverflow.net/users/4144 | 24801 | 16,264 |
https://mathoverflow.net/questions/24780 | 5 | Is there a general method for determining the domain of dependence of (higher-order) PDEs? I would be plenty happy with a reference to a paper or textbook I could look at; despite much reading around, I couldn't find this problem addressed for anything besides first- and second-order PDEs.
If there's a simple answer to this question, then I probably don't need to write anything else. But for completeness, I'll provide background on the subject and on my particular interest.
Domain of dependence
--------------------
The **domain of dependence** of point $(\vec{x}\_0,t\_0)$ on the solution $U(\vec{x},t)$ of a hyperbolic$^\dagger$ PDE is the subset of the initial conditions which uniquely determine the value $U(\vec{x}\_0,t\_0)$. For the Advection equation
$U\_t + c U\_x = 0$
with initial conditions $U(x,0)=f(x)$, the domain of dependence of $(x,t)$ is just the point $x-c t$. For the wave equation
$U\_{tt} - c^2 U\_{xx} = 0$
the domain of dependence of $(x,t)$ is the line $[x-ct,x+ct]$. In every book I've read, the domain of dependence is only given for PDEs which have been explicitly solved (like these two examples) and no domain is given for a higher-order PDE without explict solutions.
Background for my problem
-------------------------
I'm trying to simulate the evolution of the Wigner function (a pseudo probability distribution over phase space) for a point particle moving in a chaotic potential. The PDE governing the Wigner function $W(x,p,t)$ can be approximated as
$\partial\_t W = -\frac{p}{m} \partial\_x W + V^\prime (x) \partial\_p W - \frac{\hbar^2}{24} V^{\prime \prime \prime} (x) \partial\_p^3 W$
or, in the dimensionless PDE notation,
$W\_t = -p W\_x + f(x) W\_p - f^{\prime \prime}(x) W\_{ppp} .$
[It appears that an instability in my simulation is due to my time steps not satisfying the CFL condition](https://mathoverflow.net/questions/24272/numerical-instability-using-only-heuns-method-on-a-simple-pde) and I'm trying to rigorously derive the CFL condition for this PDE.
This may not be possible, in which case I'll use the CFL condition under the approximation that $f''(x)=0$.
---
$^\dagger$ Here I am using "Hyperbolic" in the sense that the Cauchy problem is well defined *without* restricting to second-order PDEs.
| https://mathoverflow.net/users/5789 | Is there a general method for determing the domain of dependence of (higher-order) PDEs? | While I am aware of some facts about domain of dependence properties for hyperbolic PDEs, I don't think most of them will be useful for you. The problem is that what you consider as hyperbolic (in your footnote) is too large of a class of equations for the notion to be useful: an illustration is the Heat equation. It is usually classified as a parabolic equation, but it does admit a well-posed initial value problem. So by your definition is hyperbolic. Now it is well known that the heat equation has infinite smoothing properties and infinite speed of propagation.
Now, in the special case of symmetric hyperbolic systems, even in higher orders, one can generally describe the domain of dependence by considering the characteristic cones (see, eg. <https://encyclopediaofmath.org/wiki/Characteristic> ) of the system. Essentially the "fattest cone" gives you the maximum speed of propagation (which may depend quasilinearly on the solution) and integrating back this cone gets you the domain of dependence.
The domain of dependence properties are really closely associated to a priori energy estimates (see Courant and Hilbert, Methods of Mathematical Physics).
But I don't think this will solve your problem since I don't believe your question can be recast in a form in which such estimates are available.
In particular, looking directly at your equation, on the spatial side it has potentially infinite speed of propagation since the spatial propagation is essentially just a transport equation. So if $W(x,p,t)$ has non-compact support, then the spatial propagation can have arbitrary large speeds. So if your potential vanishes or if your initial data is homogeneous in momentum, your solution will have, as its spatial domain of dependence given by the largest and smallest momentum at which $W$ is supported.
Assuming $f'' = 0$, then you equation can be solved by the method of characteristics: $\partial\_t W = v\cdot\nabla W$ where $v(x,p) = (p,f(x))$ is a vector field. The domain of dependence for this problem can be easily found by integrating the vector field independently of the function $W$. I don't know how to deal with your third order term.
Like I said, in general there are only two ways to study domain of dependence properties that are well established, the first is via explicit notion of the Green's function, the second is energy estimates.
| 7 | https://mathoverflow.net/users/3948 | 24802 | 16,265 |
https://mathoverflow.net/questions/24309 | 8 | I decided to spend this summer working through exercises in Hartshorne, and I found myself frustrated by the way I was solving one of them, specifically IV.2.3 on pp. 304-305. No, I'm not asking for a solution to the problem --- I almost have the whole thing solved, as far as I can tell --- what worries me is that I think I'm thinking about this stuff the wrong way.
The problem is about the map from a projective plane curve in characteristic 0 to its dual curve, and it has 8 parts, which ask you to prove things like the fact that bitangents on the original curve correspond to nodes on the dual. I managed to solve all of them except the one about ordinary inflection points on X giving ordinary cusps on X\*, but all my solutions involved picking affine charts, finding an ugly formula for the map in question, and computing first and second partial derivatives, and it's all very long and messy and doesn't give any clue as to what's going on until you get to the end and see that the thing you have is 0 or whatever if and only if the condition in the problem is met for some magical reason.
I feel like there must be a more "high-brow" way to approach this object, and the fact that I haven't been able to come up with one seems to speak to the sort of backward way I've been learning about the subject (I just took a class that was very good and covered a lot but was almost completely devoid of examples). There must be some approach to this that actually uses all the machinery that's developed in the rest of the book. Is there a satisfying, pretty way to deal with this thing that I'm missing, something that would tell me *why* these relationships hold and not just *that* they hold?
| https://mathoverflow.net/users/5281 | Dual Curves in Fancy Language | You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve. For example, try to identify the line bundle on your curve which embeds it into the dual plane. E.g.: let $X \subset \mathbb{P}^2$ be your curve. Then define $X' \subset X \times \mathbb{P}^{2\*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$. When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N\_{X/\mathbb{P}^2}$.
Then think about the map $X' \rightarrow X \times \mathbb{P}^{2\*} \rightarrow \mathbb{P}^{2\*}$. This is the dual curve embedding. The morphism comes from the inclusion of $N^\* \subset \Omega\_{\mathbb{P}^2}|\_X \subset \mathcal{O}(-1)^3$.
At some point, you have to do computations in local coordinates - but setting it up like this may help a little.
Example of local computation:
Locally, the curve looks like $(x(t), y(t), z(t))$. The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$). If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.
| 3 | https://mathoverflow.net/users/397 | 24809 | 16,269 |
https://mathoverflow.net/questions/24755 | 3 | One aspect of category theory that caught my eye is that it can give simultaneously prove the 1st isomorphism theorem for groups/rings/fields/vector spaces/... Yet whenever I look up works on category theory, it takes hundreds of pages to reach this theorem. Is it possible to prove this result using substantially less theory, and if so, is there any work which provides such a proof?
| https://mathoverflow.net/users/4692 | Reference request for category theory works which quickly prove the theorem which generalises the 1st isomorphism theorem for groups/rings/... | If you want a categorical proof that encompasses nonabelian groups and rings-without-identity, then you need to work with concepts such as "normal subobject" that are (I think) not really part of the category theory that most mathematicians routinely encounter. You may also find that a proof is not particularly enlightening at this level of generality. I think this might explain why a simultaneous generalization does not appear early in textbooks.
I'll consider the formulation of the first isomorphism theorem which states that for any homomorphism $f: A \to B$, $ker(f): K \to A$ is a normal subobject, and the image of $f$ in $B$ is isomorphic to the quotient of $A$ by $K$. The formalism is roughly as follows: We work in a category in which every morphism has a kernel and a cokernel. Then any morphism $f: A \to B$ factors into a composition of $coker(ker(f)): A \to A/K$, $m: A/K \to N$, and $ker(coker(f)): N \to B$, and the diagram is defined up to unique isomorphism by the universal properties of kernel and cokernel. For the first part of the theorem, one typically defines normal subobject to be a kernel (although there are alternative definitions involving congruences), so we reduce to the question of the image. Here we have a conflict of terminology, since the "image" is often defined in categories as the kernel of the cokernel, and the first isomorphism theorem interprets image set-theoretically. Given this alternative interpretation, the theorem amounts to the assertion that $m$ and $ker(coker(f))$ are monomorphisms. Here we encounter another problem, since as far as I know, $m$ does not need to be a monomorphism in general.
At this point, we need to check that the categories we like have kernels and cokernels. The kernel of a group homomorphism $f:A \to B$ is (the inclusion of) the preimage of the identity, and the cokernel of $f$ is given by taking the normal hull $N$ of $f(A)$ in $B$, and taking the quotient group $B/N$. We get our theorem because the maps of $f(A)$ into $N$ and $N$ into $B$ are inclusions, hence monomorphisms. A similar argument works for rings-without-identity. For abelian groups and vector spaces, the map $m: f(A) \to N$ is an isomorphism. For fields, there are no kernels, so the first isomorphism theorem doesn't make sense.
| 6 | https://mathoverflow.net/users/121 | 24810 | 16,270 |
https://mathoverflow.net/questions/24669 | 2 | I'm going to ask a very vague question, and then give specifics for the version I particularly care about. I'm interested in answers at all levels of vagueness.
At the most vague version, I am in the following situation. I have some particular (smooth) function $A(t\_0,q\_0,t\_1,q\_1)$, where $t\_0,t\_1$ are real variables and $q\_0,q\_1$ vary over some manifold $M$, i.e. $A: \mathbb R \times M \times \mathbb R \times M \to \mathbb R$. I have another (smooth) function $Q(t\_0,q\_0,t\_1,q\_1;t) : \mathbb R \times M \times \mathbb R \times M \times \mathbb R \to M$.
I am interested in (smooth) solutions $F(t\_0,q\_0,t\_1,q\_1)$ to the "composition" formula:
$$ A(t\_0,q\_0,t\_1,q\_1)\,F(t\_0,q\_0,t,Q(t\_0,q\_0,t\_1,q\_1;t))\,F(t,Q(t\_0,q\_0,t\_1,q\_1;t),t\_1,q\_1) = F(t\_0,q\_0,t\_1,q\_1)$$
I know one such (nonzero) solution explicitly. I am hoping that, presumably with slightly more data (some sort of "initial data"), I can conclude that my solution is uniquely determined. So my question is:
>
> What extra data for $F$, and what restrictions on $A$, are needed for a functional equation of the above type to have a unique solution?
>
>
>
---
For those interested, I can give you more precise information. I have some "physics" encoded by a function $S(t\_0,q\_0,t\_1,q\_1)$, which I am thinking of as a "Hamilton-Jacobi function". This function defines $Q$ via:
$$ \left. \frac{\partial}{\partial q} \bigl[ S(t\_0,q\_0,t,q) + S(t,q,t\_1,q\_1) \bigr] \right]\_{q = Q(t\_0,q\_0,t\_1,q\_1;t)} = 0 $$
and satisfies:
$$ S(t\_0,q\_0,t,q) + S(t,q,t\_1,q\_1) \bigr|\_{q = Q(t\_0,q\_0,t\_1,q\_1;t)} = S(t\_0,q\_0,t\_1,q\_1)$$
The function $A$ is given by
$$ \left| \det \left[ \frac{\partial^2}{\partial q^2} \bigl[ S(t\_0,q\_0,t,q) + S(t,q,t\_1,q\_1) \bigr] \right]\_{q = Q(t\_0,q\_0,t\_1,q\_1;t)} \right|^{-1/2} $$
Then the function $F$ is:
$$ F(t\_0,q\_0,t\_1,q\_1) = \left| \det \frac{\partial^2}{\partial q\_0\partial q\_1} \bigl[ S(t\_0,q\_0,t\_1,q\_1) \bigr] \right|^{1/2}$$
which as defined makes sense only on $\mathbb R^n$. Actually, $A$ makes sense as a volume form, and $F$ as a half density in each variable, and then everything transforms correctly under changes of coordinates.
Anyway, I would really like to conclude that this $F$ is the only solution, but I don't know if I can.
| https://mathoverflow.net/users/78 | What kind of uniqueness can I conclude for solutions to a simple functional equation? | Here's a vague answer then: there's [an old 1978 paper in Aequationes Mathematicae](https://doi.org/10.1007/BF01818567 "Baron, K., Sablik, M. On the uniqueness of continuous solutions of a functional equation of n-th order. Aequat. Math. 17, 295–304 (1978). https://zbmath.org/?q=an:0393.39002") mentioning the problem of uniqueness for the functional equation $\varphi (x)=h(x,\varphi [f\_1(x)],\dots, \varphi [f\_n(x)] )$ under some hypotheses. This seems to be of the same form as your equation (identifying $x=(q\_0,t\_0,q\_1,t\_1)$ and so on). Maybe what you're looking for is in there, or in later papers quoting that one.
Having said that, since your inspiration is Hamiltonian mechanics, I would imagine at least some variant of your problem to have been investigated already. Obviously you're looking at the stationary phase approximation to the quantum propagator, such as in [equation (33.32) of a chapter about the semiclassical propagator](https://chaosbook.org/chapters/VanVleck.pdf) in a [great physics textbook](https://chaosbook.org/). Now there's been a lot of mathematical work on the topic which may, at least implicitly, help you with your specific question, namely [this paper of Meinrenken](https://lib-extopc.kek.jp/preprints/PDF/1992/9201/9201019.pdf "Meinrenken, E. Semiclassical principal symbols and Gutzwiller's trace formula. Rep. Math. Phys. 31, No. 3, 279–295 (1992). doi:10.1016/0034-4877(92)90019-W. zbMATH review at https://zbmath.org/?q=an:0794.58046") in particular page 7 and beyond, and [that paper](https://web.ma.utexas.edu/mp_arc-bin/mpa?yn=97-599 "Combescure, M., Ralston, J. & Robert, D. A Proof of the Gutzwiller Semiclassical Trace Formula Using Coherent States Decomposition. Comm Math Phys 202, 463–480 (1999). doi:10.1007/s002200050591. https://zbmath.org/?q=an:0939.58031") of Combescure–Ralston–Robert.
| 1 | https://mathoverflow.net/users/469 | 24817 | 16,272 |
https://mathoverflow.net/questions/24824 | 1 | I stumbled upon [this](http://haroonsaeed.wordpress.com/2006/04/25/self-describing-numbers-and-sequence/) number sequence while surfing the web. And I generated the next terms with my pc, and I was amazed to see, only the numbers 1 to 3 come up in megabytes of output. The sequence describes the previous number. The first term is one one, so 11 is the second one. The second consist 2 ones, so the third element is 21.
```
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
…
…
```
So now I'm wondering, if there's any good explanation why only the numbers 1 to 3 come up? Or does any higher number comes up later on?
| https://mathoverflow.net/users/4338 | Self-describing number sequence | If a 4 existed in the sequence, as such:
... 41 ...
That would mean that there were 4 one's in the previous sequence, as such:
... 1111 ...
So what does THAT describe? Basically it's saying "there's 1 occurrence of 1, followed by 1 occurrence of 1." Since you're describing consecutive numbers, that would ACTUALLY translate to "two occurrences of 1", or "21".
Hope that helps.
| 6 | https://mathoverflow.net/users/3275 | 24826 | 16,278 |
https://mathoverflow.net/questions/24842 | 3 | If the group of rational points of $E$, which is finitely generated by the Mordell-Weil Theorem, has $g$ generators of infinite order, then the Birch-Swinnerton-Dyer conjecture gives
$L\_E(s)$ has a zero of order $g$ at $s=1$.
Assuming the BSD conjecture, is it possible to (and if so how) to construct such $L\_E(s)$? Specifically, if we want $g=3$ or $4$?
| https://mathoverflow.net/users/695 | BSD conjecture and L functions with zeroes of order g | I'm not entirely sure what you mean by your question. Here are two remarks:
1. If you *assume* BSD, then to "construct" $L\_E$ you just need to give the curve $E$. There are (many) elliptic curves /$\mathbb{Q}$ whose ranks have been computed, and are (say) equal to 3 or 4.
2. If one wants an example without assuming BSD, then you are in trouble - for given $E$, you can compute $L^{(n)}(s)$ to any desired degree of accuracy, but proving that it vanishes computationally is impossible.
However, two things help you. If the sign in the functional equation is -1, then you have that the order of vanishing is also odd. The Gross-Zagier formula can be used to check the vanishing of the first derivative. For example, this is used in the following paper to exhibit an elliptic curve $E$ whose $L\_E$ provably vanishes to order 3.
On the Conjecture of Birch and Swinnerton-Dyer for an Elliptic Curve of Rank 3
Author(s): Joe P. Buhler, Benedict H. Gross, Don B. Zagier
Source: Mathematics of Computation, Vol. 44, No. 170 (Apr., 1985), pp. 473-481
| 2 | https://mathoverflow.net/users/1594 | 24846 | 16,289 |
https://mathoverflow.net/questions/24845 | 14 | Let $G$ be a simple Lie group and let $G(\mathbb{C}((t)))$ be its loop group.
The Lie algebra $\mathfrak{g}[[t]][t^{-1}]$ has a well known central extension
(see e.g. [Wikipedia](http://en.wikipedia.org/wiki/Affine_Lie_algebra)) given by the cocycle
$c(f,g) = Res\_0\langle f,dg\rangle$. Here, $\langle\ ,\ \rangle \colon \mathfrak{g}\otimes \mathfrak{g}\to\mathbb{C}$ denotes some invariant bilinear form on $\mathfrak{g}$, and $f dg$ is the $\mathfrak{g}\otimes \mathfrak{g}$-valued differential given by multiplying $f$ and *dg*.
>
> **Question:** It there a similarly concrete cocycle for the central extension of $G(\mathbb{C}((t)))$ by $\mathbb{C}^\ast$?
>
>
>
To give you an idea of what I'm looking for, let me show
you a cocycle for central extension by $S^1$ of the smooth loop group $LG = \mathop{Map} \_ {C^\infty} (S^1,G)$ of a compact Lie group $G$.
Pick a bounding disc $D\_\gamma$ : $D^2 \to G$ for each element $\gamma\in LG$. The cocycle is then given by
$$
c(\gamma,\delta) = \exp\left(i\int \langle D\_\gamma^\*\theta\_L,D\_\delta^\*\theta\_R\rangle
+i\int H^\*\eta\right)
$$
where $\theta\_L,\theta\_R\in\Omega(G,\mathfrak{g})$ are the Maurer-Cartan 1-forms, $\eta\in\Omega^3(G)$ is the Cartan 3-form,
and $H:D^3\to G$ in a homotopy between $D\_\gamma D\_\delta$ and $D \_ {\gamma\delta}$.
---
**References:**
The cocycle for the smooth loop group can be found on page 19 of the paper
[From Loop groups to 2-groups](http://arxiv.org/pdf/math/0504123v2), by Baez, Crans, Schreiber, and Stevenson,
and also on page 8 of Mickelsson's paper [From Gauge anomalies to Gerbes and Gerbal actions](http://arxiv.org/pdf/0812.1640v1).
| https://mathoverflow.net/users/5690 | Explicit cocycle for the central extension of the algebraic loop group G(C((t))) | For $SL\_2$ a cocycle is given by
$$
\sigma(g,h)=\left( \frac{x(gh)}{x(g)} , \frac{x(gh)}{x(h)} \right)
$$
where for $g=\left(\begin{array}{ll} a & b \\ c & d\end{array}\right)\in SL\_2(\mathbb{C}((t)))$, we define $x(g)=c$ unless $c=0$ in which case $x(g)=d$. $(\cdot,\cdot)$ is the tame symbol.
I'll see if I can come up with a good reference. I've seen this stuff over local fields, where this is attributed to Kubota, and [Kazhdan-Patterson's paper on Metaplectic Forms](https://eudml.org/doc/103999) has this formula in it (actually for $GL\_2$). I would be suprised if there was a usable formula for higher rank groups.
| 8 | https://mathoverflow.net/users/425 | 24847 | 16,290 |
https://mathoverflow.net/questions/24841 | 1 | Does there exists a connected graph G which is a subgraph of two graphs H and H' for which
* G, H and H' have the same vertex set,
* H is minimally 2-connected (i.e. deleting any edge from H makes is not 2-connected),
* H' is 3-connected and
* H is not a subgraph of H'?
I arrived at this question when thinking about how grant applications with "anti-terrorism" in the title seem to get much more funding than those without. I came out with this problem: A terrorist wants to disconnect G by deleting edges. To prepare for this attack, I want to add edges to G to make it 2-connected (i.e. H) at minimal cost, while at the same time preparing to make it 3-connected (i.e. H') at a later date. Does the choice of H matter?
| https://mathoverflow.net/users/2264 | A minimally 2-connected graph H and a 3-connected graph H' both contain G, does H' contain H? | Fedja's comment definitely earns the grant. A minimal example is to let G be a star with four leaves {1,2,3,4}, let H be the bowtie obtained from G be adding the edges 13 and 24, and let H' be the wheel obtained from G by adding the cycle 1234.
| 4 | https://mathoverflow.net/users/2233 | 24848 | 16,291 |
https://mathoverflow.net/questions/24719 | 28 | Recently I tried to learn class field theory, but I find it is difficult. I have read the book "Algebraic Number Theory" by J. W. S. Cassels and A. Frohlich. In the book, the approach to class field theory is cohomology of groups. Although I have learned cohomology of groups, I find that those theorems in the book are complicated and can not form a system.
I'm wondering what are people's opinions of the book above, can you give me some suggestions on learning class field theory, and could you recommend some good books on class field theory?
| https://mathoverflow.net/users/6104 | Suggestions for good books on class field theory | When you are first learning class field theory, it helps to start by getting some idea of what the fuss is about. I am not sure if you have already gotten past this stage, but if not, I recommend B. F. Wyman's article "What is a Reciprocity Law?" in the American Mathematical Monthly, Vol. 79, No. 6 (Jun. - Jul., 1972), pp. 571-586. I also highly recommend David Cox's book *Primes of the Form $x^2 + ny^2$* (mentioned by Daniel Larsson). Cox's book will show you what class field theory is good for and will get you to the statements of the main theorems quickly in a very accessible way. (You can safely skim through most the earlier sections of the book if your goal is to get to the class field theory section quickly.) As a bonus, the book will also give you an introduction to complex multiplication on elliptic curves.
However, Cox's book does not prove the main theorems of class field theory. You will need to look elsewhere for the proofs. There are several different approaches and someone else's favorite book may be unappealing to you and vice versa. You will have to dip into several different books and see which approach appeals to you. One book that has not been mentioned yet is Serge Lang's *Algebraic Number Theory*. Even if you ultimately choose not to use Lang's book as your main text, there is a short essay by Lang in that book, summarizing the different approaches to class field theory, that is worth its weight in gold.
| 21 | https://mathoverflow.net/users/3106 | 24852 | 16,293 |
https://mathoverflow.net/questions/24836 | 4 | Four red vectors are given, one per quadrant, $[0,90^\circ)$,
$[90^\circ,180^\circ)$, etc.
A rigid *star* of six green vectors separated by $60^\circ$
can be positioned at
$(\theta,
\theta+60^\circ,
\theta+120^\circ,
\theta+180^\circ,
\theta+240^\circ,
\theta+300^\circ)$.
The goal is to spin the green star so that the red vectors are centralized
in green sectors as much as possible.
Define the *deviation* $\delta(r)$ of a red vector $r$ as the larger of the
(absolute value of the) two
angles from the red vector to the boundaries of the green sector in which it lies.
I want to minimize the largest red deviation.
For example, let the red vectors be at
$(0^\circ,
90^\circ,
180^\circ,
270^\circ)$.
Then choosing $\theta=15^\circ$ yields a deviation of $45^\circ$ for all
red vectors.
For example, $\delta(0^\circ) = \max \{ 15^\circ, 45^\circ \}$.
My question is: What is the largest deviation of any four red vectors?
I thought it might approach $60^\circ$,
but it seems that perhaps $52.5^\circ$ is the worst
($52.5^\circ = 7 \pi / 24$).
The problem generalizes to $k$ red vectors and $m > k$ green sectors. Likely the logic to establish
the answer for $(k,m)=(4,6)$ will work for any $(k,m)$.
| https://mathoverflow.net/users/6094 | Centralizing four red vectors in six green sectors | Replace each of your red vectors by its value modulo 60. You are then seeking to find a choice for $\theta$ that is as far as possible (mod 60) from any of the red vectors. The best choice for theta is to put it into the largest gap (mod 60) between red vectors, so the worst choice for the red vectors is for them to be equally spaced 15 degrees apart. With this choice, you get theta at distance 7.5 degrees from its nearest red vector or (as you already calculated) $\delta = 52.5 = 60 - 7.5$.
More generally, if you have k and m, the worst case is when the k vectors are equally spaced modulo $2\pi/m$, in which case the gaps between them have size $2\pi/km$, the distance from $\theta$ to the nearest red vector when it's placed in the middle of a gap will be $2\pi/2km$, and $\delta=\frac{2\pi}{m}-\frac{2\pi}{2km}=(1-\frac{1}{2k})\frac{2\pi}{m}$.
I don't understand why you want to restrict $m>k$ since the answer doesn't depend on that.
| 4 | https://mathoverflow.net/users/440 | 24867 | 16,304 |
https://mathoverflow.net/questions/24864 | 28 | This is to do with high dimensional geometry, which I'm always useless with. Suppose we have some large integer $n$ and some small $\epsilon>0$. Working in the unit sphere of $\mathbb R^n$ or $\mathbb C^n$, I want to pick a large family of vectors $(u\_i)\_{i=1}^k$ which is almost orthogonal in the sense that $|(u\_i|u\_j)| < \epsilon$ when $i\not=j$. I guess I'm interested in how the biggest choice of $k$ grows with $n$ and $\epsilon$.
For example, we can let $\{u\_1,\cdots,u\_n\}$ be the usual basis, and then choose $u\_{n+1} = (1,1,\cdots,1)/n^{1/2}$, which works if $n^{-1/2} < \epsilon$. Then you can let $u\_{n+2} = (1,\cdots,1,-1,\cdots,-1)/n^{1/2}$ and so forth, but it's not clear to me how far you can go.
| https://mathoverflow.net/users/406 | Almost orthogonal vectors | Matt, to get $k$ points, you only need $n\ge C \epsilon^{-2} \log k$. See <http://en.wikipedia.org/wiki/Johnson%E2%80%93Lindenstrauss_lemma> or Google "Johnson-Lindenstrauss lemma".
| 23 | https://mathoverflow.net/users/2554 | 24873 | 16,309 |
https://mathoverflow.net/questions/24874 | 8 | Suppose one tries to formalize first-order logic. How much "strength" is required to do this?
Strength can mean in various senses:
1. The fragment of ZFC needed to codify first-order logic.
2. Which system of 2nd-order arithmetic is needed to codify first-order logic. (reverse mathematics)
3. The fragment of PA needed to codify first-order logic.
4. Categorial logic?
Codify means be able to (i) represent formulae and sentences, (ii) recognize proofs, (ii) represent structures and models.
(I'm also interested to the answer for this question in the case of propositional logic.)
| https://mathoverflow.net/users/nan | What is the reverse mathematics of first-order logic and propositional logic? | In general, "reverse mathematics" refers to work with subsystems of second-order arithmetic only; it does not include ZFC. Assuming this is actually what you meant, everything is in Stephen Simpson's book *Subsystems of second-order arithmetic*.
All the basic syntactics of formulas can be done in a primitive recursive way, and can be directly formalized into RCA0. It is well known that Goedel's incompleteness theorems can be formalized in PRA, and so they can also be formalized in RCA0.
The restriction of Goedel's completeness theorem to countable theories is equivalent to WKL0 over RCA0, as is the similar restriction of the compactness theorem. The corresponding theorems for propositional logic are also equivalent to WKL0 over RCA0.
The underlying reason that WKL0 is needed is that RCA0 is not strong enough to prove that the deductive closure of an arbitrary set of sentences exists. If one adds a hypothesis that the input theory is already deductively closed, RCA0 will do. This is all made precise in Simpson's book.
There has been some interesting work [1] on the reverse mathematics strength of the atomic model theorem of elementary model theory, which turns out to be very weak in the sense of reverse mathematics while still being stronger than RCA0.
[1] Denis R. Hirschfeldt; Richard A. Shore; Theodore A. Slaman. "The atomic model theorem and type omitting". Trans. Amer. Math. Soc. 361 (2009), 5805-5837.
| 11 | https://mathoverflow.net/users/5442 | 24882 | 16,316 |
https://mathoverflow.net/questions/24889 | 1 | More precisely, let $M$ be a subspace $\mathbb R^n$ with the following properties:
* $M$ is a topological manifold of dimension $n-1$.
* M is compact.
Does there exist a homological characterization of when the following happens:
* $\mathbb R^n \backslash M$ has two components, the bounded one being "inside" and the other one "outside". Both are $n$-dimensional manifolds.
If the above is not possible, is there a different formulation of the question which would allow a nice characterization?
The motivation of this question is of course the realization that the solution for $n = 3$ seems to be that $M$ is an oriented surface.
| https://mathoverflow.net/users/6031 | Does there exists a (possibly homological) characterization of the Jordan curve property in all dimensions? | More genreally, the number (finite or not) of the connected components of the complement set of a compact subset $M\subset {\mathbb R}^n$, which is the rank of $H\_0({\mathbb R}^n\setminus M)$, is a homotopic invariant for compact subspaces of ${\mathbb R}^n$, by duality in homology.
| 0 | https://mathoverflow.net/users/6101 | 24892 | 16,322 |
https://mathoverflow.net/questions/24711 | 14 | The beginning of homological algebra involves lots of diagram chasing, for proving most of the theorems. This gets repetitive after a while. To make things more interesting and satisfactory, one would like to remove the mechanical dredge work as far as possible.
One possible approach, as advocated in Lang's Algebra book, is to first prove the snake lemma by diagram chasing, and then reduce everything else as much possible to the snake lemma. This is mostly satisfactory. But some proofs, such as that of five lemma(as far as I am able to reconstruct now), require a small amount of diagram chasing in addition to using snake lemma.
There seems to be an alternate approach using the salamander lemma, as given in the references in [this question](https://mathoverflow.net/questions/6749/). Salamander lemma can be used to prove the other important lemmas. But according to the description in Anton Geraschenko's post, it is like chasing salamanders around, rather than chasing elements.
Are there other possible approaches to founding homological algebra avoiding extensive diagram chasing? That is, is it possible to base homological algebra on some other lemma/proposition, than the two possibilities mentioned above?
| https://mathoverflow.net/users/6031 | Ways of formulating homological algebra without diagram chasing | One other way you can prove all diagram chasing results (that I know of) is to first prove that there is a spectral sequence associated with a double complex (actually, there are two!). This involves some diagram chasing, but you only do it once. I guess you could also argue that you still have to chase the spectral sequences.
As an example, here's how you'd prove that a short exact sequence of chain complexes induces a long exact sequence in homology. We start with the double complex with exact rows
```
↑ ↑ ↑
0 → Ai+1 → Bi+1 → Ci+1 → 0
↑ ↑ ↑
0 → Ai → Bi → Ci → 0
↑ ↑ ↑
```
(I'm assuming the columns are bounded below, but it doesn't matter since this result is "local": you can truncate above and below the point you're interested in and then prove the result)
We can regard this as the $E\_0$ page of two different spectral sequences, depending on whether we decide to start with the vertical or horizontal arrows as our differential. Both spectral sequences have to abut to the homology of the total complex associated to this double complex.
First, use the horizontal arrows. Since the rows are exact, when we take homology, we get zero everywhere, so the $E\_1$ page is identically zero. Nothing interesting is going to happen now: we've gotten to $E\_\infty$. So the homology of the total complex is zero.
Now what if we use the vertical arrows? We get the $E\_1$ page
```
0 → Hi+1(A) → Hi+1(B) → Hi+1(C) → 0
0 → Hi(A) → Hi(B) → Hi(C) → 0
```
We'd like to prove that the rows are exact in the middle and that kernel of $H^{i+1}(A)\to H^{i+1}(B)$ is isomorphic to the cokernel of $H^i(B)\to H^i(C)$. Let's flip to the $E\_2$ page and see what happens. Let $K^i=\ker(H^{i+1}(A)\to H^{i+1}(B))$, $M^i=\mathrm{cok}(H^i(B)\to H^i(C))$, and let $L^i$ be the homology of the above row at $H^i(B)$. We get the $E\_2$ page
```
0 Ki+1 Li+1 Mi+1 0
↘ ↘ ↘
↘ ↘ ↘
0 Ki Li Mi 0
```
(pardon my ascii art: those are meant to be arrows going two spots to the right and one spot down)
Note that on the $E\_{\ge 3}$ pages, the differentials will be too long to connect anything in these three columns to anything other than zero. Since we must abut to zero, this $E\_2$ page is the "last chance" for the homology to vanish. It follows that the sequences $0\to L^i\to 0$ and $0\to K^{i+1}\to M^i\to 0$ must be exact, so $L^i=0$ and $K^{i+1}\cong M^i$, which is exactly what we wanted to show.
---
Note that I was able to completely ignore the question of what the differentials on higher pages were; I just had to know that they exist.
Diagram-chasing results almost always assume that the rows are exact and then make assertions about the homology groups you get from the columns, so you can always run one spectral sequence and immediately get that the homology of the total complex is zero, then run the other spectral sequence and note when the spectral sequence has it's "last chance" to cancel all homology groups.
| 11 | https://mathoverflow.net/users/1 | 24907 | 16,330 |
https://mathoverflow.net/questions/24903 | 5 | The Lakes of Wada partitions the unit square in to three regions, all of whom share a common boundary. The Wikipedia entry (<http://en.wikipedia.org/wiki/Lakes_of_Wada>) gives a construction approach, and a picture of a first few steps of the construction.
Is there a good algorithm available somewhere to explicitly list the partition up to some specific iteration? Or a closed form expression for the boundary in the limit of the process?
I'm hoping for something I can implement myself, or a piece of software that already does it, and in the end get a picture to arbitrary high levels of detail and arbitrary high iterations of the construction.
| https://mathoverflow.net/users/102 | Algorithms for the Lakes of Wada | Wada lakes can be obtained following a recipe given by Plykin.
An algorithm is explained in an online [article](http://www.ams.org/notices/200601/fea-coudene.pdf) in the Notices of the AMS. You only need to iterate a single explicit function to get the picture. So, this can be done using any fractal generator (`chaospro`,`fractint`,...). Note that zooming on
the boundary of the lake is not so interesting. You only get straight interlaced color bands after a few zooms.
The result can be seen [online](http://www.math.univ-brest.fr/perso/yves.coudene/dyn3.html) (scroll to the section about Wada Lakes). There is also a [movie](http://www.math.univ-brest.fr/perso/yves.coudene/lakes.avi).
Actually there are four regions on the pictures/movie. You can merge two of them if three is enough for your need.
Cheers
| 8 | https://mathoverflow.net/users/6129 | 24911 | 16,334 |
https://mathoverflow.net/questions/24891 | 3 | Ok, imagine having a finite line segment from point (a) to point (b) in $R^2$ . I'm not familiar with mathematical terminology of this kind, but let me state that the line we began with is the geometric interpretation of $A^1$. The geometric interpretation of $A^2$ is a square with sides $A^1$. We could go on by saying that $ A^3$ is a cube with edge length $ A^{1}$ again.
I wonder what the geometric interpration of $ \sqrt[2]{A}=A^{1/2}$ would look like. Is it a (straight) line? Is it constructible? Of course, we could extend this question by asking ourselves what $A^{k} $ would look like, in which $k \in \mathbb{R}$ or even $\mathbb{C}$ . When $k>2$, $A^k$ probably isn't constructible anymore on a sheet of paper, but one can still think about how these constructions of $A$ would 'look like'.
Thanks in advance,
Max Muller
P.S. I realize I ask more than one question now, which is an indictation I don't know a lot about this (yet) and I'd like to know more about this subject. Should this be a community wiki? Feel free to modify the Tags, I don't know how to classify this question exactly.
| https://mathoverflow.net/users/93724 | The Root of a Line | Here's my proposal for the square root of a line segment.
It's the Cantor set obtaining by repeatedly
splitting the intervals into four, and removing the two middle pieces.
When you take the cartesian square of that space, you obtain something whose projection
is exactly an interval:
[alt text http://www.staff.science.uu.nl/~henri105/drawing.pdf](http://www.staff.science.uu.nl/~henri105/drawing.pdf)
| 4 | https://mathoverflow.net/users/5690 | 24924 | 16,343 |
https://mathoverflow.net/questions/24932 | 3 | Apparently, there is the following fact:
>
> The set of homeomorphism classes of connected manifolds has the same cardinality $c$ as that of $\mathbb R$.
>
>
>
I find it to be interesting; but would be happier to see a proof, and would be grateful for a reference somewhere.
| https://mathoverflow.net/users/6031 | Counting connected manifolds | Upper bound (assuming the manifolds are second countable): every manifold admits a complete metric, and the "set" of isometry classes of complete separable metric spaces is of cardinality continuum. Indeed, every such a space is a completion of a countable metric space, and there is only continuum of metrics on a countable set.
Lower bound: there is a continuum of non-isomorphic fundamental groups of manifolds. E.g. any countable sum of cyclic groups is a first homology group of a connected manifold: just take a connected sum of an appropriate countable collection of lens spaces.
| 13 | https://mathoverflow.net/users/4354 | 24939 | 16,352 |
https://mathoverflow.net/questions/24936 | 11 | This question arose out of my attempts to understand [another question](https://mathoverflow.net/questions/24453/). The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.
But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.
So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?
Re to Greg Kuperberg: The details go roughly like the following. An $n$-disc is the unit ball in $\mathbb R^n$. The boundary map is $D^n$ going to $S^{n-1}$ on the boundary, but so that this boundary is a union of two discs $D^{n-1}$ but equipped with opposite orientation, ie the parts above and below the equator. Orientation for an $n$-disc is a choice of direction into the center, or going away from the center. Orientation for a $1$-disc is just a choice of direction. A bit more checking is needed to fix the orientation, but I am hopeful that it can be done. Assuming this, and after extending to chains, clearly the images are contained in kernels. Thus homology can be defined. It will be more difficult to prove that this is (after sorting out minor discrepancies) essentially the same as the homology given by simplexes. There is no nice barycentric subdivision, which was a problem. But, since the $n$-disc is homeomorphic to the $n$-simplex, I was hoping that at least an ugly proof of equivalence could be constructed.
| https://mathoverflow.net/users/6031 | Why the choice of the simplex for defining homology? | There is a specific reason that singular homology with simplices is simpler than singular homology with cubes. You would like the homology of a point to be "trivial" according to the Eilenberg-Steenrod axioms. (That is, $H^0 = \mathbb{Z}$ and the others are trivial.) However, if you look carefully at the chain complex of maps from cubes to a point, it is not true. The homology groups have to be corrected by hand by annihilating certain degenerate chains. If instead you carefully use the standard ordered simplex --- not just an abstract simplex floating in space but one with numbered vertices --- then it works automatically. On the other hand, products of chains look a bit simpler at first with cubes. They are only barely simpler though, because of the magic of simplicial sets. So on balance simplices are nicer than cubes.
Likewise, at the entirely rigorous level, defining singular homology with "disks" is not really a complete proposal, and not necessarily a simple proposal if you were to flesh it out. Do you mean maps from "all" balls, or some specific collection of balls, or just one standard ball? What is the boundary operator?
You can think of singular homology as a conversion from topology to combinatorics, attained by building a simplicial set (which is a generalization of a simplicial complex) from a topological space X, and then taking the simplicial homology of the combinatorial object. There is a specific way to do that with simplices, and an analogue with cubes that gives you a cubical complex. For instance, in the case of simplices, even if you start with a point, you get an infinite-dimensional complex with one simplex in each dimension. Presumably you have in mind bulding some CW complex from a topological space X, and then taking its CW homology. But what do you really want to connect to what?
It is also true that once you have a definition of homology, you can use a mapped-in disk to define a cycle or a relative cycle, which is then unique up to homology. But that is not the same as defining homology.
---
The question was extended to include a more precise description of the boundary operator in the proposed disk-based definition of homology. Namely, the proposal is that the boundary a mapped-in disk $D^n$ is a formal sum of two mapped-in disks $D^{n-1}$, whose restriction to the equator is the same mapped-in sphere $S^{n-2}$. I don't know what it means to equip these maps with an orientation; one thing that it could mean is to take the formal difference rather than the formal sum of the two hemispherical maps.
Either way, I don't think that the homology groups that result are correct. In this theory, a mapped-in line segment is only homologous to another mapped-in line segment with the same endpoints; there is no way to split the line segment into two line segments. I think that if the space is $\mathbb{R}^n$, then the resulting CW complex is weakly homotopy equivalent to a complete graph whose vertex set is $\mathbb{R}^n$ with the discrete topology. This isn't what you want.
| 22 | https://mathoverflow.net/users/1450 | 24944 | 16,356 |
https://mathoverflow.net/questions/24930 | 9 | This is for the sake of completeness(for references, understanding).
I ask for references for proofs that:
-There is exactly one differentiable(ie $C^\infty$) structure on $\mathbb R$, upto diffeomorphism.
-Ditto for $\mathbb R^2$.
-Ditto for $\mathbb R^3$.
| https://mathoverflow.net/users/6031 | Differentiable structures on R^3 | UPDATE : I found some precise references that answer the OP's question and fill in some details in my original answer. See the end for them.
I don't have a precise reference for the cases of $n=2$ or $n=3$, though I suspect that they could be found in Moise's "Geometric Topology in Dimensions 2 and 3".
However, I do have a nice reference for the amazing theorem (due to Stallings) that $\mathbb{R}^n$ has a unique $C^{\infty}$ structure for $n \geq 5$. It is written up very nicely in Steve Ferry's [Geometric Topology Notes](http://www.math.rutgers.edu/~sferry/ps/geotop.pdf). See Chapter 10 starting on page 56. What he proves is that $\mathbb{R}^n$ has a unique PL structure, but that is the key to the result. The rest of the notes are also wonderful.
This theorem is surprising for a number of reasons. For instance, it says that if $\Sigma$ is an exotic $n$-sphere for $n \geq 5$, then $\Sigma \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^n$ for any point $p \in \Sigma$. In other words, the "exoticness" is "concentrated at a point". Of course, this also follows from the usual proof of the high dimensional Poincare conjecture using the $h$-cobordism theorem, which constructs a homeomorphism between a homotopy sphere and the usual sphere which is differentiable except at one point (that one point giving trouble due to the "Alexander trick").
Another remark that should be made is that Freedman and Donaldson proved that $\mathbb{R}^4$ has uncountably many $C^{\infty}$ structures.
UPDATE : OK, here are some precise references. In
MR0121804 (22 #12534)
Munkres, James
Obstructions to the smoothing of piecewise-differentiable homeomorphisms.
Ann. of Math. (2) 72 1960 521--554.
it is proven in Corollary 6.6 that two PL-isomorphic differentiable manifolds that are homeomorphic to $\mathbb{R}^n$ are actually diffeomorphic. To make sense of this, recall that Cairns proved that differentiable manifolds have canonical PL structures; the original reference for this is his paper
MR0017531 (8,166d)
Cairns, Stewart S.
The triangulation problem and its role in analysis.
Bull. Amer. Math. Soc. 52, (1946). 545--571.
I think JHC Whitehead might have also proven this, but I don't have a reference for that.
This reduces us to showing that $\mathbb{R}^n$ has a unique PL structure. The cases $n=2$ and $n=3$ can be found in Moise's book "Geometric Topology in Dimensions 2 and 3". As far as the original results go, the 2-dimensional case is classical, while the 3-dimensional case was originally proven by Moise in his paper
MR0048805 (14,72d)
Moise, Edwin E.
Affine structures in $3$-manifolds. V. The triangulation theorem and Hauptvermutung.
Ann. of Math. (2) 56, (1952). 96--114.
For dimensions at least $5$, the original reference is the following
MR0149457 (26 #6945)
Stallings, John
The piecewise-linear structure of Euclidean space.
Proc. Cambridge Philos. Soc. 58 1962 481--488.
I hope this is helpful!
| 17 | https://mathoverflow.net/users/317 | 24950 | 16,361 |
https://mathoverflow.net/questions/24849 | 12 | Let $N \in \mathfrak{M}\_n(\mathbb{C})$ nilpotent, such that there exists $X \in \mathfrak M\_n(\mathbb{C})$ with $X^2=N$ (take for instance $n>2$ and $N(1,n)=1$; $N(i,j)=0$ otherwise).
Denote by $\mathcal{S}\_N$ the set of $X \in \mathfrak M\_n(\mathbb{C})$ such that $X^2=N$.
Is $\mathcal{S}\_N$ connected or path-connected ? What happens when we change $2$ by $3,4,\ldots $?
| https://mathoverflow.net/users/3958 | (Path) connected set of matrices? | This is a memorial to an incorrect solution that used to be here. Unfortunately, I can't delete it since it was accepted.
| 2 | https://mathoverflow.net/users/2349 | 24956 | 16,364 |
https://mathoverflow.net/questions/24392 | 7 | Sorry the title is a bit vague. Let A be a C\*-algebra, and let x and y be positive elements in A. Is it true that $$ \|x-y\|^2 \leq \|x^2-y^2\|? $$
Well, yes. But the proof I have is a bit of a hack, so I wonder if anyone has a "nice" proof, or a reference?
Aside: if $A=C\_0(X)$ then this reduces to the inequality $(a-b)^2 \leq |a^2-b^2|$ for non-negative real numbers a and b.
**Update:** Jonas points me to <http://www.springerlink.com/content/j4756m418220644r/> where Kittaneh has a proof pretty similar to what I had in mind (unpack the proof of Theorem 1). I guess I was interested in whether this sort of thing was standard (if I looked in the right textbook) or if it was a bit of a curiosity. I think the latter seems more likely...
| https://mathoverflow.net/users/406 | Simple inequality in C*-algebras | Theorem 1.5 of [this 1987 paper](https://dspace.library.uvic.ca:8443/dspace/handle/1828/1506) by J. Phillips says that if $f:[0,\infty)\to [0,\infty)$ is a continuous operator monotone function and $a$ and $b$ are positive operators on a Hilbert space, then $\|f(a)-f(b)\|\leq f(\|a-b\|)-f(0)$. I think that the proof is nice. Corollary 1.6 says that $\|a^{1/n}-b^{1/n}\|\leq\|a-b\|^{1/n}$, $n\geq1.$ Of course your inequality follows from taking $a=x^2$, $b=y^2$, and $n=2$.
Apparently Kittaneh and Kosaki have a similar approach in "Inequalities for the Schatten p-norm. V." Publ. Res. Inst. Math. Sci. 23 (1987), no. 2, 433--443 ([MR link](http://www.ams.org/mathscinet-getitem?mr=890926)). I haven't read any of this article.
Perhaps I should add the following for a more general audience. A continuous function $f:[0,\infty)\to [0,\infty)$ is operator monotone if whenever $x$ and $y$ are positive operators such that $y-x$ is positive, it follows that $f(y)-f(x)$ is positive. The functions $t\mapsto t^\alpha$ are operator monotone for $0<\alpha\leq1$ (but not for $\alpha>1$).
| 9 | https://mathoverflow.net/users/1119 | 24963 | 16,368 |
https://mathoverflow.net/questions/24970 | 38 | This was going to be a comment to [Differentiable structures on R^3](https://mathoverflow.net/questions/24930/differentiable-structures-on-r3), but I thought it would be better asked as a separate question.
So, it's mentioned in the previous question that $\mathbb{R}^4$ has uncountably many (smooth) differentiable structures. This is a claim I've certainly heard before, and I have looked a little bit at the construction of exotic $\mathbb{R}^4$s, but it's something that I really can't say I have an intuitive understanding of.
It seems reasonable enough to me that a generic manifold can have more than one differentiable structure, just from the definition; and is in fact, a little surprising to me that manifolds have only one differentiable structure for dimension $d \le 3$.
But it's very odd to me that $\mathbb{R}^d$ has *exactly* one differentiable structure, unless $d=4$, when it has way too many!
Naively, I would have thought that, since $\mathbb{R}^4 = \mathbb{R}^2 \times \mathbb{R}^2$, and $\mathbb{R}^2$ has only one differentiable structure, not much can happen. Although, we know [$\text{Diff}(M\times N)$ cannot generically be reasonably decomposed in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$](https://mathoverflow.net/questions/18034/can-we-decompose-diffmxn) in general, I would not have expected there to be obstructions for this to happen in this case.
I would have also thought, that since $\mathbb{R}^5$ has only one differentiable structure, and $\mathbb{R}^4$ is a submanifold of $\mathbb{R}^5$, and $\mathbb{R}^3$ is a submanifold of $\mathbb{R}^4$ with only one differentiable structure, this would be fairly restrictive on the differentiable structures $\mathbb{R}^4$ can have.
Although it seems that this only restricts the "inherited" differentiable structure to be a unique one, it still seems odd to me that the there are "non-inherited" structures in $d=4$ from $d=5$, and somehow all of these non-inhereted structures are identical on the submanifold $\mathbb{R}^3$!
Anyway, can anyone provide a intuitively sensible explanation of why $\mathbb{R}^4$ is so screwed up compared to every other dimension? Usually I would associate multiple differentiable structures with something topologically "wrong" with the manifold. Is something topologically "wrong" with $\mathbb{R}^4$ compared to every other dimension? Or is this a geometric problem somehow?
| https://mathoverflow.net/users/3329 | Exotic differentiable structures on R^4? | I once heard Witten say that topology in 5 and higher dimensions "linearizes". What he meant by that is that the geometric topology of manifolds reduces to algebraic topology. Beginning with the Whitney trick to cancel intersections of submanifolds in dimension $d \ge 5$, you then get the [h-cobordism theorem](http://en.wikipedia.org/wiki/H-cobordism_theorem), the solution to the Poincare conjecture, and surgery theory. As a result, any manifold in high dimensions that is algebraically close enough to $\mathbb{R}^d$ is homeomorphic or diffeomorphic to $\mathbb{R}^d$.
By the work of Freedman and others using Casson handles, there is a version of or alternative to the Whitney trick in $d=4$ dimensions, but only in the continuous category and not in the smooth category. Otherwise geometric topology does not "linearize" in Witten's sense. But in $d \le 3$ dimensions, the dimension is too low for the smooth category to separate from the continuous category, at least for the question of classification of manifolds.
What you have in 3 dimensions is examples such as the [Whitehead manifold](http://en.wikipedia.org/wiki/Whitehead_manifold), which is contractible but not homeomorphic to $\mathbb{R}^3$. In 4 dimensions you ~~instead~~ get open manifolds that are homeomorphic to $\mathbb{R}^4$ (because they are contractible ~~I'm not sure if other conditions are needed~~ and simply connected at infinity), but not diffeomorphic to $\mathbb{R}^4$. You have to be on the threshold between low dimensions and high dimensions to have the phenomenon. I would say that these exotic $\mathbb{R}^4$s don't really look that much like standard $\mathbb{R}^4$, they just happen to be homeomorphic. The homeomorphism has fractal features, and so does the Whitehead manifold.
Meanwhile 2 dimensions is too low to have non-standard contractible manifolds. In the smooth category, the Riemann uniformization theorem proves that smooth 2-manifolds are very predictable, or you can get the same result in the PL category with a direct combinatorial attack on planar graphs. And as mentioned, smooth, PL, and topological manifolds don't separate in this dimension.
---
Also, concerning your question about Cartesian products: Obviously the famous results imply that there is a fibration of standard $\mathbb{R}^5$ by exotic $\mathbb{R}^4$. The Whitehead manifold cross $\mathbb{R}$ is also homeomorphic to $\mathbb{R}^4$. (I don't know if it's diffeomorphic.) These fibrations are also fractal or have fractal features.
| 36 | https://mathoverflow.net/users/1450 | 24973 | 16,374 |
https://mathoverflow.net/questions/24974 | 3 | Of course if two morphisms of complexes are homotopic their induced maps coincide, but I'm wondering about the converse: if the induced maps on the cohomologies coincide, when does that imply that the morphisms are homotopic?
I've played around with it a bit and I think it might be true for complexes of projective modules? But I'm not sure... are there any well-known results regarding this?
| https://mathoverflow.net/users/2503 | Coinciding induced maps | The answer is no. Consider the complex over the integers $A$ which is $\mathbb Z$ in degree $0$ and $1$ and the only non-trivial differential being multiplication by $2$ and let $B$ be the same complex shifted once to the left (so that it is $\mathbb Z$ in degrees $-1$ and $0$). We have a map of complexes $A \to B$ which is the identity in degree $0$ and (necessarily) zero in all other degrees. This induces zero in cohomology (for trivial reasons) but is not null homotopic. (This is easily seen by an explicit calculation. Abstractly however it has to do with the fact that it realises the non-zero element of $\mathrm{Ext}^1\_{\mathbb Z}(\mathbb Z/2,\mathbb Z/2)$.)
| 7 | https://mathoverflow.net/users/4008 | 24977 | 16,377 |
https://mathoverflow.net/questions/24967 | 17 | The physical meaning of a Lie group symmetry is clear: for example, if you have a quantum system whose states have values in some Hilbert space $H$, then a Lie group symmetry of the system means that $H$ is a representation of some Lie group $G$. So you want to understand this Lie group $G$, and generally you do it by looking at its Lie algebra. At least initially, this is understood as a way to make the problem of classifying Lie groups and their representations easier.
But there are Lie algebras which are not the Lie algebra of a Lie group, and people are still interested in them. One possible way to justify this perspective from a physical point of view is that Lie algebras might still be viewable as ("infinitesimal"?) symmetries of physical systems in some sense. However, I have never seen a precise statement of how this works (and maybe I just haven't read carefully enough). Can anyone enlighten me?
| https://mathoverflow.net/users/290 | What is the physical meaning of a Lie algebra symmetry? | I'm not sure which sort of examples of Lie algebras without the corresponding groups you have in mind, but here is a typical example from Physics.
Many physical systems can be described in a hamiltonian formalism. The geometric data is usually a symplectic manifold $(M,\omega)$ and a smooth function $H: M \to \mathbb{R}$ called the *hamiltonian*. If $f \in C^\infty(M)$ is any smooth function, let $X\_f$ denote the vector field such that $i\_{X\_f}\omega = df$. If $f,g \in C^\infty(M)$ we define their *Poisson bracket*
$$\lbrace f, g\rbrace = X\_f(g).$$
It defines a Lie algebra structure on $C^\infty(M)$. (In fact, a Poisson algebra structure once we take the commutative multiplication of functions into account.)
In this context one works with the Lie algebra $C^\infty(M)$ (or particular Lie subalgebras thereof) and not with the corresponding Lie groups, should they even exist.
Symmetries in this context are functions which Poisson commute with the hamiltonian, hence the centraliser of $H$ in $C^\infty(M)$. They define a Lie subalgebra of $C^\infty(M)$.
**Added**
Another famous example occurs in two-dimensional conformal field theory. For example, the Lie algebra of conformal transformations of the Riemann sphere is infinite-dimensional: any holomorphic or antiholomorphic function defines an infinitesimal conformal transformation. On the other hand, the group of conformal transformations is finite-dimensional and isomorphic to $\mathrm{PSL}(2,\mathbb{C})$.
| 9 | https://mathoverflow.net/users/394 | 24979 | 16,378 |
https://mathoverflow.net/questions/24960 | 53 | Over the years, I have heard two different proposed answers to this question.
1. It has something to do with parabolic elements of $SL(2,\mathbb{R})$. This sounds plausible, but I haven't heard a really convincing explanation along these lines.
2. "Parabolic" is short for "para-Borelic," meaning "containing a Borel subgroup."
Which answer, if either, is correct?
A related question is who first introduced the term and when. Chevalley perhaps?
| https://mathoverflow.net/users/3106 | Why are parabolic subgroups called "parabolic subgroups"? | It appears that neither of the answers is fully correct. There is a great book, "Essays in the history of Lie groups and algebraic groups" by Armand Borel, when it comes to references of this type. To quote from chapter VI section 2:
>
> ...There was no nice terminology for the subgroups $P \_I$ with lie algebra the $\mathfrak p \_I$ until R. Godement suggested calling them parabolic subgroups. I shall therefore anachronistically call them that...
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"The geometry of the finite simple groups" by F. Buekenhout is on the other hand the only paper that came up in a search for paraborelic, and the author mentions he is using this term instead of parabolic to distinguish from parabolic subgroups of Chevalley groups.
| 25 | https://mathoverflow.net/users/2384 | 24987 | 16,385 |
https://mathoverflow.net/questions/24984 | 3 | According to Serre's book 'Galois cohomology', Galois chomology group are always torsion, but it seems to me that H^1(k, End\_{Z\_l}(T\_l(A)))=coker(Frob-1) on End\_{Z\_l}(T\_l(A)), which has the same Z\_l rank as End\_{k}(T\_l(A))
So maybe End\_{Z\_l}(T\_l(A)) is not a discrete galois module. And why is the Tate module a discrete galois module?
waht are the Galois cohomology groups of the Tate module of some abelian variety over a finite field or a number field?
| https://mathoverflow.net/users/3848 | what is the first Galois cohomology group of the Galois module End(T_l(A)) for some abelian variety A over a finite field k and l some prime number different from the characteristic of the base field? | In general, if $G$ is a profinite group and $M$ a continuous discrete $G$--module, then $H^i(G,M)$ is torsion for $i>0$. This applies in particular to Galois cohomology, i.e. when $G$ is a Galois group.
Tate modules are *not* discrete Galois modules, and their cohomology will usually not be torsion. The same goes for $\mathrm{End}(T\_\ell A)$.
Over finite or local fields the cohomology of $T\_\ell A$ is more or less well understood. Not so over global fields.
| 0 | https://mathoverflow.net/users/5952 | 24993 | 16,390 |
https://mathoverflow.net/questions/17508 | 5 | As I mentioned before, I'm a novice at deformation theory. I was wondering if the definition of versality in deformation theory is related to the versality in moduli spaces:
### Deformation theory
"Moduli of Curves" defines a versal deformation space as a deformation $\phi: X \rightarrow Y$ such that for any other deformation $\xi: X \rightarrow Z$ and for every point in $Z$ there exists an open set (in the complex topology) $U$ such that the pullback of $\phi$ via $f: U \rightarrow Y$ is $\xi$ restricted to $U$. (I imagine that in general instead of an open set one takes open etale covers - is this true?)
### Moduli Spaces
In moduli spaces, versality has always meant a space such that instead of the geometric points being in 1-1 correspondence with the objects we're interested in (over the field of the geo. point), each object is going have several geometric points in the versal space corresponding to it.
### Question
Are these two notions related? If so - how?
| https://mathoverflow.net/users/3238 | Versality in deformation theory vs. versality in moduli spaces | There is an easy answer to your question (without stacks) which has not been given yet: Yes.
The deformation space of a curve $X\_0$ is just a local model of a moduli space of all curves near a special member $X\_0$.
Your definition of versality for deformations says that, $Y$ is over-parametrizing
local deforamtions, i.e. every (germ of) a family of curves with central fiber $X\_0$
factorizes **non-uniquely** over $Y$.
Thus every nearby curve $X\_1$ might be represented by several elements $y \in Y$.
| 7 | https://mathoverflow.net/users/5714 | 25000 | 16,396 |
https://mathoverflow.net/questions/25003 | 3 | The construction of tangent bundle on a $C^\infty$ manifold, as for example in the book of Warner, uses the existence of double derivatives. Of course the tangent space for a point is first constructed in case of an open set in a Euclidean space and then the whole setup is glued up. But then in the neighborhood of a point the second derivatives might be different for a different $C^1$-chart around the point. So I suppose the tangent space/tangent bundle construction is for $C^2$-manifolds.
Question is:
>
> Since $C^2$ is apparently the minimum condition for existence of tangent bundle, is the tangent bundle just $C^1$, or it is again $C^2$?
>
>
>
| https://mathoverflow.net/users/6031 | What is the order of differentiability of the tangent bundle of a C^2- manifold? | The tangent bundle of a $C^1$ manifold exists: it's a $C^0$ manifold.
Similarly, for every $n$, the tangent bundle of a $C^n$ manifold is a $C^{n-1}$ manifold.
| 5 | https://mathoverflow.net/users/5690 | 25005 | 16,399 |
https://mathoverflow.net/questions/25008 | 10 | Given a finite field $K$, what are the possible degrees of a polynomial $p\in K[x]$ such that $x\longmapsto p(x)$ is one-to-one?
Such a polynomial has clearly not degree $0$ and it cannot be of degree two except for
$x\longmapsto (\alpha(x))^2$ for $\alpha$ an affine bijection of a field of characteristic $2$.
Are there many examples of degree $3$ (except for the stupid $x\longmapsto (\alpha(x))^3$
with $\alpha$ an affine bijection of a field of characteristic $3$)?
I guess that the degrees of such polynomials (except for affine bijections and their
composition with the Frobenius map)
are generically fairly high (the interpolation polynomial for a "random" permutation
of a finite field with $q$ elements should typically be of degree $q-1$).
What can for instance be said on the smallest degree $>1$ of a non-affine polynomial
inducing a bijection of $\mathbb Z/p\mathbb Z$?
| https://mathoverflow.net/users/4556 | Degrees of a polynomial $p$ such that $x\mapsto p(x)$ is one-to-one | Such things are referred to as ‘permutation polynomials’ and if you do a search, you'll find a whole menagerie of non-stupid classes which is constantly expanding. One simple result going back to Dickson provides something converse to damiano's [observation](https://mathoverflow.net/questions/25008/given-a-finite-field-k-what-are-the-possible-degrees-of-a-polynomial-p-in-k#comment52602_25008) — there are no (non-linear) permutation polynomials of degree dividing $q-1$.
Something backing up your guess that the degrees are 'generically' fairly high is a conjecture of Carlitz:
Fix even degree $n$, then the cardinality $q$ (with $q$ odd) of a field having a degree $n$ permutation polynomial is bounded from above.
This has been proved in cases for $n$ up to 14 by Dickson, Hayes and then Daqing Wan in [*On a conjecture of Carlitz*](https://doi.org/10.1017/S1446788700029657)
J. Austral. Math. Soc. Ser. A 43 (1987), no. 3, 375–384
but as far as I can tell, the general case is completely open.
Edit: a quick search provides more:
S. Cohen [*Permutation polynomials and primitive permutation groups.*](https://doi.org/10.1007/BF01246737)
Arch. Math. (Basel) 57 (1991), no. 5, 417–423
proves the above conjecture for $n\leq 1000$ and for n begin 2 times an odd prime.
And in fact the whole conjecture is implied by the result proved by Fried, Saxl and Guralnick in
[*Schur covers and Carlitz's conjecture.*](https://doi.org/10.1007/BF02808112) Israel J. Math. 82 (1993), no. 1-3, 157–225.
| 17 | https://mathoverflow.net/users/3143 | 25017 | 16,406 |
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