parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/19788 | 4 | Let $G=GL(\mathbb{R})$, $P$ be the subgroup of $G$ consisting of elements with the last row $(0,0,...,1)$. Then Kirillov conjecture states that for any irreducible unitary representation of $G$, its restriction to $P$ remains irreducible. This conjecture has been proved (not only over $\mathbb{R}$, but also over $\mathbb{C}$ and p-adic fields). Here I'm wondering if we consider irreducible smooth representations in Hilbert space(or Banach, Frechet space), does this conjecture remains true?
Another related question is generally, how to prove the irreducibility for a smooth representation besides the definition?
| https://mathoverflow.net/users/1832 | a question about irreducibility of representations and Kirillov conjecture | I think it's best to look at the relatively recent paper of Moshe Baruch, Annals of Math., "A Proof of Kirillov's Conjecture" -- in the introduction of his paper, he discusses the basic techniques of proof, and a bit of the history (Bernstein proved this conjecture in the p-adic case, for example).
Baruch and others (e.g. Kirillov, in the original conjecture, I think) consider unitary representations. This is necessary for the methods which they use. From the beginning, they use the "converse of Schur's lemma", i.e., if $Hom\_P(V,V)$ is one-dimensional then $V$ is irreducible. This converse of Schur's lemma requires one to work in the unitary (or unitarizable) setting.
Now, to address Kevin Buzzard's point, consider $G = GL\_2(F)$ for a $p$-adic field, and a unitary principal series representation $V = Ind\_B^G \chi \delta^{1/2}$, where $\chi = \chi\_1 \boxtimes \chi\_2$ is a unitary character of the standard maximal torus, and $\delta$ is the modular character for the Borel $B$.
Restricting $V$ back down to $B$, one gets a short exact sequence of $B$-modules:
$$0 \rightarrow V(BwB) \rightarrow V \rightarrow V(B) \rightarrow 0,$$
where $V(X)$ denotes a space of functions (compactly supported modulo $B$ on the left) on the ($B$-stable) locus $X$. On can check that these spaces are nonzero using the structure of the Bruhat cells, and hence the restriction of $V$ to $B$ is reducible as Kevin suggests.
*But*, if one considers the Hilbert space completion $\hat V$ of $V$, with respect to a natural Hermitian inner product, one finds that $\hat V$ is an irreducible unitary representation of $G$ which remains irreducible upon restriction to $B$ (and to the even smaller "mirabolic" subgroup of Kirillov's conjecture). Here it is important to note that "irreducibility" for unitary representations on Hilbert spaces refers to *closed* subspaces. The $B$-stable subspace $V(BwB)$ of $V$ is not closed, and its closure is all of $\hat V$ I think.
So - I think that Kirillov's conjecture is false, in the setting of smooth representations of $p$-adic groups (and most probably for smooth representations of moderate growth of real groups).
However, the techniques still apply in the smooth setting to give weaker (but still useful) results. After all, it is still useful to know that $Hom\_P(V,V)$ is one-dimensional! This can be used to prove multiplicity one for certain representations, for example.
The general technique to prove $Hom\_P(V,V)$ is one-dimensional involves various forms of Frobenius reciprocity and characterization of distributions. Without explaining too much (you should look at old papers of Bernstein, perhaps), and being sloppy about dualities sometimes,
$$Hom\_P(V,V) \cong Hom\_P(End(V), C) \cong Hom\_G(End(V), Ind\_P^G C).$$
Some sequence of Frobenius reciprocity and linear algebra (I don't think I have it quite right above) identifies $Hom\_P(V,V)$ with a space of functions or distributions: $f: G \rightarrow End(V)$, which are $(P,V)$-bi-invariant. In other words,
$$f(p\_1 g p\_2) = \pi(p\_1^{-1}) \circ f(g) \circ \pi(p\_2),$$
or something close.
So in the end, one is led to classify a family of $P$-bi-quasi-invariant $End(V)$-valued distributions on $G$. This leads to two problems: one geometric, involving the $P$-double cosets in $G$. This is particularly easy for the "mirabolic" subgroup $P$. The second problem is often more difficult, analyzing distributions on each double coset, and proving most of them are zero or else have very simple properties.
Hope this clarifies a little bit... you might read more on the Gelfand-Kazhdan method (Gross has an exposition in the Bulletin) to understand this better.
| 13 | https://mathoverflow.net/users/3545 | 19844 | 13,192 |
https://mathoverflow.net/questions/19809 | 10 | There is of course the standard construction of the reals by considering the set of sequences that are Cauchy with respect to the standard metric and taking the quotient by sequences that converge to 0, but in many other cases of the completion of a ring or group, we can complete by taking an inverse limit of factor rings/groups corresponding to some predetermined ideals/subgroups. For example, the $\mathfrak{p}$-adic integers are constructed using precisely this type of inverse limit construction, where the ideals are the powers of $\mathfrak{p}$, where $\mathfrak{p}$ is the ideal $(p)$ for $p$ a prime.
Is there any way to construct the reals using a similar inverse limit construction?
It seems like this might be related to the so-called "infinite prime", but I don't know if there is really a way to complete with respect to an "infinite prime", which doesn't seem like it is strictly a prime integer or ideal, so that doesn't really seem to help..
(If anyone can think of better tags, please go ahead and change the ones I've used. I wasn't really sure how to classify this).
| https://mathoverflow.net/users/1353 | Completion of the rationals to the reals as an inverse limit construction? | Although your question is not at all vague, there are a few completely different ways to interpret what a good answer would be. Knowing a bit about your personal preferences, I suspect the following is not at all what you wanted, but it is still of interest to the community.
The fact that $\mathbb{Q}\_p$, for example, can be seen as an inverse limit boils down to the fact that the series representations
$$\sum\_{n \in \mathbb{Z}} a\_n p^{-n}$$
give a homeomorphism between $\mathbb{Q}\_p$ and the subspace of $\{0,1,\dots,p-1\}^{\mathbb{Z}}$ of eventually zero sequences (on the positive side). At first glance, it would seem that the same works for $\mathbb{R}$ since every element of $[0,\infty)$ has a very similar binary expansion
$$\sum\_{n \in \mathbb{Z}} a\_n 2^n$$
where $(a\_n)\_{n \in \mathbb{Z}}$ is again in the space of all eventually zero sequences in $\{0,1\}^{\mathbb{Z}}$. Unfortunately this is not a bijection. In fact, there is no way to select such a sequence $(a\_n)\_{n\in\mathbb{Z}}$ continuously — binary rationals are always points of discontinuity no matter how the selection is made.
There is a funny way to remedy this which is commonly used in Computable Analysis and Reverse Mathematics. The idea is to represent real numbers again with series of the form
$$\sum\_{n \in \mathbb{Z}} a\_n 2^n$$
but where $(a\_n)\_{n \in \mathbb{Z}}$ is now an eventually zero sequence from the extended set $\{-1,0,1\}^{\mathbb{Z}}$. The obvious cost is that no number has a unique representation in this form, but the side benefit is that the binary rationals are no longer 'special' in this way. In the end, this representation is extremely well behaved compared to ordinary binary expansions. This can be seen from the fact that any *invariant* continuous function between representations (see note) gives rise to a continuous function $\mathbb{R}\to\mathbb{R}$, and every continuous function $\mathbb{R}\to\mathbb{R}$ admits such a lifting.
The conclusion to draw from this is that, up to some very nice blurring, $\mathbb{R}$ is indeed an inverse limit just like $\mathbb{Q}\_p$.
Note: The topology on eventually zero sequences is not exactly the product topology, it is given by the metric
$$d(\vec{x},\vec{y}) = \inf\{2^n : \forall m \geq n\,({x\_m = y\_m})\}.$$ The *invariance* of a function $f$ between eventually zero sequences from $\{-1,0,1\}^{\mathbb{Z}}$ means that if $\vec{x}$ and $\vec{y}$ represent the same real number then so do $f(\vec{x})$ and $f(\vec{y})$.
| 9 | https://mathoverflow.net/users/2000 | 19854 | 13,197 |
https://mathoverflow.net/questions/19857 | 58 | Note: I have modified the question to make it clearer and more relevant. That makes some of references to the old version no longer hold. I hope the victims won't be furious over this.
---
### Motivation:
Recently Pace Nielsen asked the question "How do we recognize an integer inside the rationals?". That reminds me of this question I had in the past but did not have chance to ask since I did not know of MO.
There seems to be a few evidence which suggest some possible relationship between decidability and prime numbers:
1) Tameness and wildness of structures
One of the slogan of modern model theory is " Model theory = the geography of tame mathematics". Tame structure are structures in which a model of arithmetic can not be defined and hence we do not have incompleteness theorem. A structure which is not tame is wild.
The following structures are tame:
* Algebraic closed fields. Proved by Tarski.
* Real closed fields e.g $\mathbb{R}$. Proved by Tarski.
* p-adic closed fields e.g $ \mathbb{Q}\_p$. Proved by Ax and Kochen.
Tame structures often behave nicely. Tame structures often admits quantifier elimination i.e. every formula are equivalent to some quantifier free formula, thus the definable sets has simple description. Tame structures are decidable i.e there is a program which tell us which statements are true in these structure.
The following structures are wilds;
* Natural number (Godel incompleteness theorem)
* Rational number ( Julia Robinson)
Wild structure behaves badly (interestingly). There is no program telling us which statements are true in these structures.
One of the difference between the tame structure and wild structure is the presence of prime in the later. The suggestion is strongest for the case of p-adic field, we can see this as getting rid of all except for one prime.
2) The use of prime number in proof of incompleteness theorem
The proof of the incompleteness theorems has some fancy parts and some boring parts. The fancy parts involves Godel's Fixed point lemma and other things. The boring parts involves the proof that proofs can be coded using natural number. I am kind of convinced that the boring part is in fact deeper. I remember that at some place in the proof we need to use the Chinese Remainder theorem, and thus invoke something related to primes.
3) Decidability of Presburger arithmetic and Skolem arithmetic ( extracted from the answer of Grant Olney Passmore)
Presburger arithmetic is arithmetic of natural number with only addition. Skolem arithmetic is arithmetic of natural number with only multiplication.
Wishful thinking: The condition that primes or something alike definable in the theory will implies incompleteness. Conversely If a theory is incomplete, the incompleteness come from something like primes.
---
### Questions:
(following suggestion by François G. Dorais)
Forward direction: Consider a bounded system of arithmetic, suppose the primes are definable in the system. Does it implies incompleteness.
Backward direction: Consider a bounded system of arithmetic, suppose the system can prove incompleteness theorem, is primes definable in the system? is the enumeration of prime definable? is the prime factoring function definable?
---
### Status of the answer:
For the forward direction: A weak theory of prime does not implies incompleteness. For more details, see the answer of Grant Olney Passmore and answer of Neel Krishnaswami
For backward direction: The incompleteness does not necessary come from prime. It is not yet clear whether it must come from something alike prime. For more details, see the answer of Joel David Hamkins.
---
Since perhaps this is as much information I can get, I accept the first answer by Joel David Hamkins. Great thanks to Grant Olney Passmore and Neel Krishnaswami who also point out important aspects.
Recently, Francois G. Dorais also post a new and interesting answer.
| https://mathoverflow.net/users/2701 | Has decidability got something to do with primes? | Goedel did indeed use the Chinese remainder theorem in his proof of the Incompleteness theorem. This was used in what you describe as the `boring' part of the proof, the arithmetization of syntax. Contemporary researchers often agree with your later assessment, however, that the arithmetization of syntax is profound. This is the part of the proof that reveals the self-referential nature of elementary number theory, for example, since in talking about numbers we can in effect talk about statements involving numbers. Ultimately, we arrive in this way at a sentence that asserts its own unprovability, and this gives the Incompleteness Theorem straight away.
But there are other coding methods besides the Chinese Remainder theorem, and not all of them involve primes directly. For example, the only reason Goedel needed CRT was that he worked in a very limited formal language, just the ring theory language. But one can just as easily work in a richer language, with all primitive recursive functions, and the proof proceeds mostly as before, with a somewhat easier time for the coding part, involving no primes. Other proofs formalize the theory in the hereditary finite sets HF, which is mutually interpreted with the natural numbers N, and then the coding is fundamentally set-theoretic, also involving no primes numbers especially.
| 39 | https://mathoverflow.net/users/1946 | 19858 | 13,200 |
https://mathoverflow.net/questions/19840 | 33 | My question is fairly simple, and may at first glance seem a bit silly, but stick with me. If we are given the rationals, and we pick an element, how do we recognize whether or not what we picked is an integer?
Some obvious answers that we might think of are:
A. Write it in lowest terms, and check the denominator is 1.
B. Check that the p-adic valuation is non-negative, for all p.
C. Decide whether the number is positive (or negative) and add 1 to itself (or -1 to itself) until it is bigger than the rational you picked. (If these multiples ever equaled your rational, then you picked an integers.)
Each of these methods has pluses and minuses. For example, in option A we presuppose we know how to write an arbitrary rational number q as a quotient of integers and reduce. In C, we have issues with stopping times. etc..
To provide some context for my question: We know, due to the work of Davis, Putnam, Robinson, and Matijasevic, that the positive existential theory of $\mathbb{Z}$ is undecidable. The same question for $\mathbb{Q}$ is not entirely answered. One approach to this new question is to show that that, using very few quantifiers, one can describe the set of integers inside the rationals; and then reduce to the integer case. For example, see Bjorn Poonen's paper "Characterizing integers among rational numbers with a universal-existential formula." There, he finds a way to describe the p-valuation of a rational number (i.e. he finds a way to encode option B in the language of quantifiers and polynomials on the rationals). I'm wondering if there are other characterizations of the integers which would follow suit.
| https://mathoverflow.net/users/3199 | How do we recognize an integer inside the rationals? | Here's one way: Show that the rational number is an algebraic integer. This may sound like a silly idea, but it has non-trivial applications. A rational number is an integer if it has an expression as a sum of products of algebraic integers. See for example, Prop. 5 in the appendix of *Groups and Representations* by J.L. Alperin, and Rowen B. Bell, where this is used to prove that, for an irreducible character $\chi$ of a finite group $G$, $\chi(1)$ divides $|G|$.
| 29 | https://mathoverflow.net/users/2604 | 19860 | 13,202 |
https://mathoverflow.net/questions/19826 | 3 | Let A be an $m\times n$ matrix and $k$ be an integer. Assume that $A$ is non-negative. We want to find a scalar $\epsilon$ and an $n\times n$ matrix $B$ such that $A\leq A(\epsilon I + B)$ (where $\leq$ is an element-wise comparison). The goal is to minimize $\epsilon$ and we have the following restrictions on $B$: 1) $B$ is non-negative. 2) Each column of $B$ has $L\_1$ norm at most 1. 3) There are at most $k$ rows of $B$ that are non-zero (i.e., at least $n-k$ rows are zero vectors).
In case it helps, we may assume that $n>>k>>m$.
My goal is to get an algorithm for computing $B$ to minimize $\epsilon$ (either exactly or approximately) and my general question is whether you know of anything related. (I'm not familiar with this area. It's not even clear to me if the problem is NP-hard or not.) Another thing is whether it is possible to bound $\epsilon$ in terms of $k$, $m$ and $n$?
| https://mathoverflow.net/users/5016 | Matrix approximation | I'll address the last question (about an a priori bound for $\epsilon$).
If $n\gg k\gg m$, the worst-case bound for $\epsilon$ is between $c(m)\cdot k^{-2/(m-1)}$ and $C(m)\cdot k^{-1/(m-1)}$ (probably near the former but I haven't checked this carefully). Note that the bound does not depend on $n$.
*Proof*.
The columns of $A$ form a set $S$ of cardinality at most $n$ in $\mathbb R^m$. For a given $\epsilon$, a suitable $B$ exists if and only if there is a subset $T\subset S$ of cardinality at most $m$ such that the convex hull $conv(T)$ majorizes the set $(1-\epsilon)S$ in the following sense: for every $v\in S$, there is a point in $conv(T)$ which is component-wise greater than $(1-\epsilon)v$. And this majorization is implied by the following: $conv(sym(T))$ contains the set $(1-\epsilon)S$, or equivalently, the set $(1-\epsilon)conv(sym(S))$, where by $sym(X)$ denotes the minimal origin-symmetric set containing $X$, that is, $sym(X)=X\cup -X$.
Consider the polytope $P=conv(sym(S))$. We want to find a subset of its vertices of cardinality at most $k$, such that their convex hull approximates $P$ up to $(1-\epsilon)$-rescaling. This problem is invariant under linear transformations, and we may assume that $P$ has nonempty interior. Then Fritz John's theorem asserts that there is a linear transformation of $\mathbb R^m$ which transforms $P$ to a body contained in the unit ball and containing the ball of radius $1/\sqrt m$. For such a set, $(1-\epsilon)$-scaling approximation follows from $(\epsilon/\sqrt m)$-approximation in the sense of Hausdorff distance. So it suffices to choose $T$ to be an $(\epsilon/\sqrt m)$-net in $S$. Then a standard packing argument gives the above upper bound for $\epsilon$.
On the other hand, if $S$ is contained in the unit sphere and separated away from the coordinate hyperplanes, you must choose $T$ to be a $\sqrt\epsilon$-net in $S$. This gives the lower bound; the "worst case" is a uniformly packed set of $n=C(m)\cdot k$ points on the sphere.
**UPDATE**.
Fritz John theorem, also known as John Ellipsoid Theorem, says that for any origin-symmetric convex body $K\subset\mathbb R^m$, there is an ellipsoid $E$ (also centered at the origin) such that $E\subset K\subset\sqrt m E$. (There is a non-symmetric variant as well but the constant is worse.) The linear transformation that I used just sends $E$ to the unit ball. There are lecture notes about John ellipsoid [here](http://www.math.sc.edu/~howard/Notes/john.ps.gz) and probably in many other sources.
Comparing scaling distance (see also Banach-Mazur distance) and Hausdorff distance between convex bodies is based on the following. The scaling distance is determined by the worst ratio of the support functions of the two bodies, and the Hausdorff distance is the maximum difference between the support functions. Once you captured the bodies between two balls, you can compare relative and absolute difference. This should be explained in any reasonable textbook in convex geometry; unfortunately I'm not an expert in textbooks, especially English-language ones.
By "packing argument" I mean variants of the following argument showing that for any $\epsilon$, any subset $S$ of the unit ball in $\mathbb R^m$ contains an $\epsilon$-net of cardinality at most $(1+2/\epsilon)^m$. Take a maximal $\epsilon$-separated subset $T$ of $S$, it is always an $\epsilon$-net. Since $T$ is $\epsilon$-separated, the balls of radius $\epsilon/2$ centered at the points of $T$ are disjoint, hence the sum of their volumes is no greater than the volume of the $(1+\epsilon/2)$-ball that contains them all. Writing the volume of an $r$-ball as $c(m)\cdot r^m$ yields the result. This argument gives a rough estimate
$$
\epsilon \le (2\sqrt m+1) k^{-1/m}
$$
in the original problem (up to errors in my quick computations). To improve the exponent one can consider the $(m-1)$-dimensional surface of $P$ rather that the whole ball.
| 4 | https://mathoverflow.net/users/4354 | 19876 | 13,214 |
https://mathoverflow.net/questions/19829 | 12 | For any set $S$ one can consider the free abelian group $\mathbb{Z}[S]$ generated by this set. Now suppose, there is a topology on $S$ given. Is it possible to find a topology on $\mathbb{Z}[S]$ in such a way, that:
i) The map $S\rightarrow \mathbb{Z}[S]$ is a homeomorphism onto the image.
ii) The addition and the inverse map are continuous
And if it is possible, is this topology unique?
| https://mathoverflow.net/users/3969 | Topologizing free abelian groups | I don't know if such a topology is unique, but it exists if and only if $S$ is [completely regular](http://en.wikipedia.org/wiki/Completely_regular_space). This includes locally compact hausdorff spaces and CW complexes.
With Freyd's Adjoint Functor Theorem, it can be shown that the forgetful functor from abelian top. groups to top. spaces has a left adjoint. This is essentially the same proof as in the discrete case. Explicitely, $\mathbb{Z}[S]$, the free abelian top. group over the top. space $S$, is the usual free abelian group endowed with the weak topology for all homomorphisms $\mathbb{Z}[S] \to A$, such that $S \to \mathbb{Z}[S] \to A$ is continuous. Here, $A$ is an arbitrary abelian top. group. In order to show that this topology exists, we may assume that $A$ is, as a group, a quotient of $\mathbb{Z}[S]$, so that these $A$ form a set. But the description of the topology does not change and even without Freyd's Theorem it is easy to see that $\mathbb{Z}[S]$ thus becomes an abelian top. group satisfying the desired universal property.
Now I claim that the three assertions
1. $S \to \mathbb{Z}[S]$ is a homeomorphism onto its image.
2. $S$ is a subspace of an abelian top. group.
3. $S$ is completely regular.
are actually equivalent!
1) implies 2), that's clear. Now assume 2), thus $S \subseteq A$ for some top. abelian group. Extend the inclusion $S \to A$ to a continuous homomorphism $\mathbb{Z}[S] \to A$. Every open subset of $S$ can be extended to an open subset of $A$. Pull it back to $\mathbb{Z}[S]$. This is an open subset of $\mathbb{Z}[S]$ which restricts to the given oben subset of $S$. This proves 1).
2) implies 3), this follows from the fact that every topological group is completely regular and subspaces of completely regular spaces are obviously completely regular.
Finally assume 3), i.e. $S$ carries the initial topology with respect to all continuous functions $S \to \mathbb{R}$. Endow $\mathbb{R}^{(S)}$ with the initial topology with respect to all homomorphisms $\mathbb{R}^{(S)} \to \mathbb{R}$, such that the restriction $S \to \mathbb{R}$ is continuous. Then $\mathbb{R}^{(S)}$ is an abelian topological group and $S \to \mathbb{R}^{(S)}$ is an embedding, thus 2).
I also believe that (but cannot prove)
* If $S$ is hausdorff and completely regular, $\mathbb{Z}[S]$ is hausdorff.
In another comment, it was suggested to endow $\mathbb{Z}[S]$ with the final topology with respect to $S \to \mathbb{Z}[S]$. But this does not even yield a translation invariant topology: If $S=\{a,b\}$ with the only nontrivial open subset $\{a\}$, then $\{a\}$ is open in $\mathbb{Z}[S]$, but $\{b\}$ is not.
>
> But maybe, if $S$ is a completely regular space, the topology of $\mathbb{Z}[S]$ used above is the final topology?
>
>
>
| 9 | https://mathoverflow.net/users/2841 | 19882 | 13,217 |
https://mathoverflow.net/questions/19802 | 8 | From Weil conjecture we know the relation between the zeta-function and the cohomology of the variety, however it appears that there are certainly more information containing in the zeta-function, and the question remains whether they can be used to compute some more geometric invariants of the variety, such as the Chern classes. For instance, can one spot a Calabi-Yau manifold just by looking at the zeta-function? Is the zeta-function a birational invariant, or stronger?
And consider the roots and poles of the zeta-function. Their absolute values are determined by the Riemann Hypothesis, nevertheless the "phases" still appear to be very mysterious. Are that any good explanations for them, e.g. in the case of elliptic curves?
| https://mathoverflow.net/users/4782 | How much complex geometry does the zeta-function of a variety know | Although the question is phrased a bit sloppily, there is a standard interpretation: Given a smooth complex proper variety $X$, choose a smooth proper model over a finitely generated ring $R$. Then one can reduce modulo maximal ideals of $R$ to get a variety $X\_m$ over a finite field, and ask what information about $X$ it retains. As has been remarked, the zeta function of $X\_m$ gives back the Betti numbers of $X$.
I believe Batyrev shows in [this paper](http://arxiv.org/PS_cache/alg-geom/pdf/9710/9710020v2.pdf) that the zeta function of a Calabi-Yau is a birational invariant and deduces from this the birational invariance of Betti numbers for Calabi-Yau's. And then, Tetsushi Ito showed [here](http://arxiv.org/PS_cache/math/pdf/0209/0209269v2.pdf) that knowledge of the zeta function at all but finitely many primes contains info about the Hodge numbers. (He did this for smooth proper varieties over a number field, but a formulation in the 'general' situation should be possible.)
For an algebraic surface, once you have the Hodge numbers, you can get the Chern numbers back by combining the fact that
$c\_2=\chi\_{top},$
the topological Euler characteristic, and Noether's formula:
$\chi(O\_X)=(c\_1^2+c\_2)/12.$
I guess this formula also shows that if you know a priori that $m$ is a maximal ideal of ordinary reduction for both $H^1$ and $H^2$ of the surface, then you can recover the Chern numbers from the zeta functions, since $H^1(O\_X)$ and $H^2(O\_X)$ can then be read off from the number of Frobenius eigenvalues of slope 0 and of weights one and two.
You might be amused to know that the *homeomorphism class* of a simply-connected smooth projective surface can be recovered from the isomorphism class of $X\_m$. (One needs to formulate this statement also a bit more carefully, but in an obvious way.) However, not from the zeta function. If you compare $P^1\times P^1$ and $P^2$ blown up at one point, the zeta functions are the same but even the rational homotopy types are different, as can be seen from the cup product in rational cohomology. See [this paper](http://arxiv.org/PS_cache/math/pdf/0204/0204066v4.pdf).
Added: Although people can see from the paper, I should have mentioned that Ito even deduces the birational invariance of the Hodge (and hence Betti) numbers for smooth minimal projective varieties, that is, varieties whose canonical classes are nef. Regarding the last example, I might also point out that this is a situation where the real homotopy types are the same.
Added again: I'm sorry to return repeatedly to this question, but someone reminded me that Ito in fact does not need the zeta function at 'all but finitely many primes.' He only needs, in fact, the number of points in the residue field itself, not in any extension.
| 20 | https://mathoverflow.net/users/1826 | 19883 | 13,218 |
https://mathoverflow.net/questions/19886 | 3 | I've been dealing with the following situation:
Let $R\subseteq S$ be an extension of Dedekind rings, where $Quot(R)=:L \subseteq E:=Quot(S)$ is a $G$-Galois extension. Let $\mathfrak{p}$ be a prime of $R$, and $\mathfrak{q}$ a primes of $S$ above $\mathfrak{p}$. Let $D\_{\mathfrak{q}}$ denote the decomposition group, and $I\_{\mathfrak{q}}$ the inertia group, of $\mathfrak{q}$ over $\mathfrak{p}$.
However, unlike in the classic case, I allow the residue fields of $\mathfrak{p}$ to be infinite, with positive characteristic. So the extension of residue fields may be inseparable.
It seems that the paper I'm reading implicitly assumes:
$|I\_{\mathfrak{q}}|=e[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]\_ i $ (the ramification index times the inseparability degree of the residue extension)
$|D\_{\mathfrak{q}}|=e[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]$
$|G|=re[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]$ (where $r$ is the number of primes above $\mathfrak{p}$)
Is that right? I keep hitting walls when I try to prove it.
| https://mathoverflow.net/users/3238 | Decomposition of primes, where the residue field extensions are allowed to be inseparable | I believe that's right, at least when $S$ is finitely generated over $R$. See Serre's Local Fields page 21-22 (in the English translation); he states his assumptions on page 13.
| 3 | https://mathoverflow.net/users/949 | 19893 | 13,224 |
https://mathoverflow.net/questions/19924 | 8 | I know the definition a spectral triple and that it is some kind of non-commutative generalisation of (the ring of functions on) a compact spin manifold.
But, why is it called *spectral* triple?
| https://mathoverflow.net/users/1291 | Why is it called *spectral* triple? | Well, it uses the spectral properties of the Dirac operator $D$ in the spectral triple quite extensively.
Also, in the article where he (essentially) introduces the notion of spectral triples ( <http://www.alainconnes.org/docs/reality.pdf> ) Alain Connes writes about the naming:
*"We shall need for that purpose to adapt the tools of the differential and integral calculus
to our new framework. This will be done by building a long dictionary which relates the usual
calculus (done with local differentiation of functions) with the new calculus which will be done
with operators in Hilbert space and spectral analysis, commutators.... The first two lines of the
dictionary give the usual interpretation of variable quantities in quantum mechanics as operators in
Hilbert space. For this reason and many others (which include integrality results) the new calculus
can be called the quantized calculus’ but the reader who has seen the word “quantized” overused
so many times may as well drop it and use “spectral calculus” instead."*
| 10 | https://mathoverflow.net/users/3897 | 19926 | 13,244 |
https://mathoverflow.net/questions/19880 | 10 | ### Background
In the course of answering another question ([Infinite collection of elements of a number field with very similar annihilating polynomials](https://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating)) I found myself with a curve, that if it had a positive genus, then I could prove something about an interesting number field. Specifically, I was trying to show that, in the terminology of the above question, all biquadratic fields over the rationals have $r=2$.
The curve was defined by a single affine equation with large degrees in $x$ and in $y$ (23 and 25). Giving sage, or to be more precise Singular, the task of computing the genus, it broke down after an hour. The degree is just too much.
### Question
But really, all I want is to bound the genus from below. What effective ways are there for doing so?
### Possible solution
Compute the number of points in a random small finite field, hoping for no bad reduction, and checking this is much easier than finding all algebraic singular points. If it is not true that $q-2g\sqrt{q}+1 \le |C(\mathbb{F}\_q)| \le q+2g\sqrt{q}+1$, then the genus is greater than $g$. Is this correct?
| https://mathoverflow.net/users/2024 | Computationally bounding a curve's genus from below? | If you can check that the curve is geometrically irreducible, then you may try using the Hurwitz formula (you may use the formula in any case, but you would have to be more careful with the conclusions if the curve is not irreducible). Assuming that your example is geometrically integral, projection onto one of the axis is a morphism of degree 23. If you know that the morphism ramifies in at least 2\*23+h **smooth** points, then the curve has genus at least 1+h/2. If you know the multiplicities of the ramification or if you know the behaviour of ramification at the singular points, then you can deduce more accurate information.
EDIT: here is an expanded, more computational, version. Write an equation *f* defining your curve *C* as a polynomial of degree *d* in *y* whose coefficients are polynomials in *x*. Thus the morphism $(x,y) \to x$ is a morphism of degree *d* to $\mathbb{A}^1$. We know that this induces a finite morphism from a smooth projective model *C'* of *C* to $\mathbb{P}^1$ and we use the Hurwitz formula to compute the (arithmetic) genus *g* of *C'*: $2g-2=-2d+r$, where *r* is the degree of the ramification divisor. The better you can estimate *r* from below, the better the approximation to the geometric genus of *C*.
Clearly, the points where *f=df/dy=0* but $df/dx \neq 0$ are smooth points of *C* that are ramified for the projection to $\mathbb{A}^1$ and hence contribute to the Hurwitz formula (ramification occurs in *C'* since such points are **smooth** and hence *C* and *C'* are locally isomorphic here).
Computationally we find the resultant in *y* of *(f,df/dy)* (thus you get a polynomial in *x*) and purge out of it the roots it has in common with the resultant in *y* of *(df/dx,df,dy)*. Denote this final polynomial by *R*; the number of distinct roots of *R* is a lower bound for *r*. If you are more careful, you might do a little better than this without having to worry about the singular points of *C*, but for "generic" projections this is optimal.
It seems to me that all this requires is the ability to compute resultants of polynomials in two variables of relatively small degree as well as factoring polynomials in a single variable (and computing gcd's and quotients of polynomials in one variable).
This can further be improved to a better and better lower bound by analyzing more carefully the structure of the ramification at the singular points of *C* and what happens at infinity: some of the points that we threw out could in fact contribute to the arithmetic genus of *C'*, and there could be ramification "at infinity". I leave this as an exercise!
| 3 | https://mathoverflow.net/users/4344 | 19927 | 13,245 |
https://mathoverflow.net/questions/19881 | 4 | Topological manifolds of dimension ≠4 have a [Lipschitz structure](http://en.wikipedia.org/wiki/Lipschitz_continuity#Lipschitz_manifolds). [Ed: Is this "well-known"? Is it obvious? Can somebody give a reference?] Does this imply the following result?
>
> Assume M and N are smooth Riemannian manifold, with same dimension other than 4. If M homeomorphic to N, then M is bi-Lipschitz homeomorphic to N.
>
>
>
In other words, can two manifolds (of dimension ≠4) be homeomorphic without being [bi-Lipschitz homeomorphic](http://en.wikipedia.org/wiki/Lipschitz_continuity#Lipschitz_manifolds)?
| https://mathoverflow.net/users/3922 | Can two Riemannian manifolds (dim≠4) be homeomorphic without being bi-Lipschitz homeomorphic? | If you don't assume compactness, then no. Silly example: $\mathbb R^1$ and $(0,1)$. Example with complete metrics: $\mathbb R^2$ and $\mathbb H^2$ (they have essentially different volume growths and hence are not bi-Lipschitz equivalent).
If $M$ and $N$ are closed, then yes, by Sullivan's uniqueness result pointed out to by Leonid Kovalev in comments (provided that the MR review is correct - I'm not an expert in any way and don't have access to the paper). The uniqueness means that for every two Lipschitz structures there is an isomorphism between them. And for Lipschitz structures defined by Riemannian metrics, isomorphisms are a locally bi-Lipschitz homeomorphisms of the metrics. By compactness, locally bi-Lipschitz implies bi-Lipschitz.
| 4 | https://mathoverflow.net/users/4354 | 19940 | 13,252 |
https://mathoverflow.net/questions/19801 | 4 | Let $H$ and $K$ be Hilbert spaces, and let $u$ be a partial isometry in $\mathcal{B}(H \otimes K)$ between projections $p\_0 = u^\ast u$ and $p\_1 = u u^\ast$ such that $p\_0, p\_1 \leq 1 \otimes (1-q)$ for some projection $q \in \mathcal B(K)$ equivalent to $1 \in \mathcal B (K)$. Does $p\_1 u (a \otimes 1)p\_0 = p\_1 (a \otimes 1)u p\_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ imply that $u$ can be extended to a unitary operator in the commutant $(\mathcal B(H)\otimes 1)' = 1 \otimes \mathcal B(K)$?
I've verified the implication for several simple examples, but I'm having trouble proving it in full generality.
Update: I've reworded the question for clarity thanks to Yemon's comments. By a unitary operator extending $u$, I mean a unitary operator that agrees with $u$ on $p\_0 (\mathcal H \otimes \mathcal K)$.
Update: Here's one example.
Let $K$ be infinite dimensional with basis $e\_1,e\_2,\ldots$, and $H$ be $n$-dimensional with bases $\xi\_1, \ldots, \xi\_n$ and $\zeta\_1, \ldots, \zeta\_n$. Let $\xi = \frac{1}{\sqrt n}\sum\_i \xi\_i \otimes e\_i$ and $\zeta = \frac{1}{\sqrt n}\sum\_I \zeta\_i \otimes e\_i$. Define $u\colon\eta \mapsto \langle \eta, \xi \rangle \zeta$. Note that the assumptions in the first sentence are satisfied with $q$ the projection on the span of $\{e\_{n+1}, e\_{n+2}, \ldots \}$. The condition that $p\_1 u (a \otimes 1)p\_0 = p\_1 (a \otimes 1)u p\_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ is here equivalent to $\langle u (a \otimes 1) \xi, \zeta\rangle = \langle (a \otimes 1) u \xi , \zeta \rangle $ for all $a \in \mathcal B(H)$, which is in turn equivalent to $\langle(a \otimes 1)\xi, \xi \rangle = \langle (a \otimes 1) \zeta, \zeta \rangle$ for all $a \in \mathcal B(H)$. The latter equation holds because both sides are equal to the trace of $a$. Thus, all conditions are satisfied.
To see that the conclusion holds, let $K\_n$ be the span of $e\_1,\ldots,e\_n$, and let $c\_{ij}$ be the unitary $n \times n$ matrix such that $\zeta\_i = \sum\_j c\_{ij} \xi\_j$. Then $$u \xi = \zeta = \frac{1}{\sqrt n}\sum\_i \zeta\_i \otimes e\_i = \frac{1}{\sqrt n} \sum\_{ij} c\_{ij}\xi\_j \otimes e\_i = \frac{1}{\sqrt n}\sum\_j \xi\_j \otimes we\_j = (1 \otimes w) \xi,$$ where $w\colon K\_n \rightarrow K\_n$ is the unitary defined by $we\_j = \sum\_i c\_{ij}e\_i$. We extend $w$ to all of $K$ by setting $\tilde{w} = w \oplus 1$. Then $1 \otimes \tilde w$ is a unitary operator in $1 \otimes B(K)$ that extends $u$.
I realized that the other examples I was thinking about are trivial.
| https://mathoverflow.net/users/2206 | When can a partial isometry $u$ in $\mathcal B(H \otimes K)$ be extended to a unitary in $1 \otimes \mathcal B(K)$? | If I get it right this time, the following Zorn argument proves the implication.
Denote by $q\_0=\bigvee\_{v\in \mathcal{U}(H)} (v\otimes 1)p\_0(v^\ast\otimes 1)$ and similarly $q\_1=\bigvee\_{v\in \mathcal{U}(H)} (v\otimes 1)p\_1(v^\ast\otimes 1)$. Apply Zorn's lemma to the set $I=\left\{w\in \mathcal{B}(H\otimes K)\left\vert \begin{array}{l}\,\,w\text{ is a partial isometry, }\\w^\ast w\leq q\_0,\\ww^\ast\leq q\_1,\\w\text{ extends u, and }\\\!\!\!ww^\ast(v\otimes 1)w=w(v\otimes 1)ww^\ast\text{ for all }v\in\mathcal{U}(H)\end{array}\right\}\right.$
ordered by extension: $w\_1\leq w\_2$ iff $w\_2 w\_1^\ast=w\_1 w\_1^\ast$ (in other words: $w\_2$ has larger initial support and agrees with $w\_1$ on the initial support of $w\_1$)
A maximal element $w\in I$ must have $w^\ast w=q\_0$, for otherwise there would have been a unitary $v\in\mathcal{B}(H)$ such that $f=(v^\ast\otimes 1)w^\ast w(v\otimes 1)\wedge (q\_0-w^\ast w)\not=0$. Then $(v^\ast\otimes 1)w(v\otimes 1)f + w$ is a partial isometry in $I$ that extends $w$. Since $1-q\_0$ and $1-q\_1$ are equivalent projections in $1\otimes \mathcal{B}(K)$, we can extend $w$ to a unitary.
| 3 | https://mathoverflow.net/users/2055 | 19941 | 13,253 |
https://mathoverflow.net/questions/19932 | 49 | Hi,
to completely describe a classical mechanical system, you need to do three things:
-Specify a manifold $X$, the phase space. Intuitively this is the space of all possible states of your system.
-Specify a hamilton function $H:X\rightarrow \mathbb{R}$, intuitivly it assigns to each state its energy.
-Specify a symplectic form $\omega$ on $X$. What is $\omega$ intuitively? What kind of information about physics does it capture?
| https://mathoverflow.net/users/2837 | What is a symplectic form intuitively? | To elaborate on a comment of Steve Huntsman: the symplectic form turns a form $d H$ into a flow $X\_H$ with a number of properties, but other types of forms can do a similar job. Indeed, there are a number of situations in physics where the relevant $\omega$ is not symplectic, for example for the following reasons:
* $\omega$ might be degenerate in the sense that $i\_X \omega = 0$ for certain $X \ne 0$. This occurs for instance when you pull back $\omega$ to a constraint surface in phase space. Or you might be working on the Lagrangian side, taking the pull-back of $\omega$ along a non-invertible Lagrangian.
* In non-holonomic mechanics, $\omega$ is sometimes not closed, with the derivative $d \omega$ being related to the non-integrability of the constraint distribution.
The point is that such forms all lead to valid generalizations of Hamilton's equations, so using a symplectic form to write down Hamilton's equations is to a large extent motivated by the fact that it "just works". The physical properties offered by using a symplectic form instead of an arbitrary two-form are the following:
* Non-degeneracy: the evolution vector field $X\_H$ is determined uniquely by the Hamiltonian $H$. By contrast, if you have gauge freedom, there will typically be constraints in the phase space, hence a degenerate symplectic form (see above), leading to a non-unique evolution (which is what gauge freedom is --- several mathematically distinct evolutions being physically the same).
* Closedness: the system preserves the symplectic form
$$
L\_{X\_H} \omega = d i\_{X\_H} \omega + i\_{X\_H} d \omega = 0
$$
if $\omega$ is closed. In the classical literature, this gives rise to a series of conservation laws called the "Poincare invariants". Again, non-holonomic systems typically don't exhibit this property, leading to all sort of weirdness.
| 27 | https://mathoverflow.net/users/3909 | 19943 | 13,254 |
https://mathoverflow.net/questions/19942 | 0 | I have a n-by-k matrix A and a n-by-n matrix B, where B is positive definite. I can form the matrix $M = A^t B A$. Playing around, I always found $rk(M) = rk(A)$ but I can't prove this.
| https://mathoverflow.net/users/737 | Rank of ABA where B is positive definite | $A^T BAx = 0 \implies (Ax)^TB(Ax)=0 \implies Ax=0$ by positve definiteness of $B$. So $ker(M)=ker(A)$ and hence $rk(M)=rk(A)$.
| 5 | https://mathoverflow.net/users/3041 | 19954 | 13,263 |
https://mathoverflow.net/questions/19957 | 63 | My son is one year old, so it is perhaps a bit too early to worry about his mathematical education, but I do. I would like to hear from mathematicians that have older children: *What do you wish you'd have known early? What do you think you did particularly well? What do you think would be particularly bad? Is there a book (for children or parents) that you recommend?*
(This a community wiki, so please give *one* advice per answer, as usual.)
### Background
I ask here because I believe that the challenges a mathematician faces in educating a child are special. For example, at least [some websites](http://www.bbc.co.uk/schools/parents/work/primary/numeracy_and_science/maths_at_home_primary.shtml) and [books](http://books.google.com/books?id=6GxoPgAACAAJ) address the parents' fear of not knowing how to solve homework, which keeps them from becoming involved. On the contrary, I fear I might get too involved and either bore my son or make him think he likes math when in fact his skills are elsewhere.
Christos Papadimitriou [said](http://linkinghub.elsevier.com/retrieve/pii/S1574013709000185) in an interview that, even though his father was teaching math in high-school, they never discussed math. I wonder if that means his father didn't teach him how to count and I wonder if it's a good strategy. (It certainly turned out well in one case.)
Timothy Gowers (in [Mathematics, a very short introduction](http://books.google.com/books?id=DBxSM7TIq48C)) says that it was inappropriate to explain to his son, who was six, the concept 'zero' using the group axioms. (Or something to this effect, I don't have the book near to check.) That was surprising to me, because I wouldn't have thought that I need to restrain myself from mentioning abstract concepts. (**Update.** Here's the quote: "[The non-abstract] way of thinking makes it hard to answer questions such as the one asked by my son John (when six): how can nought times nought be nought, since nought times nought means that you have *no* noughts? A good answer, though not one that was suitable at the time, is that it can be deduced from the [field axioms] as follows. [...]")
There is a [somewhat related Mathoverflow question](https://mathoverflow.net/questions/5139/how-have-mathematicians-been-raised). This one is different, because I'm looking for advice (rather than statistics/anecdotes) and because my goal is to give my son a good math education (rather than to make him a mathematician). I also found an [online book](http://www.math.com/parents/articles/helpmath.html) that seems to give particularly good generic advice. Here I'm looking more for advice geared towards parents that are mathematicians.
In short, I'm looking for *specific* advice on how a mathematician should approach his/her child's math education, especially for the 1 to 10 age range.
| https://mathoverflow.net/users/840 | How do you approach your child's math education? | I would recommend a great book by [Alexandre Zvonkine](http://www.labri.fr/perso/zvonkin/), "Math for little ones", but it is only available in Russian ([here](http://www.mccme.ru/free-books/zvonkine/zvonkine.html)); however, two articles by Zvonkine which were published earlier are translated into English, see [here](http://www.de.ufpe.br/~toom/others_articles/engeduc/Zvon-eng.pdf). (You might also want to check other materials [linked](http://www.de.ufpe.br/~toom/others_articles/engeduc/index.htm) on Andrei Toom's webpage.)
And, as a tiny bit of more general advice, you know about Piaget's works, right? They are highly relevant when trying to teach kids anything at all, I think.
| 19 | https://mathoverflow.net/users/1306 | 19968 | 13,273 |
https://mathoverflow.net/questions/19914 | 6 | It's well known that if E is a vector bundle with Chern roots $a\_1,\ldots, a\_r$,
then the Chern roots of the $p$th exterior power of E consist of all sums of $k$ distinct $a\_i$'s. I would like to say the same is true if E is just a torsion-free coherent sheaf on $P^n$. It seems non-obvious, though, maybe because an exterior power isn't generally an additive functor.
Presumably this is either false or also well known, but I can't find a reference.
| https://mathoverflow.net/users/5045 | A reference: the splitting principle for exterior powers of coherent sheaves? | My guess would be that the formula you want does not extend to the case of coherent sheaves. As indicated in Mariano and David answers (which has unfortunately been deleted), the best hope to compute is via a resolution $\mathcal F$ of $E$ by vector bundles. In general, for 2 perfect complexes $\mathcal F, \mathcal G$ of vector bundles, there is a formula for the localized chern classes
$$ch\_{Y\cap Z}(\mathcal F \otimes \mathcal G) = ch\_Y(\mathcal F)ch\_Z(\mathcal G)$$, with $Y,Z$ being the respective support. Unfortunately, this only gives the right formula for the "derived tensor product".
So to mess up the formula, one can pick $E$ such that $Tor^i(E,E)$ are non-trivial. I think an ideal sheaf of codimension at least 2 would be your best bet for computation purpose.
| 3 | https://mathoverflow.net/users/2083 | 19979 | 13,281 |
https://mathoverflow.net/questions/19422 | 3 | Suppose $X$ is a simplicial space of dimension $M$ (i.e. all simplices above dimension $M$ are degenerate). The claim is:
$|X|$ is compact. iff $X\_n$ is compact for each $n$.
Suppose each $X\_n$ is compact. Then $|X|$ is by definition a quotient of a compact space (you don't have to include the simplices above dimension $M$ in the realization). I wonder, whether the converse is true.
Here is one motivating example. Equip the unit interval with the structure of a simplicial space in the following way:
Let $X\_0$ be the Cantor-set and let the nondegenerate simplices in $X\_1$ are just all the intervals, that get removed in the construction of the cantor set.
One can regard $X\_1$ as a subspace of $[0;1]$ using the map (of sets) $X\_1\rightarrow [0;1]$ , that sends every point in the Cantor set to the corresponding point in the unit interval and that sends each of the small intervals to its barycenter. Equip $X\_1$ with the subspace topology using this map.
The geometric realization of this space is the unit interval, which is compact and $X\_0,X\_1$ are also compact ($X\_1$ is a closed subset of the unit interval).
This question arose in the context of this [question](https://mathoverflow.net/questions/19232/is-the-realization-of-a-proper-map-of-simplicial-spaces-proper). I realized, that I don't have a good criterium to say, when a subspace of the geometric realization of a simplicial space is compact.
| https://mathoverflow.net/users/3969 | When is the realization of a simplicial space compact ? | Ok I checked the idea mentioned in the comments above.
As $X\_n$ is a closed subspace of $X\_M$ (use any degeneracy; it has a left inverse; composing both the other way round yields a projection and projections have closed images in the Hausdorff setting), it is enough to show, that $X\_M$ is compact.
So consider the "generalized midpoint map"
$i: X\_M\rightarrow \coprod\_{n\le M} X\_n\times \Delta^n\rightarrow |X|\qquad x\mapsto [x,c]$, where $c\in\Delta^M$ denotes the barycenter.
For example, if $M=2$ and $x$ is a nondegenerate one simplex in $X$, there are two degeneracies $s\_0,s\_1:X\_1\rightarrow X\_2$. Then one gets in |X|:
$i(s\_0(x))=[s\_0(x),\frac{1}{3},\frac{1}{3},\frac{1}{3}]=[x,\frac{2}{3},\frac{1}{3}]$
$i(s\_1(x))=[s\_1(x),\frac{1}{3},\frac{1}{3},\frac{1}{3}]=[x,\frac{1}{3},\frac{2}{3}]$
So in the realisation one really picks several midpoints of a simplex, that doesn't have maximal dimension.
But the map $i$ is still injective. Any simplex $x\in X\_M$ might be written in a unique way as $x=s(y)$, where $y\in X\_n$ is nondegenerate and $s$ is a degeneracy map (it corresponds to a surjective map $[M]\rightarrow [n]$ in $\Delta^{Op}$, $\Delta^{Op}$ can be identified with the category whose objects are the sets $\{0,\ldots,n\}$ and whose morphisms are nondecreasing maps).
Then $i(x)=[y,s(c)]=[y,\frac{s^{-1}(1)}{3},\ldots,\frac{s^{-1}(1)}{3}]$. Now the right side has normal form (There is a normal form for points in $|X|$. This can be found in every book about simplicial sets and it works the same way for simplicial spaces). Given any other $x'=s'(y')$ with $i(x)=i(x')$, we get
$[y,\frac{|s^{-1}(1)|}{M+1},\ldots,|\frac{s^{-1}(1)|}{M+1}]=[y',\frac{|s'^{-1}(1)|}{M+1},\ldots,\frac{|s'^{-1}(n)|}{M+1}]$. As this is in normal form, we get $y=y'$ and $|s'^{-1}(i)|=|s^{-1}(i)|$. But such nondecreasing maps are characterized by the size of the preimages, so $s=s'$ and hence $x=x'$. So the map $i$ is injective.
We still have to show, that $i$ is a homeo onto its image and that the image of $i$ is closed. Then $X\_M$ is a closed subspace of a compact space and hence compact. Both will follow from the next claim:
For any closed subset $A\subset X\_M$ the saturation $A'$ of $A\times \{c\}$ is still a closed subset of $\coprod\_{n\le M} X\_n\times \Delta^n$. This is an even worse calculation than the last one. One has to consider for each degeneracy map $s:X\_N\rightarrow X\_M$ the preimage $s^{-1}(A)\times \{s(c)\}$ and the set $B\_s$ of all points that can be simplified to one of those points. Then one can show, that $B\_s$ is closed and $A'=\bigcup\_s B\_s$. As there are only finitely many degeneracy maps (below dim $M$) this is a finite union and hence closed.
| 0 | https://mathoverflow.net/users/3969 | 19981 | 13,282 |
https://mathoverflow.net/questions/19984 | 23 | My question is as follows.
Do the Chern classes as defined by Grothendieck for smooth projective varieties coincide with the Chern classes as defined with the aid of invariant polynomials and connections on complex vector bundles (when the ground field is $\mathbf{C}$)?
I suppose GAGA is involved here. Could anybody give me a reference where this is shown as detailed as possible? Or is the above not true?
Some background on my question:
Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any integer $r$, let $A^r X$ be the group of cycles of codimension $r$ rationally equivalent to zero. Let $AX=\bigoplus A^r X$ be the Chow ring.
Grothendieck proved the following theorem on Chern classes.
There is a unique "theory of Chern classes", which assigns to each locally free coherent sheaf $\mathcal{E}$ on $X$ an $i$-th Chern class $c\_i(\mathcal{E})\in A^i(X)$ and satisfies the following properties:
**C0**. It holds that $c\_0(\mathcal{E}) = 1$.
**C1**. For an invertible sheaf $\mathcal{O}\_X(D)$ on $X$, we have that $c\_1(\mathcal{O}\_X(D)) = [D]$ in $A^1(X)$.
**C2**. For a morphism of smooth quasi-projective varieties $f:X\longrightarrow Y$ and any positive integer $i$, we have that $f^\ast(c\_i(\mathcal{E})) =c\_i(f^\ast(\mathcal{E}))$.
**C3**. If $$0\longrightarrow \mathcal{E}^\prime \longrightarrow \mathcal{E} \longrightarrow \mathcal{E}^{\prime\prime} \longrightarrow 0$$ is an exact sequence of vector bundles on $X$, then $c\_t(\mathcal{E}) = c\_t(\mathcal{E}^\prime)c\_t(\mathcal{E}^{\prime\prime})$ in $A(X)[t]$.
So that's how it works in algebraic geometry. Now let me sketch the complex analytic case.
Let $E\longrightarrow X$ be a complex vector bundle. We are going to associate certain cohomology classes in $H^{even}(X)$ to $E$. The outline of this construction is as follows.
**Step 1**. We choose a connection $\nabla^E$ on $E$;
**Step 2**. We construct closed even graded differential forms with the aid of $\nabla^E$;
**Step 3**. We show that the cohomology classes of these differential forms are independent of $\nabla^E$.
Let us sketch this construction. Let $k= \textrm{rank}(E)$. Let us fix an invariant polynomial $P$ on $\mathfrak{gl}\_k(\mathbf{C})$, i.e. $P$ is invariant under conjugation by $\textrm{GL}\_k(\mathbf{C})$.
Let us fix a connection $\nabla^E$ on $E$. We denote denote its curvature by $R^E = (\nabla^E)^2$. One shows that $$R^E \in \mathcal{C}^\infty(X,\Lambda^2(T^\ast X)\otimes \textrm{End}(E)).$$ That is, $R^E$ is a $2$-form on $X$ with values in $\textrm{End}(E)$. Define $$P(E,\nabla^E) = P(-R^E/{2i\pi}).$$ (This is well-defined.)
The Chern-Weil theorem now says that:
The even graded form $P(E,\nabla^E)$ is a smooth complex differential form which is closed. The cohomology class of $P(E,\nabla^E)$ is independent of the chosen connection $\nabla^E$ on $E$.
Choosing $P$ suitably, we get the Chern classes of $E$ (by definition). These are cohomology classes. In order for one to show the equivalence of these "theories" one is forced to take the leap from the Chow ring to the cohomology ring.
How does one choose $P$? You just take $P(B) = \det(1+B)$ for a matrix.
Motivation: If one shows the equivalence of these two theories one gets "two ways" of "computing" the Chern character.
| https://mathoverflow.net/users/4333 | GAGA and Chern classes | See [this question](https://mathoverflow.net/questions/13813/construction-of-the-stiefel-whitney-and-chern-classes).
Also, read Milnor-Stasheff or Hatcher's book "Vector Bundles and K-Theory". In particular, Milnor-Stasheff and Hatcher prove that there is a unique "theory of Chern classes" for complex vector bundles over topological spaces satisfying axioms totally analogous to the C0, C1, C2, C3. Milnor-Stasheff constructs Chern classes using the Thom isomorphism theorem; Hatcher constructs them using the Leray-Hirsch theorem. Hatcher's construction (in topology) is essentially the same as Grothendieck's construction (in algebraic geometry). I think if you study the two constructions (Hatcher's and Grothendieck's) carefully, their equivalence should follow fairly easily. I did this once a while ago.
I don't think you need any GAGA theorem. I think you just need the fact that there is an analytification functor.
Appendix C of Milnor-Stasheff then proves the equivalence with the Chern-Weil theory.
| 23 | https://mathoverflow.net/users/83 | 19988 | 13,286 |
https://mathoverflow.net/questions/19999 | 19 | Is it possible, for an arbitrary polynomial in one variable with integer coefficients, to determine the roots of the polynomial in the Complex Field to arbitrary accuracy? When I was looking into this, I found some papers on homotopy continuation that seem to solve this problem (for the Real solutions at least), is that correct? Or are there restrictions on whether homotopy continuation will work? Does the solution region need to be bounded?
| https://mathoverflow.net/users/5068 | Finding all roots of a polynomial | This argument is problematic; see Andrej Bauer's comment below.
---
Sure. I have no idea what an efficient algorithm looks like, but since you only asked whether it's possible I'll offer a terrible one.
**Lemma:** Let $f(z) = z^n + a\_{n-1} z^{n-1} + \cdots + a\_0$ be a complex polynomial and let $R = \max(1, |a\_{n-1}| + \cdots + |a\_0|)$. Then all the roots of $f$ lie in the circle of radius $R$ centered at the origin.
*Proof.* If $|z| > R$, then $|z|^n > R |z|^{n-1} \ge |a\_{n-1} z^{n-1}| + \cdots + |a\_0|$, so by the triangle inequality no such $z$ is a root.
Now subdivide the disk of radius $R$ into, say, a mesh of squares of side length $\varepsilon > 0$ and evaluate the polynomial at all the lattice points of the mesh. As the mesh size tends to zero you'll find points that approximate the zeroes to arbitrary accuracy.
There are also lots of specialized algorithms for finding roots of polynomials [at the Wikipedia article](http://en.wikipedia.org/wiki/Root-finding_algorithm#Finding_roots_of_polynomials).
| 16 | https://mathoverflow.net/users/290 | 20002 | 13,292 |
https://mathoverflow.net/questions/19910 | 9 | As will become clear, this is in some sense a follow up on my earlier question [Why should I prefer bundles to (surjective) submersions?](https://mathoverflow.net/questions/17826/). As with that one, I hope that it's not too open-ended or discussion-y. If y'all feel it is too discussion-y, I will happily close it.
Let $\rm Man$ be the category of smooth (finite-dimensional) manifolds. I can think of (at least) two natural "smooth structures" on $\rm Man$, which I will outline. My question is whether one of these is the "right" one, or if there is a better one.
I should mention first of all that there many subtly different definitions of "smooth structure" — see e.g. [n-Lab: smooth space](http://ncatlab.org/nlab/show/smooth+space) and [n-Lab: generalized smooth space](http://ncatlab.org/nlab/show/generalized+smooth+space) and the many references therein — and I don't know enough to know which to prefer. Moreover, I haven't checked that my proposals match any of those definitions.
In any case, the definition of "smooth structure" that I'm happiest with is one where I only have to tell you what all the **smooth curves** are (and these should satisfy some compatibility condition). So that's what I'll do, but I'm not sure if they *do* satisfy the compatibility conditions. Without further ado, here are two proposals:
1. A **smooth curve** in $\rm Man$ is a fiber bundle $P \to \mathbb R$.
2. A **smooth curve** in $\rm Man$ is a submersion $Y \to \mathbb R$.
Then given a manifold $M$, we can make it into a category by declaring that it has only identity morphisms. Then I believe that the **smooth functors** $M \to {\rm Man}$ under definition 1 are precisely the fiber bundles over $M$, whereas in definition 2 they are precisely the submersions over $M$.
(Each of these claims requires checking. In the first case, it's clear that bundles pull back, so all bundles are smooth functors, and so it suffices to check that if a surjective submersion to the disk is trivializable over any curve, then it is trivializable. In the second case, it's clear that if a smooth map restricts to a submersion over each curve, then it is a submersion, so any smooth functor in a submersion, and so one must check that submersions pull back along curves.)
I can see arguments in support of either of these. On the one hand, bundles are cool, so it would be nice if they were simply "smooth functors". On the other hand, we should not ask for smooth functions (i.e. 0-functors) to be necessarily "locally trivializable", as then they'd necessarily be constant. Maybe the correct answer is definition 2, and that bundles are "locally constant smooth functors", or something.
Anyway, thoughts? Or am I missing some other good definition?
### Addendum
In the comments, folks have asked for applications, which is very reasonable. The answer is that I would really like to have a good grasp of words like "smooth functor", at least in the special case of "smooth functor to $\rm Man$". Of course, Waldorf and Shreiber have explained these words in certain cases in terms of local gluing data (charts), but I expect that a more universal definition would come directly from a good notion of "smooth structure" on a category directly.
Here's an example. Once we have a smooth structure on $\rm Man$, we can presumably talk about smooth structures on subcategories, like the category of $G$-torsors for $G$ your favorite group. Indeed, for the two definitions above, I think the natural smooth structure on $G\text{-tor}$ coincide: either we want fiber bundles where all the fibers are $G$-torsors, or submersions where all the fibers are $G$-torsors, and in either case we should expect that the $G$ action is smooth. So then we could say something like: "A principle $G$-bundle on $M$ is (i.e. there is a natural equivalence of categories) a smooth functor $M \to G\text{-tor}$", where $M \rightrightarrows M$ is the (smooth) category whose objects are $M$ and with only trivial morphisms. (Any category object internal to $\rm Man$ automatically has a smooth structure.) And if I understood the [path groupoid mod thin homotopy](https://mathoverflow.net/questions/17803/what-is-the-infinite-dimensional-manifold-structure-on-the-space-of-smooth-paths) $\mathcal P^1(M) \rightrightarrows M$ as a smooth category, then I would hope that the smooth functors $\mathcal P^1(M) \to G\text{-tor}$ would be the same as principle $G$-bundles on $M$ with connection. Functors from the groupoid of paths mod "thick" homotopy should of course be bundles with flat connections. Again, [Schreiber and Waldorf](http://arxiv.org/abs/0705.0452) have already defined these things categorically, but their definition is reasonably long, because they don't have smooth structures on $\rm Man$ that are strong enough to let them take advantage of general smooth-space yoga.
Here's another example. When I draw a bordism between manifolds, what am I actually drawing? I would like to say that I'm drawing something close to a "smooth map $[0,1] \to \rm Man$". I'm not quite, by my definitions — if you look at the pair of pants, for instance, at the "crotch" it is not a submersion to the interval. So I guess there's at least one more possible definition of "smooth curve in $\rm Man$":
* A **smooth curve** in $\rm Man$ is a smooth map $X \to \mathbb R$.
But this, I think, won't be as friendly a definition as those above: I bet that it does not satisfy the compatibility axioms that your favorite notion of "smooth space" demands.
| https://mathoverflow.net/users/78 | What's the "correct" smooth structure on the category of manifolds? | I think that the most interesting part of your question is the part you put in parentheses!
>
> (and these should satisfy some compatibility condition)
>
>
>
What are your compatibility conditions? That is **everything** here. If you specify the correct conditions, you may find that all your definitions collapse to just one.
I have an issue with Konrad's answer (which I doubt very much that he will be surprised to hear me express!). Whenever I heard words like "Grothendieck topology" or "sheaves" or encoding similar ideas then I feel that something's been lost. I don't like the idea that "smooth" is just "really nice continuity". "Smooth" sits alongside continuity and can be expressed in a different way which is extremely simple: takes smooth curves to smooth curves.
Of course, I would say that, as everyone by now presumably knows that I prefer [Frolicher spaces](http://ncatlab.org/nlab/show/Froelicher+space) to the other variants (like Chen spaces or diffeological spaces, see [generalized spaces](http://ncatlab.org/nlab/show/generalized+smooth+space) for links). It is interesting that Chen's third definition (by my count) was stronger than his eventual sheaf condition and was more along the lines of "a map is smooth if enough tests say that it is smooth".
But Frolicher spaces have a problem, which is that it is extremely difficult to prise them away from being a set-based theory. The compatibility condition is so strong that it forces an underlying set. I'd really like to figure out how to make this extension, and I know that Urs would as well. If I could just encourage you and Konrad over to the nLab to play around with these ideas to see how they could work ...
If you want to study The Smooth, The Whole Smooth, and Nothing But The Smooth, then you should do so and not flirt continually with continuity. The stronger compatibility condition means that more stuff is smooth than you first thought (witness my recent question on this) and **that makes it interesting**! The unexpected happens, so *study it*!
This isn't much of an answer so far, it's more of a commentary on your question which (as is usual for me) is too long for an actual comment. So let me end with an actual answer (which I freely confess that I stole from a rabbi):
>
> That is such a great question, *why on earth would you want an answer*?
>
>
>
| 5 | https://mathoverflow.net/users/45 | 20004 | 13,294 |
https://mathoverflow.net/questions/20020 | 17 | I apologize that this is vague, but I'm trying to understand a little bit of the historical context in which the zoo of quantum invariants emerged.
For some reason, I have in my head the folklore:
The discovery in the 80s by Jones of his new knot polynomial was a shock because people thought that the Alexander polynomial was the only knot invariant of its kind (involving a skein relation, taking values in a polynomial ring, ??). Before Jones, there were independent discoveries of invariants that each boiled down to the Alexander polynomial, possibly after some normalization.
Is there any truth to this? Where is this written?
| https://mathoverflow.net/users/813 | Who thought that the Alexander polynomial was the only knot invariant of its kind? | The skein relation approach to knot invariants was not very popular before the Jones polynomial. The Alexander polynomial was thought of as coming from homology (of the cyclic branched cover); Conway had found the skein relation, but it was not well-known. Of course once you start investigating skein relations systematically, you rapidly find the Jones, Kauffman, and HOMFLY relations.
Basically, people had been looking for invariants using their standard tools like homology, and had trouble constructing interesting invariants that way. The idea of just looking for a skein relation was new. The notion of "polynomial invariants" by itself is too vague to give a place to look.
| 24 | https://mathoverflow.net/users/5010 | 20022 | 13,305 |
https://mathoverflow.net/questions/20025 | 21 | ### Background
Let $A$ be a subset of a topological space $X$. An old problem asks, by applying various combinations of closure and complement operations, how many distinct subsets of $X$ can you describe?
The answer is 14, which follows from the observations that $Cl(Cl(A))=Cl(A)$, $\neg(\neg (A)=A$ and the slightly harder fact that
$$ Cl(\neg (Cl(\neg (Cl(\neg (Cl(A))))=Cl(\neg (Cl(A))$$
where $Cl(A)$ is the closure of $A$ and $\neg (A)$ is the complement. This makes every expression in $Cl$ and $\neg$ equivalent to one of 14 possible expressions, and all that remains is to produce a specific choice of $A$ which makes all 14 possiblities distinct.
This problem goes by the name [Kuratowski's closure-complement problem](http://en.wikipedia.org/wiki/Kuratowski%2527s_closure-complement_problem), since it was first stated and solved by Kuratowski in 1922.
### The Problem
A very similar problem recently came up in a discussion that was based on a topological model for modal logic (though the logical connection is unrelated to the basic question). The idea was to take a subset $A$ of $\mathbb{R}$, and to consider all possible expressions on $A$ consisting of closure, complement, and intersection. To be clear, we are allowed to take the complement or closure of any subset we have already constructed, and we are allowed to intersect any two subsets we have already constructed.
The question: **Is this collection of subsets always finite?**
A potentially harder question: **If there are multiple starting subsets $A\_1$, $A\_2$, $A\_3$... (finite in number), is this collection of subsets always finite?**
The first question is essentially the Kuratowski question, with the added operation of intersection. There is also the closely related (but slightly stronger) question of whether there are a finite number of *formally* distinct expressions on an indeterminant subset $A$ (or a collection of subsets $A\_1$, $A\_2$...).
### Some Thoughts
My guess to both questions is yes, but the trick is showing it. I can take an example of a set $A$ which realizes all 14 possibilities from Kuratowski's problem, and show that the collection of distinct subsets I can construct from it is finite. However, just because such a set captures all the interesting phenomena which can happen when closing and complementing, doesn't mean this example is missing a property that is only important when intersecting.
It also seems difficult to approach this problem formally. The problem is that there are many non-trivial intersections of the 14 expressions coming from Kuratowski's theorem. Then, each of these intersections could potentially have its own new set of 14 possible expressions using closures and complements. In examples, the intersections don't contribute the full number of 14 new sets, but its hard to show this aside from case-by-case analysis.
| https://mathoverflow.net/users/750 | The Closure-Complement-Intersection Problem | From one set, you can generate infinitely many sets.
Let A be a closed set of infinite [Cantor-Bendixon rank](http://en.wikipedia.org/wiki/Derived_set_(mathematics)). That is, the successive finite Cantor-Bendixon derivatives A', A'', and so on are all distinct. The rank of an element x in A is the least n such that x becomes isolated in An, where A0 = A and An+1 = A', the set of limit points of An. Thus, the rank 0 elements are the isolated points of A, and the rank 1 elements are the limits of these points which are not limits of limit points of A, and so on.
Now, let B be the subset of A consisting of the elements of A having even rank. This includes all isolated points of A, but not their limits that are not limits of limits, but does include the limits of limits (as long as they are not limits-of-limits-of-limits) and so on. Define the operation B+ = cl(B) - B, which is obtainable from your operations. Note that cl(B) = A, and that B+ consists of all elements of A having odd rank in A. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in cl(B+) = A', which is exactly those elements of A having even rank at least 2 in A. And so on. The set B+n will consist of all elements of A having rank at least n+1, which have even/odd rank depending on the parity of n. Since A has infinite rank, these sets will all be distinct.
Thus, the set B generates infinitely many distinct sets B, B+, B++, B+++, etc.
| 12 | https://mathoverflow.net/users/1946 | 20026 | 13,308 |
https://mathoverflow.net/questions/20019 | 5 | Say I have $X\_{ij}$, $j \le i$ with the property that $X\_{ij}$ are centered and identically distributed and $E(X\_{ij} X\_{ij'}) = o(\exp(-i)))$. Then does $\sum\_j X\_{ij}$ have Gaussian domain of attraction?
As a related question, if $X\_1, X\_2, X\_3$ are identically distributed and centered and $E(X\_i X\_j) = c$, what bound can I get for $E(X\_1 X\_2 X\_3)$ in terms of $c$?
| https://mathoverflow.net/users/4923 | Does central limit theorem hold for general weakly dependent variables? | Not necessarily. One has to impose more restrictive mixing and moment conditions. A classical book is:
Ibragimov I.A., Linnik Yu.V. *Independent and stationary sequences of random variables*
There is a long-standing question asked by Ibragimov: is $\phi$-mixing and finiteness of second moment sufficient for CLT to hold for a stationary sequence?
Also, there are various concepts of dependence. For example, if your r.v.'s are associated (i.e. satisfy FKG inequalities) and the covariance decays as you describe, then CLT holds.
UPD. As for the second part of your question: you cannot estimate higher-order moments in terms of lower-order ones unless the joint distributions have some special structure.
| 6 | https://mathoverflow.net/users/2968 | 20030 | 13,311 |
https://mathoverflow.net/questions/19989 | 7 | My primary motivation for asking this question comes from the discussion taking place in the comments to [What is a symplectic form intuitively?](https://mathoverflow.net/questions/19932/).
Let $M$ be a smooth finite-dimensional manifold, and $A = \cal C^\infty(M)$ its algebra of smooth functions. A *derivation* on $A$ is a linear map $\{\}: A \to A$ such that $\{fg\} = f\{g\} + \{f\}g$ (multiplication in $A$). Recall that all derivations factor through the de Rham differential, and so: **Theorem:** Derivations are the same as vector fields.
A *biderivation* is a linear map $\{,\}: A\otimes A \to A$ such that $\{f,-\}$ and $\{-,f\}$ are derivations for each $f\in A$. By the same argument as above, biderivations are the same as sections of the tensor square bundle ${\rm T}^{\otimes 2}M$. *Antisymmetric* biderivations are the same as sections of the exterior square bundle ${\rm T}^{\wedge 2}M$. A *Poisson structure* is an antisymmetric biderivation such that $\{,\}$ satisfies the Jacobi identity.
Recall that sections of ${\rm T}^{\otimes 2}M$ are the same as vector-bundle maps ${\rm T}^\*M \to {\rm T}M$. A *symplectic structure* on $M$ is a Poisson structure such that the corresponding bundle map is an isomorphism. Then its inverse map makes sense as an antisymmetric section of ${\rm T^\*}^{\otimes 2}M$, i.e. a differential 2-form, and the Jacobi identity translates into this 2-form being closed. So this definition agrees with the one you may be used to of "closed nondegenerate 2-form".
>
> **Question:** Is there a "purely algebraic" way to test whether a Poisson structure is symplectic? I.e. a way that refers only to the algebra $A$ and not the manifold $M$?
>
>
>
For example, it is necessary but not sufficient that $\{f,-\} = 0$ implies that $f$ be locally constant, where I guess "locally constant" means "in the kernel of every derivation". The easiest way that I know to see that it is necessary is to use Darboux theorem to make $f$ locally a coordinate wherever its derivative doesn't vanish; it is not sufficient because, for example, the rank of the Poisson structure can drop at points.
Please add tags as you see fit.
| https://mathoverflow.net/users/78 | How can I tell whether a Poisson structure is symplectic "algebraically"? | In the purely algebraic setting, Daniel Farkas proved in his beautiful paper [Farkas, Daniel R. Characterizations of Poisson algebras. Comm. Algebra 23 (1995), no. 12, 4669--4686. [MR1352562](http://www.ams.org/mathscinet-getitem?mr=MR1352562)] that a Poisson-simple linear Poisson algebra over an algebraically closed field is a regular symplectic domain, a partial converse of the much easier fact that a commutative regular affine domain which is symplectic is Poisson-simple. There are examples of non symplectic Poisson-simple polynomial algebras, though.
| 3 | https://mathoverflow.net/users/1409 | 20032 | 13,313 |
https://mathoverflow.net/questions/20028 | 1 | This question comes from the 4th line of the proof of Theorem E of Halmos' "Measure Theory", in page 25, which says that **C** is a sigma-ring. Because this website does not allow new users to link images, I rephrase it as follows: Suppose A is any subset of the whole space X, E is any collection of subsets of X, S(E) denotes the sigma-ring generated by E, $E\cap A$ means the collection formed by all intersections of elements from E with A. Then the collection of all sets of the form $B\cup(C-A)$ where B is from S($E\cap A$) and C is from S(E) is a sigma-ring.
I just can not prove this because I can not make up the difference $[B1\cup(C1-A)]-[B2\cup(C2-A)]$ into a form of $B\cup(C-A)$. Could you please help me prove this statement? Thanks!
| https://mathoverflow.net/users/5072 | A question about sigma-ring | How about choosing $B = B\_1 - B\_2 $ and $C = C\_1 -C\_2$, we have $ B \in S(E \cap A)$ by property of $ S(E \cap A) $, and $ C \in S(E) $, by property of $S(E)$.
The fact that this works can be seen by observing that both $B1 \cup (C1−A)$, and $B2\cup (C2−A)$ are in fact both disjoint unions. $B\_1,B\_2$ has everything to do with $A$ and $C\_1 -A, C\_2 -A$ has nothing to do with $A$ at all.
ps: I don't know why you use the and and or symbol here instead of intersection and union symbol.
| 1 | https://mathoverflow.net/users/2701 | 20038 | 13,319 |
https://mathoverflow.net/questions/19997 | 1 | Let $f\in L^1(R)$ such that $F(f)$ is odd, where $F$ is the Fourier transform. Can I then say that $f$ is odd?
If $F(f)$ is odd, then
$\int \cos(x\xi) f(x) dx = 0 \:\:\forall \xi\in R$
Can I deduce from it that $f$ is odd?
| https://mathoverflow.net/users/4928 | If the fourier transformed of f is odd, is f odd? | Write $f^-(x)=f(-x)$ etc. Then $F(f^-)=F(f)^-$. If $F(f)$ is odd then
$$0=F(f)+F(f)^-=F(f+f^-).$$
The only $L^1$ function with zero Fourier transform is $0$ so that $f+f^-=0$,
that is, $f$ is odd.
| 12 | https://mathoverflow.net/users/4213 | 20042 | 13,322 |
https://mathoverflow.net/questions/20039 | 1 | I recently tried to prove the following characterization of [chordal graphs](http://en.wikipedia.org/wiki/Chordal_graph), attributed to Fulkerson & Gross:
"A graph $G$ is chordal if and only if it has an ordering such that for all $v \in G$, all the neighbors of $v$ that precede it in the ordering form a clique."
I believe the wikipedia page calls this (or its reverse) a "perfect elimination ordering." In any case, my proof was harder than I expected and I ended up with the following strengthening:
"$G$ is chordal if and only if for any $v \in G$, we can find such an ordering of $G$ that starts with $v$."
I would like to know if this strengthening is:
1) known / obvious from the weaker theorem / obvious from folklore, or
2) wrong. =)
I can of course give a proof to anyone who is interested - it is just too long to fit here.
Best,
-Yan
| https://mathoverflow.net/users/81883 | characterization of chordal graphs | It's known/obvious from the next few lines of the Wikipedia article stating that a LexBFS ordering (or its reverse, depending on your conventions) gives a perfect elimination ordering, since LexBFS can be made to start at any vertex.
| 2 | https://mathoverflow.net/users/440 | 20043 | 13,323 |
https://mathoverflow.net/questions/20046 | 4 | Do paracompact non-Hausdorff spaces admit partions of unity? I'm just curious.
| https://mathoverflow.net/users/4528 | Paracompact but not Hausdorff | The answer is no.
Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.
EDIT: I think the question is a kind of ["duplicate"](https://mathoverflow.net/questions/19219/smooth-manifolds-that-dont-admit-a-partition-of-unity) .
Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.
First the definition:
The line with two origins is the quotient space of two copies of the real line
$\mathbb{R} \times {a}$ and $\mathbb{R} \times {b}$.
with equivalence relation given by
$(x,a) \sim (x,b)\text{ if }x \neq 0$.
Since all neighbourhoods of $0\_a$ intersect all neighbourhoods of $0\_b$, it is non-Hausdorff.
However, this space is paracompact, since $\mathbb{R}$ is paracompact.
For the non-existence of a partition of unity: take the open covering $ U = (-\infty,0) \cup \{ 0\_a \} \cup (0,\infty)$ and $\tilde{U} = (-\infty,0) \cup \{ 0\_b \} \cup (0,\infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )
| 7 | https://mathoverflow.net/users/675 | 20051 | 13,328 |
https://mathoverflow.net/questions/20045 | 31 | When I was teaching calculus recently, a freshman asked me the conditions of the Riemann integrability of composite functions.
For the composite function $f \circ g$, He presented three cases:
1) both $f$ and $g$ are Riemann integrable;
2) $f$ is continuous and $g$ is Riemann integrable;
3) $f$ is Riemann integrable and $g$ is continuous.
For case 1 there is a counterexample using Riemann function. For case 2 the proof of the integrability is straight forward. However, for case 3, I can neither give a proof nor construct any counterexample. Even under the condition that $g$ is differentiable, I cannot work out anything. How to reply my student?
| https://mathoverflow.net/users/3926 | About the Riemann integrability of composite functions | Let $f$ be bounded and discontinuous on exactly the Cantor set $C$ (for example, the characteristic function of $C$). Let $g$ be continuous increasing on $[0,1]$ and map a set of positive measure (for example a fat Cantor set) onto $C$. Then $f \circ g$ is discontinuous on a set of positive measure. So $f$ is Riemann integrable, $g$ is continuous, and $f \circ g$ is not Riemann integrable. Of course, a Freshman calculus student wont know about "measure zero" so this example is not good for an elementary course.
| 29 | https://mathoverflow.net/users/454 | 20063 | 13,338 |
https://mathoverflow.net/questions/20059 | 2 | This is from an exercise in Koch, Vainsencher - An invitation to quamtum cohomology.
**Background**
The exercise asks to compute the 3-points Gromov-Witten invariants of the Grassmannian $G = \mathop{Gr}(1, \mathbb{P}^3) = \mathop{Gr}(2, 4)$ via the enumerative interepretation. In particular my problem is with computing the invariant $I\_2(p \cdot p \cdot p)$, where $p$ is the class of a point on $G$. This is the number of rational curves of degree $2$ through $3$ generic points on $G$. Here we see $G$ as embedded by the Plucker map, and the degree is defined accordingly.
A rational curve $C \subset G$ of degree $d$ will sweep out a rational ruled surface $S$ of degree $d$ in $\mathbb{P}^3$; up to here I agree with the hints of the book. The problem is the following hint:
>
> Show that the condition on $C$ of passing through a point $q \in G$ corresponds to the condition on $S$ of containing the line in $\mathbb{P}^3$ corresponding to $q$.
>
>
>
This seems to me plain false. Of course one implication is true, but is absolutely possible that $S$ contains a line without $C$ passing through the corresponding point.
For instance, when $d = 1$, $C$ is a line on the Grassmannian, and it is well-known that these have the form $\{ \ell \mid a \in \ell \subset A \}$, where $a$ is a point and $A$ a plane of $\mathbb{P}^3$. In this case $S$ is the plane $A$, so it contains many lines which do not pass through $a$, hence these lines are not parametrized by $C$.
Similarly, when $d = 2$, the surface $S$ can be a smooth quadric, which has two distinct rulings of lines; one will correspond to lines parametrized by $C$, but the other one will not. To see that a smooth quadric can actually arise, just invert the construction. Starting from a smooth quadric $S$ take any line $\ell \subset S$. There is a natural map $\ell \to G$ given by sending a point $q \in \ell$ to the unique line in the other ruling passing through $q$. The image of this map is a curve $C \subset G$, such that the associated surface is $S$ itself.
Given the hint, the book goes on to say
>
> Show that $I\_2(p \cdot p \cdot p) = 1$, by interpreting this number as a count of quadrics containing three lines.
>
>
>
Now I certainly agree that given three generic lines in $\mathbb{P}^3$ there is a unique quadric containing them. To see this, just choose $3$ points on each line: a quadric will contain the lines iff it contains the $9$ points, and it can be shown that these $9$ points give $9$ independent conditions.
Still I do no see how this implies the count $I\_2(p \cdot p \cdot p) = 1$. What I guess happens is that generically we will have two lines in one ruling and one line in the other, so that the curve $C \subset G$ which sweeps $S$ will only pass through one or two of the assigned points.
**Question**
What is the right count? Is there something wrong in what I said above? Is it even true that $I\_2(p \cdot p \cdot p) = 1$?
| https://mathoverflow.net/users/828 | Computing 3 points Gromov-Witten invariants of the Grassmannian | Reading more carefully your question, I think that in the hint, the ruled surface is meant to only contain the lines that are part of the ruling, not the spurious ones that may come when you look at the scroll in P^3. Thus in the case of degree 1, you only get the lines through the point, in the case of degree two, you only get the lines in one ruling. I thus suspect that the three lines will be in the same ruling (otherwise they would intersect, which is not very generic) and hence the unique quadric containing them will be the 1 you need. You still need to make sure it is a reduced point.
(Si, sono il damiano che conosci!)
| 3 | https://mathoverflow.net/users/4344 | 20067 | 13,341 |
https://mathoverflow.net/questions/20057 | 26 | Chas and Sullivan constructed in 1999 a Batalin-Vilkovisky algebra structure on the shifted homology of the loop space of a manifold: $\mathbb{H}\_\*(LM) := H\_{\*+d}(LM;\mathbb{Q})$. This structure includes a product which combines the intersection product and Pontryagin product and a BV operater $\Delta: \mathbb{H}\_\*(LM) \to \mathbb{H}\_{\*+1}(LM)$.
I was wondering about the applications of this structure. Has it even been used to prove theorems in other parts of mathematics? A more concrete question is the following: Usually, considering a more complicated structure on topological invariants of a space allows you to prove certain non-existince results. For example, the cup product in cohomology allows you to distinguish between $ S^2 \vee S^1 \vee S^1 $
and
$T^2$. Is there an example of this type for string topology?
| https://mathoverflow.net/users/798 | Applications of string topology structure | Hossein Abbaspour gave an interesting connection between 3-manifold topology and the string topology algebraic structure in [arXiv:0310112](https://arxiv.org/pdf/math/0310112v2). The map $M \to LM$ given by sending a point $x$ to the constant loop at $x$ allows one to split
$\mathbb{H}\_\*(LM)$ as $H\_\*(M) \oplus A\_M$.
He showed essentially that the restriction of
the string product to the $A\_M$ summand is nontrivial if and only if $M$ is hyperbolic. There are some technical details in the statements in his paper, but it was written pre-Perelman and I believe the statements can be made a bit more elegant in light of the Geometrization Theorem.
Philosophically, Sullivan has said that he his goal in inventing string topology was to try to find new invariants of smooth structures on manifolds. His original idea was that if you have to use the smooth structure to smoothly put chains into transversal positions to intersect them then you might hope that the answer will depend on the smooth structure. Unfortunately, we now know that the string topology BV algebra depends only on the underlying homotopy type of the manifold (there are now quite a few different proofs of various parts of this statement).
The string topology BV algebra is only a piece of a potentially much richer algebraic structure. Roughly speaking, $\mathbb{H}\_\*(LM)$ is a homological conformal field theory. This was believed to be true for quite some time but it took a while before it was finally produced by Veronique Godin [arxiv:0711.4859](https://arxiv.org/abs/0711.4859). She constructed an action of the PROP made from the homology of moduli spaces of Riemann surfaces with boundary. Restricting this action to pairs of pants recovers the original Chas-Sullivan structure.
Unfortunately, for degree reasons, nearly all of the higher operations vanish. In particular, any operation given by a class in the Harer stable range of the homology of the moduli space must act by zero. Hirotaka Tamanoi has a paper that spells out the details, but it is nothing deep.
Furthermore, it seems that the higher operations are homotopy invariant as well. For instance Lurie gets this as a corollary of his work on the classification of topological field theories.
Last I heard, Sullivan, ever the optimist, believes that there is still hope for string topology to detect smooth structures. He says that one should be able to extend from the moduli spaces of Riemann surfaces to a certain piece of the boundary of the Deligne-Mumford compactification. I've heard that the partial compactification here is meant to be that one allows nodes to form, but only so long as the nodes collectively do not separate the incoming boundary components from the outgoing boundary. Sullivan now has some reasons to hope that operations coming from homology classes related to the boundary of these moduli spaces might see some information about the underlying smooth structure of the manifold.
| 31 | https://mathoverflow.net/users/4910 | 20072 | 13,344 |
https://mathoverflow.net/questions/20070 | 2 | Is Hironaka's resolution of singularities functorial? I know that the resolution is not unique, there are flips etc. But if we have a rational map f:X---> Y, can we chose resolutions X'->X and Y'->Y and a map $f\_\*:X'\to Y'$ that makes the relevant diagram commute?
| https://mathoverflow.net/users/nan | Functoriality of Hironaka's resolution of singularities | In addition to what Damiano says, characteristic zero resolution is canonical in a stronger sense: a readable account is given [in a paper by Hauser.](http://www.ams.org/bull/2003-40-03/S0273-0979-03-00982-0/S0273-0979-03-00982-0.pdf) In particular, resolution commutes with smooth morphisms and commutes with group actions. Recently, this has also been shown to be true for all excellent reduced schemes of dimension at most two by [Cossart, Jannsen, and Saito](http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.2191v1.pdf)
| 3 | https://mathoverflow.net/users/397 | 20075 | 13,347 |
https://mathoverflow.net/questions/19772 | 6 | In real analysis one can define something known as the approximative derivative of a function. [See here eg](http://eom.springer.de/a/a012850.htm) Roughly speaking one asks that the limit of the difference quotient exists as long as h goes to zero while only taking values in some subset that is sufficiently dense.
Does anyone know if this concept has been studied for complex-valued valued functions of a complex variable? The basic definition should go through without problem so it should make sense to speak of an approximately holomorphic function as one that has an approximate complex derivative at every point of some open set. It would be interesting how much of the classical complex analysis one could generalize. Even knowing whether there exists functions that are approximately holomorphic but not holomorphic in the normal sense would be interesting.
| https://mathoverflow.net/users/2888 | Approximately holomorphic functions | Men'shov proved in 1936 that if $f\colon D\to\mathbb C$ is continuous and approximately differentiable outside of a countable set, then it is holomorphic in $D$ [1](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=5370&option_lang=rus) (Russian original with French summary). In the same paper he gives an example, attributed to Lusin, which shows that continuity cannot be dropped entirely even if the function is approximately differentiable at every point. Let $\varphi\colon\mathbb C\to\mathbb C$ be an entire function that tends to $0$ as $z\to\infty$ within any sector $\{|\arg z|<\pi-\epsilon\}$ (approximation theory can be used to create such examples). Then $f(z)=z\varphi(1/z)$ has an approximate derivative everywhere in $\mathbb C$, but of course it is not differentiable or even continuous at $0$.
Men'shov asked if continuity can be replaced with boundedness. This was answered affirmatively by Telyakovskii 50 years later [2](http://www.ams.org/mathscinet-getitem?mr=837136). In fact, boundedness can be weakened to logarithmic integrability.
Brodovich [3](http://www.ams.org/mathscinet-getitem?mr=1241165) proved that an injective function with an approximate derivative at every point is holomorphic. (MR review omits the injectivity assumption). A survey of results in this area was written by Dolzhenko [4](http://www.ams.org/mathscinet-getitem?mr=1211822), but it predates the work of Brodovich.
| 6 | https://mathoverflow.net/users/2912 | 20084 | 13,351 |
https://mathoverflow.net/questions/20082 | 8 | Let $Gr$ be the affine Grassmannian of $G=G((t))/G[[t]]$, and let $Perv(Gr)$ be the category of perverse sheaves on $Gr$. We have action of $G((t))$ on the left-hand side of $Perv(Gr)$, also we have action of the tensor category $Rep(G^\vee)$ on the right-hand side, through geometric Satake correspondence.
It is clear that we have the action of $G((t))$ on $Perv(Gr)$ as functors.
Follow Gaitsgory's paper "the notion of category over stack", he claimed that an action of algebraic group scheme $H$ on a category $\mathcal{C}$ is actually equivalent to a category $\mathcal{C}$ with the action of the tensor category "Rep(H)".
Go back to the original set-up, that means we also have an action of $G^\vee$ on $Perv(Gr)$.
I think, on this $Perv(Gr)$, the actions of $G((t))$ and $G^\vee$ should have different meaning, but why they gave them the same name?
Am I confused?
| https://mathoverflow.net/users/5082 | A question on group action on categories | Yes, you are confused. What is claimed by Gaitsgory is that datum of category with
action of $H$ is equivalent to datum of of another category with action of $Rep(H)$.
You go back and forth between these two categories using constructions of "equivariantization"
and "de-equivariantization".
| 13 | https://mathoverflow.net/users/4158 | 20090 | 13,355 |
https://mathoverflow.net/questions/20080 | 4 | I don't know whether this question is a bit too vague for MO or not, so feel free to delete it if you see fit.
The p-adic integer is defined by taking the inverse limit $\ldots \mathbb{Z} / p^2 \rightarrow \mathbb{Z}/p $.One way to see the p-adic integers is to see it as dealing with $ \mathbb{Z} / p, \mathbb{Z} / p^2, \ldots $ at the same time. So $p$-adic integers allow us to see the structure of the ring of integers at the prime $p$. Taking the fractional field we obtain the $ p$-adic rational field $\mathbb{Q}\_p$.
This construction is useful in study the arithmetic of the ring (field). For example, in the theory of class field theory, we study the question in $\mathbb{Q}\_p$ first and glue them together and do something more to get the solution for $\mathbb{Q}$.
I want to ask why it is not possible for us to construct a structure that will allow us to see the ring of integer at two primes $p,q$ together and see how do they interacts? The analog of the above inverse limit construction seems to still work though not an intergral domain. However, we can still localize where it is possible.
Here is the motivation for the above question. We know that the global question are not solve by simply gluing together solution for local question. I attribute that to the fact that the primes does not play alone but interact with one another. An illustration of this can be seen through the fact that the product of all normalized absolute value is 1. So my question is why not isolate two primes to understand how they are interacting with one another instead of looking at all of them at once. I think there is some complicated issue that will arise from this. Just want to know what they are.
A more particular question may be like this: Let call the construction obtained above $\mathbb{Q}\_{p,q} $. Is there something in the same vein of class field theory for this object. What is the obstacle in having such a theory. I am vaguely know that we have a more general Galois theory not only for fields but for rings also.
| https://mathoverflow.net/users/2701 | Why isn't there a structure with two primes? | As stankewicz said, it is a general principle in number theory that whenever only finitely many primes are involved, they act "independently" in the sense that analyzing what is happening locally at each prime separately is enough to understand what all the finitely many primes are doing. One example of this is the Chinese Remainder Theorem.
Here is another: if you want a version of the integers with two primes $p$ and $q$, start with $\mathbb{Z}$ and invert every prime $\ell \not \in \{p,q\}$. This gives a semilocal ring with $(p)$ and $(q)$ as the nonzero primes. Similarly, you can do this with any two prime ideals $\mathfrak{p}, \mathfrak{q}$ in a Dedekind domain $R$. But the ring you get is not very interesting: it is a semi-local Dedekind domain, hence its class group is trivial, very likely the unit group $R^{\times}$ will be infinitely generated, etc. This domain is the intersection of the two DVRs $R\_{\mathfrak{p}}$, $R\_{\mathfrak{q}}$, and everything you want to know about it can be reduced to the DVRs. The same with two replaced by any finite set...
| 5 | https://mathoverflow.net/users/1149 | 20096 | 13,357 |
https://mathoverflow.net/questions/20008 | 4 | The basic question is whether there is a notion of chief factor of a connected solvable algebraic group that matches my intuition. A few smaller assertions are sprinkled through the explanation, and the implicit question is if these are correct (are unipotent groups nilpotent, are their chief factors all isomorphic to subgroups of Ga, etc.).
In J.S. Milne's course notes on the basic theory of algebraic groups, theorem 14.30:
A connected solvable smooth group over a perfect field has a connected unipotent normal subgroup whose quotient is of multiplicative type.
In other words, the derived subgroup is contained in the unipotent radical.
Now, a connected group acts on a group of multiplicative type trivially, by 13.21.
I did not see it mentioned, but I believe unipotent groups are nilpotent in both the group theoretic sense and whatever fancy definition one might cook up for these functors. I think that the lower central factors should be direct products of subgroups of the additive group Ga.
By the Jordan decomposition or 13.13 or 13.15, I think any action of Gm on (Ga)^n is diagonal.
It looks like a connected solvable affine algebraic group over an algebraically closed field has a chief series consisting of subgroups of Ga and Gm, all of which I would describe as being one-dimensional.
The analogy with finite groups takes the unipotent radical to be O\_p(G), the p-core, and so it appears that a finite group of solvable algebraic type always has [G,G] <= O\_p(G), so that not only is G ~~supersolvable~~ nilpotent-by-abelian, it is p-closed and its eccentric chief factors are all for the same prime p. In the algebraic case, the central chief factors would be the subgroups of Gm, and the eccentric chief factors would be the subgroups of Ga.
In other words, connected solvable affine algebraic groups over algebraically closed fields are very dissimilar from finite solvable groups in that the representations they define on their own chief factors are all one-dimensional. More briefly, connected solvable affine algebraic groups over algebraically closed fields are supersolvable.
**Edit:** I think the answer to my question is relatively simple: "supersolvable" is a little tricky to directly generalize, but "nilpotent-by-abelian" is quite easy and true, and still implies that any chief factors will be one-dimensional. In Jim's answer, it appears Borel-Serre-Mostow (at least by the time they are translated into Russian) also considered these groups to be "supersolvable", so the name is reasonable, even if the correct definition is just "nilpotent-by-abelian".
| https://mathoverflow.net/users/3710 | Are all connected solvable affine algebraic groups supersolvable? | Properties of solvable linear algebraic groups have been explored sporadically for over
50 years, with various assumptions on the base field. There seems to be no single
viewpoint about how this subject interacts with finite groups or with general
algebraic groups. But to get perspective it may be helpful to consult older work
originating with Lie groups (Borel, Mostow, Serre) as well as the long paper by
Platonov, V. P.,
The theory of algebraic linear groups and periodic groups. (Russian)
Izv. Akad. Nauk SSSR Ser. Mat. 30 1966 573--620. This appears in Series 2, Volume 69
of the AMS Translations. For example, in 3.1 a solvable algebraic group is defined
to be "supersolvable" if its finite component group is supersolvable. This may or may not
be everyone's preferred definition, but given this history one has to be careful about the underlying assumptions when asking new questions about solvable algebraic groups.
| 3 | https://mathoverflow.net/users/4231 | 20100 | 13,360 |
https://mathoverflow.net/questions/20077 | 6 | Hello,
I am the 3rd year undegraduate student of mathematics.
After I obtain a bachelor degree I want to study maths at graduate level, especially algebraic geometry and complex analysis.
This fields of mathematics are well-represented at my univeristy, so at the first glance this plan looks fine.
Unfortunately almost all of my collegues are not interested in this fields (they think that AG and CA are too technically involved; most of them are interested mainly in functional analysis and topology).
It will have such unpleasant effect on my studies, that most probably the most (or all) of courses in AG and CA in the upcoming year won't start at my university.
So I'll end up learning alone, from books.
It's not that it's a big problem to learn from the book, it's not a problem at all. But I think such learning have no comparison with the regular course, where I could discuss problems and see another approches of other (more gifted) students.
To avoid such situation I may try to convince my collegues to study CA and AG, but I don't have many arguments since I'm still an ignorant in this fields.
And here arise my questions:
>
> 1. How would You encourage graduate students to learn algebraic geometry?
> 2. How would You encourage graduate students to learn complex analysis?
>
>
>
| https://mathoverflow.net/users/5080 | How would You encourage graduate students to learn algebraic geometry and/or complex analysis? | This feels really like an "Ask Professor Nescio" question.
Let me ask you a question: if you feel like you cannot learn what you want to learn, why are you staying at the same university? I see from your profile that you are studying a Jagielloni, and you are Polish, so I understand somewhat that, if you want to remain in Poland, you feel that you should stay. But mathematics being the international field as it is, I would recommend going to a different university in a different country. (For example, in the US there seems to be no lack of graduate students who share your interest. I'm sure if you ask around a bit you can find out about other places in Europe.)
Now, with that said, if you decide that you want to stay:
1. Don't over sell it. Being too pushy will have a negative effect on the other students.
2. Don't be evangelical. You should not tell them why they *should* be interested or why they *ought* to study the subject with you. That'll have the opposite effect.
3. From your descriptions you need to first dispel the myth that complex analysis and algebraic geometry is too technical. I guess the best thing to do is to introduce them to partial differential equations. (That's a joke.) But you need to be able to show them some examples of how sometimes, things becomes much more clear when viewed in the right framework. Show them a nice theorem or two with relatively simple proofs. A nice forum for this could be an informal seminar organized by the students for themselves: try to run a seminar where each student presents a result (not due to himself) that he finds interesting (don't just hijack it for your ulterior motives). When it is your turn talk about something really pretty from algebraic geometry. It may win you some converts.
4. Find out what your fellow students like to do. You said functional analysis and topology. Anything else? You need to sell to your audience. For the topologists, at least some introductory complex analysis and algebraic geometry should be that hard to sell: tell them about (Hirzebruch-)Riemann-Roch! Tell them about the works of Kodaira! Complex analysis in one-variable is basically just topology anyway. (Can't help you with the functional analysts there.)
5. An extension of the above: convince your fellow students that those subjects are useful for *them*. So a good idea is to find some theorems in *their* field that was first proven, or has nice interpretations, using the tools of complex analysis or algebraic geometry.
If all else fail, and you cannot get another person to study with you, you can always ask questions here or on sci.math.
| 5 | https://mathoverflow.net/users/3948 | 20102 | 13,362 |
https://mathoverflow.net/questions/20089 | 11 | I figured out the first part of this years ago, but completely forget how I did it. I looked at the second, but don't think I figured it out.
This I am sure is true, but don't remember why. Suppose that G is a finite group of size $n$, and H is a normal subgroup with |G/H| = $k$. Then at least $\frac{1}{k}$ of the conjugacy classes of G are within H.
This I don't know the answer to. If H is allowed to be an arbitrary subgroup, must H intersect at least $\frac{1}{k}$ of the conjugacy classes?
An example of a simple consequence of the first statement is that if we look at $S\_n$ and $A\_n$ is that at least half the partitions of $n$ have an even number of even parts.
| https://mathoverflow.net/users/5023 | Conjugacy classes intersecting subgroups of finite groups | Let $r$ be the number of conjugacy classes in $G$.
The action of $G$ on itself by conjugation gives, via the Cauchy-Frobenius formula, $r=\frac{1}{|G|}\sum |C(g)|$, where $g$ ranges over $G$. This action restricted to $H$ gives the number of $G$-conjugacy classes in $H$ as $\frac{1}{|G|}\sum |C\_H(g)|$, and from the fact $|C\_H(g)|\ge \frac{1}{k}|C(g)|$ the first part of your problem follows.
**EDIT**: As pointed out by Sergei Ivanov below, this argument also shows the second part is true as well.
| 13 | https://mathoverflow.net/users/1446 | 20104 | 13,364 |
https://mathoverflow.net/questions/20105 | 13 | Dear all,
Sorry if the question is naive: any nice example of such a ring or, better, of a class of such rings?
| https://mathoverflow.net/users/5087 | A ring on which all finitely generated projectives modules are free but not all projectives are free? | Cher Michel, these rings are uncommon.
1. Over a local ring ALL projective modules are free : this is a celebrated theorem due to Kaplansky.
2. If $R$ is commutative noetherian and $Spec(R)$ is connected, every NON-finitely generated projective module is free. This is due to Bass in his article "Big projective modules are free" which you can download for free here
[Link](https://web.archive.org/web/20161229203318/http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ijm/1255637479)
And now for the good news: the rings you are after are uncommon but they exist. Bass in the article just quoted shows that the ring $R=\mathcal C([0,1])$ of continuous functions on the unit interval has all its finitely generated projective modules free. Nevertheless the ideal consisting of functions vanishing in a neighbourhood of zero (depending on the function) is projective, not finitely generated and not free. Bass attributes the result to Kaplansky.
| 34 | https://mathoverflow.net/users/450 | 20109 | 13,366 |
https://mathoverflow.net/questions/12425 | 10 | The ring $S=\mathbb{C}[x\_1,x\_2,\dots,x\_n]^{S\_n}$ of symmetric polynomials has a number of commonly used bases, but the undisputed world champion of these is the basis consisting of [Schur polynomials](http://en.wikipedia.org/wiki/Schur_polynomial) $s\_\lambda$, where $\lambda$ ranges over non-increasing sequences $\lambda\_1 \geq \lambda\_2 \geq \cdots \geq \lambda\_n \geq 0$ of non-negative integers. For a partition $\mu$ of $n$, let $V\_\mu$ be the corresponding irreducible $S\_n$-module, and let $M(\mu)=(\mathbb{C}[x\_1,x\_2,\dots,x\_n] \otimes V\_\mu)^{S\_n}$ be the ($S$-module of) $S\_n$-invariant polynomial functions on $\mathbb{C}^n$ with values in $V\_\mu$. Is there a $\mathbb{C}$-basis of $M(\mu)$ that deserves top billing?
(A bit of background: the dimensions of the homogeneous components of $M(\mu)$ can be computed from the *exponents* of $V\_\mu$, that is, the degrees in which it appears in the coinvariant algebra. There are combinatorial expressions known for these numbers---see e.g. Stembridge's paper "On the eigenvalues of representations of reflection groups and wreath
products", Pacific J. Math. 140 (1989), 353--396 and the references therein, but they are not obtained by writing down a particularly nice basis.)
| https://mathoverflow.net/users/nan | Canonical bases for modules over the ring of symmetric polynomials | In my paper Cyclage, catabolism, and the affine Hecke algebra
<http://arxiv.org/abs/1001.1569>
I exhibit a canonical basis for $\mathbb{C}[x\_1,x\_2,\dots,x\_n]$ and more generally a canonical basis for $\mathbb{C}[x\_1,x\_2,\dots,x\_n] \otimes V\_\mu$ coming from the extended affine Hecke algebra of type A. The subset of this canonical basis corresponding to cells of shape $(n)$ is a basis for the $S\_n$-invariants in this module. For the special case $M(n)$, these canonical basis elements do correspond to Schur functions -- they are Schur functions in the Bernstein generators times $e^+$, where $e^+$ spans the trivial representation for the finite Hecke algebra (see Theorem 6.1).
I have not thought too much about the combinatorics of the canonical basis for $M(\mu)$, but it may be possible to work this out explicitly, including an explicit description of the $S\_n$ invariants (see Example 9.21 for a little bit about this). This paper is long and may take some time to get through. Feel free to contact me at the email address at the very bottom of the paper if you have any questions or want to discuss this in detail.
| 3 | https://mathoverflow.net/users/3318 | 20111 | 13,368 |
https://mathoverflow.net/questions/20113 | 9 | I would like to have an estimate for the infinite series
$$
\sum\_{k=B}^\infty \frac{A^k}{k!},
$$
where $A$ is a large positive quantity and $B$ is just a little bit bigger than $A$, namely, $B = A + C \sqrt A$ for some fixed large positive constant $C$. (In my application, $A$ and thus $B$ are increasing functions of some other variable, but $C$ really will stay fixed.)
I expect that the answer should look something like
$$
?\ \sum\_{k=B}^\infty \frac{A^k}{k!} \ll e^{-C^2/2} \ \ ?
$$
uniformly in $A$, $B$, and $C$. (Possibly there should even be an asymptotic formula.) It would be great to be able to just quote such an estimate "off the shelf". I've only been able to find such estimates when $B$ is substantially larger than $A$, such as $B > 5A$.
Does anyone know of a bound of this type in the literature? Many thanks.
| https://mathoverflow.net/users/5091 | Estimate for tail of power series of exponential function? | Let's instead consider the sum
$$ \sum\_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!} $$
which of course differs from yours just by a factor of $e^{-A}$.
Then this sum is the probability that a Poisson random variable of mean $A$ is at least $A + C\sqrt{A}$.
A Poisson with mean $A$ has standard deviation $\sqrt{A}$, and as $A \to \infty$ the Poissons become asymptotically normal. So we have
$$ \sum\_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!} \to \Phi(C) $$
as $A \to \infty$, where $\Phi$ is the CDF of the standard normal. So your sum is asymptotic to $e^A \Phi(C)$.
Alternatively, if you'd like an explicit inequality, your sum can be bounded above by the geometric series with first term $A^B/B!$ and common term ratio $A/B$. Therefore, your sum is less than
$$ {A^B \over B!} {1 \over 1-A/B} $$
and this can be rewritten as
$$ {A^B \over B!} \left( 1 + {\sqrt{A} \over C} \right) $$
The product $A^B/B!$ is, as $A \to \infty$ with $B = A + C \sqrt{A}$,
$$ {1 \over \sqrt{2\pi}} e^{-C^2/2} A^{-1/2} e^A (1+o(1))$$
by Stirling's formula. In the factor $1 + \sqrt{A}/C$ we can neglect $1$ as $A \to \infty$, so we get that
$$ \sum\_{k = A+C\sqrt{A}}^\infty {e^{-A} A^k \over k!} \le {1 \over \sqrt{2\pi}} C e^{-C^2/2} e^A (1 + o(1)) $$
By, say, [the double inequality (26) here](http://mathworld.wolfram.com/StirlingsApproximation.html) it should be possible to get explicit bounds.
| 16 | https://mathoverflow.net/users/143 | 20118 | 13,373 |
https://mathoverflow.net/questions/17369 | 12 | Let $f: R \to S$ be a morphism of Noetherian rings (or more generally $S$ can just be an $R-R$ bimodule with a bimodule morphism $R \to S$). Suppose $f$ is faithfully flat on both sides, so $M \to M \otimes\_R S$ is injective for any right $R$-module $M$, and $N \to S \otimes\_R N$ is injective for any left $R$-module $N$.
Is it then true that $M \otimes\_R N \to M\otimes\_R S \otimes\_R N$ is injective for any pair $(M, N)$ of a right and a left $R$-module? This seems way too optimistic, but I can't seem to find a counterexample.
| https://mathoverflow.net/users/2481 | Tensor products and two-sided faithful flatness | Here's an example. Let $R = {\mathbb C}[x]$ and let $S = {\mathbb C}\langle x,y\rangle/(xy-yx-1)$, i.e. the first Weyl algebra $A\_1$. Then $S$ is free as both a left and right
$R$-module, and comes equipped with the natural ($R$-bimodule) inclusion of $R$. On the
other hand, if you take $M = {\mathbb C}[x]/(x) = N$, you'll get for $M\otimes\_R S\otimes\_R M$ the zero module. Indeed, any element of $S$, i.e. any differential operator with polynomial coefficients (writing $\partial = \partial/\partial x$ in place of $y$) can be written in the form $\sum\_i p\_i(x) \partial^i$, so any element in $M\otimes\_R S = {\mathbb C}[x]/(x) \otimes S$ is represented by an expression $\sum\_i c\_i \partial^i$ where the $c\_i$ are constants, and now an induction on $k$ shows (I believe, my brain is a little fuzzy at this hour) that, for the right $R$-module structure on $S/xS$, one has $\partial^k \cdot x = k \partial^{k-1}$. One can conclude that $M\otimes\_R S\otimes\_R M = 0$, whereas of course $M\otimes\_R M\cong M$ in this example...
| 8 | https://mathoverflow.net/users/2628 | 20136 | 13,386 |
https://mathoverflow.net/questions/20132 | 13 | Let $f:X\to B$ be a family of curves of genus $g$ over a smooth curve $B$. Let $F\_0$ be a singular fiber.
If $F\_0$ is a semistable fiber, the monodromy matrix can be gotten by the classical Picard-Lefschetz formula.
If $F\_0$ is non-semistable, I don't know how to compute its monodromy matrix. For example, in Namikawa and Ueno's paper[1], they can compute the Picard-Lefschetz monodromy matrix for each type of singular fiber of genus 2. It's not clear to me how they did that.
[1] Namikawa, Y. and Ueno, K., The complete classification of fibres in pencils of curves of genus two, Manuscripta math., Vol. 9 (1973), 143-186.
| https://mathoverflow.net/users/5093 | How to compute the Picard-Lefschetz monodromy matrix of a non-semistable fiber? | One approach (I don't know how effective it is in the genus 2 case you asked about)
is to explicitly apply the semi-stable reduction theorem, and so reduce to the semi-stable
case.
To achieve semi-stable reduction, you have to alternately blow-up singular points in
the special fibre, and then make ramified base-changes. The latter operation just extracts
a root of the monodromy operator (i.e. if $\gamma$ is a generator of $\pi\_1$ of the punctured
$t$-disk, and we set $t = s^n$, then $\gamma = \tau^n,$ where $\tau$ is a generator of
$\pi\_1$ of the punctured $s$-disk), so it is easy to see how the monodromy matrix
changes. And blowing up a point in the special fibre doesn't change the monodromy action
around the puncture at all.
So using this process, one can relate the original (unknown) monodromy matrix
to the corresponding matrix in the semi-stable context, where it is known thanks to the
Picard--Lefshcetz formula.
| 5 | https://mathoverflow.net/users/2874 | 20137 | 13,387 |
https://mathoverflow.net/questions/20112 | 56 | On [another thread](https://mathoverflow.net/questions/20077/how-would-you-encourage-graduate-students-to-learn-algebraic-geometry-and-or-comp) I asked how I could encourage my final year undergraduate colleagues to take an algebraic geometry or complex analysis courses during their graduate studies.
Willie Wong proposed me following idea - to show them some interesting results in this fields with relatively simply proofs and some consequences in other fields.
Thus by 'interesting' result in algebraic geometry I here mean the result which may convince 3rd year undergraduate student to study algebraic geometry.
In fact I'm supposed to give some talk during the seminar dedicated to final year undergraduates, and I can propose my own topic, so I thought that it could work.
But my problem is that I'm just wanna-be student of algebraic geometry and I don't have enough insight and knowledge to find a topic which 'could work'.
Also I doubt whether it is possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field.
So in short my first question is as above:
>
> Is it possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field?
>
>
>
To be more precise - the talk should take a one or two meetings, 90 minutes each one.
The audience will be, as I said 3rd year undergraduate students, all of them after two semester course in algebra, some of them after one or two semesters in commutative algebra. All taking the course in one complex variable, and after one semester introductory course in differential geometry.
Some of them may be interested in number theory.
The second, related question is as follows:
>
> If You think that the answer to the previuos question is positive, please try to give an example of idea/theorem/result which would be accessible in such time for such audience, and which You find interesting enough to make them consider possibility of studying algebraic geometry. Try to think about the results which shows connections of AG with some other fields of mathematics.
>
>
>
| https://mathoverflow.net/users/5080 | Interesting results in algebraic geometry accessible to 3rd year undergraduates | If you want to teach something intriguing, you should do something that introduces a new geometric idea while also involving algebra in an essential way. I recommend that you give an introduction to the projective plane, showing the other students that it is a natural extension of ordinary space which makes some geometric properties more uniform (such as intersection properties of curves), gives a fruitful new way to think about old topics (like asymptotes), and lets you do things that are impossible to conceive without it (reducing rational points mod $p$). There should be substantial interplay between algebra and geometry, but make sure to draw pictures to emphasize the geometric aspects.
1. In algebra, we can conceive of the quadratic formula in a uniform manner, but the ancient Greeks [Edit: Babylonians, not Greeks] couldn't do this because they didn't have the idea of negative numbers. So they had several quadratic formulas on account of not being able to write something as simple as $ax^2 + bx + c = 0$ at one stroke (for any signs on $a, b$, and $c$, with $a$ nonzero). Our extended skill at algebra lets us work with one case where the ancients had to take multiple cases. We can also say with complex numbers that *any* quadratic equation has two roots, allowing for a double root to count as one root with multiplicity two.
The thrust of what comes next is to extend the plane so that geometric properties become nicer in a similar way the algebra is becoming nicer when we use more general number systems.
2. Consider the intersection properties of lines in the plane. There is a dichotomy: usually two lines in the plane meet in one point, but some pairs of lines (the parallel ones) meet in no points. Let's see what this looks like under stereographic projection. Lines in the plane become circles through the north pole, but not including the north pole itself. It's natural to close up the image and take that whole circle as a substitute for the original line. So we can see that lines in the plane naturally close up into circles through the north pole. Under stereographic projection, the old dichotomy between parallel and non-parallel lines takes on a new appearance: a pair of non-parallel lines corresponds under stereographic projection to a pair of circles intersecting in two different points, one of which is the north pole, while a pair of parallel lines corresponds under stereographic projection to a pair of circles which are tangent at the north pole. It is natural to think of two tangent circles as having their point of tangency be an intersection point of multiplicity two, much like a quadratic polynomial can have a double root. So after stereographic projection we can "see" two points of intersection for any pair of lines. This geometric construction is something like the algebraic use of more general number systems to find roots to all quadratic equations. The moral to take from this example is that in a larger space, curves that used to not intersect may now intersect (or rather, their natural closures in the new space intersect) with a *uniform* count of the number of intersection points. If the students agree that enlarging number systems to create solutions to polynomial equations is good, they should agree that enlarging space to make intersection properties more uniform is good too. Another important feature is that the sphere, like the plane, is a homogeneous object: we can transform (rotate) the space to carry one point to any other point. On the sphere as a space in its own right, there is truly nothing special about the north pole.
3. An even better geometric extension of the plane is the projective plane, although at first it will feel unfamiliar and strange because you can't see it all at once.
You should introduce it in a uniform manner as points described with homogeneous coordinates $[x,y,z]$ where $x$, $y$, and $z$ are not all 0 and, say,
$$
[3,6,2] = [1,2,2/3] = [1/2,1,1/3] = [3/2,3,1] \text{ and } [0,5,0] = [0,1,0].
$$
Although it is impossible to see the whole projective plane at once, we can get glimpses of large parts of it using three different charts: $U\_0$ is the points where $x \not= 0$, $U\_1$ is the points where $y \not= 0$ and $U\_2$ is the points where $z \not= 0$. These three charts together cover the projective plane. Any nonzero coordinate can be scaled to 1 and that fixes the other two homogeneous coordinates of the point, e.g., $[x,y,1] = [x',y',1]$ if and only if $x = x'$ and $y = y'$. This means we can identify each of $U\_0$, $U\_1$, and $U\_2$ with the usual plane (e.g., identify $U\_2$ with ${\mathbf R}^2$ by identifying $[x,y,1]$ with $(x,y)$). This means the projective plane locally looks like the plane, much like the sphere does, except we can't see all of it at the same time as we can with the sphere.
(In case you want to show students that the projective plane is a really natural model of something they have known in another context, think about nonzero ideals in ${\mathbf R}[x]$. Any ideal has a generator, but the polynomial generator is only defined up to a nonzero scaling factor. Usually we normalize the generator to be monic, but if we don't want to insist on a particular choice of generator then the right model for the generator is a point in projective space. In particular, for any nonzero ideal $(f(x))$ where $\deg f(x) \leq 2$, write $f(x) = ax^2 + bx + c$; the coefficients $a, b, c$ are only defined up to an overall scaling factor, so the point $[a,b,c]$ is one way to think about that ideal.)
Next introduce curves in the projective plane as solutions to *homogeneous* polynomial equations in $x$, $y$, and $z$ and explain what the algebraic process of homogenization and dehomogenization of polynomials is, e.g., it makes $y = 2x + 1$ into $y = 2x + z$ or $x^2 - y^2 = x+ 1$ into $x^2 - y^2 = xz + z^2$. In particular a *line* in the projective plane is the solution set to any equation $ax + by + cz = 0$ where the coefficients are not all 0.
Now let's look at what a point on a specific curve in the projective plane looks like in each of the three standard charts, carry out the same kind of calculus computation in each chart, and compare the results with each other. We will use the curve $C : x^2 + y^2 = z^2$ in the projective plane (not to be confused with a surface in 3-space given by the same equation) and the points $P = [3,4,5]$ and $Q = [1,0,1]$ which lie on $C$. How do $C$, $P$, and $Q$ appear in each of the charts $U\_0$, $U\_1$, and $U\_2$?
a) In $U\_0$, which is identified with the plane by $[x,y,z] \mapsto (y/x,z/x)$, $C$ becomes the hyperbola $z^2 - y^2 = 1$, $P$ becomes $(4/3,5/3)$, and $Q$ becomes $(0,1)$. Here we identify $U\_0$ with the usual $yz$-plane. By calculus, the tangent line to $z^2 - y^2 = 1$ at the point $(4/3,5/3)$ is $z = (4/5)y + 3/5$ and the tangent line at $(0,1)$ is $z = 1$.
Note that we actually miss two points from $C$ when we look at the intersection of it with $U\_0$: $[0,1,\pm 1]$.
b) In $U\_1$, $C$ becomes the hyperbola $z^2 - x^2 = 1$ in the $xz$-plane, $P$ becomes the point $(3/4,5/4)$ with tangent line $z = (3/5)x + 4/5$, and $Q$ doesn't actually live in this chart (kind of like the north pole under stereographic projection not going to anything the in the plane). Here two points from $C$ are missing: $[1,0,\pm 1]$.
c) In $U\_2$, $C$ becomes the circle $x^2 + y^2 = 1$, $P$ becomes $(3/5,4/5)$ with tangent line $y = (-3/4)x + 5/4$, and $Q$ becomes $(1,0)$ with tangent line $x = 1$. Every point from $C$ lies in $U\_2$, so no points are missing here. We see the "complete" curve in this chart.
It is *essential* to draw three pictures here (of the $yz$-plane, $xz$-plane, and $xy$-plane) and mark $P$ and $Q$ in each (except you don't see $Q$ in the $xz$-plane).
Now comes the beautiful comparison step: in all three charts the homogenization of the tangent line at $P$ is exactly the same equation: $3x + 4y = 5z$. The tangent line at $Q$ in $U\_0$ and $U\_2$ homogenizes in both cases back to $x = z$. This suggests there should be an intrinsic concept of tangent line in the projective plane to the curve $C$ at the points $P$ and $Q$, and you can compute the tangent line by looking at any chart containing the relevant point of interest, doing calculus there, and then homogenizing back. The homogenization of your ordinary linear equation to a homogenuous linear equation will always be the same, and its solutions in the projective plane define the tangent line to the projective curve at that point.
As further evidence of the consistency of this new space and the geometry in it, look at the intersections of the two tangent lines at $P$ and $Q$: in $U\_0$ -- the $yz$-plane -- the tangent lines meet in $(1/2,1)$ while in $U\_2$ -- the $xy$-plane -- the tangent lines meet in $(1,1/2)$. These points both homogenize back to the same point $[2,1,2]$, which is the unique (!) point in the projective plane satisfying $3x + 4y = 5z$ and $x = z$.
Remember that $Q$ went missing in the chart $U\_1$? Well, its tangent line did not go missing: the projective line $x = z$ in the projective plane meets the chart $U\_1$ in the ordinary line $x = z$ of the $xz$-plane, which is an *asymptote* to the piece of $C$ we can see in $U\_1$. This is really amazing: asymptotes to (algebraic) curves in the usual plane are "really" the tangent lines to missing points on the complete version of that curve in the projective plane. To see this from another point of view, *move* around $C$ clockwise in the chart $U\_2$ (where it's a circle) and figure out the corresponding motion along the piece of $C$ in the chart $U\_0$ (where it's a hyperbola): as you pass through the point $Q = (1,0)$ in $U\_2$, what happens in the chart $U\_0$ is that you jump off one branch of the hyperbola onto the other branch by skipping through an asymptote, sort of. (There is a second point on $C$ in $U\_2$ that you don't see in $U\_0$ -- the point $R = [-1,0,1]$ is $(-1,0)$ in $U\_2$ -- and paying attention to that point may help here.)
The conic sections -- parabolas, hyperbolas, and ellipses -- which look quite different in ${\mathbf R}^2$, simplify in the projective plane because they all look like the same kind of curve (once you close them up): $y = x^2$ becomes $yz = x^2$, $xy = 1$ becomes $xy = z^2$, and $x^2 + y^2 = 1$ becomes $x^2 + y^2 = z^2$, which is the same as $x^2 = (z-y)(x+y) = z'y'$, where $z' = z-y$ and $y' = z+y$. I like to think about this as a fancy analogue of the Greek [Edit: Babylonian] use of many forms of the quadratic formula because they didn't have the right algebraic technique to realize there is one quadratic formula. Using the projective plane we see there is really one conic section.
You might want to show by examples the nicer intersection properties of lines in the projective plane: any two lines in the projective plane meet in exactly one point. This is just a glimpse of the fact that curves in the projective plane have nicer *intersection properties* than in the ordinary plane, but to get the correct theorem in that direction for curves other than lines, you need to (a) work over the complex numbers and (b) introduce an appropriate concept of intersection multiplicity for intersection points of curves, somewhat like the idea of tangent circles intersecting in a point of multiplicity two which I mentioned earlier. The relevant theorem here is Bezout's theorem, but to state it correctly is complicated precisely because it is technical to give a good definition of what the intersection multiplicity is for two curves meeting at a common point.
For the student who wants to be a number theorist, compare reduction mod $p$ in the usual plane and the projective plane. In the study of Diophantine equations (e.g., to show $y^2 = x^3 - 5$ has no integral solutions), it is very useful to reduce mod $p$, and there is a natural way to reduce a point in ${\mathbf Z}^2$ modulo $p$ However, there's no reasonable way to reduce all points in ${\mathbf Q}^2$ modulo $p$: when the rational numbers have denominator divisible by $p$, you can't make sense of them mod $p$: we can reduce $(-7/4,51/8)$ mod 5, for example, but not mod 2. In the projective plane, however, we can reduce rational points mod $p$ by the idea of choosing a set of primitive integral coordinates, where the homogeneous coordinates are relatively prime. For example, $[-7/4,51/8,1] = [-14,51,4]$ in ${\mathbf P}^2({\mathbf Q})$, and this can be reduced mod $p$ for any $p$ at all. For example, in ${\mathbf P}^2({\mathbf F}\_2)$ it becomes $[0,1,0]$.
(There is another primitive set of homogeneous coordinates for the point, namely $[14,-51,-4]$, but that reduces mod $p$ to the same thing as before, so this reduction mod $p$ process is well-defined.) This suggests that the projective plane has better mapping properties than the usual plane, in some sense.
| 61 | https://mathoverflow.net/users/3272 | 20143 | 13,391 |
https://mathoverflow.net/questions/20138 | 16 | I have seen it stated that Proj of any graded ring $A$, finitely generated as an $A\_0$-algebra, is isomorphic to Proj of a graded ring $B$ such that $B\_0 = A\_0$ and $B$ is generated as a $B\_0$-algebra by $B\_1$.
Could someone either supply a reference for or a sketch a proof of this statement?
Note: An obvious approach to this question is to make $B$ a Veronese subring of $A$. However, when I try this approach, I end up getting a terrible combinatorics problem that I do not know how to approach.
| https://mathoverflow.net/users/5094 | Why is Proj of any graded ring isomorphic to Proj of a graded ring generated in degree one? | Bourbaki Commutative Algebra Chapter 3:
Let $A$ be a non-negatively graded algebra. Assume that $A$ is finitely generated over $A\_0$. There exists $e \geq 1$ such that $A^{me} = A\_0[A^{me}\_1]$ for any $m \geq 1$.
Here $A^{e} = \oplus\_{n \in\mathbb Z} A^{e}\_n$, where $A^{e}\_n := A\_{ne}$.
The desired result follows because replacing $A$ with $A^{me}$ does not change Proj.
| 17 | https://mathoverflow.net/users/5097 | 20145 | 13,392 |
https://mathoverflow.net/questions/20153 | 6 | Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasi-projective homogeneous G-variety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$-variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$-variety is already quasi-projective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical *abstract* isomorphism $\pi:X \rightarrow G/G \_x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counter-examples or is any (quasi-projective) homogeneous $G$-variety up to isomorphism of the form $G/H$?
| https://mathoverflow.net/users/717 | Is every homogeneous G-variety of the form G/H? | It depends on what you mean by "closed subgroup". If you mean a Zariski closed subset which forms a subgroup then the answer is no. If you mean a closed subgroup scheme, then the answer is yes. An example where you need to use the second definition is the Frobenius map $F\colon G \to G^{(p)}$. If we let $G$ act on $G^{(p)}$ through $F$ then the action is transitive and indeed $G^{(p)}$ is isomorphic to $G/Ker F$. However, unless $G$ is zero-dimensional $Ker F$ is a non-trivial finite group scheme whose $k$-points consist of just the identity.
Note however that $G/H$ is always quasi-projective even when $H$ is a subgroup scheme so all homogeneous $G$-spaces are quasi-projective.
| 16 | https://mathoverflow.net/users/4008 | 20155 | 13,398 |
https://mathoverflow.net/questions/20142 | 1 | Fluorescence correlation spectroscopy (FCS) is a common technique used by physicists, chemists, and biologists to experimentally characterize the dynamics of fluorescent species.
The key of the technique is the auto correlation function. The (temporal) autocorrelation function is the correlation of a time series with itself shifted by time τ, as a function of τ:
The formula is given by
$$ G(\tau) = \frac{\langle \delta I(t) \delta I (t+\tau)\rangle}{\langle I(t)\rangle^2}
= \frac{\langle I(t)I(t+\tau)\rangle}{\langle I(t)\rangle^2} - 1$$
where
$$ \delta I(t) = I(t)-\langle I(t)\rangle $$
is the deviation from the mean intensity.
My question is:
Given a row vector of finite elements, what is the right way to calculate G(tau).
The matlab code below shows two methods giving two different result. Which one is the correct one?
```
rawdata = [1 2 3 4 5 6 7 8];
%the regular method.Result shown in the next line
%0.259259259 0.2 0.151515152 0.111111111 0.076923077 0.047619048 0.022222222
count = rawdata;
Ntime = length(count);
G = [];
for t = 0:Ntime-1
top = [];
bottom = [];
ai = [];
bi = [];
ai = count(1:end-t);
bi = count(t+1:end);
top = mean( (ai-mean(ai)) .* (bi-mean(bi)) );
bottom = mean(ai) * mean(bi);
%bottom = mean(count)^2;
G = [G, top/bottom];
end
%The typical FFT method. Results shown in the next line
%0.259259259 0.086419753 -0.037037037 -0.111111111 -0.135802469 -0.111111111 -0.037037037 0.086419753
%NFFT euqals to the length of the input.
count = rawdata;
NFFT = 8;
tmpGfft = length(count) * ifft( fft(count,NFFT).*conj(fft(count,NFFT)))/sum(count)^2 - 1;
plot(G,'*');hold on; plot(tmpGfft,'o')
```
More reference on the technique itself, <http://en.wikipedia.org/wiki/Fluorescence_correlation_spectroscopy>
| https://mathoverflow.net/users/5096 | Which way is the right way to calculate auto correlation function | I am not sure about my answer, as I am not completely fluent in MatLab. The "regular" method is self-evident enough that I can parse it. The FFT method, if I am interpreting it correctly, is by evaluating the mean $\langle I(t+\tau)I(t)\rangle $ via convolution as $I\*I(\tau)$?
So of course the two answers are different: the main difference is that using the FFT method, you are extending the raw data to a periodic (in time) data of period 8, whereas using the regular data, you are extending the raw data to a pulsed-data so that before $t = 0$ the data is all 0, and after $t = 7$ the data is also all 0.
So in essence, you are feeding into the algorithm two different sets of data. So you get two different answers. If you change ai and bi in your regular method so that, instead of sub-arrays, you shift the array (fix ai = count, but bi = pop the first t elements off of count and push them to the end), you'll probably get then the same answer for the two methods.
I am also a bit troubled by the fact that you computed bottom as mean(ai) \* mean(bi), that seems wrong to me. Your original mean(count)^2 which you commented out feels more right.
Of course, this all depends on what you mean by the auto-correlation of a finite time-sequence (whether you should extend the data periodically or impulsively). If the raw data you are processing is roughly periodic, then you should extend periodically, and then you should just use the FFT method. If the raw data you are processing is not periodic, personally I don't think the auto-correlation function makes much sense unless $\tau$ is much much smaller than your sample time-span (in which case you may as well assume the data is periodic since the error terms will be small). In that case, the easiest thing to do is to pre- and post-pad your data with as many zeroes as there are data points (if there were N data points to start with, pad N zeroes in front and N zeroes after), and do the FFT and read off the answer using positions N through 2N - 1 (okay, maybe up to a normalization...)
In any case, the MatLab code you wrote for the regular method is almost certainly wrong: all the places you write mean(ai) and mean(bi) should really be mean(count) for the expressions to make sense compared with the mathematical formula you wrote.
| 1 | https://mathoverflow.net/users/3948 | 20157 | 13,400 |
https://mathoverflow.net/questions/20154 | 7 | In this question: [What is the definition of "canonical"?](https://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical)
, people gave interesting "philosophical" takes on what the word "canonical" means. Moreover I percieved an underlying opinion that there was no formal mathematical definition.
Whilst looking for something else entirely, I just ran into Bill Messing's post
<http://www.cs.nyu.edu/pipermail/fom/2007-December/012359.html>
on the FOM (Foundations of Mathematics) mailing list. I'll just quote the last paragraph:
"It is my impression that there is very little FOM discussion of either
Hilbert's epsilon symbol or of Bourbaki formulation of set theory. In
particular the chapitre IV Structures of Bourbaki. For reasons,
altogether mysterious to me, the second edition (1970) of this book
supressed the appendix of the first edition (1958). This appendix gave
what is, as far as I know, the only rigorous mathematical discussion of
the definition of the word "canonical". Given the fact that Chevalley
was, early in his career, a close friend of Herbrand and also very
interested in logic, I have guessed that it was Chevalley who was the
author of this appendix. But I have never asked any of the current or
past members of Bourbaki whom I know whether this is correct."
It's a 4-day weekend here in the UK and I'm very unlikely to get to a library to find out what this suppressed appendix says. Wouldn't surprise me if someone could find this appendix on the web somewhere though! Is there really a *mathematical* definition of "canonical"??
NOTE: if anyone has more "philosophical" definitions of the word, they can put them in the other thread. I am hoping for something different here.
| https://mathoverflow.net/users/1384 | candidate for rigorous _mathematical_ definition of "canonical"? | Although the Bourbaki formulation of set theory is very seldom used in foundations, the existence of a definable Hilbert $\varepsilon$ operator has been well studied by set theorists but under a different name. The hypothesis that there is a definable well-ordering of the universe of sets is denoted V = OD (or V = HOD); this hypothesis is equivalent to the existence of a definable Hilbert $\varepsilon$ operator.
More precisely, an [ordinal definable set](http://en.wikipedia.org/wiki/Ordinal_definable_set) is a set $x$ which is the unique solution to a formula $\phi(x,\alpha)$ where $\alpha$ is an ordinal parameter. Using the reflection principle and syntactic tricks, one can show that there is a single formula $\theta(x,\alpha)$ such that for every ordinal $\alpha$ there is a unique $x$ satisfying $\theta(x,\alpha)$ and every ordinal definable set is the unique solution of $\theta(x,\alpha)$ for some ordinal $\alpha$. Therefore, the (proper class) function $T$ defined by $T(\alpha) = x$ iff $\theta(x,\alpha)$ enumerates all ordinal definable sets.
The axiom V = OD is the sentence $\forall x \exists \alpha \theta(x,\alpha)$. If this statement is true, then given any formula $\phi(x,y,z,\ldots)$, one can define a Hilbert $\varepsilon$ operator $\varepsilon x \phi(x,y,z,\ldots)$ to be $T(\alpha)$ where $\alpha$ is the first ordinal $\alpha$ such that $\phi(T(\alpha),y,z,\ldots)$ (when there is one).
The statement V = OD is independent of ZFC. It implies the axiom of choice, but the axiom of choice does not imply V = OD; V = OD is implied by the axiom of constructibility V = L.
---
When I wrote the above (which is actually a reply to Messing) I was expecting that Bourbaki would define canonical in terms of their $\tau$ operator (Bourbaki's $\varepsilon$ operator). However, I was happily surprised when reading the 'état 9' that Thomas Sauvaget found, they make the correct observation that $\varepsilon$ operators do not generally give canonical objects.
A term is said to be 'canonically associated' to structures of a given species if (1) it makes no mention of objects other than 'constants' associated to such structures and (2) it is invariant under transport of structure. Thus, in the species of two element fields the terms 0 and 1 are canonically associated to the field F, but $\varepsilon x(x \in F)$ is not since there is no reason to believe that it is invariant under transport of structures. They also remark that $\varepsilon x(x \in F)$ is actually invariant under automorphisms, so the weaker requirement of invariance under automorphisms does not suffice for being canonical.
---
To translate 'canonically associated' in modern terms:
1) This condition amounts to saying that the 'term' is definable without parameters, without any choices involved. (Note that the language is not necessarily first-order.)
2) This amounts to 'functoriality' (in the loose sense) of the term over the [core groupoid](http://ncatlab.org/nlab/show/core) of the concrete category associated to the given species of structures.
So this seems to capture most of the points brought up in the answers to the [earlier question](https://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical).
| 12 | https://mathoverflow.net/users/2000 | 20159 | 13,402 |
https://mathoverflow.net/questions/20149 | 11 | I just skimmed a bit of [this fresh-off-the-press paper on homological mirror symmetry for general type varieties](http://arxiv.org/abs/1004.0129).
One thing that intrigued me was statement (ii) of Conjecture 3.3. It suggests that, just as there are Fourier-Mukai functors $DCoh(X) \to DCoh(Y)$ associated to objects of $DCoh(X \times Y)$, there are also "Fourier-Mukai" functors $DFuk(M) \to DFuk(N)$ associated to objects of $DFuk(M \times N)$. How does this work, or how might this work? The paper does not seem to explain it.
| https://mathoverflow.net/users/83 | "Fourier-Mukai" functors for Fukaya categories? | I can't speak for these authors, but what I understand by a "Fourier-Mukai" transform between Fukaya categories is *the functor between extended Fukaya categories associated with a Lagrangian correspondence*. I expect these will appear a good deal in the next few years, in symplectic topology and its applications to low-dim topology and mirror symmetry (cf. these papers of [Auroux](http://arxiv.org/abs/1003.2962) and [Abouzaid-Smith](http://arxiv.org/abs/0903.3065)).
A *Lagrangian correspondence* from $X$ to $Y$ is an embedded Lagrangian submanifold of $X\times Y$ with symplectic form $(-\omega\_X)\oplus\omega\_Y$. One can take the graph of a symplectomorphism, for instance, or the vanishing moment-map locus $\mu^{-1}(0)$ as a correspondence from a Hamiltonian $G$-manifold $M$ to the quotient $\mu^{-1}(0)/G$. According to Wehrheim-Woodward, a *generalised Lagrangian* in $X$ is a sequence of symplectic manifolds $X\_0=pt., X\_1, X\_2,\dots,X\_d=X$ and Lagrangian correspondences $L\_{i,i+1}$ from $X\_i$ to $X\_{i+1}$. Generalised Lagrangians (subject to the usual sorts of restrictions and decorations) form objects in the *extended Fukaya category* $F^{\sharp}(X)$, whose $A\_\infty$-structure is under construction by Ma'u-Wehrheim-Woodward. If it happens that two adjacent Lagrangian correspondences have a smooth, embedded composition, say $L' = L\_{i+1,i+2}\circ L\_{i,i+1}$, then deleting $X\_{i+1}$ from the chain and substituting $L'$ for its two factors results in an isomorphic object; see [arxiv:0905.1368](http://arxiv.org/abs/0905.1368).
The geometric mechanism behind this is the idea of "pseudo-holomorphic quilts", see [arXiv:0905.1369](http://arxiv.org/abs/0905.1369), [arxiv:math.SG.0606061](http://arxiv.org/abs/math/0606061).
Whilst $F^\sharp(X)$ seems even less tractable than the Fukaya category $F(X)$, we expect that in many cases the embedding $F(X)\to F^\sharp(X)$ induces a quasi-isomorphism of module categories.
A Lagrangian correspondence from $X$ to $Y$ induces a "Fourier-Mukai" $A\_\infty$-functor between extended Fukaya categories. Even better, we expect that in this way one gets an $A\_\infty$-functor $F(X\_-\times Y)\to \hom(F^{\sharp}(X),F^{\sharp}(Y))$.
*Interpolated paragraph, in response to Kevin's query*: The definition of the $A\_\infty$-functor associated with a correspondence $C$ from $X$ to $Y$ is very simple, at least on objects. An object of $F^\sharp(X)$ is a chain of Lagrangian correspondences, beginning at the 1-point manifold and ending at $X$. We just tack $C$ to the end of this sequence. The clever thing about this formalism is that when there's a nice geometric way to pass Lagrangians through a correspondence (for instance taking the preimage of $L\subset \mu^{-1}(0)/G$ in $\mu^{-1}(0)\subset M$) the geometric and formal approaches give quasi-isomorphic objects in $F^\sharp(Y)$. (Usually you can't pass a Lagrangian submanifold through a correspondence without making a terrible mess - hence the formal approach.)
For me, all this is exciting because we can at last compute Floer cohomology using its naturality properties, rather than by direct attack on the equations.
[In the new paper that inspired your question, the authors suggest that their F-M kernels should be coisotropic branes in the sense of Kapustin-Orlov. Floer cohomology for such branes is supposed to be some weird mixture of pseudo-holomorphic discs and Dolbeault cohomology over a holomorphic foliation - but there is no concrete proposal on the table
and for the moment this is just an intriguing idea. For the purposes of homological mirror symmetry, idempotent endomorphisms in the Fukaya category apparently provide the enlargement that K-O observed was necessary.]
| 13 | https://mathoverflow.net/users/2356 | 20162 | 13,404 |
https://mathoverflow.net/questions/20164 | 11 | Recently, I found a paper by Schilling <http://www.jstor.org/pss/2371426>, which mentions that for certain infinite field of algebraic numbers there is an analog of class field theory. By infinite field of algebraic number we mean an infinite extension of $\mathbb{Q}$. The paper cite a previous paper by Moriya which was the origin of the idea. I could not read the later since it is in German. Since the first paper is quite old (1937), I believe there must have been a lot of development in the mean time.
My question: Do we have an analog of class field theory over an arbitrary infinite field of algebraic number?
An even more general question: Do we have an analog of class field theory over an arbitrary field. This seems a bit greedy, but since we know that an algebraic closed field of characteristic 0 is totally characterized by its trancendence degree so if the answer to the previous question is positive the answer to this is perhaps not too far. Am I making sense?
| https://mathoverflow.net/users/2701 | Is there an analog of class field theory over an arbitrary infinite field of algebraic numbers? | Iwasawa theory studies abelian extensions of fields $K$ where $K$ is a $\mathbb{Z}\_p$-extension of $\mathbb{Q}$, that is the Galois group of $K/\mathbb{Q}$ is $\mathbb{Z}\_p$. The corresponding Galois groups (of extensions of $K$) and class groups (of really subfields) of $K$, suitably interpreted, become $\mathbb{Z}\_p$-modules and there are interesting relations between these modules and $p$-adic $L$-functions. It is a vast subject. Washington's book, Introduction to Cyclotomic Fields, is a good entry point.
| 8 | https://mathoverflow.net/users/2290 | 20167 | 13,408 |
https://mathoverflow.net/questions/20168 | 5 | More generally, I'm interested in the situation of Lagrangian mechanics. And actually my question is local, so you can work on $\mathbb R^n$ if you like. I will begin with some background on Lagrangian mechanics, and then recall the notion of Morse index in the positive-definite case. My final question will be whether there is a similar story in the indefinite case.
Background on Lagrangian mechanics
----------------------------------
Let $\mathcal N$ be a smooth manifold. A **Lagrangian** on $\mathcal N$ is a function $L: {\rm T}\mathcal N \to \mathbb R$, where ${\rm T}\mathcal N$ is the tangent bundle to $\mathcal N$. I will always suppose that the Lagrangian is **nondegenerate**, in the sense that when restricted to any fiber ${\rm T}\_q\mathcal N$, the second derivative $\frac{\partial^2 L}{\partial v^2}$ is everywhere invertible (for $v\in {\rm T}\_q\mathcal N$, $\frac{\partial^2 L}{\partial v^2}(v)$ makes sense as a map ${\rm T}\_v({\rm T}\_q\mathcal N) \to {\rm T}\_v^\*({\rm T}\_q\mathcal N)$). In this case, the **Euler-Lagrange equations**
$$ \frac{\partial L}{\partial q}\bigl(\dot\gamma(t),\gamma(t)\bigr) = \frac{\rm d}{{\rm d}t}\left[ \frac{\partial L}{\partial v}\bigl(\dot\gamma(t),\gamma(t)\bigr)\right]$$
define a nondegenerate second order differential equation for $\gamma$ a parameterized path in $\mathcal N$.
An important example is when $\mathcal N$ is equipped with a (semi-)Riemannian metric, in which case the Euler-Lagrange equations specify that $\gamma$ is an arc-length-parameterized geodesic.
By nondegeneracy, a solution to the Euler-Lagrange equations is determined by its initial conditions, a point $\bigl(\dot\gamma(0),\gamma(0)\bigr) \in {\rm T}\mathcal N$. Thus, the Euler-Lagrange equations determine a smooth function $\text{flow}: {\rm T}\mathcal N \times \mathbb R \to \mathcal N \times \mathcal N \times \mathbb R$, which sends an initial condition and a duration to the triple (initial location, final location, duration). Actually, the function is only defined on an open neighborhood of the zero section of ${\rm T}\mathcal N \times \mathbb R$.
I will use the shorthand **path** to mean a smooth function $[0,T] \to \mathcal N$, i.e. a parameterized path of finite duration. A path is **classical** if it solves the Euler-Lagrange equations, so that classical paths are in bijection with points in (that open neighborhood of) ${\rm T}\mathcal N \times \mathbb R$. A classical path is **nonfocal** if near it the function $\text{flow}: {\rm T}\mathcal N \times \mathbb R \to \mathcal N \times \mathcal N \times \mathbb R$ is a local diffeomorphism. Thus a choice of nonfocal classical path of duration $T$ determines a function $\gamma: \mathcal O \times [0,T] \to \mathcal N$, where $\mathcal O \subseteq \mathcal N \times \mathcal N$, such that for each $(q\_0,q\_1) \in \mathcal O$, the path $\gamma(q\_0,q\_1,-)$ is classical with $\gamma(q\_0,q\_1,0) = q\_0$ and $\gamma(q\_0,q\_1,T) = q\_1$. Given a nonfocal classical path $\gamma$, the corresponding **Hamilton principal function** $S\_\gamma: \mathcal O \to \mathbb R$ is given by:
$$S\_\gamma(q\_0,q\_1) = \int\_{t=0}^T L\left( \frac{\partial \gamma}{\partial t}(q\_0,q\_1,t), \gamma(q\_0,q\_1,t)\right){\rm d}t$$
Finally, given a classical path $\gamma: [0,T] \to \mathcal N$, there is a well-defined second-order linear differential operator $D: \gamma^\*{\rm T}\mathcal N \to \gamma^\*{\rm T}^\*\mathcal N$, given by:
$$ D\_\gamma[\xi] = -\frac{\rm d}{{\rm d}t} \left(\frac{\partial^2 L}{\partial v^2} \frac{{\rm d}\xi}{{\rm d}t}\right) - \frac{\rm d}{{\rm d}t}\left( \frac{\partial^2 L}{\partial q \partial v}\xi\right) + \frac{\partial^2 L}{\partial v \partial q} \frac{{\rm d}\xi}{{\rm d}t} + \frac{\partial^2 L}{\partial q \partial q}\xi$$
The second derivatives of $L$ are evaluated at $(\dot\gamma(t),\gamma(t))$ and act as "matrices"; in particular, $\frac{\partial^2 L}{\partial v \partial q}$ and $\frac{\partial^2 L}{\partial q \partial v}$ are transpose to each other. These individual matrices require local coordinates to be defined, but $D\_\gamma$ is well-defined all together if $\gamma$ is classical.
Then $\gamma$ is nonfocal iff $D\_\gamma$ has trivial kernel among the space of sections of $\gamma^\*{\rm T}\mathcal N \to [0,T]$ that vanish at $0,T$.
Of course, really what's going on is that for $(q\_0,q\_1) \in \mathcal N \times \mathcal N$, the space of paths of duration $T$ that start at $q\_0$ and end at $q\_1$ is an infinite-dimensional manifold. Using the Lagrangian $L$ we can define an **action** function on this manifold. The Euler-Lagrange equations assert that a path $\gamma$ is a critical point for this function, $S\_\gamma$ is the value of the function, and the operator $D\_\gamma$ is the Hessian at that point.
The Morse index of a classical path
-----------------------------------
Recall the following fact. Let $\mathcal V$ be a vector space and $D: \mathcal V\otimes \mathcal V \to \mathbb R$ a symmetric bilinear form on $\mathcal V$. Then any subspace $\mathcal V\_- \subseteq \mathcal V$ that is maximal with respect to the property that $D|\_{\mathcal V\_-}$ is negative-definite has the same dimension as any other such subspace. This dimension is the **Morse index** $\eta$ of the operator $D$ acting on $\mathcal V$.
Recall that there is a canonical pairing between sections of $\gamma^\*{\rm T}\mathcal N$ and sections of $\gamma^\*{\rm T}^\*\mathcal N$ (pairing the vectors and covectors for each $t\in [0,T]$ gives a function on $[0,T]$, which we then integrate). By composing with this pairing, we can think of the operator $D\_\gamma$ as a bilinear form on $\gamma^\*{\rm T}\mathcal N$, and it is symmetric on the space of sections of $\gamma^\*{\rm T}\mathcal N$ that vanish at the endpoints $0,T$.
Given a nonfocal classical path $\gamma$, its **Morse index** $\eta(\gamma)$ is the Morse index of $D\_\gamma$ acting on such endpoint-zero sections.
Let $L$ be a Lagrangian on $\mathcal N$, and assume moreover that the matrix $\frac{\partial^2 L}{\partial v^2}$ is not just everywhere invertible but actually everywhere positive-definite (this is a coordinate-independent statement, even though the value of the matrix depends on coordinates). Then the Morse index of any nonfocal classical path is finite. (And conversely: if $\frac{\partial^2 L}{\partial v^2}$ is not positive-definite along $\gamma$, then the Morse index as defined above is infinite.)
Moreover, suppose that $\gamma: [0,T] \to \mathcal N$ is classical and nonfocal and choose $T' \in [0,T]$ such that the obvious restrictions $\gamma\_1: [0,T'] \to \mathcal N$ and $\gamma\_2: [T',T] \to \mathcal N$ are both nonfocal. Let $S\_{\gamma\_1}$ and $S\_{\gamma\_2}$ be the corresponding Hamilton-principle functions. Then $q = \gamma(T')$ is a nondegenerate critical point for $S\_{\gamma\_1}(\gamma(0),-) + S\_{\gamma\_2}(-,\gamma(T))$. Define its Morse index of $\eta(q)$ to be the number of negative eigenvalues of the Hessian of $S\_{\gamma\_1}(\gamma(0),-) + S\_{\gamma\_2}(-,\gamma(T))$ at $q$. Then the following is a fact:
>
> $\eta(\gamma) = \eta(\gamma\_1) + \eta(q) + \eta(\gamma\_2)$
>
>
>
My question
-----------
Is there a similar story when $\frac{\partial^2 L}{\partial v^2}$ is invertible but indefinite? More precisely:
>
> Suppose that you are given a nondegenerate (but not convex on fibers) Lagrangian $L$ on a manifold $\mathcal N$. Is there a way to assign a (finite) number $\eta(\gamma)$ to each classical nonfocal path $\gamma$ such that $\eta(\gamma) = \eta(\gamma\_1) + \eta(q) + \eta(\gamma\_2)$, where $\gamma,\gamma\_1,\gamma\_2$ are as above, and $\eta(q)$ is the usual Morse index of $S\_{\gamma\_1}(\gamma(0),-) + S\_{\gamma\_2}(-,\gamma(T))$?
>
>
>
The starting idea would be to recall the fact that in the positive-definite case, $\eta(\gamma)$ counts with multiplicity the number of times $T' \in [0,T]$ such that the restriction $\gamma|\_{[0,T']}$ is focal. (The multiplicity is given by the rank of the differential of the flow map.) This counting still makes sense in the indefinite case. So perhaps it can be used, and the signature of $S\_{\gamma\_1}(\gamma(0),-) + S\_{\gamma\_2}(-,\gamma(T))$ can be added in by hand?
| https://mathoverflow.net/users/78 | Is there a notion of "Morse index" for geodesics in a manifold with indefinite metric that is well-behaved under cutting and gluing? | Yes, there is: [The Maslov index and a generalized Morse index theorem for non-positive definite metrics](http://www.ams.org/mathscinet-getitem?mr=1784919).
| 2 | https://mathoverflow.net/users/394 | 20169 | 13,409 |
https://mathoverflow.net/questions/20176 | 20 | The Eilenberg-Maclane space $K(\mathbb{Z}/2\mathbb{Z}, 1)$ has a particularly simple cell structure: it has exactly one cell of each dimension. This means that its "Euler characteristic" should be equal to
$$1 - 1 + 1 - 1 \pm ...,$$
or Grandi's series. Now, we "know" (for example by analytic continuation) that this sum is morally equal to $\frac{1}{2}$. One way to see this is to think of $K(\mathbb{Z}/2\mathbb{Z}, 1)$ as infinite projective space, e.g. the quotient of the infinite sphere $S^{\infty}$ by antipodes. Since $S^{\infty}$ is contractible, the "orbifold Euler characteristic" of the quotient by the action of a group of order two should be $\frac{1}{2}$.
More generally, following John Baez $K(G, 1)$ for a finite group $G$ should be "the same" (I'm really unclear about what notion of sameness is being used here) as $G$ thought of as a one-object category, which has groupoid cardinality $\frac{1}{|G|}$. In particular, $K(\mathbb{Z}/n\mathbb{Z}, 1)$ should have groupoid cardinality $\frac{1}{n}$. I suspect that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ has $1, n-1, (n-1)^2, ...$ cells of each dimension, hence orbifold Euler characteristic
$$\frac{1}{n} = 1 - (n-1) + (n-1)^2 \mp ....$$
Unfortunately, I don't actually know how to construct Eilenberg-Maclane spaces...
**Question 1a:** How do I construct $K(\mathbb{Z}/n\mathbb{Z}, 1)$, and does it have the cell structure I think it has? (I've been told that one can write down the cell structure of $K(G, 1)$ for a finitely presented group $G$ explicitly, but I would really appreciate a reference for this construction.)
**Question 2:** $K(\mathbb{Z}/2\mathbb{Z}, 1)$ turns out to be "the same" as the set of all finite subsets of $(0, 1)$, suitably interpreted; the finite subsets of size $n$ form the cell of dimension $n$. Jim Propp and other people who think about combinatorial Euler characteristic would write this as $\chi(2^{(0, 1)}) = 2^{\chi(0, 1)}$. Is it true more generally that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ is "the same" as the set of all functions $(0, 1) \to [n]$, suitably interpreted?
**Question 3:** What notion of "sameness" makes the above things I said actually true?
**Question 4:** Let $G$ be a finite group and let $K(G, 1)$ be constructed using the standard construction I asked about in Question 1. If $c\_n$ denotes the number of cells of dimension $n$, let $f\_G(z) = \sum\_{n \ge 0} c\_n z^n$. Can $f\_G$ always be analytically continued to $z = -1$ so that $f\_G(-1) = \frac{1}{|G|}$?
| https://mathoverflow.net/users/290 | What's the cell structure of K(Z/nZ, 1)? Does it let me sum this divergent series? What about other finite groups? | There are multiple possible cell structures on K(Z/n,1).
One is generic. For any finite group G there is a model for BG that has (|G|-1)k new simplices in each nonzero degree k. This is the standard simplicial bar construction of K(G,1). This gives you that BG has Euler characteristic 1/|G|, if you like.
One is more specific. There is another cell structure on K(Z/n,1), viewing it as a union of generalized lens spaces, that has exactly one cell in each degree. This is a topological avatar of the "simple" resolution of Z by free Z[Z/n]-modules. Obviously this doesn't give you the Euler characteristic argument you're seeking - one needs to keep track of more intricate information about the cell attachments in order to extract something.
| 14 | https://mathoverflow.net/users/360 | 20177 | 13,412 |
https://mathoverflow.net/questions/20174 | 9 | I'd like a name for an augmented algebra $A = \langle 1\rangle \oplus A\_+$ for which there is an $N$ so that any product of more than $N$ elements in the augmentation ideal is $0$, i.e., $(A\_+)^N = 0$. Is there a name? It seems related to nilpotence, and it implies that all elements in $A\_+$ are nilpotent, but is stronger than that. There is a uniform bound on the degree of nilpotence, but that's not enough either, as the example of the exterior algebra in infinitely many variables over $\mathbb{Z}/2$ shows.
MathWorld defines a nilpotent algebra or nilalgebra to be one where every element is nilpotent. (They are therefore not considering unital algebras, contrary to an [earlier discussion](https://mathoverflow.net/questions/15107/algebra-unital-associative-algebra-better-terminology) here.) Is this standard? Is there a better term?
| https://mathoverflow.net/users/5010 | Terminology: Algebras where long strings of products are 0? | I believe the terminology you want is that "the augmentation ideal is locally nilpotent." See, e.g. [Link](https://planetmath.org/NilAndNilpotentIdeals), although there are surely places in actual literature where this is used. I think an ideal being nilpotent means that the powers of the ideal are eventually zero (and is stronger than just requiring it to be a nil-ideal, meaning it's comprised of individually nilpotent elements). So a nilpotent ideal is one where there is a uniform bound, i.e. I^N=0 for some sufficiently large N. Locally nilpotent means that for any finitely generated subalgebra there exists such an N.
Is that the notation you want? I don't know a name for an augmented algebra whose ideal is locally nilpotent other than just that.
| 6 | https://mathoverflow.net/users/1040 | 20178 | 13,413 |
https://mathoverflow.net/questions/20172 | 8 | I come across the following problem in my study.
Let $x\_i, y\_i\in \mathbb{R}, i=1,2,\cdots,n$ with $\sum\limits\_{i=1}^nx\_i^2=\sum\limits\_{i=1}^ny\_i^2=1$, and $a\_1\ge a\_2\ge \cdots \ge a\_n>0 $. Is it true $$\left(\frac{\sum\limits\_{i=1}^na\_i(x\_i^2-y\_i^2)}{a\_1-a\_n}\right)^2\le 1-\left(\sum\limits\_{i=1}^nx\_iy\_i\right)^2~~?$$
Has anyone seen this inequality before, or can you give a counterexample?
| https://mathoverflow.net/users/3818 | A plausible inequality | [Wrong ounter-example deleted]
This it true for all $n$. The case $n=2$ is handled by Hailong Dao, let's reduce the general case to $n=2$.
First, we may assume that $a\_1=1$ and $a\_n=0$ as others mentioned. So remove the denominator in LHS. Then forget the condition that $a\_i$ are monotone, let's only assume that they are in $[0,1]$ (we can rearrange the indices anyway). Also we may assume that the sum under the square in the LHS is nonnegative - otherwise swap $x$ and $y$.
Then, for every $i$ such that $x\_i^2-y\_i^2>0$, set $a\_i=1$, otherwise $a\_i=0$. The LHS grows, the RHS stays. So it suffices to prove the inequality for $a\_i\in\{0,1\}$. Rearrange indices so that the first $k$ of $a\_i$'s are 1. We arrive to
$$
\left(\sum\_{i=1}^k x\_i^2-\sum\_{i=1}^ky\_i^2 \right)^2 \le 1 - \left(\sum\_{i=1}^n x\_iy\_i\right)^2 .
$$
Define $X\_1,X\_2,Y\_1,Y\_2\ge 0$ by
$$
X\_1^2 = \sum\_{i=1}^k x\_i^2, \ \ \ X\_2^2 = \sum\_{i=k+1}^n x\_i^2, \ \ \
Y\_1^2 = \sum\_{i=1}^k y\_i^2, \ \ \ Y\_2^2 = \sum\_{i=k+1}^n y\_i^2.
$$
Then the LHS equals $(X\_1^2-Y\_1^2)^2$ and $X\_1^2+X\_2^2=Y\_1^2+Y\_2^2=1$. By Cauchy-Schwarz,
$$
\left|\sum x\_iy\_i\right| \le X\_1Y\_1+X\_2Y\_2 ,
$$
so the RHS is greater or equal to $1-(X\_1Y\_1+X\_2Y\_2)^2$. Now the inequality follows from
$$
(X\_1^2-Y\_1^2)^2 \le 1-(X\_1Y\_1+X\_2Y\_2)^2
$$
which is the same inequality for $n=2$.
| 13 | https://mathoverflow.net/users/4354 | 20179 | 13,414 |
https://mathoverflow.net/questions/20181 | 5 | I need the year of death of the following mathematicians all of whom are written up in R.C.Archibald's book Mathematical Table Makers.
Carl Burrau, b. 1867, d. ???? - Danish, astronomer and actuary
Herbert Bristol Dwight, b. 1885, d. ???? - American, tables of integrals
Alexander John Thompson, b. 1885, d. ????, British, statistician, BAASMTC
Thanks for any insight.
Cheers, Scott
| https://mathoverflow.net/users/4111 | Query: Year of Death of Three Mathematicians | Burrau died in 1947 ([source](http://linkinghub.elsevier.com/retrieve/pii/S0315086085710154))
Dwight died in 1975 ([source](http://ru.wikisource.org/wiki/%D0%93%D0%B5%D1%80%D0%B1%D0%B5%D1%80%D1%82_%D0%91%D1%80%D0%B8%D1%81%D1%82%D0%BE%D0%BB%D1%8C_%D0%94%D0%B2%D0%B0%D0%B9%D1%82))
I could not find info on A.J. Thompson's year of death.
| 9 | https://mathoverflow.net/users/4925 | 20185 | 13,417 |
https://mathoverflow.net/questions/20184 | 38 | I am currently trying to get my head around flatness in algebraic geometry. In particular, I'm trying to relate the notion of flatness in algebraic geometry to the notion of fibration in algebraic topology, because they do formally seem quite similar. I'm guessing that the answers to my questions are "well-known", but I am struggling to find anything decent in the literature. Any help/references will be most useful.
The set up is this: Let $E,B$ be smooth projective complex algebraic varieties, and let $\pi:E \to B$ be a surjective flat map such that the fibres $E\_b:=\pi^{-1}(b), b \in B$ are smooth projective complex algebraic varieties.
I am aware that each fibre has the same Hilbert polynomial, so cohomologically they are quite similar. But each fibre can certainly be non-isomorphic as algebraic varieties (e.g. moduli spaces). However:
1. Using GAGA type methods, we can consider $E$ and $B$ as complex manifolds. Is it true that $(\pi,E,B)$ is a fibration? That is, satisfies the homotopy lifting property with respect to any topological space?
2. Again considering $E,B$ and each fibre $E\_b$ as a complex manifold, is it true that each fibre $E\_b$ is homotopy equivalent to each other? What about homeomorphic/diffeomorphic?
Thanks,
Dan
| https://mathoverflow.net/users/5101 | Flatness in Algebraic Geometry vs. Fibration in Topology | A surjective flat (equals faithfully flat) map with smooth fibres is in fact a smooth morphism, and hence induces a submersion on the underlying manifolds obtained by passing
to complex points. Since the fibres are projective, it is furthermore proper (in the sense of algebraic geometry) [see the note added at the end; this is not a logical deduction from the given condition on the fibres, but nevertheless seems to be a reasonable reinterpretation of that condition], and hence proper (in the sense of topology). A theorem
of Ehresmann states that any proper submersion of smooth manifolds is a fibre bundle.
In particular, it is a fibration in the sense of homotopy theory, and the fibres are
diffeomorphic (thus also homeomorphic, homotopic, ... ).
Note: Your specific question is really about smooth morphisms (these are flat morphisms
with smooth fibres, although there are other definitions too, which are equivalent
under mild hypotheses on the schemes involved, and in particular, are equivalent for maps
of varieties over a field). One point about the notion of flat map is that it allows one
to consider cases in which the fibres over certain points degenerate, but still vary continuously (in some sense). It may well be a special feature of algebraic geometry
(and closely related theories such as complex analytic geometry) that one can have such a reasonable notion, a feature related to the fact that one can
work in a reasonable manner with singular spaces in algebraic geometry, because the singularities are so mild compared to what can occur in (say) differential topology.
[Added: I should add that I took a slight liberty with the question, in that I interpreted the condition that the fibres are projective stronger than is literally justified, in so far as I replaced it with the condition that the map is proper. As is implicit in Chris Schommer-Pries's comment below, we can find non-proper smooth surjections whose fibres are projective varieties: e.g. if, as in his example, we consider the covering of $\mathbb P^1$ by two copies of $\mathbb A^1$ in the usual way, then the fibres consists of either one or two points (one point for $0$ and the point at $\infty$, two points for all the others), and any finite set of points is certainly a projective variety.
Nevertheless, my interpretation of the question seems to have been helpful; hopefully, with the addition of this remark, it is not too misleading.]
| 38 | https://mathoverflow.net/users/2874 | 20187 | 13,419 |
https://mathoverflow.net/questions/20192 | 8 | I have several questions about Steinberg group and K2 for symplectic group:
1. Can I extend the definition of Steinberg symbols to symplectic case? Will they generate the center of Steinberg group?
2. Does the center of symplectic Steinberg group coincide with K2 (the kernel of $\mathrm{SpSt}\rightarrow\mathrm{Sp}$ as usual)?
3. Is there an analogue for Matsumoto's theorem?
I tryed to read "Sur les sous-groupes arithmetiques des groupes semi-simples deployes" by Hideya Matsumoto and all I got to know about symplectic case is that there is some problems with long roots in Cl. Also, is it written in english anywhere about non-Al K-theory? There is "The Classical groups and K-theory" by Hahn and O'Meara, but it tells about SL and about unitary groups only.
| https://mathoverflow.net/users/5018 | Symplectic Steinberg group | There is useful information about the symplectic analogues of the Steinberg group, the Steinberg symbols, and the $K\_2$ functor in some of Michael Stein's papers from the '70's. In particular, see his papers "Generators, relations and coverings of Chevalley groups over commutative rings" and "Surjective stability in dimension $0$ for $K\_{2}$ and related functors" and "Injective stability for $K\_{2}$ of local rings".
| 6 | https://mathoverflow.net/users/317 | 20198 | 13,423 |
https://mathoverflow.net/questions/20203 | -2 | Please forgive me if this is inappropriate for MathOverflow. I've been working/playing with generating functions for a little while and may have stumbled upon a new technique or methodology.
The problem is that it's incomplete, and I don't have a lot to show to someone to prove its effectiveness. I believe I need some help fine tuning this method to get it to work. I'm unsure if I can solve it, at least easily, on my own. I've tried contacting a few big name professors in the field, but most seemed to busy. I was wondering if there's someone who could spare some time to look into what I'm working on.
Are there any suggestions of people that I could try contacting? If it's not advisable to go forward with contacting people at this time, what else should/could I try to help get answers? I realize I haven't described my method here; it's a pretty complicated one, but one that can probably be explained in a few pages.
| https://mathoverflow.net/users/3647 | Where do I turn for help with generating functions? | Your best bet is to clarify what you have first, in line with accepted current terminology. Then type up what you have in a few pages with something like Latex. I note your question uses none, and that may be one reason people did not reply to your queries.
To be specific, there are two free books available online, one by Wilf called generatingfunctionology, one by Flajolet and Sedgewick called something like Analytic Combinatorics, both titles are given correctly with links in the answers to this:
[Free, high quality mathematical writing online?](https://mathoverflow.net/questions/1722/free-high-quality-mathematical-writing-online)
If you can get a fairly clean description of what you do have, you may contact me, just click on my name and there will be instructions for how to find one of my email addresses. It is not that I am expert in this particular field, of course. But I answer email.
| 7 | https://mathoverflow.net/users/3324 | 20204 | 13,427 |
https://mathoverflow.net/questions/19337 | 30 | I hope these are not to vague questions for MO.
Is there an analog of the concept of a Riemannian metric, in algebraic geometry?
Of course, transporting things literally from the differential geometric context, we have to forget about the notion of positive definiteness, cause a bare field has no ordering. So perhaps we're looking to an algebro geometric analog of *semi-* Riemannian geometry.
Suppose to consider a pair $(X,g)$, where $X$ is a (perhaps smooth) variety and $g$ is a nondegenerate section of the second symmetric power of the tangent bundle (or sheaf) of $X$.
What can be said about this structure? Can some results of DG be reproduced in this context? Is there a literature about this things?
| https://mathoverflow.net/users/4721 | Algebraic (semi-) Riemannian geometry ? | Joel Kamnitzer had a very similar question a couple years ago, that prompted a nice discussion at the [Secret Blogging Seminar](http://sbseminar.wordpress.com/2007/11/15/algebraic-riemannian-manifolds/). I'm afraid no one ended up citing any literature, and I have been unable to find anything with a quick Google search, but that doesn't rule out the possibility of existence.
| 9 | https://mathoverflow.net/users/121 | 20211 | 13,430 |
https://mathoverflow.net/questions/20205 | 1 | Let $U\_\infty$ be a compact space, and let $U\_r$ be an increasing family of compact subspaces whose closure is all of $U\_\infty$. That is, $U\_r \subseteq U\_{r'}$ if $r \le r'$ and $U\_\infty = \overline{\bigcup U\_r}$.
For $r \in [1,\infty]$, let $Y\_r = C(U\_r,\mathbb R)$ be the Banach space of real-valued continuous functions over $U\_r$ with the supremum norm. For $r \le r'$, let $\phi\_{r,r'} : Y\_{r'} \to Y\_r$ be the restriction maps, so that $Y\_\infty$ is the inverse limit of the spaces $Y\_r$. Write $\phi\_r : Y\_\infty \to Y\_r$ for the restriction map $\phi\_{r,\infty}$.
Suppose there exists a family of continuous linear operators $m\_r : Y\_r \to Y\_\infty$ such that $\|m\_r\| \le M$ for all $r$, and $\phi\_r \circ m\_r$ is the identity map on $Y\_r$.
**Question:** Suppose $\Gamma \subseteq Y\_\infty$ is compact. Does $m\_r \circ \phi\_r$ converge strongly to the identity operator on $\Gamma$? That is, for all $\epsilon > 0$, does there exist $R > 0$ such that if $r \ge R$, then $$\sup\_{y \in \Gamma} \left\| (m\_r \circ \phi\_r)(y) - y \right\|\_{Y\_\infty} < \epsilon?$$
| https://mathoverflow.net/users/238 | Convergence of operators to the identity on Banach spaces | Contrary to my original muddled guess, the answer is no: the problem is that your `extension operators' don't give enough control over what happens in the gap between $U\_\infty$ and $U\_r$.
For a concrete example, take $U\_r$ to be the closed interval $[r^{-1},2]$ (for $1\leq r\leq\infty$), which clearly satisfies the conditions of the question. Now define the extension operator $m\_r$ as follows: given $f$ continuous and real-valued on $U\_r$, extend it to all of $[0,2]$ by putting $m\_r(f)(0)=0$ and interpolating linearly, i.e.
$$ m\_r(f)(t) = r^{-1}t f(1/r) \hbox{ if } 0\leq t\leq r \hbox{ and } m\_r(f)(t)=f(t)
\hbox{ if } 1/r\leq t \leq 2.$$
Clearly each $m\_r$ is a linear extension operator with norm $1$.
Now let ${\bf 1}$ be the function on $[0,2]$ with constant function $1$. Then
$$ \Vert m\_r\circ\phi\_r({\bf 1}) - {\bf 1} \Vert \geq
\vert m\_r\circ\phi\_r({\bf 1})(0) - {\bf 1}(0) \vert = 1 $$
for all $r<\infty$.
| 3 | https://mathoverflow.net/users/763 | 20212 | 13,431 |
https://mathoverflow.net/questions/20144 | 10 | Let $f: \mathbb R^n \to \mathbb R$ be a smooth function. Then the first derivative $f^{(1)}$ makes sense as a function $\mathbb R^n \to \mathbb R^n$, and the second derivative makes sense as a function $f^{(2)}: \mathbb R^n \to \{\text{symmetric }n\times n\text{ matrices}\}$. I would like either a proof or a counterexample to the following claim:
>
> **Claim:** Suppose that for every $x\in \mathbb R^n$, $f^{(2)}(x)$ is invertible as an $n\times n$ matrix. Then $f^{(1)}: \mathbb R^n \to \mathbb R^n$ is one-to-one.
>
>
>
Some comments:
1. If you replace $\mathbb R$ by $\mathbb C$ and "smooth" by "algebraic", then the only such functions are (inhomogeneous) quadratic in $x$, since over $\mathbb C$ all non-constant functions have zeros. Then $f^{(1)}$ is (inhomogeneous) linear, and so one-to-one.
2. If you replace "invertible" by "positive (or negative) definite", then $f$ is convex, and so the claim follows. In particular, the claim is true when $n=1$.
3. This is a special case of the following more general question. If $g: \mathbb R^n \to \mathbb R^n$ is smooth, then its derivative makes sense as a function $g^{(1)}: \mathbb R^n \to \{n\times n\text{ matrices}\}$. If $g^{(1)}(x)$ is invertible for all $x\in \mathbb R^n$, is $g$ necessarily one-to-one? Of course, then $g$ is *locally* a diffeomorphism, but I don't know if it is globally. I don't think it is.
Oh, and I have no idea what the best tags are.
| https://mathoverflow.net/users/78 | If the second derivative of a function on $\mathbb R^n$ is everywhere nondegenerate, does it follow that the first derivative is an injection? | The counter example given in the comments by Brian Conrad (and Dylan Thurston) is very nice. However, it oscillates wildly at $\infty$, and I believe that if you assume some nice properties at $\infty$ you will get a possitive answer.
Translating $f$ by a linear function does not change the assumptions on $f$ and so the claim is equivalent to the fact that all such $f$ has no or a unique critical points.
If we add assumptions such that e.g. the Conley index (or homotopy index) is well-defined and a sphere then this modified claim would follow. Indeed, if two or more critical points existed they would by assumptions be non-degenerate with same Morse index and a small pertubation would yield a Conley index which is a vedge of two or more spheres - a contradiction.
| 1 | https://mathoverflow.net/users/4500 | 20223 | 13,440 |
https://mathoverflow.net/questions/20200 | 7 | This is an attempt to extend the current full fledged random matrix theory to fields of positive characteristics. So here is a possible setup for the problem: Let $A\_{n,p}$ be an $n \times n$ matrix with entries iid taking values uniformly in $F\_p$. Then one should be able to find its eigenvalues together with multiplicities, which might lie in some finite extension of the field $F\_p$. To ensure diagonalizability, one might even take $A\_{n,p}$ to be symmetric or antisymmetric (I am not so sure if that guarantees diagonalizability in $F\_p$ but I have no counterexamples either). Now the question is if we associate to each eigenvalue $\lambda$ the degree of its minimal polynomial $d(\lambda)$, then does the distribution of $d(\lambda)$ as $n$ goes to infinite converge to some law upon normalization (say maybe Gaussian)? I am very curious whether others have studied this problem before. Maybe it's completely trivial.
| https://mathoverflow.net/users/4923 | distribution of degree of minimum polynomial for eigenvalues of random matrix with elements in finite field | The survey article
[Jason Fulman, Random matrix theory over finite fields, *Bulletin of the AMS* **39** (2002), 51-85](http://www.ams.org/bull/2002-39-01/S0273-0979-01-00920-X/S0273-0979-01-00920-X.pdf)
and the references therein should answer your questions to the extent that the answers are currently known. See in particular Example 3 in Section 2.2. Roughly, the distribution of the degrees of the factors of the characteristic polynomial of a random matrix over a finite field is close to the distribution of the degrees of the factors of a random polynomial over the same finite field, which is close to the distribution of the cycle lengths of a random element of a symmetric group.
| 12 | https://mathoverflow.net/users/2757 | 20237 | 13,449 |
https://mathoverflow.net/questions/19632 | 5 | Hi to all!
I'm studying complex geometry from Huybrechts book "Complex Geometry"
and i have problems with an exercise, please can anyone help me?
I define the kahler cone of a compact kahler manifold X as the set
$K\_X \subseteq H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$
of kahler classes. I have to prove that $K\_X$ doesn't contain any line
of the form $\alpha + t \beta$ with $\alpha , \beta\in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$
and $\beta\neq 0$ (i identify classes with representatives).
This is what i thought: i know that a form $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$
that is positive definite (locally of the form $\frac{i}{2}\sum\_{i,j} h\_{ij}(x)dz^i\wedge d \overline{z}^{j}$ and $(h\_{ij}(x))$ is a positive definite hermitian matrix $\forall x\in X$) is the kahler form associated to a kahler structure. Supposing $\alpha$ a kahler class i want to show that there is a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not a kahler class. Since $\beta\neq0$ i can find a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not positive definite any more, now i want to prove that there is no form $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ such that $\omega=d\lambda$ with $\lambda$ a real 1-form and $\omega=\overline{\partial}\mu$ with $\mu$ a complex (1,0)-form (what i'd like to prove is: correcting representatives of cohomology classes with an exact form i don't get a kahler class). From $\partial\overline{\partial}$-lemma and a little work i know that $\omega=i\partial\overline{\partial}f$ with f a real function. And now (and here i can't go on) i want to prove that i can't have a function f such that $\alpha + t \beta+i\partial\overline{\partial}f$ is positive definite.
Please, if i made mistakes, or you know how to go on, or another way to solve this, tell me.
Thank you in advance.
| https://mathoverflow.net/users/4971 | question about kahler cone of a compact kahler manifold | Here is another try:
WLOG, we assume $\alpha$ is kahler, fix it as a metric on $M$.
Assume $\alpha+t\beta$ is kahler for every $t$. So
$\int(\alpha+t\beta)\wedge \alpha^{n-1}=\int
\alpha^n+t\int\beta\wedge\alpha^{n-1}>0$ for every $t$. It then
follows $\int\beta\wedge\alpha^{n-1}=0$. In a same manner, by considering $\int(\alpha+t\_1\beta)\wedge(\alpha+t\_2\beta)\wedge\alpha^{n-2}$, we have
$\int \beta^2\wedge\alpha^{n-2}=0$. By Lefschetz decomposition, we
can write $\beta=\beta\_1+c\alpha$, where $\beta\_1$ is a primitive
cohomology class. Then $\beta\_1\wedge\alpha^{n-1}=0$. By the fact
$\int\beta\wedge\alpha^{n-1}=0$, we conclude $c=0$ and $\beta$
itself is primitive class. Then it is a contradiction that $\int
\beta^2\wedge\alpha^{n-2}=0$ unless $\beta=0$ by Hodge-Riemann
bilinear relation for primitive classes.
| 5 | https://mathoverflow.net/users/1947 | 20256 | 13,459 |
https://mathoverflow.net/questions/19169 | 9 | In the Wikipedia article it states that Ramanujan's tau conjecture was shown to be a consequence of Riemann's hypothesis for varieties over finite fields by the efforts of
Michio Kuga, Mikio Sato, Goro Shimura, Yasutaka Ihara, and Pierre Deligne. Do their papers consist of the only published proof of this result? And is this proof of a similar level of difficulty to Deligne's proof of Riemann's hypothesis?
| https://mathoverflow.net/users/4692 | Reference request for a proof of Ramanujan's tau conjecture | Following the discussion at [meta.MO](http://mathoverflow.tqft.net/discussion/328/should-we-do-anything-if-a-question-is-answered-well-in-the-comments/), I'm going to post a good answer from the comments (made by JT) as a "community wiki" answer. I should mention that the Rogawski article mentioned by Tommaso says almost nothing about the proof of Ramanujan's conjecture, but it seems to be a very nice introduction to Jacquet-Langlands.
Deligne reduced Ramanujan's conjecture about the growth of tau to the Weil conjectures (in particular, the Riemann hypothesis) applied to a Kuga-Sato variety, in his paper [Formes modulaires et representations l-adiques](http://www.google.com/search?q=formes+modulaires+et+representations+l-adiques), Seminaire Bourbaki 355. I believe Jay Pottharst has made an English translation available.
Deligne then proved the Weil conjectures in his paper [La conjecture de Weil. I](http://www.numdam.org/item?id=PMIHES_1974__43__273_0).
As far as I know, all known proofs of this conjecture involve the use of cohomology of varieties over finite fields in an essential way.
Added by Emerton: One point to make is that the Weil conjectures (in their basic form,
saying that the eigenvalues of Frobenius on the $i$th etale cohomology of a variety over
$\mathbb F\_q$ have absolute value $q^{i/2}$) apply only to smooth proper varieties. On the other hand, the Kuga-Sato variety is the symmeteric power of the universal elliptic curve over a modular curve, which is not projective. Thus one has to pass to a smooth compactification in order to apply the Weil conjectures, and then hope that this does not mess anything up in the rest of the argument. A certain amount of Deligne's effort in
his Bourbaki seminar is devoted to dealing with this issue. If you don't worry about this
(i.e. you accept that it all works out okay) then the proof is essentially just Eichler--Shimura theory (i.e. the relation between modular forms and cohomology of modular curves), but done with etale cohomology, combined with the Eichler--Shimura congruence relation that connects the $p$th Hecke operator to Frobenius mod $p$. (The latter was treated in the
[following question](https://mathoverflow.net/questions/19390).)
| 7 | https://mathoverflow.net/users/121 | 20259 | 13,462 |
https://mathoverflow.net/questions/20246 | 18 | **Background:** One says that continuous maps $f: X \to X, g: Y \to Y$ are [topologically conjugate](http://en.wikipedia.org/wiki/Topological_conjugacy) if there exists a homeomorphism $h: X \to Y$ such that $h \circ f = g \circ h$. There are many ways one can see that two maps are *not* topologically conjugate. For instance, if $f$ has fixed points and $g$ does not (more generally if the same power $f$ and $g$ have different numbers of fixed points), they cannot be topologically conjugate. [Topological entropy](http://en.wikipedia.org/wiki/Topological_entropy) provides a fancier invariant in terms of coverings (assuming $X$ and $Y$ are compact spaces).
I see also that there are many general theorems of that allow one to conclude that two maps are topologically conjugate (e.g. the [Hartman-Grobman theorem](http://en.wikipedia.org/wiki/Hartman-Grobman_theorem)).
However, I am curious:
>
> Given two discrete dynamical systems, is there a trick one can use to *construct* a topological conjugacy between them?
>
>
>
I suppose an analogy would be a comparison between the Brouwer and Banach fixed point theorems. I'm curious if there is an iterative process as in the proof of the latter.
(Full disclosure: I was motivated to ask this question because I got stuck on what should be a simple exercise in a book on dynamical systems. The exercise was to prove that any two $C^1$ maps $f,g: \mathbb{R} \to \mathbb{R}$ such that $0<f'(0)<1, 0 < g'(0)<1$ are locally topologically conjugate.)
| https://mathoverflow.net/users/344 | How to construct a topological conjugacy? | Let me throw in some speculations based on my limited involvement in dynamical systems.
The conjugation formula $f=h^{-1}gh$ is in general not a type of a functional equation that can be solved by iterative approximations or a clever fixed-point trick. The problem is that you cannot determine how badly a particular $h$ fails by looking at the difference between LHS and RHS of this equation. The obstructions are not local and you don't see them until you consider all iterations of $f$ and $g$. Sometimes you can do approximations (e.g. Anosov system more or less survive under perturbations) but this works only in special types of systems (some kind of "hyperbolicity" is needed).
It seems that the only "general" way for constructing a topological conjugation is to study the orbits of $f$ and $g$ and send each orbit of $f$ to a similar orbit of $g$ so that the whole map is continuous. (A dense set of orbits is sufficient, e.g. the set of periodic points of an Anosov system.) The problem is, of course, that the structure of orbits can be *really* complicated. But if it is simple, one can hope to construct a conjugation directly.
For example, consider two homeomorphisms $f,g:\mathbb R\to\mathbb R$ satisfying $f(x)>x$ and $g(x)>x$ for all $x$. They are conjugate. To see this, consider an orbit $\dots,x\_{-1},x\_0,x\_1,x\_2,\dots$ of a point $x\_0$ under the iterations of $f$. This is an increasing sequence and the intervals $[x\_i,x\_{i+1}]$ cover $\mathbb R$. Every other orbit "interleaves" with this one: for example, if $y\_0\in(x\_0,x\_1)$, then $y\_i:=f^i(y\_0)$ lies between $x\_i$ and $x\_{i+1}$. So every orbit has a unique member in the interval $[x\_0,x\_1)$. In a sense, this interval (or rather the closed one with the endpoints glued together) naturally represents the set of all orbits.
So take any orbit $(x\_i')$ of $g$ and let $h\_0$ be *any* order-preserving bijection from $[x\_0,x\_1]$ to $[x\_0',x\_1']$. This defines a unique conjugacy map $h:\mathbb R\to\mathbb R$ such that $h|\_{[x\_0,x\_1]}=h\_0$: the orbit $\{f^i(y)\}$ of a point $y\in [x\_0,x\_1]$ is mapped to the $g$-orbit $\{g^i(h\_0(y))\}$ of the point $h\_0(y)$. And all conjugations can be obtained this way.
Already in this simple example, you can see how fragile things can be. Even if $f$ and $g$ are smooth and have bounded derivatives, you have no control over how big the derivatives of $h$ can grow. (And you actually lose smoothness if you do the same on a closed interval rather than $\mathbb R$.)
If you vary the map $g$, it remains conjugate to $f$ while the condition $g(x)>x$ holds true. But it suddenly stops being conjugate once a fixed point appears. However trivial this fact is, is shows that limit of conjugacy maps does not make sense in general.
The exercise you mention can be solved in a similar fashion as my toy example; the orbits are not much more complicated. In fact, given any two homeomorphisms $\mathbb R\to\mathbb R$, it is easy to understand whether they are conjugate or not (just study the intervals between fixed points). But the next step - homeomorphisms of the circle - is much more difficult: there are beautiful theorems, unexpected conterexamples, connections to number theory and other signs of a rich theory around such a seemingly trivial object. See Denjoy theorem and especially its smoothness requirements to get an idea how interesting these things can be.
| 14 | https://mathoverflow.net/users/4354 | 20262 | 13,465 |
https://mathoverflow.net/questions/19892 | 10 | A complete, simply connected Riemannian manifold has no conjugate points if and only if every geodesic is length-minimizing. I just realized that I don't know whether the same is true for a locally CAT(k) space (i.e. a geodesic space with curvature bounded above in the Alexandrov sense).
Thanks to Alexander and Bishop, there is a developed "geodesic analysis" in these spaces, including Jacobi fields and conjugate points. And there is a Cartan-Hadamard theorem for spaces without conjugate points: if it is simply connected, then every pair of points is connected by a unique geodesic.
In the Riemannian case, the converse statement follows from the basic fact that a geodesic beyond a conjugate point is no longer minimizing. This is proved by constructing a length-decreasing variation, or something similar, from a vanishing Jacobi field. Unfortunately, this argument uses a lower curvature bound. Well, not quite that, because it also works in Finsler geometry, but anyway it fails for CAT(k): on a bouquet of two spheres there are geodesics that remain minimizing beyond a conjugate point.
However this does not disprove the converse Cartan-Hadamard theorem. Hence the question:
Let $X$ be a space with curvature locally bounded above. Let's not talk about monsters: the space is complete, locally compact, all geodesics are extensible (otherwise one can play dirty tricks with a boundary). Suppose that every geodesic in $X$ is minimizing. Or even better: every pair of points is connected by a unique geodesic. Does this imply that the geodesics have no conjugate points?
**UPDATE**. Thanks to Henry Wilton, I've found that there is no standard definition of a conjugate point. In fact, some definitions are designed so as to imply the affirmative answer to my question immediately. When I asked the question, I meant the following (maybe not the best possible) definition.
Fix a point $p\in X$ and consider the space $X\_p$ of geodesic segments emanating from $p$. The segments are parametrized by $[0,1]$ proportionally to arc length. The space $X\_p$ is regarded with the $C^0$ metric. The exponential map $\exp\_p:X\_p\to X$ is defined by $\exp\_p(\gamma)=\gamma(1)$. A point $q=\gamma(1)$ is conjugate to $p$ along $\gamma$ iff $\exp\_p$ is not bi-Lipschitz near $\gamma$.
| https://mathoverflow.net/users/4354 | In a locally CAT(k) space, does uniqueness of geodesics imply the lack of conjugate points? | Consider a surface of revolution with an equator $\ell$ of lenght $2{\cdot}\pi$ such that its Gauss curvature $$K=1/\left(1+\sqrt[5]{\mathrm{dist}\_ \ell}\right).$$
Choose $z\in \ell$ and let $\Sigma=B\_{\pi/2}(z)$.
Clearly $\Sigma$ is a $\mathrm{CAT}(1)$-space it has just one pair of conjugate points (say $p$ and $q$ --- the ends of $\Sigma\cap\ell$) and it has unique geodesics between each pair.
It remains to make geodesics extensible.
To do this, we take $\Lambda=(S^1\times [0,\infty), d)$ with flat metric and concave boundary $\partial \Lambda=\partial\Sigma$.
Then we glue $\Lambda$ and $\Sigma$ along the boundary.
The metric on $\Lambda$ is completely described by curvature $k(u)$ of its boundary [$u\in \partial \Lambda=\partial \Sigma$].
We only need to choose a function $k$ which is
* on one had is large enough so that the glued surface still has unique geodesics between each pair (in particular $k(p)=k(q)=\infty$).
* on the other hand is $\int\_{S^1} k<\infty$, so that glued space is locally compact.
I believe it is possible...
| 5 | https://mathoverflow.net/users/1441 | 20269 | 13,468 |
https://mathoverflow.net/questions/20275 | 21 | Today my fellow grad student asked me a question, given a map f from X to Y, assume $f\_\*(\pi\_i(X))=0$ in Y, when is f null-homotopic?
I search the literature a little bit, D.W.Kahn
[Link](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-15/issue-2/Maps-which-induce-the-zero-map-on-homotopy/pjm/1102995805.full)
And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces.
<http://www.jstor.org/stable/pdfplus/2037939.pdf>
However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint?
or, maybe even some examples arising from manifolds?
| https://mathoverflow.net/users/1877 | Maps inducing zero on homotopy groups but are not null-homotopic | Consider ordinary singular cohomology with varying coefficients. You can look at the short exact sequence of abelian groups:
$$0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$$
This gives rise, for any space X, to a short exact sequence of chain complexes:
$$0 \to C^i(X;\mathbb{Z}/2) \to C^i(X;\mathbb{Z}/4) \to C^i(X;\mathbb{Z}/2) \to 0$$
and hence you get a long exact sequence in cohomology. Thus we get an interesting boundary map known as the [Bockstein](http://en.wikipedia.org/wiki/Bockstein_homomorphism)
$$H^i(X; \mathbb{Z}/2) \to H^{i+1}(X; \mathbb{Z}/2).$$
This is natural in X and so is represented by a (homotopy class of) map(s) of Eilenberg-Maclane spaces:
$$K(i, \mathbb{Z}/2) \to K(i+1, \mathbb{Z}/2)$$
This map is necessarily zero on homotopy groups. To show that this map is not null-homotopy, you just need to find a space for which the Bockstein is non-trivial. There are lots of examples of this. Rather then explain one, I suggest you look up "Bockstein homomorphism" in a standard algebraic topology reference, e.g. Hatcher's book.
| 34 | https://mathoverflow.net/users/184 | 20278 | 13,473 |
https://mathoverflow.net/questions/20277 | 6 | What is the best asymptotic approximation of the inverse $x=g(y)$ of $y = x^x$ for large $x$? [Clearly, if $x>e$, then $f(x) > e^x$ implies $g(x) < \log x$.]
| https://mathoverflow.net/users/5122 | Approximately Invert x^x | I don't know how accurate you want to be, but a quick and dirty approximate inversion of $x\log x$ is $x/\log x$. So if $y=x^x$ then $\log y\approx x\log x$, so $x\approx\log y/\log\log y$. But perhaps you want something better than this.
| 16 | https://mathoverflow.net/users/1459 | 20279 | 13,474 |
https://mathoverflow.net/questions/20283 | 17 | What is the "correct" pronunciation of Robin Hartshorne's last name? Mostly I hear it pronounced "Har-shorn" although I've also heard "Harts-orn" and maybe a few other variations.
| https://mathoverflow.net/users/1148 | How do you pronounce "Hartshorne"? | He prefers it be pronounced as in Hart's Horn. I asked him a few years ago, our brief common ground being assisting Marvin Jay Greenberg with revisions for the fourth edition of his book on Euclidean and non-Euclidean geometry. That is not to say that I have ever heard anyone else say it that way. But then few people get my name right.
| 28 | https://mathoverflow.net/users/3324 | 20284 | 13,476 |
https://mathoverflow.net/questions/20281 | 19 | I'm reading Terry Gannon's [Moonshine Beyond the Monster](http://books.google.com/books?id=ehrUt21SnsoC&printsec=frontcover&dq=terry+gannon+moonshine&source=bl&ots=7m8tyu7_0n&sig=yOYPV3kAm_eEimKFIGWcidHyR6M&hl=en&ei=3wm4S9HYHMH88Aa7_OXhBw&sa=X&oi=book_result&ct=result&resnum=3&ved=0CA4Q6AEwAg#v=onepage&q=&f=false), and in section 2.4.3 he hints at (but does not explicitly describe) a way to extend the action of $SL\_2(\mathbb{Z})$ on modular forms to an action of the braid group $B\_3$. Here is what he says about this action:
First, we lift modular forms $f : \mathbb{H} \to \mathbb{C}$ to functions $\phi\_f : SL\_2(\mathbb{R}) \to \mathbb{C}$ as follows: let
$$\phi\_f \left( \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right] \right) = f \left( \frac{ai + b}{ci + d} \right) (ci + d)^{-k}.$$
Thinking of $f$ as a function on $SL\_2(\mathbb{R})$ invariant under $SO\_2(\mathbb{R})$, we have now exchanged invariance under $SO\_2(\mathbb{R})$ for invariance under $SL\_2(\mathbb{Z})$. ($SO\_2(\mathbb{R})$ now acts by the character corresponding to $k$.) In moduli space terms, an element $g \in SL\_2(\mathbb{R})$ can be identified with the elliptic curve $\mathbb{C}/\Lambda$ where $\Lambda$ has basis the first and second columns (say) of $g$, and $\phi\_f$ is a function on this space invariant under change of basis but covariant under rotation.
Second, $SL\_2(\mathbb{R})$ admits a universal cover $\widetilde{SL\_2(\mathbb{R})}$ in which the universal central extension $B\_3$ of $SL\_2(\mathbb{Z})$ sits as a discrete subgroup. Unfortunately, Gannon doesn't give an explicit description of this universal cover (presumably because it's somewhat complicated).
**Question:** What is a good explicit description of this universal cover and of how $B\_3$ sits in it (hence of how it acts on modular forms)? In particular, does it have a moduli-theoretic interpretation related to the description of $B\_3$ as the fundamental group of the space $C\_3$ of unordered triplets of distinct points in $\mathbb{C}$? (These triplets $(a, b, c$) can, of course, be identified with elliptic curves $y^2 = 4(x - a)(x - b)(x - c)$.)
| https://mathoverflow.net/users/290 | Details for the action of the braid group B_3 on modular forms | You can think of the space of positively oriented covolume-one bases of $\mathbb{R}^2$ as a torsor under $SL\_2(\mathbb{R})$, i.e., it is a manifold with a simply transitive action of the group. If you choose a preferred basepoint, such as $(\mathbf{i},\mathbf{j})$, you get an identification with the group. You can think of elements of the universal cover as positively oriented area-one bases equipped with a homotopy class of paths in $\mathbb{R}^2 - \{0\}$ from $\mathbf{i}$ to the first element of the basis. There is a reasonably straightforward composition law that involves multiplying matrices and composing paths.
Gannon's description of lifting to $SL\_2(\mathbb{R})$ implies the lifts of even weight modular forms to $\widetilde{SL\_2(\mathbb{R})}$ are invariant under the action of $B\_3$. In particular, classical modular forms are rather boring from the perspective of the braid group. In order to detect the central extension, you need to consider modular forms of fractional weight. When the weight is not an integer, you don't get an action of $SO(2)$, but instead, an action of the universal cover $\mathbb{R}$. The resulting action of $B\_3$ is necessarily nontrivial, since the restriction to the center is by a nontrivial character of $\mathbb{Z}$. I don't know many explicit constructions of fractional weight forms, other than half-integer weight forms like $\eta$ and theta functions, and vector-valued forms constructed from them. However, you can generate a family of examples by choosing powers of the cusp form $\Delta$, which admits a logarithm since it is globally regular and nonvanishing.
My understanding of the explicit relationship to configurations of points and elliptic curves is the following: Given a path of triples of distinct points $(a\_1(t),a\_2(t),a\_3(t))$, we get a path on the space of elliptic curves of the form $y^2 = (x-a\_1(t))(x-a\_2(t))(x-a\_3(t))$, but this will throw away an action of real translations and dilations (irrelevant) together with the central extension and the circle action (important). If we just look at the isomorphism types (i.e., the $j$-invariants) of the curves, we get a path through the quotient of the upper half plane by $SL\_2(\mathbb{Z})$. We need to choose a discrete structure to remove the quotient by the center, and a one dimensional continuous structure to promote our space to three dimensions. To retain the angular information that we lost by passing to elliptic curve isomorphism, we fix a tangent direction at infinity to remove rotational symmetry. This tangent direction is manifested when we choose our discrete structure: a homotopy class of nonintersecting paths from the three points to infinity, because we demand that the paths asymptotically approach infinity in that direction. The elliptic curve is a double cover of the complex projective line, ramified at the three points and infinity. We can choose once and for all a uniform convention for lifting the three paths to primitive homology cycles, such that any pair generates $H\_1$, and one cycle is the sum of the other two, so those two form a preferred basis. To get the parametrization of $\widetilde{SL\_2(\mathbb{R})}$ from the first paragraph, we choose a preferred basepoint configuration of three points with paths from infinity and asymptotic direction, and consider a triple $(a\_1(t),a\_2(t),a\_3(t))$ that starts at the basepoint. By uniqueness of homotopy lifting, we get a family of tangent vectors at infinity together with a family of elliptic curves with oriented bases of homology. By rescaling the bases in $\mathbb{R}^2$ to have unit covolume, we get the parametrization in the first paragraph.
| 8 | https://mathoverflow.net/users/121 | 20287 | 13,478 |
https://mathoverflow.net/questions/20288 | 8 | The *punctual Hilbert scheme* in dimension $d$ parameterizes ideals $I$ of codimension $n$ in $k[x\_1,\dots, x\_d]$ which are contained in some power of the ideal $(x\_1,\dots, x\_d)$. In other words, it is the Hilbert scheme of $n$ points supported at the origin in $\mathbb A^d$.
Can anybody give me a reference for this object? In particular, I'd like to know if it is irreducible for arbitrary $d$.
---
**A related question:** the *curvilinear Hilbert scheme* parameterizes ideals $I$ of codimension $n$ in $k[x\_1,\dots, x\_d]$ such that $I\not\subset(x\_1,\dots, x\_d)^{n-1}$. This object is irreducible; in fact, it's covered by dense open subschemes isomorphic to $\mathbb A^m$ for some $m$. Does anybody know of a reference that calculates what these spaces are? For $n=2$, it's $\mathbb P^{d-1}$. For $n=3$, I think it's the total space of $\mathcal O\_{\mathbb P^{d-1}}(1)$.
| https://mathoverflow.net/users/1 | Reference request: is the punctual Hilbert scheme irreducible? | It is certainly not irreducible if *n=8* and *d>3*. This is analyzed nicely in the paper
Hilbert schemes of 8 points, Dustin A. Cartwright, Daniel Erman, Mauricio Velasco, Bianca Viray available at <http://arxiv.org/abs/0803.0341> (and I think published in ANT).
From there you can look at the references, especially I think the first paper studying in detail this problem was
Anthony Iarrobino. Reducibility of the family of 0-dimensional schemes on a variety. Inventiones Math., 15:72–77, 1972.
Note that, at least the arXiv reference above, deals with subschemes of $\mathbb{A}^d$ not necessarily supported at the origin. On the other hand, since for $n \leq 8$ the above is the only non-irreducible example, the "new" component must be supported at the origin. Indeed in the case *n=8* and *d=4*, the extra component is a product of a Grassmannian and an affine space.
EDIT: Let me expand on my initial answer. First a general remark: if a point on a scheme is non-singular, then it lies on a unique irreducible component of that scheme. Now, in the usual (not the punctual) Hilbert scheme of 8 points in A^4 there are two components, one being the closure of the component consisting of 8 distinct points, and one supported entirely at a single point (the point being allowed to vary; by translating the support, we can assume the point is the origin). In the arXiv paper they show that there are points supported at a single point that are non-singular and **smoothable** in the full Hilbert scheme (Section 4.4). It follows that such points cannot lie in the component whose points consist entirely of subschemes supported at the origin. We conclude that the punctual Hilbert scheme contains the "non-smoothable" component (necessarily supported at the origin, this one being a minimal example of a reducible Hilbert scheme), as well as more points, corresponding to the points described above (the ones appearing in Section 4.4). In particular, the punctual Hilbert scheme is not irreducible. Hope this clears up the doubts (or that it exposes my mistake!).
| 9 | https://mathoverflow.net/users/4344 | 20290 | 13,480 |
https://mathoverflow.net/questions/20228 | 3 | Once upon a time I asked whether $\omega\_1 \times \beta \mathbb{N}$ is normal. I got the answer no and a fairly convincing proof of this [here](http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=1582.0002)
However I'm currently in a situation where I have three plausible proofs of plausible results at most two of which can be true, and of the three this is the one of which I'm currently the least sure (there are a bunch of details I haven't yet fully checked as I'm in the process of dusting off some of my recently rather unused knowledge about general topology), so was hoping someone could confirm.
| https://mathoverflow.net/users/4959 | Is ω1 × βN normal? | I have reread the proof and it's completely correct.
The idea is that $\omega\_1 \times \beta\mathbb{N}$ maps perfectly onto
a non-normal space, and normality is preserved under perfect maps.
Tamano's theorem says that $X$ is paracompact Hausdorff iff $X \times \beta X$ is normal,
and we use that $\omega\_1$ is not paracompact and $\beta \omega\_1 = \omega\_1 + 1$. But the direct proof as sketched is also correct (we can use the pushdown lemma to prove it).
All building blocks can be found in Engelking, e.g.
| 7 | https://mathoverflow.net/users/2060 | 20294 | 13,484 |
https://mathoverflow.net/questions/20267 | 7 | Let $G$ be a finite group, and $k$ be a field of characteristic zero (not necessarily algebraically closed!). Let $\rho : G \to \mathrm{End}\_k \left(k^n\right)$ be a irreducible representation of $G$ over $k$. Consider the vector space
$S=\left\lbrace H\in \mathrm{End}\_k\left(k^n\right) \mid \rho\left(g\right)^T H\rho\left(g\right)=H\text{ for any }g\in G\right\rbrace$
$=\left\lbrace \sum\limits\_{g\in G}\rho\left(g\right)^T H\rho\left(g\right)\mid H\in \mathrm{End}\_k\left(k^n\right)\right\rbrace$
and its subspace
$T=\left\lbrace H\in S\mid H\text{ is a symmetric matrix}\right\rbrace$.
It is easy to show that, if we denote our representation of $G$ on $k^n$ by $V$, then the elements of $S$ uniquely correspond to homomorphisms of representations $V\to V^{\ast}$ (namely, $H\in S$ corresponds to the homomorphism $v\mapsto\left(w\mapsto v^THw\right)$), while the elements of $T$ uniquely correspond to $G$-invariant quadratic forms on $V$ (namely, $H\in T$ corresponds to the quadratic form $v\mapsto v^THv$).
**(1)** In the case when $k=\mathbb C$, Schur's lemma yields $\dim S\leq 1$, with equality if and only if $V\cong V^{\ast}$ (which holds if and only if $V$ is a real or quaternionic representation). Thus, $\dim T\leq 1$, and it is known that this is an equality if and only if $V$ is a real representation. (Except of the equality parts, this all pertains to the more general case when $k$ is algebraically closed of zero characteristic).
**(2)** In the case when $k=\mathbb R$, it is easily seen that $T\neq 0$ (that's the famous nondegenerate unitary form, which in the case $k=\mathbb R$ is a quadratic form), and I think I can show (using the spectral theorem) that $\dim T=1$. As for $S$, it can have dimension $>1$.
**(3)** I am wondering what can be said about other fields $k$; for instance, $k=\mathbb Q$. If $k\subseteq\mathbb R$, do we still have $\dim T=1$ as in the $\mathbb R$ case? In fact, $T\neq 0$ can be shown in the same way.
| https://mathoverflow.net/users/2530 | Invariant quadratic forms of irreducible representations | There are certainly examples over $k=\mathbb{Q}$ where $\dim T\ge2$.
Let's take the cyclic group $G$ of order $5$ and the representation
space
$$V=\{(a\_0,\ldots,a\_4)\in\mathbb{Q}^5:a\_0+\cdots +a\_4=0\}$$
where $G$ acts by cyclic permutation. Two linearly independent
elements of $T$ are given by
$$\left(\begin{array}{rrrrr}
2&-1&0&0&-1\\\
-1&2&-1&0&0\\\
0&-1&2&-1&0\\\
0&0&-1&2&-1\\\
-1&0&0&-1&2\end{array}\right)
$$
and
$$\left(\begin{array}{rrrrr}
2&0&-1&-1&0\\\
0&2&0&-1&-1\\\
-1&0&2&0&-1\\\
-1&-1&0&2&0\\\
0&-1&-1&0&2\end{array}\right)
$$
(these define quadratic forms on $V$ since they annihilate the all-one
vector).
The point here is that this representation splits into two over
$\mathbb{R}$. I think the dimension of $T$ in general will be
the number of irreducible representations it splits into over
$\mathbb{R}$.
| 4 | https://mathoverflow.net/users/4213 | 20296 | 13,486 |
https://mathoverflow.net/questions/20272 | 9 | In the English translation of *The Gamma Function* by Emil Artin (1964 - Holt, Rinehart and Winston) there appears to be a mistake in the formula given for the gamma function on page 24:
$$\Gamma(x) = \sqrt{2\pi}x^{x-1/2}e^{-x+\mu(x)}$$
$$\mu(x)=\sum\_{n=0}^\infty(x+n+\frac{1}{2})\text{log}(1+\frac{1}{x+n})-1=\frac{\theta}{12x},\ \ \ \ \ 0 < \theta < 1$$
and on page 22 where this is derived, it is noted that '$\theta$ is a number independent of $x$ between 0 and 1'.
This sounds incorrect, as $\theta$ does depend on $x$, but since the wording is a little ambiguous it may just be an unclear translation. The original German might have meant that $0< \theta(x) < 1$ for any $x$. That the variable $x$ is suppressed from $\theta$ could be just confusing notation, or someone's misunderstanding (possibly mine.)
The preface does mention that a (different) formula had to be corrected for the English reprint.
I would like to know if there are mistakes in this book, and if so, whether they exist in the German edition. Is there an available list of errata?
| https://mathoverflow.net/users/2604 | Errata for Emil Artin's 'The Gamma Function'? | It seems clear that $\theta$ can indeed be chosen to be a number independent of $x$ as stated, to get Stirling's formulas for the gamma function when
$x$ is *large*. The wording, at least in English, is not too helpful
in this section.
But I'm less clear about where in the formula on page 24 there is supposed to be
a mistake. Here as in any mathematics book (especially a translation) one has to be
wary about misprints or errors. Probably there is no publicly available list
of errata for this small monograph published originally in 1931 in German and
later republished in 1964 in an English translation by
Michael Butler. This English version is included in the 2007 AMS softcover
book *Exposition by Emil Artin: A Selection* edited by Michael Rosen. (There is an older 1965 book
*The Collected Papers of Emil Artin* published by Addison-Wesley and edited
by Lang & Tate. This contains Artin's research papers, in the original
German or English.) As Zavosh observes, the 1964 preface by Edwin Hewitt reprinted here does indicate one formula corrected in the translation:
" ... a small error following formula (59) (this edition) was corrected..."
However, the formula seems to be the one actually numbered (5.9). Caveat lector.
| 4 | https://mathoverflow.net/users/4231 | 20305 | 13,493 |
https://mathoverflow.net/questions/20311 | 0 | Here you are another question in basic measure theory...
Let $f\_k$ be a measurable sequence of functions on $(X,M,\mu)$ measure space. Suppose that $f\_k$ *does not go to 0 a.e*.. Can I then find a set $A\subseteq X$ with positive measure and a subsequence $f\_{k\_j}$ and an $\varepsilon > 0$ such that $\liminf\_j |f\_{k\_j}(x)| > \varepsilon$ foreach $x\in A$?
| https://mathoverflow.net/users/4928 | If $f_k \not\to 0$ a.e., does there exist a subsequence, a set of positive measure, and $c > 0$, on which $\liminf |f_{k_j}| > c$? | That's not true. For example, in $(0,1)$ take
$f\_1 =1$,
$f\_2=1\_{(0,1/2)}$, $f\_3= 1\_{(1/2,1)}$
$f\_4=1\_{(0,1/3)}$, $f\_5= 1\_{(1/3,2/3)}$, $f\_6= 1\_{(2/3,1)}$
and so on. $f\_k(x)$ does not go to 0 a.e. (the limit does not exist, for each x), but we can't find any succession that satisfies the statement, because $m(supp f\_k)$ goes to zero
| 3 | https://mathoverflow.net/users/4928 | 20317 | 13,500 |
https://mathoverflow.net/questions/8622 | 18 | Under what conditions can a Riemannian manifold be embedded isometrically as a submanifold of a complete one of the same dimension? There should some kinds of necessary conditions. For instance, any ball in $M$ (considered as a metric space) must be totally bounded. Is this sufficient?
I am curious because it seems that many theorems are stated and proved only for the complete case, and I was wondering how to what extent they could be generalized using a completion tool (if it existed).
Also, is there any kind of uniqueness (there is for $C^{\omega}$ manifolds--implied by the Myers-Rinow theorem)?
| https://mathoverflow.net/users/344 | When is a Riemannian manifold an open subset of a complete one? | I take the opportunity to advertise the work of a colleague Charles Frances, which is somehow related.
There are counter-examples to a more flexible question: given a (pseudo-)riemannian manifold, is it always possible to *conformally*, non-trivially embed it into another?
A counter-example to this question gives a counter-example to yours since a conformal class on a (non-compact !) manifold contains a non-complete riemannian metric.
Details can be found at the following adress: <http://www.math.u-psud.fr/~frances/boundary-frances.pdf>
| 5 | https://mathoverflow.net/users/4961 | 20320 | 13,503 |
https://mathoverflow.net/questions/20314 | 30 | Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.
| https://mathoverflow.net/users/4928 | Good books on theory of distributions | Grubb's recent *Distributions And Operators* is supposed to be quite good.
There's also the recommended reference work, Strichartz, R. (1994), *A Guide to Distribution Theory and Fourier Transforms*
The comprehensive treatise on the subject-although quite old now-is Gel'fand, I.M.; Shilov, G.E. (1966–1968), *Generalized functions, 1–5,*.
A very good,though quite advanced,source that's now available in Dover is Trèves, François (1967), *Topological Vector Spaces, Distributions and Kernels* That book is one of the classic texts on functional analysis and if you're an analyst or aspire to be,there's no reason not to have it now. But as I said,it's quite challenging.
That should be enough to get you started.And of course,if you read French,you really should go back and read Schwartz's original treatise.
| 23 | https://mathoverflow.net/users/3546 | 20322 | 13,505 |
https://mathoverflow.net/questions/20330 | 7 | Upon thinking about [this question](https://mathoverflow.net/questions/20308), I have a feeling that there is an interesting general problem like that, but I cannot verbalise it. Here is an approximation.
The question is: given a finitely generated group $G$ and a finite set $S\subset G$, we want to find out whether the subgroup generated by $S$ is the whole group. Is there an algorithm deciding this?
One has to specify how $G$ and $S$ are presented to a machine. As for $S$, its elements are given as words in some generating set for $G$. The question how to represent a group is more delicate. Obviously not by a set of relations, since the identity problem is undecidable. There are two ways I see.
1) Let's say that $G$ is *nice* if there is an algorithm deciding the above problem for this group only. For example, $\mathbb Z^n$ is nice, [edit:] free groups are nice, and probably many others. Are all groups nice? What about some reasonable classes, like lattices in Lie groups?
2) This one is the best approximation of my intuitive picture of a question. Suppose we have an algorithm $A$ deciding the identity problem in $G$. (This algorithm tells us whether two elements of $G$, presented as words, are equal.) Or maybe it's an oracle rather than algorithm, I don't feel the difference. Can we decide the above problem using $A$?
**Update.** It turns out that even $F\_6\times F\_6$ is not nice in the above sense (see John Stillwell's answer). This kills the second part too. The only question that remains unanswered is the one with the least motivation behind it: Is "generating problem" solvable in lattices of Lie groups?
| https://mathoverflow.net/users/4354 | Is it decidable whether a given set generates the whole group? | The answer to the title question is that the problem is unsolvable. See p. 194 of Lyndon and Schupp's *Combinatorial Group Theory*, where it is called the "generating problem." It is unsolvable even when $G$ is the direct product of free groups of rank at least 6.
| 9 | https://mathoverflow.net/users/1587 | 20334 | 13,513 |
https://mathoverflow.net/questions/19906 | 14 | Is there some sort of monad whose algebras are monads? How about if we are internal to a bicategory B? Are internal monads in B monadic? Certainly not always, as otherwise free T-multicategories a la Leinster would always exist. What is known?
| https://mathoverflow.net/users/4528 | Are monads monadic? | In the book "Toposes, Triples and Theories", Barr and Wells study the question when a *particular* endofunctor admits a free monad. This is the case if the underlying category is complete and cocomplete and if the endofunctor preserves filtered colimits (we say that such a functor is *finitary*).
The question remains whether the resulting adjunction between monads on $C$ and endofunctors on $C$ is monadic. If $C$ is [locally finitely presentable](http://ncatlab.org/nlab/show/locally+presentable+category) (lfp), this is true: Steve Lack showed in
"On the monadicity of finitary monads", Journal of Pure and Applied Algebra, Volume 140, Number 1, 21 July 1999 , pp. 65-73(9) (available [here](http://www.maths.usyd.edu.au/res/Catecomb/Lack/1997-29.html))
that the forgetful functor $\mathrm{Mnd}\_f(C) \rightarrow \mathrm{End}\_f(C)$ from finitary monads on a lfp category $C$ to finitary endofunctors is monadic. Note that both these categories are again lfp categories, and we don't need universes to make sense of them, essentially because a finitary endofunctor is determined by what it does on a set of objects.
The result remains true if we consider categories and functors enriched in a complete and cocomplete symmetric monoidal closed category $V$ which is lfp as a closed category.
| 16 | https://mathoverflow.net/users/1649 | 20335 | 13,514 |
https://mathoverflow.net/questions/20239 | 2 | With this question I try to build up a systematization of different kinds of integrals. The following table differentiates between deterministic and stochastic integrals, the summation processes ("from left" or "average between left and right") and different kinds of variations (bounded and quadratic).
The following integrals are already present:
(A) [Riemann integral](http://en.wikipedia.org/wiki/Riemann_integral)
(F) [Ito integral](http://en.wikipedia.org/wiki/Ito_integral)
(G) [Stratonovich integral](http://en.wikipedia.org/wiki/Stratonovich_integral)
Unfortunately I didn't find references for the remaining fields. Is this because it doesn't make any sense to define these? And then, why not?
In any case: I would very much appreciate your help here. Could you please reference the field (X) and give some further information regarding
* type of integral
* examples
* refences
* links
* if it doesn't make any sense: why not
Every little bit of information helps - thank you very much and I promise to consolidate the information given here into a proper form.
**Addendum**: For the cases (D) and (E) [perhaps even (H) and (I)?] there seem to be some possible connections to *deterministic fractal functions* (like e.g. [Weierstrass function](http://en.wikipedia.org/wiki/Weierstrass_function)) but I can't find any further references on definitions for integrals/integration here.
[alt text http://www.freeimagehosting.net/uploads/f5c860660a.png](http://www.freeimagehosting.net/uploads/f5c860660a.png)
| https://mathoverflow.net/users/1047 | Systematization of deterministic and stochastic integrals | Hi,
I think that you can define for some integrands with restrictive conditions, integrals with respect to q-variational integrator (q>0). You can google for Young Integrals.
You could also have a look at Rough Path theory also where you solve equations with respect to path of infinite variations.
Moreover there is a way to define a path-by-path Itô integral for continuous integrators with Itô's formula also being derived in a path-by-path way. This is done (in french) in an article "Calculd'Itô sans probabilité" by Hans Föllmer in volume XV of the Séminaire of Probablity of Strasbourg LNM 850 . To my knowledge it hasn't been translted in english.
All this is very partial but are leads that you may exlpore yourself
Regards
| 3 | https://mathoverflow.net/users/2642 | 20337 | 13,515 |
https://mathoverflow.net/questions/20341 | 2 | In the development of constructible sets via the Gödel operations, one of the main point seems to be that for $\Delta\_0$ formulas, one can construct the sets of of elements verifying them with a finite number of Gödel operations, called $G\_1$,...,$G\_{10}$.
My questions are : does this means that Set theory with separation restricted to $\Delta\_0$ formulas is finitely axiomatizable since a set is closed under a finite number of these operations iff it is closed under $\Delta\_0$ formulas? Also since $\Delta\_0$ formulas are absolute in transitive models, does this mean that if we consider $Th(M)$ with $M$ the fragment generated only by $\Delta\_0$ formulas, then it is finitely axiomatizable? Now if you had $\Delta\_1$ formulas is it still finitely axiomatizable? Finally, does this imply that a reflection theorem can't hold in Set Theory with $\Delta\_0$ separation?
I hope my questions were accurate.
| https://mathoverflow.net/users/3859 | Finite axiomatizability and constructible sets | You might be interested in looking at [Kripke-Platek Set Theory](http://en.wikipedia.org/wiki/Kripke%E2%80%93Platek_set_theory).
Yes, closure under the Gödel operations is equivalent to Δ0-comprehension (plus union, pairing, and cartesian products). Over this base theory, Σ1-reflection is equivalent to Δ0-collection. Note that Σ1-reflection does not prevent finite axiomatizability when the axioms are not Σ1.
| 2 | https://mathoverflow.net/users/2000 | 20356 | 13,528 |
https://mathoverflow.net/questions/20362 | 14 | (if any?)
I understand that in the natural numbers, the sum of two numbers can be readily thought of as the disjoint union of two finite sets.
John Baez even spent a week talking about how you can extend this idea to thinking about the integers here: [TWF 102](http://math.ucr.edu/home/baez/week102.html). This led into a discussion of the homotopy groups of spheres.
But then you have to pass to a certain colimit to get the rationals, and take a certain completion to get the reals. It all gets very complicated.
One place where we rely on a correspondence between a sum of real numbers and a certain coproduct is in measure theory-- perhaps in analogy to the relation between finite sets and natural numbers, we should think of some measure space as the categorification of the real numbers. But this sounds unpromising-- what space would be in any sense a canonical categorification? Moreover, what I was really hoping for originally was a precise sense in which the $\sigma$-additivity of a measure states that it preserves coproducts or something, so I was hoping there might be more to it than sigma-algebras.
| https://mathoverflow.net/users/2362 | In what category is the sum of real numbers a coproduct? | None (except trivially).
It's an elementary (though maybe not obvious) lemma that if $X$ and $Y$ are objects of a category and their coproduct $X + Y$ is initial, then $X$ and $Y$ are both initial.
Suppose there is some category whose objects are the real numbers, and such that finite coproducts of objects exist and are the same as finite sums of real numbers. In particular (taking the empty sum/coproduct), the real number $0$ is an initial object. Now for any real number $x$ we have $x + (-x) = 0$, so by the lemma, $x$ is initial. So every object is initial, so all objects of the category are uniquely isomorphic, so the category is equivalent to the terminal category **1**.
If you just want non-negative real numbers then this argument doesn't work, and I don't immediately see an argument to take its place. But I don't think it's too likely that an interesting such category exists.
I wonder if it would be more fruitful to ask a slightly different question. Product and coproduct aren't the only interesting binary operations on a category. You can *equip* a category with binary operations (as in the concept of *monoidal category*). Sometimes this is a better thing to do.
For example, there is on the one hand the concept of *distributive category*, which is something like a rig (=semiring) in that it has finite products $\times$ and finite coproducts $+$, with one distributing over the other. On the other hand, there is the concept of *rig category*, which is a category *equipped* with binary operations $\otimes$ and $\oplus$, with one distributing over the other. Distributive categories are examples of rig categories. Any rig, seen as a category with no morphisms other than identities, is a rig category. Any ordered rig can be regarded as a rig category (just as any poset can be regarded as a category): e.g. $[0, \infty]$ is one, with its usual ordering, $\otimes = \times$, and $\oplus = +$.
| 16 | https://mathoverflow.net/users/586 | 20365 | 13,532 |
https://mathoverflow.net/questions/20347 | 3 | DeMorgan authored the entry under TABLES in both the Penny and English Cyclopedias. Copies of a number of the tables DeMorgan discusses were in the library of Charles Babbage, a resource to which DeMorgan seems to have had access.
Questions:
1) Is there a list of the books in Babbage's Library somewhere?
2) What happened to the library upon Babbage's death?
Thanks for any insight.
Cheers, Scott
| https://mathoverflow.net/users/4111 | Mathematical Tables in Babbage's Library | See: M.R. Williams, "The Scientific Library of Charles Babbage," *IEEE Annals of the History of Computing*, vol. 3, no. 3, pp. 235-240, July-Sept. 1981, doi:10.1109/MAHC.1981.10028
Abstract: "In the early nineteenth century, Charles Babbage compiled a large scientific library that included many rare works. This paper describes the history, contents, and present location of the library. The contents are classified under twenty-one headings."
| 5 | https://mathoverflow.net/users/965 | 20370 | 13,537 |
https://mathoverflow.net/questions/20375 | 0 | Hi.
My question is probably very simple to some of you that have experience in Convex Optimization.
The dual function is defined as the infimum of the lagrangian $L(x,\lambda, \nu)$ over all $x\ $ in the domain. The lagrangian is:
$f\_0(x)+\sum \lambda\_i f\_i(x)+\sum \nu\_i h\_i(x)$
My question is, if $x\ $ is in the domain, it satisfies the equality constraints $h\_i(x)$ and in that case, $h\_i(x)=0$. So why do we even have to mention the equality constraints if they zero-out anyway?
Thanks a lot, I hope I wrote my question clearly.
| https://mathoverflow.net/users/5138 | A question about the lagrangian $L(x,\lambda, \nu)$ in the dual function in Convex Optimization | The domain in question is the intersection of all the domains of the functions $f\_i,h\_i$. Not all the points in the domain satisfy the conditions (such points constitute what's called the feasible set). Also keep in mind that the Lagrangian dual is often a relaxation of the original convex optimization and only gives you a lower bound, unless you have strong duality.
| 0 | https://mathoverflow.net/users/2384 | 20377 | 13,541 |
https://mathoverflow.net/questions/20374 | 13 | I should admit the question below does not have a serious motivation. But still I found it somehow natural.
Let $G$ be a finite group of order $n$ with $h$ conjugacy classes. If $c\_1,\ldots,c\_h$ are the orders of the conjugacy classes of $G$, then clearly
$n=c\_1+c\_2+\ldots+c\_h$.
Let now $\pi\_1,\ldots,\pi\_h$ be the pairwise non-isomorphic, irreducible complex representations of $G$. It is well known that another partition of $n$ of length $h$ is given by the squares of the degrees $d\_i$'s of the $\pi\_i$'s:
$n=d\_1^2+d\_2^2+\ldots+d\_h^2$.
Question: Assume that, up to reordering, the two partitions of $n$ described above are the same. Then what can we say about $G$? Is $G$ forced to be abelian?
| https://mathoverflow.net/users/4800 | When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group? | My standard rant about "what can we say about $G$": what we can say about $G$ is that the two partitions are the same. If the questioner doesn't find that a helpful answer then they might want to consider the possibility that they asked the wrong question ;-)
But as to the actual question: "is $G$ forced to be abelian?", the answer is no, and I discovered this by simply looping through magma's database of finite groups. Assuming I didn't make a computational slip, the smallest counterexample has order 64, is the 73rd group of order 64 in magma's database, which has 8 representations of degree 1, 14 representations of degree 2, 8 elements in the centre and 14 more conj classes each of order 4.
Letting the loop go further, I see counterexamples of size 64, 128, 192 (I guess these are just the counterexamples of size 64 multiplied by Z/3Z) and then ones of order 243 (a power of 3). So I guess all examples I know are nilpotent. Are they all nilpotent? That's a question I don't know the answer to.
| 19 | https://mathoverflow.net/users/1384 | 20378 | 13,542 |
https://mathoverflow.net/questions/20381 | 20 | I am trying to learn a little bit about crystalline cohomology (I am interested in applications to ordinariness). Whenever I try to read anything about it, I quickly encounter divided power structures, period rings and the de Rham-Witt complex. Before looking into these things, it would be nice to have an idea of what the cohomology that you construct at the end looks like.
The l-adic cohomology of abelian varieties has a simple description in terms of the Tate module. My question is: is there something similar for crystalline cohomology of abelian varieties?
More precisely, let $X$ be an abelian scheme over $\mathbb{Z}\_p$. Is there a concrete description of $H^1(X\_0/\mathbb{Z}\_p)$? (or just $H^1(X\_0/\mathbb{Z}\_p) \otimes\_{\mathbb{Z}\_p} \mathbb{Q}\_p$?) I think that this should consist of three things: a $\mathbb{Z}\_p$-module $M$, a filtration on $M \otimes\_{\mathbb{Z}\_p} \mathbb{Q}\_p$ (which in the case of an abelian variety has only one term which is neither 0 nor everything) and a Frobenius-linear morphism $M \to M$.
I believe that the answer has something to do with Dieudonné modules, but I don't know what they are either.
| https://mathoverflow.net/users/1046 | Crystalline cohomology of abelian varieties | To add a bit more to Brian's comment: the crystalline cohomology of an abelian variety (over a finite field of characteristic p, say) is canonically isomorphic to the Dieudonné module of the p-divisible group of the abelian variety (which is a finite free module over the Witt vectors of the field with a semi-linear Frobenius). If you start with an abelian scheme over the Witt vectors of this field then the crystalline cohomology of the special fibre is canonically isomorphic to the algebraic de Rham cohomology of the thing upstairs, hence receives a Hodge filtration also.
A good starting place to understand Dieudonné modules is Demazure's 'Lectures on p-divisible groups', which appears in Springer LNM. In particular, he gives a nice description of the analogy with Tate modules (and the relation between the various Frobenii that appear). For a general picture of crystalline cohomology, and the various structures that can be placed on it, I would look at Illusie's survey in the Motives volumes (this is a little out of date now, but gives a good description of the basic theory).
| 12 | https://mathoverflow.net/users/1594 | 20385 | 13,544 |
https://mathoverflow.net/questions/20392 | 11 | The following question came up in the class I'm teaching right now. There definitely exist groups $G$ with subgroups $H$ such that there exists some $g \in G$ such that $g H g^{-1}$ is a *proper* subgroup of $H$. For instance, let $G$ be the (big) permutation group of $\mathbb{Z}$ (by the big permutation group, I mean that elements of $G$ can move infinitely many elements of $\mathbb{Z}$). Let $H \subset G$ be the big permutation group of $\mathbb{N}$ and let $g \in G$ be the permutation that takes $n \in \mathbb{Z}$ to $n+1 \in \mathbb{Z}$. Then $g H g^{-1}$ is a proper subgroup of $H$.
The same sort of trick produces many examples like this. However, a feature of all of them is that $G$ is "big" in some way -- for instance, $G$ is not finitely presentable. By [Higman's embedding theorem](http://en.wikipedia.org/wiki/Higman%27s_embedding_theorem), you can embed such a $G$ into a finitely presentable group, so there exist examples where $G$ is finitely presentable. However, in all the examples I can come up with, the subgroup $H$ is not finitely presentable. I'm pretty sure that there exist examples in which $H$ is finitely presentable. Does anyone know of one? Even better, are there examples in which both $G$ and $H$ are of finite type (ie have compact $K(\pi,1)$'s)?
| https://mathoverflow.net/users/317 | Conjugating a subgroup of a group into a proper subgroup of itself | There are very simple examples with $H\cong\mathbb{Z}$. For instance
let $G$ be the affine linear group over $\mathbb{Q}$ consisting
of all maps $x\mapsto ax+b$ where $a\in\mathbb{Q}^\*$ and $b\in\mathbb{Q}$.
Let $H$ be the set of maps $x\mapsto x+b$ with $b\in\mathbb{Z}$.
Then $x\mapsto 2x$ conjugates $H$ into the proper subgroup $2H$.
In this example we could restrict $a$ to be a power of $2$
and $b$ to be a dyadic rational. Then the new $G$ has generators $a:x\mapsto 2x$
and $b:x\mapsto x+1$ and is defined by the single relation
$h^2=ghg^{-1}$.
| 18 | https://mathoverflow.net/users/4213 | 20394 | 13,549 |
https://mathoverflow.net/questions/20391 | 14 | I am teaching an advanced undergraduate class on topology. We are doing introductory knot theory at the moment. One of my students asked how do we know to use this [skein relation](http://en.wikipedia.org/wiki/Skein_relation) to compute all these wonderful polynomial invariants of knots.
I was not trained as a knot theorist, so I was at loss. Intuitively it does not seem to be that powerful, because it is not clear (to me at least) that you can always pick a recursive relation involving simpler knots. Could somebody help me motivating my students here? In other words,
**Why are the skein relations so useful for computing polynomial invariants of knots, and when did people realize that is the case?**
Thanks.
| https://mathoverflow.net/users/2083 | How to motivate the skein relations? | Regarding "when", it was Alexander, in his paper on what we call the Alexander polynomial. Conway was the first to popularize them, I believe.
Why are they useful? I'm not sure I believe they are so useful. Sometimes I'm interested in computing Alexander invariants but the knots and links that I'm looking at do not have easy-to-compute diagrams associated to them. Say you have a homology 3-sphere and you apply the JSJ-decomposition to it. This produces some knots and links in homology spheres but working out diagrams can be a pain. So sometimes it's far easier to use the covering space definition to get at the Alexander polynomials. Moreover, Skein relations don't give you the Alexander module and how Poincare duality works on the module, or naturality (when you have maps between 3-manifolds), etc.
| 7 | https://mathoverflow.net/users/1465 | 20401 | 13,555 |
https://mathoverflow.net/questions/20386 | 42 | I would like to know if practicing mathematics, constituting a hobby for some of you who are neither academics nor (advanced) mathematics, is an important part of your career. How do you go and learn a new mathematical field on your own?
Do you just pick up a book and go over all proofs and do all exercises on your own? Is there any technique would you like to share?
| https://mathoverflow.net/users/5144 | Mathematics as a hobby | I try to learn and understand as many facts as I can. Of course, many people would like to benefit from the opposite, that is, digging into a certain branch as deep as they can.
I try to do the opposite, which I see as my main advantage, as the opposite to professional mathematicians. This is because they have their own careers and has their professional criteria to fulfill (writing articles in journals, gaining citation points, etc.).
As an amateur I am not obliged to do so, and this is a great freedom. If you want to be creative, you may try to dig here and there, and probably you will be lucky to find certain problems which are not penetrated, or you may find just something interesting enough (for example, your own point of view on a well-known area, maybe you find a surprising connection and, even if it is well known, it is funny to discover it once more, etc.) to write it somewhere, maybe on a blog.
In summary: I read as much as I can, I learn as much as I can, and I ask as much as I can.
As regards to low-level entry (you need of course to be genius to discover it, but nothing more;-), an example is [Feigenbaum's famous discovery about chaos](http://mathworld.wolfram.com/FeigenbaumConstant.html), etc. As far as I know, he used only a programmable calculator to discover it. He was just inquisitive, nothing more, nothing less.
| 15 | https://mathoverflow.net/users/3811 | 20403 | 13,557 |
https://mathoverflow.net/questions/20408 | 8 | In [this](https://mathoverflow.net/questions/20392/conjugating-a-subgroup-of-a-group-into-a-proper-subgroup-of-itself) question, I asked whether there existed groups $G$ with finitely presentable subgroups $H$ such that $gHg^{-1}$ is a proper subgroup of $H$ for some $g \in G$. Robin Chapman pointed out that the group of affine automorphisms of $\mathbb{Q}$ contains examples where $H \cong \mathbb{Z}$.
This leads me to the following more general question. A group $\Gamma$ is "coHopfian" if any injection $\Gamma \hookrightarrow \Gamma$ is an isomorphism. To put it another way, $\Gamma$ does not contain any proper subgroup isomorphic to itself. The canonical example of a non-coHopfian group is a free group $F\_n$ on $n$ letters. Chapman's example exploits the fact that $F\_1 \cong \mathbb{Z}$ contains proper subgroups $k \mathbb{Z}$ isomorphic to $\mathbb{Z}$.
Now let $\Gamma$ be a non-coHopfian group and let $\Gamma' \subset \Gamma$ be a proper subgroup with $\Gamma' \cong \Gamma$. Question : does there exist a group $\Gamma''$ such that $\Gamma \subset \Gamma''$ and an automorphism $\phi$ of $\Gamma''$ such that $\phi(\Gamma) = \Gamma'$? How about if we restrict ourselves to the cases where $\Gamma$ and $\Gamma''$ are finitely presentable? I expect that the answer is "no", and I'd be interested in conditions that would assure that it is "yes".
If such a $\Gamma''$ existed, then we could construct an example answering my linked-to question above by taking $G$ to be the semidirect product of $\Gamma''$ and $\mathbb{Z}$ with $\mathbb{Z}$ acting on $\Gamma''$ via $\phi$. This question thus can be viewed as asking whether Chapman's answer really used something special about $\mathbb{Z}$.
| https://mathoverflow.net/users/317 | Automorphisms of supergroups of non-coHopfian groups | Let $\alpha: \Gamma\to\Gamma$ be an injection sending $\Gamma$ to $\Gamma'$. Then the $\Gamma''$ you're looking for is the infinite amalgamated product
$\cdots \*\_{\Gamma}\Gamma \*\_{\Gamma}\*\Gamma\*\_{\Gamma}\cdots$
where, at each stage, $\Gamma$ maps to the left by the identity and to the right by $\alpha$. Now the 'shift' automorphism has the property that you want, and the semidirect product with $\mathbb{Z}$ that you suggest is just the ascending HNN extension of $\Gamma$ via $\alpha$.
| 11 | https://mathoverflow.net/users/1463 | 20409 | 13,562 |
https://mathoverflow.net/questions/20346 | 10 | This is likely an elementary question to logicians or theoretical computer scientists, but I'm less than adequately informed on either topic and don't know where to find the answer. Please excuse the various vague notions that appear here, without their appropriate formal setting. My question is exactly about what I should have said instead of what follows:
When we make a definition, of either a property, or the explicit value of a symbol, it seems that we are somehow changing the language. Prescribing meaning to a word might be viewed as a kind of transformation of the formal language akin to taking a quotient, where we impose further relations on a set of generators.
I don't know how to describe a 'semantic' object, but am assuming an ad-hoc definition could be as a class of words under an equivalence relation supplied by an underlying logic. If the complexity of such an object is the size of the smallest word in the language that describes it, then making a definition lowers the complexity of some objects (and doesn't change the rest.) The obvious example is that if I add the word *group* to my language, then saying *G is a group* is a lot shorter than listing its properties.
It seems that lowering complexity is a main point of making definitions. Further, that one reason mathematical theory-building works, is a compression effect through which I am able to use less resources to describe more complex objects, at the cost of the energy it takes to cram definitions from a textbook.
Likely there is some theory out there that describes this process, but I've not been able to google it. I would appreciate being pointed towards the right source, even if it's a wikipedia link. Specifically:
*Where can I find a theory of formal logic or complexity theory that studies the process of adding definitions to a mathematical language, viewed as a transformation that changes complexity?*
| https://mathoverflow.net/users/2604 | Is there a formal notion of what we do when we 'Let X be ...'? | Kieffer, Avigad, & Frideman, 2008 [A language for mathematical knowledge management](http://arxiv.org/abs/0805.1386), which I mentioned in the [Proof formalization](https://mathoverflow.net/questions/16386/proof-formalization/16387#16387) thread, discusses DZFC, an extension of ZFC with definitions of terms and partial terms. Theorem 1 proves conservativity over ZFC, with respect to which the paper says:
>
> the usual method of eliminating defined function symbols and
> relation symbols by replacing them by their definiens can result in an exponential
> increase in length.
>
>
>
Which is roughly in line with some analogous results for other formalisms. Neel mentioned the doubly expontential blow-up for normalisation in the simply typed lambda calculus, and more drastic blowups are possible with higher type systems.
It's unclear to me what notion of complexity is sought, but the expansion in size of terms under expansion will typically be the same as the time complexity of the expansion algorithm.
| 9 | https://mathoverflow.net/users/3154 | 20411 | 13,564 |
https://mathoverflow.net/questions/20414 | 1 | Consider a stochastic process $X\_t$ , $t \in 1,2,3,..,N $.
$X\_t$ is a Bernoulli variable and $\Pr(X\_t=1) = p$ for all $t$.
The Autocovariance function $\gamma(|s-t|)= E[(X\_t - p)(X\_s -p)]$ is given
$
\gamma(k) = \frac{1}{2} (|k-1|^{2H} - 2|k|^{2H} + |k+1|^{2H}).
$
For a constant $H\in (0,1)$ This is the same autocovariance as for fractional gaussian noise (increments of the fractional brownian motion), and give a autocovariance which falls like a power law when $k$ goes to infinity.
Let X and Y be process with the given properties, I am interested in the following probability distribution:
$
\Pr\left(\sum\_{i=0}^N X\_i Y\_i = k\right)
$
That is the distribution of the overlap of two such processes. For $H=1/2$ the process is not correlated and I have the simple result that $\Pr(X\_t Y\_t)=p^2$, and that
$
\Pr\left(\sum\_{i=0}^N X\_i Y\_i = k\right) = {N \choose k} p^{2k} (1-p^2)^{N-k}.
$
But for $H\neq 1/2$, I do not know how to deal with the long range correlation. Is there a way to proceed on this problem? I regret i never took a class in Stochastic Analysis, and I really hope the question makes sense. Any help or input would be highly appreciated.
| https://mathoverflow.net/users/4626 | Problem with a Long Range Correlated Time Series | One thing you should understand is that Bernoulli is not Gaussian: the autocorrelation function does not determine the process uniquely. In particular, the fact that the Bernoulli variables are not correlated doesn't mean that they are independent. For instance, the 3 step process that takes the paths (0,0,0),(0,1,1),(1,1,0),(1,0,1) with probability $1/4$ each has no autocorrelations but $\sum\_{i=0}^2 X\_iY\_i$ is never $3$ here. So, your formula fails for this process. We need to know much more than just autocorrelations to answer your question.
| 4 | https://mathoverflow.net/users/1131 | 20417 | 13,568 |
https://mathoverflow.net/questions/20422 | 4 | Let $k$ be a field, and consider an irreducible polynomial $f \in k[x,y]$. Let $S(f)$ denote the singular points of $f$ (points that are simultaneously zero on $f$, the $x$-derivative of $f$, and the $y$-derivative of $f$.)
If $k$ is algebraically closed, then I can prove $S(f)$ is finite. Also, I can prove that if the field has characteristic $0$, then $S(f)$ is finite.
But what if the field has characteristic $p$ and is not algebraically closed? Is it true that $S(f)$ is finite?
I asked this question to my algebraic geometry professor last semester and stumped him! Hopefully one of you can think of a counterexample or proof.
| https://mathoverflow.net/users/491 | Singular points of an irreducible polynomial | What do you mean by irreducible and what do you mean by $S(f)$?
Does irreducible mean absolutely irreducible (ie irreducible over the algebraic closure of $k$)? Is $S(f)$ considered as a scheme or as a set of rational points? If the latter, then is $S(f) := \{ (a,b) \in k^2: f(a,b) = 0 = \frac{\partial f}{\partial x}(a,b) = \frac{\partial f}{\partial y}(a,b) \}$? Or is it the set of singular points over the algebraic closure of $k$?
If by irreducible, you mean absolutely irreducible, then as Douglas Zare suggests, you can pass to the algebraic closure and prove that $S(f)$ is finite.
If irreducible is to be read over $k$, but you are considering $S(f)$ scheme theoretically or are evaluating the points in the algebraic closure of $k$, then the assertion is false. Consider for instance $k$ of characteristic $p$ with $a \in k$ a non-$p^\mathrm{th}$ power and $f(x,y) = x^p + y^p + a$.
Finally, if by $S(f)$ you mean the $k$-rational points, then if $S(f)(k)$ were infinite, then the set of $k$-rational points on the curve defined by $f$ would be infinite and $f$ would then be absolutely irreducible so that by the first case considered, $S(f)$ would be finite.
| 8 | https://mathoverflow.net/users/5147 | 20428 | 13,574 |
https://mathoverflow.net/questions/20420 | 5 | Does this 2D cellular automaton have a known name and history?
* n colors (numbered 1 to n), assigned randomly at the start.
* For each generation, every cell that has at least one neighbour cell with a color that is one higher changes its color to that "next higher" color. Additionally, the "lowest" color is considered "next higher" to the highest one.
* Emergent behaviour shows up best around n=16, disappears for much higher or much lower n
I have it [implemented on my website](http://www.brazzy.de/Muncher.php) so you can see it in action.
I saw this ages ago and always remembered it as a great example of emergent behaviour, but can't remember what it was called, and couldn't find it on Wikipedia or Wolfram Mathworld.
| https://mathoverflow.net/users/5150 | What's the name of this 2D cellular automaton? | [Cyclic cellular automaton](http://en.wikipedia.org/wiki/Cyclic_cellular_automaton)
| 10 | https://mathoverflow.net/users/440 | 20431 | 13,576 |
https://mathoverflow.net/questions/19992 | 2 | Given a pair of distributions $x,y\in(0,1]^{n\times 1}$, so that $1^Tx=1$ and $1^Ty=1$,
I want to build a matrix $C$ (change matrix) that satisfy **at least** the following properties:
i) $C$ is diagonal if and only if $x=y$
ii) $C1 = x$
iii) $C^T1 = y$
iv) $C$ has nonnegative entries
How to build a $C$ that satisfy i)-iv)?
If $\Lambda\_x = diag(x)$ and $\Lambda\_y = diag(y)$ conditions ii) and iii) can be also written as:
(1) $C\Lambda\_y^{-1}y = x$
(2) $C^T\Lambda\_x^{-1}x = y$
respectivelly. Replacing (2) in (1) results in:
(3) $C\Lambda\_y^{-1}C^T\Lambda\_x^{-1}x = x$
And replacing $x$ by $\Lambda\_x1$ results the matricial Equation:
(4) $\(C\Lambda\_y^{-1}C^T-\Lambda\_x\)1 = 0$
or alternativelly (if 1 is replaced in 2),
(5) $\(C^T\Lambda\_x^{-1}C-\Lambda\_y\)1 = 0$
| https://mathoverflow.net/users/5066 | Matrix decomposition problem | This is exactly the measure transportation problem in finite setting. Try to google "optimal measure transportation" for references and various algorithms. (ii) and (iii) are just the definition of transport (one also usually wants the entries of $C$ to be non-negative) and (i) is a very weak requirement of "optimality" that follows from any transport cost minimization requrement usually used (they are many and yield different answers).
| 5 | https://mathoverflow.net/users/1131 | 20433 | 13,578 |
https://mathoverflow.net/questions/20373 | 5 | I want to know about people in researching complex (maybe differential) geometry are careing about what currently ? For example ,$L^2$ estimate inspired by Lars Hormander is a very useful tool,and how does this theory be developed currently ? As myself , i like this method very much ,but i don't know which is the next important problem be solved by this method . How far will this method go ? As well , just like the holomorphic morse inequalities , when it is proved by Demailly in 1985,twenty years passed , it seems that during thest twenty years there are no important results comes out in complex geometry ? I'm a beginner in complex geometry, i think that i can't scratch the direction of complex geometry ? I don't know people are careing about what in complex geometry ?Only doing some little questions or leave this field? So this is just the purpose of this question i asked , i want to communicate with all who are interested in this field.
| https://mathoverflow.net/users/4437 | the central issues in complex geometry | One major area of research is that of canonical metrics on Kahler manifolds. The original definition of "canonical" is due to Calabi. One considers all metrics in a fixed Kahler class and attempts to minimise the L^2-norm of the curvature tensor. The Euler-Lagrange equations say that a metric is a critical point for this functional precisely when its scalar curvature has holomorphic gradient. Such metrics are called *extremal*. Special cases include Kahler-Einstein metrics and constant scalar curvature metrics.
Enormous amounts of research have been done in this direction. In one direction, one tries to prove existence. A random selection of results: Aubin and Yau's work on Kahler-Einstein metrics of negative and zero scalar curvature, Tian's work on Kahler-Einstein surfaces of positive scalar curvature, more recently Donaldson's work on extremal metrics on toric surfaces.
In the opposite direction, one tries to find obstructions to existence. Futaki found one such obstruction, involving holomorphic vector fields, which was then vastly generalised by Tian and Donaldson. This gives examples of many manifolds and Kahler classes which can never admit extremal metrics.
There is a conjecture (due in various forms to Donaldson, Tian and Yau) which says that the known obstructions are the only obstructions. When they vanish, an extremal metric exists. This would be a beautiful result because the obstructions are purely algebro-geometric (at least when the Kahler class is integral) yet the conclusion is analytic - we can find a metric which solves a PDE. If you know of the Hitchin-Kobayashi correspondence for Hermitian Yang-Mills metrics on vector bundles, this conjecture can be seen as the analogue for Kahler metrics.
At the moment, the conjecture is known to hold for toric surfaces (Donaldson) and is close to being settled for Kahler-Einstein metrics. This is due to Aubin-Yau for negative scalar curvature, and Yau for zero scalar curvature - in these cases the metrics always exist. The positive scalar curvature case is far more delicate. Donaldson recently announced significant progress in this direction.
If you are looking for a link with Hormander's L^2 estimates, then look no further than projective embeddings via higher and higher powers of a positive line bundle L over your Kahler manifold X. For each power of L we consider the following problem: find a basis of holomorphic sections of L so that the image of X in CPn has zero centre of mass (to define centre of mass we think of CPn as a coadjoint orbit in the linear space su(n+1)\* equiped with an inner-product via the Killing form). This can be thought of as a finite dimensional analogue of the problem of finding an extremal metric representing the first Chern class of L. As the power of L tends to infinity, these projective problems converge in some sense to the problem of finding an extremal metric. Understanding the precise nature of this convergence involves difficult questions concerning the Bergman kernel and, ultimately, Hormander's estimate. (Again, this part of the story is due to Donaldson.)
To get started you could look at:
* *Extremal Kahler metrics 1 and 2*, by Calabi.
* *Canonical metrics in Kahler geomtery*, by Tian.
* *Remarks on gauge theory, complex geometry and 4-manifold topology*, by Donaldson.
* *Scalar curvature and stability of toric varieties*, by Donaldson.
* *Scalar curvature and projective embeddings 1*, by Donaldson.
| 15 | https://mathoverflow.net/users/380 | 20436 | 13,581 |
https://mathoverflow.net/questions/20438 | 11 | In the following every surface is assumed to be connected.
I've read that the commutative monoid of homeomorphism classes of closed surfaces is generated by $P$ (projective plane) and $T$ (torus) subject to the only(!) relation $P^3=PT$. Here the product is given by the connected sum. Now what about the commutative monoid of homeomorphism classes of compact surfaces (with boundary)? Does it also have a nice presentation? I think it is generated by the $P[k]$ and the $T[k]$, where $[k]$ means that $k$ holes have been inserted, and $k$ runs through the natural numbers. What relations do we need? And how do you prove that no others are needed?
Another question, which is rather informal: Do you think that it's worth to read the proofs of these classical classifications? I know the importance of the results, but I suspect that the proofs are just technical.
| https://mathoverflow.net/users/2841 | Presentation of the monoid of surfaces | I assume you want your surfaces-with-boundary to be compact? Anyway, this cannot be generated by the $P[k]$ and $T[k]$, since you are leaving out the genus 0 surfaces (spheres with holes). Since connect-sum-with-a-disk is the same as removing an open disk, I would work instead with the generators $P$, $T$, and the disk $D$; the $D$ won't interact with the other generators. Since every surface-with-boundary is a surface-minus-some-disks, it seems a presentation is $\langle P,T,D\vert P^3=PT\rangle$.
In answer to your second question: it can be very worthwhile to know the proofs of these classical facts, but that doesn't mean you need to learn the classical proofs. For the classification of surfaces, I have seen Benson Farb give a very nice proof (fitting with the "modern" perspective on mapping class groups etc.) hinging upon the fact that the sphere has the maximal Euler characteristic among surfaces. If I can find any notes of that lecture or a written version, I'll update with a link.
| 10 | https://mathoverflow.net/users/250 | 20443 | 13,584 |
https://mathoverflow.net/questions/20383 | 0 | I learn that Geometry has several categories/subfields from Wikipedia. But I am still not clear about the standards according to which they are classified.
1. It seems Euclidean Geometry, Affine Geometry and Projective Geometry are classified according some rule, while Hyperbolic Geometry, Elliptic Geometry and Riemann Geometry according to another, and Axiomatic, Analytic, Algebraic and Differential Geometry perhaps according to a different one? What rules are they?
2. Are Affine Geometry, Projective Geometry, Hyperbolic Geometry, Elliptic Geometry and Riemann Geometry all Non-Euclidean Geometry? What are their common characteristics that make them Non-Euclidean Geometry?
Really appreciate if someone could clarify these questions for me and also hope you can provide more insights into the subfields of Geometry not necessarily the specific questions I asked.
---
*Update:* Although my major is not math, I have been involved in some projects requiring quite a few mathematics and have taken a lot of courses in mathematics on undergraduate/graduate level. Now looking back, I am confused about what I have learned and heard, and would like to get a big picture.
| https://mathoverflow.net/users/5142 | Categories of Geometry | There are a couple of subtle differences. Some of the concepts are only relevant when talking about geometry from an axiomatic perspective.
When one is talking about geometry from an axiomatic perspective ( you want to talk about points, lines, planes, angles etc.) you are really looking at a model for your axioms. Here we might talk about Euclidean, Riemannian, Hyperbolic, Projective, Spherical and (perhaps) Elliptic geometries. Usually the main difference is whether or not we choose to take the parallel postulate as an axiom or one of its negations. If we take the parallel postulate then we are working in a model of Euclidean geometry, this is sort of flat geometry where things are what you expect. In Spherical geometry (sometimes called Riemannian, and maybe elliptical but i am not sure about that) is where we have no parallel lines, a model of this is the sphere where we take lines to be great circles. Note though that typically Riemannian geometry is about manifolds with a Riemannian structure, but that we can save until later. Hyperbolic geometry is when we have infinitely many parallel lines through a given point, a model of this is the [Poincare disk](http://en.wikipedia.org/wiki/Poincare_disk%20%22poincare%20disk%20model%22). Projective geometry has every two parallel lines meet... at the point at infinity. There are much better references for this stuff but so far we have just been looking at models of axiom systems that one could work with and "do geometry" in.
Algebraic, differential, and Riemannian geometry are more complicated. Here are some "one line slogans" which i am sure others can improve upon. Differential geometry is about curves, surfaces and homogeneous objects that you want to study via calculus, there is a priori some smooth structure on that object. Algebraic geometry wants to study similar objects but only when you are concerned with what you can tell about an object via rational functions. For me, this starts with the Gelfand-Naimark result. it gets much much richer. Riemannian geometry is differential geometry except you have a well behaved notion of distance between points, distance ON the hypersurface itself! All of these could be fixed by someone with more knowledge then I on the respective subjects.
I didn't mention affine geometry, but I will take a stab and say it is like a geometry in any one of the above models except you only care about "flat" things, lines, planes, etc.
The above comment suggests looking at some good books. It mentions Klein's Erlangen program, which is where Klein proposed studying a geometry by understanding the group of symmetries that preserve it. So you can think of the first big paragraph as looking at groups and the geometries you get from them. The second big paragraph you can get by looking at different types of sheaves on various topological spaces (I think) where the sheaf keeps track of the type of structure you care about. The two classes are sort of a bit different, but they are similar in that you can look at objects that act as receptacles for the geometric content: groups and sheaves.
Any suggestions are very welcome!
| 6 | https://mathoverflow.net/users/3901 | 20448 | 13,587 |
https://mathoverflow.net/questions/20445 | 18 | Let
>
> $F:A^{\mbox{op}} \to \mbox{Set}$
>
>
>
and define
>
> $G\_a:A\times A^{\mbox{op}} \to \mbox{Set}$
>
>
> $G\_a(b,c) = \mbox{hom}(a,b) \times F(c)$.
>
>
>
I *think* the coend of $G\_a$,
>
> $\int^AG\_a$,
>
>
>
ought to be $F(a)$--it's certainly true when A is discrete, since then hom is a delta function. But my colimit-fu isn't good enough to actually compute the thing and verify it's true. Can someone walk me through the computation, please?
| https://mathoverflow.net/users/756 | Coend computation | Hi Mike. This is what's often called the Density Formula, or (at the n-Lab) the coYoneda Lemma (I think), or (by Australian ninja category theorists) simply the Yoneda Lemma. (But Australian ninja category theorists call *everything* the Yoneda Lemma.) In any case, it's a kind of dual to the ordinary Yoneda Lemma.
But you asked to be walked through it. First: yes, it is $F(a)$. Another way of writing your coend
$$
\int^A G\_a
$$
is as
$$
\int^{b \in A} G\_a(b, b) = \int^b \mathrm{hom}(a,b) \times F(b).
$$
I claim this is canonically isomorphic to $F(a)$. I'll prove this by showing that for an arbitrary set $S$, the homset $\mathrm{hom}(\mathrm{this}, S)$ is canonically isomorphic to $\mathrm{hom}(F(a), S)$. The claim will then follow from the ordinary Yoneda Lemma.
So, let $S$ be a set. Then
$$
\begin{align}
\mathrm{Set}(\int^b \mathrm{hom}(a, b) \times F(b), S) &
\cong
\int\_b \mathrm{Set}(\mathrm{hom}(a, b) \times F(b), S) \\
&\cong
\int\_b \mathrm{Set}(\mathrm{hom}(a, b), \mathrm{Set}(F(b), S)) \\
&\cong
\mathrm{Nat}(\hom(a, -), \mathrm{Set}(F(-), S)) \\
&\cong
\mathrm{Set}(F(a), S)
\end{align}
$$
I don't know how much of this you'll want explaining, so I'll just say it briefly for now. If you want further explanation, just ask. The first isomorphism is kinda the definition of colimit. The second is the usual exponential transpose/currying operation. The third is maybe the most important: it's a fundamental fact about ends that if $F, G: C \to D$ are functors then
$$
\mathrm{Nat}(F, G) = \int\_c D(F(c), G(c)).
$$
The fourth and final isomorphism is the ordinary Yoneda Lemma applied to the functor $\mathrm{Set}(F(-), S)$.
| 30 | https://mathoverflow.net/users/586 | 20451 | 13,588 |
https://mathoverflow.net/questions/20454 | 12 | Consider a dihedral group of degree *n* and order *2n*. Its two-dimensional irreducible representations can be realized over the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$, with the usual action by rotations and reflections. Also, any splitting field of characteristic zero for this group must contain $\mathbb{Q}(\cos(2\pi/n))$, because this is the field generated by the characters of the irreducible representations. (Here, splitting field for a finite group means a field over which all the irreducible representations are realized).
When *n* is a multiple of 4, these two fields are the same; otherwise, they are not. My question: when *n* is not a multiple of 4, what conditions would ensure that the smaller subfield $\mathbb{Q}(\cos(2\pi/n))$ is a splitting field for the dihedral group? I think we can restrict attention to *n* odd.
For instance, when $n = 3$, the dihedral group of degree 3, order 6, has $\cos(2\pi/3) = -1/2$, $\sin(2\pi/3) = \sqrt{3}/2$. So, any splitting field must contain $\mathbb{Q}(1/2) = \mathbb{Q}$. Also, $\mathbb{Q}(1/2,\sqrt{3}/2) = \mathbb{Q}(\sqrt{3})$ is a splitting field since it contains the usual representation given by rotations and reflections.
However, $\mathbb{Q}$ is also a splitting field. To see this, we think of the group as the symmetric group of degree three and take the standard representation. So in this case, we see that $\mathbb{Q}(\cos(2\pi/n))$ works as splitting field even though it is smaller than the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$.
NOTE: It is *not* true *in general* that if the characters all take values in a field, the representation can be realized over that field. The standard counterexample is the quaternion group, whose characters all take rational values but whose irreducible representations are realized only when we go to $\mathbb{Q}(i)$. However, some weaker variant of the result may be true for the groups that we are interested in here, namely, the dihedral groups.
| https://mathoverflow.net/users/3040 | Realizability of irreducible representations of dihedral groups | The name of the concept you are looking for is the Schur index. The Schur index is $1$ iff the representation can be realized over the field of values. The Schur index divides the degree of the character.
In your case, the the Schur index is either $1$ or $2$. You can use a variety of tests to eliminate $2$, but for instance:
*Fein, Burton; Yamada, Toshihiko*, [**The Schur index and the order and exponent of a finite group**](https://dx.doi.org/10.1016/0021-8693(74)90055-6), J. Algebra 28, 496-498 (1974). [ZBL0243.20008](https://zbmath.org/?q=an:0243.20008).
shows that if the Schur index was $2$, then $4$ divides the exponent of $G$.
In other words, all of your representations are realizable over the field of values.
Isaacs's Character Theory of Finite Groups has most of this in it, and I found the rest of what I needed in Berkovich's Character Theory collections. Let me know if you want more specific textbook references.
**Edit:** I went ahead and looked up the Isaacs pages, and looks like textbook is enough here: Lemma 10.8 on page 165 handles induced irreducible characters from complemented subgroups, and shows that the Schur index divides the order of the original character. Taking the subgroup to be the rotation subgroup and the original character to be faithful (or whichever one you need for your particular irreducible when $n$ isn't prime), you get that the Schur index divides $1$. The basics of the Schur index are collected in Corollary 10.2 on page 161.
At any rate, Schur indices are nice to know about, and if Isaacs's book doesn't have what you want, then Berkovich (or Huppert) has just a silly number of results helping to calculate it.
**Edit:** Explicit matrices can be found too. If $n=4k+2$ is not divisible $4$, and $G$ is a dihedral group of order $n$ with presentation $\langle a,b \mid aa=b^n=1, ba=ab^{n-1} \rangle$, then one can use companion polynomials to give an explicit representation (basically creating an induced representation from a complemented subgroup). Send $a$ to $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, also known as multiplication by $x$. Send $b$ to $\begin{pmatrix}0 & -1\\1 & \zeta + \frac{1}{\zeta}\end{pmatrix}$, also known as the companion matrix to the minimum polynomial of $\zeta$ over the field $\mathbb{Q}(\zeta+\frac{1}{\zeta})$, where $\zeta$ is a primitive $(2k+1)$st root of unity.
Compare this to the more direct choice of $a = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ and $b = \begin{pmatrix} \zeta & 0\\0 & \frac{1}{\zeta}\end{pmatrix}$. If you conjugate this by $\begin{pmatrix}1 & \zeta \\\zeta & 1\end{pmatrix}$ then you get my suggested choice of a representation.
In general, finding pretty, (nearly-)integral representations over a minimal splitting field is hard (and there may not be a unique minimal splitting field), but in some cases you can do it nicely.
Let me know if you continue to find this stuff interesting. I could ramble on quite a bit longer, but I think MO prefers focused answers.
| 18 | https://mathoverflow.net/users/3710 | 20458 | 13,590 |
Subsets and Splits