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https://mathoverflow.net/questions/32765 | 18 | If functors are morphisms between categories, and natural transformations are morphisms between functors, what's a morphism between natural transformations? Is there ever a need for such a notion?
| https://mathoverflow.net/users/2592 | What's after natural transformations? | (Small) categories form what's called a 2-category, which is a structure that has objects, morphisms (functors), and morphisms between morphisms (natural transformations). There are also n-categories, which have a deeper morphisms structure. A google search will point you to a lot of references about n-categories. But for ordinary categories, the story ends at natural transformations.
| 13 | https://mathoverflow.net/users/4183 | 32767 | 21,271 |
https://mathoverflow.net/questions/32761 | 1 | Let $\mathbb{F}\_q$ be a finite field with characteristic $p$ and $p < q$ (i.e. not a prime field). Let $D\subseteq \mathbb{F}\_q$ be a some set with $|D|=n$. Find a non-empty subset $\{x\_1,\dots,x\_k\} \subseteq D$ such that $x\_1+\cdots+x\_k=s$ for some given $s\in\mathbb{F}\_q$. This is the definition of the subset sum problem.
What I cannot understand is how do you count the number of solutions for a given $s$. In this [paper](http://arxiv.org/abs/0708.2456), in page 2 it says
>
> heuristically should be approximately $\frac{1}{q}\binom{n}{k}$.
>
>
>
A more concrete questions is, given $s$ how many summands does it have given that we select $D$ randomly from $\mathbb{F}\_q$?
| https://mathoverflow.net/users/7692 | Number of subset sums | There are $\binom{n}{k}$ ways to choose $k$ elements from $n$ elements. If we consider their sum it is "expected" to be equal to every element of the field with the same probability. Hence we get $\frac{1}{q}\binom{n}{k}$ for the number of solutions of $x\_1 + x\_2 + \cdots + x\_k = s$. This is the first part of your question.
Now there are $\binom{q}{k}$ ways to choose $k$ different elements of the field. And $\frac{1}{q}$ of them have sum equal to $s$. In other words number of solutions of $x\_1 + x\_2 + \cdots + x\_k = s$ is equal to $\frac{1}{q}\binom{q}{k}$.
If we choose random $D$ with $|D|=n$ every solution will have all $x\_i \in D$ with the same probability $p$:
$$p = \frac{\binom{q-k}{n-k}}{\binom{q}{n}}$$
Hence the expected number of solutions is equal to $\frac{1}{q}\binom{q}{k} \cdot p$ which is equal to $\frac{1}{q}\binom{n}{k}$.
So the expected number of solutions is equal to $\frac{1}{q}\binom{n}{k}$.
| 5 | https://mathoverflow.net/users/7079 | 32768 | 21,272 |
https://mathoverflow.net/questions/32762 | 3 | This is a sequel to my previous question [colimits of spectral sequences](https://mathoverflow.net/questions/28972/colimits-of-spectral-sequences) .
I think I've found the answer in S.A. Mitchell's paper "Hypercohomology spectra and Thomason's descent theorem". There the author states a "colimit lemma" (page 42) for ss of homotopy groups of spectra, which I think can be literally translated with no harm for cohomology groups of cochain complexes and it's exactly the result I was looking for.
However, my problem is more basic and shameful. Previously, but in the same page, Mitchell says that, having a right half-plane cohomology spectral sequence (coming, say, from a double complex, or a filtered complex), that is
$$
E\_2^{pq} = 0 \qquad \mbox{if} \quad p < 0 \ , \qquad \qquad \qquad [1]
$$
then it *converges* if it is, for instance, *bounded on the right*, that is, if there exists $d$ such that
$$
E\_2^{pq} = 0 \qquad \mbox{if} \quad p > d . \qquad \qquad \qquad [2]
$$
I assume the author uses the term "converge" in the sense of Cartan-Eilenberg (otherwise, I don't understand the proof of his "colimit lemma" at all), that is:
(a) We have an *exhaustive* filtration $F$ on the limit $G$, $\bigcup\_p F^PG = G$, and also isomorphisms $E\_\infty^p = F^pG / F^{p+1}G$, and
(b) The filtration on $G$ is *Hausdorff*, that is $\bigcap\_p F^pG = 0$.
Now, I have no problem in assuming that in my particular ss the original filtration of my filtered complex is already exhaustive and hence so it is the induced one on the "limit" $G$ and the isomorphisms for $E\_\infty$ (as it is the case for both filtrations of a double complex) and I think Mitchell is assuming this too implicitely, because these conditions seems "for free".
Also, the boundness conditions [1] and [2] will imply that the filtration on $G$ is in fact *finite*. So, if we had (b), the converge would be in the strong sense. Great! :-)
So the point is the Hausdorff condition of the filtration on $G$: why would [1] and [2] imply that the filtration on $G$ should also be Hausdorff?
| https://mathoverflow.net/users/1246 | Convergence of right half-plane spectral sequence bounded on the right | It doesn't -- just filter any chain complex trivially, the resulting spectral sequence has vanishing $E\_2$ and obviously doesn't converge in any sense. So you have to look at how the spectral sequence is constructed. A very useful notion in this context is *conditional convergence*, which is probably satisfied in the example you are considering. In that case, if the derived functor of the $E\_\infty$ term vanishes, the spectral sequence converges strongly -- in particular, this happens if the filtration is finite. I'm willing to bet that the spectral sequence you're looking at is conditionally convergent. For this notion, and other convergence questions about spectral sequences, I highly recommend Boardman's paper "Conditionally convergent spectral sequences."
Unfortunately, a lot of the literature is quite sloppy on convergence questions for spectral sequences...
| 5 | https://mathoverflow.net/users/4183 | 32769 | 21,273 |
https://mathoverflow.net/questions/32766 | 23 | Let $X$ be a complex normal projective variety.
**Is there any sufficient condition to guarantee the torsion-freeness of Picard group of $X$?**
One technique I sometimes use is following:
If $X$ can be represented by GIT quotient $Y//G$ for some projective variety $Y$ with well-known Picard group, then by using Kempf's descent lemma, we can attack the computation of integral Picard group.
Of course, if we can make a sequence of smooth blow-ups/downs between $X$ and $X'$ with well known Picard group, then we can get the information of $\mathrm{Pic}(X)$ from $\mathrm{Pic}(X')$.
Is there any way to attack this problem?
| https://mathoverflow.net/users/4643 | Torsion-freeness of Picard group | [EDIT: A previous version mistakenly argued that the fundamental group of *X* was responsible for torsion in the Picard group. I hope that this is correct now! Btw, there is probably a more direct way of arguing, but I cannot find one at the moment.]
The Picard group of *X* is torsion free if and only if the group ${\rm H\_1}(X,\mathbf{Z})$ vanishes.
By the exponential sequence, the torsion in the Picard group of *X* comes from the torsion in ${\rm H^2}(X,\mathbf{Z})$ and from ${\rm H^1}(X,\mathbf{C})/{\rm H^1}(X,\mathbf{Z})$. Thus the vanishing of ${\rm H\_1}(X,\mathbf{Z})$ is equivalent (by the Universal Coefficient Theorem) to the torsion-freeness of the Picard group of *X*.
ADDED (for explicitness) To make everything more explicit, assume that *X* is non-singular. The Picard group of *X* may contain torsion coming from two different sources. There might contain torsion in the connected component of the identity, and this is recorded by the torsion free part of the first homology group. Or there might be torsion in the component group of the Picard group, and this is recorded by the torsion in the first homology group. In terms of the exponential sequence, the first kind of torsion appears in the image of ${\rm H^1}(X,\mathbf{C})$, while the second one "appears" in torsion in ${\rm H^2}(X,\mathbf{Z})$. The Universal Coefficient Theorem implies that the "combination" of these two groups is the whole first integral homology group.
An example of torsion of the first kind is already present in the case of curves of genus at least one: the Jacobian of the curve contains plenty of torsion bundles. An example of torsion of the second kind is the case of Enriques surfaces: the canonical divisor on such a surface is a torsion line bundle that is non-trivial. If the characteristic of the ground-field is different from two, the corresponding cover of *X* is a K3 surface.
| 30 | https://mathoverflow.net/users/4344 | 32773 | 21,275 |
https://mathoverflow.net/questions/32787 | 22 | Given any finitely-presented group $G$, there are a few equivalent techniques for constructing smooth/PL 4-manifolds $M$ such that $\pi\_1 M$ is isomorphic to $G$. For most constructions of these 4-manifolds, they embed naturally in $S^5$ (as the boundary of regular neighbourhoods of $2$-complexes in $S^5$.)
Question: Are there are any smooth/PL 4-dimensional submanifolds $M$ of $S^4$ such that $\pi\_1 M$ has an unsolvable word problem? $M$ would of course have to be a smooth $4$-manifold with non-empty boundary.
I'm aware there are several constructions and obstructions to $2$-complexes embedding in $S^4$. Moreover, I've heard some of the construction techniques fall into the tame topological world and may not be smoothable. The condition given by Kranjc (that $H^2$ of the 2-complex is cyclic) is generally a non-computable condition for a group with non-solvable word problem. Although, perhaps there are many groups with non-solvable word problem and $H^2$ trivial. The closest to references on the subject that I know:
M. Kranjc, "Embedding a 2-complex K in R^4 when H^2(K) is a cyclic group," Pac. J. Math. 150 (1991), 329-339.
A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.
edit: Thanks for the comments people. Now that I'm back in Canada with proper internet (+MathSciNet) access, I did a little digging and came across this:
A. Dranisnikov, D. Repovs, "Embeddings up to homotopy type in Euclidean Space" Bull. Austral. Math. Soc (1993).
They show that any finitely-presented group is the fundamental group of a 2-dimensional polyhedron in $\mathbb R^4$. This was apparently a question of Matthias Kreck's.
And yes, Sam Nead, this question was in part motivated by the concern that 2-knots could have undecidable word problems for the fundamental groups of their complements. I've been thinking about the fundamental groups of 2-knot complements recently, and this is a concern.
| https://mathoverflow.net/users/1465 | Word problem for fundamental group of submanifolds of the 4-sphere | Update:
My memory was quite blurry about this when I originally answered.
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.
They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.
Original answer:
You most likely would like a finitely presented group, but this might be of interest anyway:
Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group
$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$
This has unsolvable word problem. See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)
This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$. You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.
I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.
| 15 | https://mathoverflow.net/users/1335 | 32803 | 21,293 |
https://mathoverflow.net/questions/32782 | 4 | Under what conditions on a and b is there a distribution $f\_{a,b}$ such that the product $XY$ of two independent realizations $X$ and $Y$ from $f\_{a,b}$ has a Beta(a,b) distribution?
A standard result on deriving the distibution of the product of two variables indicates that the p.d.f. $f\_{a,b}$ needs to satisfy:
$\int\_0^1 f\_{a,b}(x) f\_{a,b}(\frac{v}{x}) \frac{1}{x} dx = \frac{v^{a-1} (1-v)^{b-1}}{B(a,b)}$, but I have no idea how to find such an $f\_{a,b}$. (B denotes the beta function).
This article ( <http://www.jstor.org/stable/2045709> ) might be relevant for the case a=b=1, but my background in the relevant maths is not strong enough to understand it at all.
| https://mathoverflow.net/users/7801 | "Square root" of Beta(a,b) distribution | A partial answer is that $f\_{a,b}$ exists for every positive integer $b$.
To see this, first recall that for every positive $s$ and $a$ the distribution Gamma$(s,a)$ has density proportional to $z^{s-1}e^{-az}$ on $z\ge0$ and that the sum of independent Gamma$(s,a)$ and Gamma$(t,a)$ is Gamma$(s+t,a)$.
If $b=1$, choose $X=e^{-Z}$ and $Y=e^{-T}$ where $Z$ and $T$ are i.i.d. and Gamma$(1/2,a)$. Then $X$ and $Y$ are i.i.d. and $XY=e^{-(Z+T)}$ where $Z+T$ is Gamma$(1,a)$, that is, exponential of parameter $a$, hence $XY$ is Beta$(a,1)$ and you are done if $b=1$.
From there, recall that the product of two independent Beta$(a,c)$ and Beta$(a+c,b-c)$ is Beta$(a,b)$ for every $c$ in $(0,b)$. Assume that $b$ is a positive integer and choose $X=e^{-Z\_1-Z\_2-\cdots-Z\_b}$ and $Y=e^{-T\_1-T\_2-\cdots-T\_b}$ where all the $Z\_k$ and $T\_i$ are independent and, for each $k$, $Z\_k$ and $T\_k$ are both Gamma$(1/2,a+k-1)$. Then, by the $b=1$ case, each $e^{-(Z\_k+T\_k)}$ is Beta$(a+k-1,1)$ hence $XY$ is the product of some independent Beta$(a,1)$, Beta$(a+1,1)$, ..., Beta$(a+b-1,1)$, hence it is Beta$(a,b)$ and you are done for every positive integer $b$.
| 5 | https://mathoverflow.net/users/4661 | 32805 | 21,295 |
https://mathoverflow.net/questions/32791 | 3 | For $C,D$ small categories, and $f : C \to D$ a functor between them, there is a precomposition, or "inverse image", functor $f^\* = (-) \circ f : Set^D \to Set^C$. It has a left and a right adjoint. What are their definitions, and in particular what is the right adjoint $f\_\*$? I couldn't find a definition in terms of functor categories, just "topological" ones.
| https://mathoverflow.net/users/7674 | How is the right adjoint $f_*$ to the inverse image functor $f^*$ described for functor categories $Set^C$, $Set^D$ and $f : C \to D$ | Given a functor $f:\mathcal{C}\to\mathcal{D}$ and any complete category $\mathcal{A}$ (e.g., take $\mathcal{A}=\text{Sets}$ to get the case you are asking about), there exists a right-adjoint $f\_{\ast}:[\mathcal{C},\mathcal{A}]\to[\mathcal{D},\mathcal{A}]$ to the "inverse image functor" $f^{\ast}$ and this is given by taking right Kan extension.
Explicitly, given a functor $X:\mathcal{C}\to\mathcal{A}$, the functor $f\_{\ast}(X):\mathcal{D}\to\mathcal{A}$ is the right Kan extension of $X$ along $f$. This can be described explicitly using the limit formula
$$f\_{\ast}(X)(d)=\text{lim}\_{d\to f(c)}X(c)$$
for $d$ an object of $\mathcal{D}$ (the action on arrows of $\mathcal{D}$ is then induced by the universal property of limits). The indexing category of the limit here is of course the comma category $(d\downarrow f)$.
When $\mathcal{A}$ is cocomplete there is a corresponding left-adjoint $f\_{!}\dashv f^{\ast}$ which is given by taking left Kan extension along $f$. This can be explicitly described by the colimit formula dual to the limit formula given above.
(I should say that all of this is described very nicely in Mac Lane's book *Categories for the Working Mathematician*.)
| 7 | https://mathoverflow.net/users/6485 | 32808 | 21,298 |
https://mathoverflow.net/questions/32752 | 1 | I was reading a paper where I came across the following argument :
For any $x$ in $M$ and for a geodesic ball $B(x; \varepsilon)$ in a compact Riemannian
manifold $M$ with injectivity radius bigger than or equal to epsilon, and for any smooth eigenfunction $f$ of Laplacian on $M$, we have :
the square of $f(x)$ is $\leq C$ times ( the square of $L^2$ norm of $f$ over $B(x;\varepsilon)$
$+$ square of $L^2$ norm of $L(f)$ over $B(x:\varepsilon)$),
where $L(f)=$ Laplacian of $f$, where $C$ is independent of the Riemannian metric on $M$.
I was unable to see, with my limited Analysis knowledge, why this is true, but they mentioned that it follows from Sobolev's and Garding's inequality, for which they referred to S. Agmon's "Lectures on Elliptic boundary value problems"... still it is unclear to me.
N.B.: the injectivity radius of a manifold is the smallest of all numbers r
such that I can have a geodesic ball of radius r around each point of M. e.g.
injectivity radius of the sphere of radius $1$ with standard metric is $\pi$, injectivity radius of $\Bbb R^n$ is infinity etc.
Any help ? Thanks in advance.
| https://mathoverflow.net/users/6953 | The comparison between the square of the functional value and the sum of squares of the $L^2$ norms of function and its Laplacian | You are working on a Riemann surface. That bit of information is rather important, as [Sobolev inequalites](http://en.wikipedia.org/wiki/Sobolev_inequality) depends rather much on the dimension of the space. The basic Sobolev inequality is
$$ \| f \|\_{L^q(\Omega)} \leq C (\| \partial f \|\_{L^p(\Omega)} + \| f\|\_{L^p(\Omega)})$$
where the condition $\frac1p \geq\frac1q \geq \frac1p - \frac1n$ is satisfied (and $\Omega$ needs to be suitably regular). and $C$ depends on the set $\Omega$ and the coefficients $p,q$. If you want the sup norm on the left hand side, you can morally speaking replace $q$ by $\infty$ (so $1/q = 0$ and ask that the second inequality be strict).
In any case, in two dimensions by iterating the derivatives, you can actually show that for smooth $f$
$$ |f| \leq C( \|f\|\_{L^2} + \|\partial^2 f\|\_{L^2})$$
using that $0 > 1/2 - 2/2$. (The 2 in the numerator is the number of derivatives. In the denominator in the first term is the Lebesgue exponent, and in the second term is the dimension.)
Now, a consequence of [Garding's inequality](http://en.wikipedia.org/wiki/G%C3%A5rding%27s_inequality) states that for an uniformly elliptic differential operator $L$ of order $k$, one has that
$$ \| \partial^k f\|\_{L^2} \leq C (\| Lf\|\_{L^2} + \| f\|\_{L^2})$$
so using that the Laplacian is uniformly elliptic of order 2, you can plug Garding's inequality into Sobolev inequality and square the whole expression to get what the authors claim.
As to the actual dependence of the constant $C$ on various parameters: off the top of my head I can't remember the details. So I suggest you look it up either in Agmon's book as the authors suggest, or in Gilbarg & Trudinger *Elliptic Partial Differential Equations of Second Order* or Adams *Sobolev Spaces*
| 4 | https://mathoverflow.net/users/3948 | 32811 | 21,300 |
https://mathoverflow.net/questions/32788 | 3 | Given a map of topological spaces $f:X\rightarrow Y$. Assume, that $X$ has finite Lebesgue dimension. I am wondering, what dim$(f(X))$ might be. Of course, if $f$ is a homeomorphism onto its image, then it's just dim$(X)$. On the other hand there are the space filling curves, that show, that the dimension might increase. So I am wondering, whether there is any nice condition for $f$ (such as open, closed, proper etc), that guarantees, that dim$(f(X))\le$dim$(X)$.
| https://mathoverflow.net/users/3969 | Lebesgue dimension of images | Some results from Engelking's dimension theory book:
If $f: X \mapsto Y$ is a closed, continuous and surjective function between normal spaces $X$ and $Y$, and $\forall y \in Y: | f^{-1}[{y}] | \le k$ for some integer $k \ge 1$, then $\dim(Y) \le \dim(X) + (k-1)$.
If $f: X \mapsto Y$ is an open, continuous and surjective function between a weakly paracompact space $X$ and a normal space $Y$ with finite fibres, then $\dim(X) = \dim(Y)$.
If $f: X \mapsto Y$ is an open, continuous and surjective function between a locally compact normal space $X$ and a weakly paracompact normal space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.
If $f: X \mapsto Y$ is an open-and-closed, continuous and surjective function between a locally compact normal space $X$ and a paracompact space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.
In the other direction:
If $f: X \mapsto Y$ is a closed, continuous function between a normal space $X$ and a paracompact space $Y$ such that the dimension of the fibres is at most 0 (this includes the empty fibres of dimension -1), then $\dim(X) \le \dim(Y)$.
weakly paracompact includes Hausdorff, and is called metacompact or point-paracompact by other authors: every open cover has a point-finite refinement.
| 6 | https://mathoverflow.net/users/2060 | 32813 | 21,301 |
https://mathoverflow.net/questions/32812 | 6 | Suppose we have a sequence, { f\_n }, of symplectic diffeomorphisms of R^{2n} converging to a function f. By f\_n converging to f I mean: f\_n converges to f in C^k(B\_r) for every r > 0, where B\_r is the ball centered at the origin of radius r. Suppose k > 0.
Question: Is f is diffeomorphism? Certainly, f is symplectic and hence a local diffeomorphism, and so my question is really: is f invertible?
Interestingly enough, this is not true if f\_n is not symplectic (e.g. take f\_n = x/n, which converges to 0 in C^k(B\_r) for every r > 0).
Thanks!
| https://mathoverflow.net/users/nan | Is the limit of symplectic diffeomophisms a diffeomorphism? | No, $f$ doesn't have to be invertible. It has to be injective, even in the volume-preserving category rather than the symplectic category. None of the singular values of $Df$ can go to 0, because some other singular value would go to $\infty$. This shows that it is locally injective, and then globally it is not possible to make the locally injective pieces overlap.
But even when $n=1$, it does not have to be surjective. In this case the symplectic condition says just that $f\_n$ is area-preserving. You can reshape larger and larger circles centered at $0$ to non-concentric circles with the same area that lie in the upper half plane. So the image of $f$ could be the upper half plane. Actually, I think that the image of $f$ can be any open topological disk with infinite area.
| 8 | https://mathoverflow.net/users/1450 | 32816 | 21,303 |
https://mathoverflow.net/questions/32815 | 7 | A $n$-th degree polynomial has precisely $n$ roots. So it is natural to ask the question
how large ( and small) can be the measure of a set where a polynomial takes small values ?
This, and other interesting variation of this must have been studied in depth.
I would really appreciate any reference to the relevant literature.
Also, if there are some interesting variation of this problem I would like to know.
Thank you.
| https://mathoverflow.net/users/6766 | How large (small) can be the measure of a set where a polynomial takes small values ? | There is first Polya's estimate that if $f$ is a monic polynomial, then
$$
|\{x\in \mathbb{R}:\quad |f(x)|\leq 2\}| \leq 4.
$$
A proof can be found in the book "Proofs from the book". One can obtain inequalities for non-monic polynomials by rescaling.
Second there is Cartan's lemma or estimate. It can for example be found in Levin's book on entire functions. The estimate even holds for analytic functions.
The basic statement is:
Let $f: G \to \mathbb{C}$ be analytic and assume that $f$ is bounded by $1$ on a disc of radius $2$. Then there are constant $C, c > 0$ such that
$$
|\{z\in \mathbb{C}:\quad |z| < 1, | f(z)| \leq e^{-s}\}| \leq C
\exp\left( - \frac{c}{\log(\varepsilon^{-1})} s \right)
$$
where $\varepsilon = |f(0)|$. In fact, this is sharper, since it provides some information on how the set looks. For a polynomial it's just the union of its degree many disks. (For analytic functions countably many).
| 8 | https://mathoverflow.net/users/3983 | 32818 | 21,304 |
https://mathoverflow.net/questions/32799 | 7 | Let *X* be a normed space and denote by *X\** the space of all bounded linear functionals on *X*. Take a linear subspace *G ≤ X\** which separates the elements of *X*, i.e., for each *x ∈ X*, there is an *f ∈ G* with *f(x) ≠ 0*. Denote by *B* the closed unit ball in *X*. Now consider a linear subspace *Y ≤ X*. The question is:
If *Y* is dense in *X* in the weak topology induced by *G*, is *Y ∩ B* necessarily dense in *X ∩ B* in that topology?
REMARKS, BACKGROUND AND MOTIVATION
----------------------------------
Without the assumption that *G* separates points, there exists a trivial counter-example.
Take *X* := ℝ2 with the supremum norm, i.e., ∥(*x*, *y*)∥ :=
max{|*x*|, |*y*|}. For *G*, take the linear span of the linear functional *f(x, y) := x + y*. Finally, take *Y* := {(*x*, 0) ; *x* ∈ ℝ}. Then *Y* is *G*-dense in *X* because (*x*, *y*) and (*x* + *y*, 0) are indistinguishable in the *G*-topology. However, the element (1, 1) ∈ *B* is not in the closure of *Y ∩ B* because *f*(1, 1) = 2
and *f(x, 0) ≤ 1* for each *x* with *(x, 0) ∈ B*.
An interesting example is to take the space *G := L∞(S)*, the space of all bounded measurable functions on a measurable space *S*, equipped with the supremum norm. Take *X := G\**, with the corresponding dual norm. The space *G* can be naturally considered as a subspace of *X\**. Clearly, it separates the points in *X*, and the *G*-topology is exactly the weak \*-topology.
An important subspace of *X* is *Y := M(S)*, the space of all real measures on *S* (with finite total variation). If *S* is large enough, let's say, ℕ, then *Y* is a proper subspace of *X*. It is well-known that *Y* is weakly \*-dense in *X*, but it is also interesting that *Y* is weakly \*-complete by sequences (see Diestel: Sequences and Series in Banach spaces, Springer-Verlag, 1984).
By the Banach-Alaoglu theorem, *B* is weakly \*-compact. One may wonder whether *X ∩ B* is also weakly \*-compact. The answer is no. However, the argument that *Y* is weakly \*-dense in *X* is insufficient; a sufficient argument is that that *Y ∩ B* is weakly \*-dense in *X ∩ B*. Though this is not difficult to prove in our particular case, it might by a non-trivial issue in more general cases. If the answer to my initial question is yes, it will be sufficient to only prove that *Y* is dense in *X*.
Many thanks in advance for any answer, reference or comment!
| https://mathoverflow.net/users/7804 | If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? | If I understand the question correctly, then maybe you are after special cases, as well as a general comment. So, as one of your examples suggests, one special case is to let G be a Banach space, considered as sitting inside its own bidual, and let $X=G^\*$. Thus G induces the usual weak\*-topology on X.
So an example of a positive answer is furnished by the Kaplansky Density Theorem: here G would be the predual of a von Neumann algebra, X would be a von Neumann algebra, and we let Y be any self-adjoint subalgebra which is weak\*-dense. Then Kaplansky Density tells us that indeed the unit ball of Y is weak\*-dense in the unit ball of X. This is an incredibly useful tool in Operator Algebra theory.
This then suggests that the result is unlikely to be true in general. Indeed, I think rpotrie's counter-example works! But here's an easier variant. Let $G=c\_0$ and $X=\ell^1$, and let Y be the span of vectors $e\_n+ne\_{n+1}$. To see that this is weak\*-dense, suppose that $\sum\_k a\_k e\_k^\* \in c\_0$ annihilates all of Y. Then $a\_n + na\_{n+1}=0$ for all $n$, so $a\_1 + a\_2=0$ and $0 = a\_2+2a\_3 = 2a\_3 - a\_1$ and $0=a\_3+3a\_4 = (1/2)a\_1+3a\_4$, so $a\_1 = -a\_2 = a\_3/2 = -a\_4/3$ and an easy induction shows $a\_1 = (-1)^{n-1}a\_n/n$. Thus $|a\_n| = n|a\_1|$ for all $n$, but as $|a\_n|\rightarrow 0$, it
follows that $a\_1=0$, and so actually $a\_n=0$ for all $n$. Hence Y is weak\*-dense. However, $e\_1$ is in the closed unit ball of X, but it's pretty clear that we can't approximate it by norm one elements in Y (to do this without a tedious calculation defeats me right now).
| 2 | https://mathoverflow.net/users/406 | 32831 | 21,311 |
https://mathoverflow.net/questions/32744 | 3 | I have the following discrete time dynamical system
$$ y(t+1) = y(t) + \frac{1}{1+\exp(z+ u f y(t))} ,\quad y(0)=0,$$
where $z$ is a real number $f$ and $u$ are non-negative reals. I know I have little hope of obtaining a closed form solution for this process. But, actually, for my application, a "better" solution involves finding (making the dependence of $y$ on $f$ explicit by writing $y(t)$ as $y(t,f)$):
$$\lim\_{n \to \infty} \frac{1}{n}\cdot y\left(nt,\frac{f}{n}\right).$$
**Update:** Initial simulations suggested that this converged to $$y(t) = \frac{t}{1+\exp(z)}.$$ But that was wrong. In conclusion, the differential equation is indeed a good approximation to the above limit. Sorry about the confusion.
How can I rigorously show this? Also appreciated are references to texts that discuss similar problems.
Thanks
| https://mathoverflow.net/users/3812 | Limit of a discrete time dynamical system | The continuous model of the problem suggests that the limit does depend on $f$ (and $u$). More precisely, it depends on how fast the parameter $f$ is suppressed in the expression whose limit you are taking; behavior of $\lim y(nt, f\_n)/n$ will depend on the limit of $nf\_n$ as $n \to \infty$. The answer will be a function of the limit of $nuf\_n$. Only when this limit is zero does one get the proposed formula.
The associated differential equation is $Y' = 1/(1+Ae^{BY})$ where $A = e^z$ and $B=uf$. Its solution vanishing at 0 is $Y(t) = H^{-1}(t)$ where $H(t) = t + (A/B)(e^{Bt} - 1)$. It does look like this matches the asymptotic behavior of your sequence for $y$ both in the large and small range. For small $t$, the expansion $Y(t) = t/(1+A) + O(t^2)$ corresponds to your formula, but I think the answer is not quite that simple: you have to establish whether the expression whose limit is calculated belongs to the small regime where $Y(t)$ is approximately linear, or the large regime where $Y(t)$ is logarithmic, $Y(t)=O(\log(t))$. The limit uses $n$ iterations so we want to know, as a function of $B \sim 1/n$, whether the transition between regimes happens at a point much larger than $1/B$. However, it's easy to calculate that the ratio $H(t)/t$ moves away from 1 (the difference is larger than some constant independent of $B$) as soon as $Bt$ is of order 1 (i.e., bounded below by a given positive constant) and this would spoil the limit if the differential equation is a good model of the difference equation.
(ADDED: for comparison of $Y$ predictions with $y$ simulations, in the phase transition where $Bt$ is of order 1, $Y(t) \sim t/C$ and $H(t) \sim Ct$, with $C = 1 + A(e^q - 1)/q$, and $q = Bt =uft$. That is, $Y$ stays approximately linear but the coefficient goes to zero, consistent with the idea that it's turning into a logarithmic function. Let $t=nt\_0, \quad f=f\_0/n$, for some constant $f\_0$ and with $u$ and $t\_0$ also held constant while $f$ varies with $n$, so that the phase transition parameter is $q=uf\_0 t\_0$ and the predicted value of the limit, if $Y$ is a good approximation for $y$, is $L\_{pred} = \lim Y(nt\_0,f\_0/n)/n = \lim nt\_0/nC = t\_0/C = t\_0(q/(q + Ae^q - A))$. In the original notation of the question, $L = t/(1+{e^z}F(uft))$ where $F(x)=(e^x-1)/x$. Does this match the simulations?)
To see the small-$y$ behavior directly in the difference equation, it can be expanded in powers of $y$.
$y(t+1) - y(t) = 1/(1+A) - (AB)/(1+A)^2)y + O(y^2)$
Your formula proposes that when $B \sim 1/n$, the effect of the $y^{\geq 1}$ terms is of order smaller than $n$ for $t \in [0,n]$. The sum of the first $t$ values of the $y^1$ term will be of order $t^2$, so one expects these corrections to be suppressed only on a short interval, $t << n^{1/2}$. The calculation with the differential equation suggests that $f\_n = f/n$ is too large a parameter ; this calculation with the truncated difference equation can be used to prove that $f\_n = f/n^k$ is small enough for any $k > 2$. Adding higher degree terms to the approximate difference equation would, I suppose, only get closer to the picture suggested by the differential equation.
To prove rigorously the predictions from the differential equation you could try to control $y$ by trapping the sequence $y(n)$ between two trajectories of the ODE. If simulations are consistent with a heuristically "wrong" formula it would be very interesting to sort out what the truth is.
| 3 | https://mathoverflow.net/users/6579 | 32839 | 21,317 |
https://mathoverflow.net/questions/32847 | 5 | Is there a non-projective flat module over a local ring?
Here I assume the ring is commutative with unit.
| https://mathoverflow.net/users/5292 | Is there a non-projective flat module over a local ring? | $\mathbb{Q}$ is flat over $\mathbb{Z}\_p$, but not projective.
| 20 | https://mathoverflow.net/users/6950 | 32850 | 21,325 |
https://mathoverflow.net/questions/32824 | 16 | Languages decidable by weak models of computation often have certain necessary characteristics, e.g. [the pumping lemma for regular languages](http://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages) or [the pumping lemma for context-free languages](http://en.wikipedia.org/wiki/Pumping_lemma_for_context-free_languages). Such characteristics are useful for determining when a language cannot be decided by that weak model of computation.
>
> Do Turing-decidable languages have any such necessary characteristics? If not, why not?
>
>
>
Such a characteristic should avoid mention of Turing machines.
Motivation: The only way I know to show that a language is not Turing-decidable is diagonalization, or reducing to a language where diagonalization has already been applied. It would be nice to have another method.
I suspect that there aren't any nice characteristics of this type known, or they'd probably be *very* well-known; but an argument for why they can't exist could be illuminating.
I'd also be interested in the same question with the phrase "Turing-decidable" replaced with any complexity class, e.g. NP. This question seems more tractable (as there are fewer languages under examination).
Edit: Given the comments, I'll be a bit more precise about what I mean by "characteristics." I would be thrilled with either of the following two possibilities:
>
> 1) A property that allowed one to show a language was not in the relevant class (Turing-decidable, NP, whatever) without using diagonalization.
>
>
>
Or
>
> 2) A property that can be stated in a language too weak to define Turing machines.
>
>
>
| https://mathoverflow.net/users/6950 | Structure theorems for Turing-decidable languages? | As far as I know, the answer is no. We don't have an argument not based (directly or indirectly) on digonalization for even proving there is a language outside Dec (The set of Turing decidable languages).
Characterizing a class means having criteria for showing when a language is in the class and when a language is not. The second part would give us tools to prove lowerbounds on such classes.
For very small classes like $AC^0$ and $AC^0[p]$ there are combinatorial arguments (also for monotone-$AC^0$ circuits), but the current state of knowledge is consistent with $AC^0[6] = NP$. The Razbarov-Rudich (RR) Natural Proofs theorem is related to why the previous combinatorial arguments haven't been successful in separating larger classes. In simple terms, there is no "natural proof" for separating the larger classes unless there is no pseudo-random number generators (many believe they do exists), and all previous proofs were natural. Tim Gowers wrote a number of interesting blog posts on the topic:
<http://gowers.wordpress.com/2009/09/22/a-conversation-about-complexity-lower-bounds/>
The reason that we have pumping lemmas for $Reg$ (regular languages) and $CFL$ (context-free languages) is that the machine models for them are very specific which gives the strings accepted by them special combinatorial structure. For classes defined by Turing machines this is not the case. If a complexity class contains some basic functions (say $AC^0$) and is closed under composition, proving lowerbounds for it becomes a rather difficult task. Note that $Reg$ does not contain $AC^0$ ($\{0^n1^n: n \in \omega \}$ is in $AC^0$ but not in $Reg$.)
---
EDIT:Comments
-------------
part 1.
I don't have a theoretical justification right now, but the history of complexity theory shows that the structure of much smaller complexity classes are at least very difficult to understand. Many experts "believe" that there are pseudo-random number generators (PRNG), so RR's Natural Proof theorem applies and there is no "natural" property of sets/functions that can be used to prove a lowerbound for the complexity classes satisfying the condition of the RR theorem. I think a person accepting the existing of PRNGs can at least claim that there are no simple combinatorial structures for these complexity classes.
part 2. You are looking for a combinatorial property of sets of strings in a complexity class. The computational complexity of checking a property is a good measure. Any property stated in the first order logic (on a finite structure) is $AC^0$. The RR theorem applies to properties which are in $P$, i.e. can be checked in polynomial time given the truth table of the functions (characteristic functions of sets restricted to inputs of size n). Encoding Turing Machines (TM) is not difficult, and I think it can be done if one goes a little above $AC^0$, e.g. $TC^0$ is probably capable of checking if a string encodes a TM. You are looking for a property that holds for all functions in a class, but not for those outside it. The negation of it would be a property of the kind considered in RR's theorem. By these considerations, I think there are no properties of the kind you are looking for unless there are no PRNGs.
definitions:
$AC^0$ is class of sets of strings accepted by (uniform) circuits of polynomial size and bounded depth, i.e. there is a constant $d$ and a polynomial $p(n)$ and a sequence of circuits $\{C\_n\}\_{n \in \omega}$ using unbounded fan-in $\land$,$\lor$,$\lnot$ gates and of depth $d$ and size $p(n)$ that accepts them.
$AC^0[n]$ is the same as $AC^0$ but allows $mod\_n$ gates. $p$ is a prime number.
$TC^0$ is the same as $AC^0$ but allows majority/threshold gates.
A property $Q$ is natural if it is:
1. useful: i.e. does not hold for simple functions (functions of low complexity) but holds for some function in the larger class,
2. constructive: i.e. can be checked in polytime from the truth table of the functions,
3. large: i.e. for some k and all large enough n, the property holds for at least $\frac{1}{n^k}$ of the functions.
| 8 | https://mathoverflow.net/users/7507 | 32856 | 21,329 |
https://mathoverflow.net/questions/32854 | 4 | I've been looking at some thick subcategories in $K^b(R-proj)$ (the homotopy category of bounded chain complexes of projective modules), and, as expected, I'm running into the question of when chain complexes split quite often. I'm wondering what sorts of useful criteria there are for determining when chain complexes split. By "split" I mean decompose as two nontrivial complexes as $A = A\_1 \oplus A\_2$.
Feel free to strengthen the hypotheses a bit if you need to- these can be complexes of free modules, if you like. I'm just trying to get a sense of how to look at a chain complex and think, "That probably splits..." or, "That probably doesn't..." Also remember that I'm working in the homotopy category, so if the question becomes easier when homotopy equivalent objects are identified, please feel free to use this hypothesis.
| https://mathoverflow.net/users/6936 | When do chain complexes decompose as a direct sum? | One result that guarantees such a decomposition comes from looking at the homological support of such complexes (assuming that $R$ is commutative so we have a tensor product). The homological support of a complex $A$ is just the union of the supports of the $H^i(A)$ as $R$-modules. Then it is a result of Balmer, in the paper Supports and Filtrations in Algebraic Geometry and Modular Representation Theory which is available on his website, that if the homological support of $A$ can be written as a disjoint union of closed subsets $Y\_1\cup Y\_2$ of $\mathrm{Spec} \;R$ whose complements are quasi-compact then $A\cong A\_1\oplus A\_2$ in $K^b(R\text{-}\mathrm{proj})$ where the homological support of $A\_i$ is $Y\_i$.
Another method which works for any triangulated category is if $f\colon A\to B$ is a morphism then the triangle one gets by completing splits giving $B\cong A\oplus \mathrm{cone}(f)$ if and only if the map $\mathrm{cone}(f) \to \Sigma A$ is zero. A reference for this is Corollary 1.2.5 of Neeman's book on triangulated categories (I think I've also put the proof on MO before but I can't remember in which answer, maybe I can hunt it down later).
| 5 | https://mathoverflow.net/users/310 | 32868 | 21,335 |
https://mathoverflow.net/questions/32862 | 1 | By using a regular hexagonal arrangement it is simple to fit 19 identical circles into a larger circle of five times the radius with no circles overlapping. This leaves an area equal to six smaller circles uncovered. Is it possible to rearrange the 19 circles to accommodate a twentieth circle of the same size into the larger one (also with no overlapping of course)?
| https://mathoverflow.net/users/7819 | Settling a circular argument: room for one more? | No, according to the information at <https://erich-friedman.github.io/packing/cirincir/>
| 4 | https://mathoverflow.net/users/3684 | 32869 | 21,336 |
https://mathoverflow.net/questions/32875 | 9 | **Background**
In [his beautifully short answer](https://mathoverflow.net/questions/30647/fibered-products-of-cyclic-groups/30656#30656) to [a previous question of mine](https://mathoverflow.net/questions/30647/fibered-products-of-cyclic-groups), Robin Chapman asserted the following.
>
> Let $m,n,r$ be natural numbers with $r$ coprime to $n$. Then there is $r' \equiv r \mod n$ which is coprime to
> $mn$.
>
>
>
Letting $C\_n$ denote the cyclic group of order $n$, the above statement is equivalent to this:
>
> Every automorphism of $C\_n$ lifts to an automorphism of $C\_{nm}$ for all $m$.
>
>
>
Since that is the context of the question I asked, I thought that this fact ought to have an elementary group-theoretical derivation, but alas I have been unable to find one. I asked a number theorist colleague of mine and he gave me this "sledgehammer proof" (his words):
Since $r$ is coprime to $n$, the arithmetic progression $r + kn$ for $k=1,2,\ldots$ contains an infinite number of primes (by [a theorem of Dirichlet's](http://tinyurl.com/l3a7lp)). Since only a finite number of those primes can divide $m$, there is some $k$ for which $r'= r+kn$ is a prime which does not divide $m$, and hence neither does it divide $nm$.
**Question**
Is there an elementary (and preferably group-theoretical) proof of this result?
| https://mathoverflow.net/users/394 | Lifting units from modulus n to modulus mn. | This can be done in an elementary way using the Chinese remainder theorem.
First of all, note $m$ only appears in the conclusion in the context of the product
$mn$. For any common prime factor of $m$ and $n$ suck that prime's contribution to $m$ into $n$ instead, which changes the meaning of $m$ and $n$ but does not alter $mn$ nor alter the meaning of $r$ being coprime to $n$. It does change the meaning of what a congruence mod $n$ is, but only by making the condition even stronger.
Thus we are reduced to the case that $m$ and $n$ are relatively prime, so
now solve $r' \equiv r \bmod n$ and $r' \equiv 1 \bmod m$.
| 13 | https://mathoverflow.net/users/3272 | 32878 | 21,343 |
https://mathoverflow.net/questions/32891 | 16 | (This question is motivated by the reading of the article *Large numbers and unprovable theorems* by Joel Spencer, which can be found at <http://mathdl.maa.org/images/upload_library/22/Ford/Spencer669-675.pdf> and that I recommend).
We all know of the game where a card of a predefined size, say 3x5 cm, is given to every contender, and whoever writes the biggest (positive) integer on his, wins. Naive answers are easily defeated by iteration of fast-growing functions; those are defeated by induction, and these by transfinite induction. However, if a system of axioms is prefixed, then we cannot pursue this strategy forever: for example, if we are only willing to accept Peano's Axioms (PA), then $f\_{\alpha}(n)$ (where $\alpha$ is an ordinal number and $f$ is defined à la Ackermann) is computable (in the sense that the axioms ensure that a program to compute it exists and will terminate in finite time) when $\alpha < \epsilon\_0$, but not if $\alpha = \epsilon\_0$.
This problem suggests two related questions to me:
* It is known that the Ackermann function is well-defined inside AP, and that other functions which grow much faster, like the one in the strong version of the finite Ramsey Theorem of Paris-Harrington, or Goodstein's function whenever it grows fast (I think), or $f\_{\epsilon\_0}(9)$, cannot be defined everywhere just by application of AP, because they "grow too fast for AP". Is there a rigorous definition of what it means for a function to grow too fast for AP (or any other arithmetical axiom system)? Can we establish in any sense a "limit" for this process? For example, can we find a "threshold function" F, depending on the axioms, such that if f dominates F then f is not computable and if F dominates f then it will be? (I'm thinking about something among the lines of the convergence of the p-series for p>0 whenever p>1 and its divergence whenever p<=1).
* Building in the exposition above Spencer observes that, between experts, this game is not funny and reduces to claims of legitimacy (over the validity of the axioms they are supposed to use), since if we allow just a fixed amount of characters for describing our number, and our axioms system is prefixed also, then THERE in fact IS a largest number computable on that system (and thus competitors would come to a draw). However, what happens if we consider the following metagame? Instead of fixing the axiom system beforehand, we allow every contender to (secretly) choose his own system of axioms for arithmetic, in the hope that his will allow faster-growing computable functions than those of the others. Doing this, the contender takes the risk, while trying to get more and more power from the axioms, of actually getting an inconsistent system! Whoever gets the biggest (computable) number in a consistent axiom system wins.
Is this game interesting, or is it "flawed" too? In adittion, inconsistency may be proved within the axiom system, but its consistency would have to be proven in a more powerful framework. Which one would you select and why? What about the metametagame of letting those frameworks to the election of the players? Is that still interesting?
| https://mathoverflow.net/users/1234 | Finding the largest integer describable with a string of symbols of predefined length | My first remark is that if you allow players to pick their own theories, but only allow consistent theories to win, then you will not be able to compute the winner of the contest. The reason is that the consistency of a theory is not in general a computable question.
To see this, suppose that we had an algorithm that could compute consistency of any given finite theory. Using it, I claim that we can solve the halting problem. Given any Turing machine program $p$ and input $n$, form the theory $T$ asserting the basic facts of arithmetic (a small fragment of PA suffices), plus the assertion that $p$ never halts on input $n$. If $p$ does halt on input $n$, then this theory is inconsistent. And if it doesn't, then the theory is consistent, being true in the standard model. So if we could check consistency, we could solve the halting problem. Since that is impossible, we cannot check consistency in a computable manner.
My second remark is that if all players work in a fixed consistent theory $T$, but play decriptions of numbers that succeed to define unique numbers provably in the theory $T$, then you still won't be able to compute the winner. For example, perhaps one player plays the definition: *the smallest size proof of a contradiction in some theory $S$, if $S$ is inconsistent, otherwise $10$*, which $T$ can prove uniquely describes a number, without setttling the question of Con(S). But if another player plays $12$, then you will seem to need to decide Con(S) in order to determine the winner, which is impossible by my previous remarks.
| 16 | https://mathoverflow.net/users/1946 | 32893 | 21,352 |
https://mathoverflow.net/questions/32098 | 12 | There are different ways of showing that a given sequence $a\_0,a\_1,a\_2,\dots$
of integers, say, is nonnegative. For example, one can show that $a\_n$ count
something, or express $a\_n$ as a (multiple) sum of obviously positive numbers.
Another recipe is manipulating with the corresponding generating series
$A(x)=\sum\_{n=0}^\infty a\_nx^n$ and showing that $A(x)\ge0$ (this is the
notation for a series which has all coefficients nonnegative, and this extends
to formal power series in as many variables as needed).
An example of criterion in this direction is
$$
(\*) \qquad
A(xy)\ge0 \iff \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)\ge0
$$
(the multiple $1/(1-x)(1-y)$ is introduced for cosmetic purposes only and, of course, both $A(x)\ge0$ and $A(xy)\ge0$ are by definition equivalent to the nonnegativity of the sequence $a\_n$).
The latter can be verified by comparing the corresponding coefficients
in the power series expansion
$$
\frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)
=\sum\_{n=0}^\infty a\_n\sum\_{k,m=0}^\infty\binom{n+k}n\binom{n+m}nx^{n+k}y^{n+m}.
$$
On the other hand, the one-dimensional version of $( \* )$,
$$
A(x)\ge0 \iff \frac1{1-x}A\biggl(\frac{x}{1-x}\biggr)\ge0,
$$
is simply false.
My question is whether it is possible to find two *nontrivial* rational functions
$p(x)\in\mathbb Q[[x]]$ and $r(x)\in x\mathbb Q[[x]]$ in *one* variable $x$
such that
$$
A(x)\ge0 \iff p(x)A\bigl(r(x)\bigr)\ge0.
$$
Although I am not supposed to put several problems in one question, I
would also ask about a more direct proof of $( \* )$ and about general ways
of constructing such $p$ and $r$ in more than one variable.
**Motivation.** Basically I am interested in proving nonnegativity of certain
$q$-series sequences $a\_0(q),a\_1(q),a\_2(q),\dots$ by manipulating with the
corresponding generating series $A\_q(x)=\sum\_{n=0}^\infty a\_n(q)x^n$. Some
of them can be "guessed" from non-$q$-versions, for example there is a neat
$q$-analogue of the criterion $( \* )$.
| https://mathoverflow.net/users/4953 | Positivity of sequences via generating series | It is impossible, and not just for rational functions. To see this, let's consider the coefficients $b\_n$ of $p(x) A(r(x))$ as functions of the $a\_n$, the coefficients of $A(x)$. Since $r(0) = 0$ (as it must be) we see that:
* Each $b\_n$ is a linear combination of $a\_0, \dots, a\_n$; i.e. we have an upper-triangular infinite matrix $F$, not depending on $a\_n$, such that (writing $\vec{a} = (a\_n), \vec{b} = (b\_n)$) $\vec{b} = F \vec{a}$. Suppose for now that $p(0), r'(0) \neq 0$; then $F$ has a nonzero diagonal, and so is invertible.
* If $\vec{b} \geq 0$ for all $\vec{a} \geq 0$, then in particular this is true of the columns of $F$, taking $\vec{a}$ to be infinite "basis" vectors. Conversely, we have equivalently that $\vec{a} = F^{-1} \vec{b}$, so if $\vec{a} \geq 0$ for all $\vec{b} \geq 0$ this must be true of the columns of $F^{-1}$. We conclude that both $F$ and $F^{-1}$ have nonnegative entries.
* Lemma in linear algebra: if $F$ is upper-triangular and both it and $F^{-1}$ have nonnegative real entries, then $F$ is diagonal. Proof by induction: true for $1 \times 1$ matrices vacuously. In general, by induction we may assume that the upper-left and lower-right $(n - 1) \times (n - 1)$ blocks of $F$ are diagonal, so only the $(1,n)$ entry of $F$ is nonzero off the diagonal. Then we have $(F^{-1})\_{1n} = -F\_{1n} F\_{nn}/F\_{11}$. Since $F\_{11}$ and $F\_{nn}$ are both positive, $F\_{1n}, (F^{-1})\_{1n} \geq 0$ implies $F\_{1n} = 0$.
* This is also true of infinite matrices, since we can compute the finite upper-left blocks independently of the rest of the matrix.
* However, if $F$ is diagonal then we see that $p(x)A(r(x)) = \sum a\_n p(x) r(x)^n = \sum F\_{nn} a\_n x^n$ for all choices of $a\_n$, so (for example, taking $a\_n = t^n$ for a new variable $t$ and rewriting both sides as power series in $t$) we have $p(x) r(x)^n = F\_{nn} x^n$ for all $n$ (and some $F\_{nn} > 0$). That is, $(r(x)/x)^n = F\_{nn} p(x)^{-1}$ for all $n$, so in fact $r(x)/x = F\_{11}/F\_{00}$ is constant, and finally, $p(x)$ is constant as well.
* Now we remove the assumptions that $p(0), r'(0) \neq 0$. If $x^k$ divides $p(x)$, then replacing $p(x)$ by $p(x)/x^k$ does not change positivity of the coefficients. Now suppose the bottom exponent of $r(x)$ is $x^m$ with positive coefficient; then in $\mathbb{R}[[x]]$ we can write $r(x) = s(x)^m$, and if we denote $A\_m(x) = A(x^m)$, we have $A(r(x)) = A\_m(s(x))$. Clearly, $A\_m$ has nonnegative coefficients if and only if $A$ does, and $s'(0) \neq 0$, so the previous proof applies and $s(x)$ is a multiple of $x$, i.e. $r(x)$ is a multiple of $x^m$, and $p(x)$ is a multiple of $x^k$.
| 3 | https://mathoverflow.net/users/6545 | 32902 | 21,359 |
https://mathoverflow.net/questions/32908 | 21 | Yesterday, in the short course on model theory I am currently teaching, I gave the following nice application of downward Lowenheim-Skolem which I found in W. Hodges *A Shorter Model Theory*:
Thm: Let $G$ be an infinite simple group, and let $\kappa$ be an infinite cardinal with $\kappa \leq |G|$. Then there exists a simple subgroup $H \subset G$ with $|H| = \kappa$.
(The proof, which is short but rather clever, is reproduced on p. 10 of [http://alpha.math.uga.edu/~pete/modeltheory2010Chapter2.pdf.](http://alpha.math.uga.edu/%7Epete/modeltheory2010Chapter2.pdf.))
This example led both the students and I (and, course mechanics aside, I am certainly still a student of model theory) to ask some questions:
$1$. The theorem is certainly striking, but to guarantee content we need to see an uncountable simple group without, say, an obvious countable simple subgroup. I don't know that many uncountable simple groups. The most familiar examples are linear algebraic groups like $\operatorname{PSL}\_n(F)$ for $F$ an uncountable field like $\mathbb{R}$ or $\mathbb{C}$. But this doesn't help, an infinite field has infinite subfields of all infinite cardinalities -- as one does not need Lowenheim-Skolem to see! (I also mentioned the case of a simple Lie group with trivial center, although how different this is from the previous example I'm not sure.) The one good example I know is supplied by the Schreier-Ulam-Baer theorem: let $X$ be an infinite set. Then the quotient of $\operatorname{Sym}(X)$ by the normal subgroup of all permutations moving less than $|X|$ elements is a simple group of cardinality $2^{|X|}$. (Hmm -- at least it is when $X$ is countably infinite. I'm getting a little nervous about the cardinality of the normal subgroup in the general case. Maybe I want an inaccessible cardinal or somesuch, but I'm getting a little out of my depth.) So:
>
> Are there there other nice examples of uncountable simple groups?
>
>
>
$2$. At the beginning of the proof of the theorem, I remarked that straightforward application of Lowenheim-Skolem to produce a subgroup $H$ of cardinality $\kappa$ which is elementarily embedded in $G$ is not enough, because it is not clear whether the class of simple groups, or its negation, is elementary. Afterwards I wrote this on a sideboard as a question:
>
> Is the class of simple groups (or the class of nonsimple groups) an elementary class?
>
>
>
Someone asked me what techniques one could apply to try to answer a problem like this. Good question!
$3$. The way I stated Hodges' result above is the way it is in my lecture notes. But when I wrote it on the board, for no particular reason I decided to write $\kappa < |G|$ instead of $\kappa \leq |G|$. I got asked about this, and was ready with my defense: $G$ itself is a simple subgroup of $G$ of cardinality $|G|$. But then we mutually remarked that in the case of $\kappa = |G|$ we could ask for a *proper* simple subgroup $H$ of $G$ of cardinality $|G|$. My response was: well, let's see whether the proof gives us this stronger result. It doesn't. Thus:
>
> Let $G$ be an infinite simple group. Must there exist a *proper* simple subgroup $H$ of $G$ with $|H| = |G|$?
>
>
>
Wait, I just remembered about the existence of [Tarski monsters](http://en.wikipedia.org/wiki/Tarski_monster_group). So the answer is no. But what if we require $G$ to be uncountable?
| https://mathoverflow.net/users/1149 | Three questions on large simple groups and model theory | The class of simple groups isn't elementary. To see this, first note that if it were, then an ultraproduct of simple groups would be simple. But an ultraproduct of the finite alternating groups is clearly not simple. (An $n$-cycle cannot be expressed as a product of less than $n/3$ conjugates of $(1 2 3)$ and so an ultraproduct of $n$-cycyles doesn't lie in the normal closure of the ultraproduct of $(1 2 3)$. )
It turns out that an ultraproduct $\prod\_{\mathcal{U}} Alt(n)$ has a unique maximal proper normal subgroup and the corresponding quotient $G$ is an uncountable simple group. This group $G$ has the property that a countable group $H$ is sofic if and only if $H$ embeds into $G$. For this reason, $G$ is said to be a universal sofic group.
As for your third question, Shelah has constructed a group $G$ of cardinality $\omega\_{1}$ which has no uncountable proper subgroups. Clearly $Z(G)$ is countable. Consider $H = G/Z(G)$. Then $H$ also has no uncountable proper subgroups.Furthermore, every nontrivial conjugacy class of $H$ is uncountable and it follows that $H$ is simple.
| 21 | https://mathoverflow.net/users/4706 | 32909 | 21,364 |
https://mathoverflow.net/questions/32892 | 67 | In Ebbinghaus-Flum-Thomas's *Introduction to Mathematical Logic,* the following assertion is made:
If ZFC is consistent, then one can obtain a polynomial $P(x\_1, ..., x\_n)$ which has no roots in the integers. However, this cannot be proved (within ZFC).
So if $P$ has no roots, then mathematics (=ZFC, for now) cannot prove it.
The justification is that Matiyasevich's solution to Hilbert's tenth problem allows one to turn statements about provable truths in a formal system to the existence of integer roots to polynomial equations. The statement is "ZFC is consistent," which cannot be proved within ZFC thanks to Gödel's theorem.
**Question:** Has such a polynomial ever been computed?
(This arose in a comment thread on the beta site math.SE.)
| https://mathoverflow.net/users/344 | Does anyone know a polynomial whose lack of roots can't be proved? | In his Master's Thesis, Merlin Carl has computed a polynomial that is solvable in the integers iff ZFC is inconsistent. A joint paper with his advisor Boris Moroz on this subject can be found at <http://www.math.uni-bonn.de/people/carl/preprint.pdf>.
| 41 | https://mathoverflow.net/users/7743 | 32914 | 21,366 |
https://mathoverflow.net/questions/32920 | 3 | Does anyone have an idea how to prove the following? It is a step in the proof of some theorem in a book about gaussian processes.
Let $f\_n$ be an orthonormal sequence of gaussian variables. Consider $\sum\_{n \geq 1} a\_n f\_n$ and assume that it is convergent in $L^2$. Show that it converges also almost everywhere.
The hint is that every orthonormal family of gaussian variables is automatically independent, but I have no idea how to proceed.
| https://mathoverflow.net/users/7827 | Convergence of a series of orthonormal gaussian variables | You can assume $f\_n$ have mean zero, independence implies
$$E[ \sum\_{n=1}^{\infty} f\_n| \mathcal{A}m ] = \sum\_{n=1}^m f\_n $$
where $\mathcal{A}\_m$ is the sigma field generated by $f\_1,..,f\_m$
This means that $Z\_m=\sum\_{n=1}^mf\_n$ is a martingal with respect to $\mathcal{A}\_m$.
You then have to apply doob martingal convergence theorem
<http://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems>
This is brutal force as the result you ask for is originally due to Kolmogorov,
Hence if you want to reproduce the direct proof of kolmogorov you have to show the following lemma (called Kolmogorov inequality):
$$ P(\max\_{k=1,..,m} |Z\_m|\geq \epsilon )\leq E[Z\_m^2]/\epsilon^2$$
You will get it by using the partition
$$( \max\_{k=1,..,m} |Z\_m|\geq \epsilon ) =\cup\_{n=1}^m B\_n$$
where $B\_n$ is the event:
$$B\_n =(|Z\_1|<\epsilon,.. |Z\_{n-1}|<\epsilon, |Z\_n|\geq \epsilon )$$
| 7 | https://mathoverflow.net/users/6531 | 32924 | 21,372 |
https://mathoverflow.net/questions/32880 | 11 | Does anybody know who was Atle Selberg's advisor?
I find it interesting to know the advisor's impact on his students.
Unfortunately, in Selberg case, this information (even his advisor's name) seems to be nowhere to be found.
| https://mathoverflow.net/users/7820 | Selberg's advisor? | I'm a student at the university of Oslo, so I thought I'd have a go at this. I just talked to Erling Størmer (Carl's grandson) who is a professor emeritus here. He said that in practice Atle had no advisor. Of course someone must have signed the papers but he doesn't know who (I don't really see what difference it makes anyway). Erling told me that according to Atle the reason why so many Norwegian mathematicians at the time worked in number theory is that they were all self-taught, and number theory is more accessible to the autodidact.
| 26 | https://mathoverflow.net/users/1123 | 32927 | 21,373 |
https://mathoverflow.net/questions/31810 | 32 | One common reason given for the circularity of manhole covers is that they can't fall through the manhole. For convex manhole covers, this property is equivalent to having [constant width](http://en.wikipedia.org/wiki/Curve_of_constant_width) — if you have different widths, just orient the cover so that the shorter width slides through the larger one. Since convex polygons can't have constant width, this rules them out for manhole covers. However, for nonconvex shapes, a longer width does not necessarily give you a longer hole.
Is it possible to have a nonconvex polygon that cannot be moved through a hole of the same shape?
Some Clarifications:
* The hole has zero thickness.
* I was thinking of the polygon being simply connected. But if this matters I would like to know the answer both ways.
| https://mathoverflow.net/users/27 | Nonconvex manhole covers | Here is a construction of a polygon that cannot fall through the hole.
Begin with a regular $MN$-gon circumscribed around a unit circle, where $M\gg N\gg 1$ and $N$ is even. For every $M$th side, draw a segment of length 1 extending this side in the clockwise direction. Take a rectangular neighborhood of width $\varepsilon\ll 1/MN$ of each of these segments. The manhole is the union of the $MN$-gon and these $N$ narrow rectangles.
Suppose that the polygon can be moved through the hole. Then there is a moment when its center is on the ground level. Consider the intersection line of the two planes: the ground and the polygon. Its intersection $S$ with the polygon consists of the central segment of length $>2$ and several pairs of short segments arising from the pairs of symmetric narrow rectangles ($N$ should be chosen large enough so that there are at least 3 such pairs for any line through the center).
The same line on the ground must intersect the hole by a set that contains $S$ in the interior. The central segment must go very close to the center of the hole because there is no other place for a segment of length 2 (we can control how close is should be by choosing $M$ large enough). Now consider a symmetric pair of short segments in $S$ somewhere in the middle (to avoid borderline effects). If the line on the ground is sufficiently close to the center (and we can ensure that), this pair of segments must fit into a pair of symmetric rectangles in the hole. But it is easy to see that it cannot fit - locally these rectangles are just parallel strips of width $\varepsilon$ separated by a fixed distance $2-\varepsilon$, and however you rotate the intersecting line, you will get either shorter segments or shorter interval between them.
| 8 | https://mathoverflow.net/users/4354 | 32928 | 21,374 |
https://mathoverflow.net/questions/32925 | 6 | We all know that Gödel showed that there, in a formal system, are true statements that are non-provable (undecidable). In ZFC, there's Kaplansky's Conjecture, the Whitehead problem, etc.
We can also agree that we're sure to find more non-provable statements in ZFC. What I'm curious about is the following:
What is the defining characteristic of a non-provable statement in ZFC? Are they all "strong" in some sense? Is it a necessary condition that they are strong, then? What future theorems might turn out to be non-provable? Is it, through the characteristics of the theorems known to be non-provable possible to make a fairly accurate guess if a theorem will turn out to be non-provable in ZFC?
| https://mathoverflow.net/users/7607 | When is a statement provable? | Not all statements that are not provable in ZFC are "strong", if by strong you mean that
ZFC + the statement in question is stronger than ZFC in the sense that it implies the consistency of ZFC.
The typical example is the Continuum Hypothesis (CH). ZFC + CH is consistent iff ZFC is,
and in this sense, CH is not strong since ZFC + CH itself does not imply the consistency
of ZFC.
However, if one wants to prove that the existence of a measurable cardinal is not provable in ZFC, the argument is that ZFC + there is a measurable cardinal implies the consistency of
ZFC and hence, by the second incompleteness thm, the existence of a measurable cardinal cannot
be proved in ZFC.
Note that the situation in set theory is different from number theory.
In order to show that something is not provable in Peano Arithmetic (something which is consistent with PA), one usually uses the Incompleteness Theorems, i.e.,
one shows that the statement implies the consistency of PA.
So in some sense, number theoretic statements known to be independent over PA
are strong (over PA), statements known to be independent over ZFC are not necessarily strong over ZFC.
| 9 | https://mathoverflow.net/users/7743 | 32929 | 21,375 |
https://mathoverflow.net/questions/32423 | 3 | All the definitions I have seen in the literature present the possibility of computing the $\epsilon$-pseudospectrum of a rectangular matrix $A^(m \* n)$ with $m < n$ but I have not seen any definitions for the case $m > n$, why ? is there a trivial answer ?
| https://mathoverflow.net/users/7728 | Pseudo-spectrum of a short and fat rectangular matrix ? | I think that in the "short and fat" case (more columns than rows), the $\epsilon$-pseudospectrum with $\epsilon>0$ is the whole complex plane. So you don't get much information out of pseudospectra.
In more detail: Suppose you define the $\epsilon$-pseudospectrum of a "short and fat" matrix $A$ as the set of all complex numbers $z$ for which there is a vector $v$ and a matrix $E$ with $\|E\| < \epsilon$ such that $(A+E-zI)v = 0$ where $I$ is the rectangular "identity matrix", whose first $n$ columns form the $n$-by-$n$ square identity matrix and whose remaining columns are zero. In other words, $z$ is an "eigenvalue" of an $\epsilon$-perturbation of $A$. However, in the "short and fat" case, almost all $z$ are eigenvalues of the rectangular matrix $A$: if you decompose $A = [ \, A\_1 \,\, A\_2 \, ]$ where $A\_1$ is square, and do the with $v$, then the eigenvalue equation $(A-zI)v = 0$ becomes $A\_1v\_1 + A\_2v\_2 = zv\_1$, with solution $v\_1 = -(A\_1-zI)^{-1} A\_2v\_2$ if $z$ is not an eigenvalue of the square matrix $A\_1$. I think that with arbitrarily small perturbations you can make every $z$ an eigenvalue of the rectangular matrix, and thus the $\epsilon$-pseudospectrum is the whole complex plane.
| 3 | https://mathoverflow.net/users/2610 | 32936 | 21,380 |
https://mathoverflow.net/questions/32912 | 6 | There are two common ways to define a series-parallel graph (or 2-terminal series-parallel graph).
Definition 1
* start with K\_2 marking both vertices as terminals
* repeatedly join two smaller 2-terminal s/p graphs either in series or in parallel
Definition 2
* start with K\_2
* repeatedly replace a single edge by two in series or two in parallel
A "decomposition tree" for an s/p graph shows how it is constructed according to Definition 1; each node of the tree is a s/p graph and the children of each node are the components from which that graph was built by series or parallel composition.
It is well known that a series-parallel graph can be recognized in linear time; the usual reference to this is Valdes, Tarjan and Lawler (Valdes, Jacobo; Tarjan, Robert E.; Lawler, Eugene L. The recognition of series parallel digraphs. SIAM J. Comput. 11 (1982), no. 2, 298--313.)
It is also frequently *stated* in the literature that the decomposition tree can be found in linear time, either just as an assertion or with a reference to the same Valdes/Tarjan/Lawler paper.
*However*, when you actually read Valdes, Tarjan and Lawler, they do *not* construct the decomposition tree in linear time, but rather they run "Definition 2" in reverse and work on *reducing* the graph to a single edge by series and parallel reductions. So they *recognize* that the graph is s/p but they do not actually give the decomposition tree.
Does anyone know if there is an explicit reference in the literature to actually *constructing* the sp-tree for a series-parallel graph in linear time?
| https://mathoverflow.net/users/1492 | Where is it shown how to construct a decomposition tree for a series-parallel graph in linear time? | It is easy enough to build the tree from a definition 2 description. Here is a sketch:
Suppose that SP graph $G$ occurs from subdividing graph $G'$ at an edge $e$, either in series or in parallel. Recursively, let $T'$ be the tree for $G'$. Then $e$ is a leaf of $T'$. Add two children below that leaf. Mark the node $e$ as either series or parallel according to whether the subdivision of $e$ was series or parallel.
I don't know what the efficiency of this is; it probably depends on your precise implementation of trees. And I don't know any references in the literature.
| 9 | https://mathoverflow.net/users/297 | 32946 | 21,386 |
https://mathoverflow.net/questions/32965 | 1 | I have heard about KKM (Knaster-Kuratowski-Mazurkiewicz) theorem in nonlinear analysis and I am trying to use that for a theorem I am working on. The original paper (1929) is in German and all the references I have found on the Net mainly deal with generalizations of this theorem or present the theorem with a lot of complex topological flavor. I was wondering if anybody knows a source which contains this theorem in its basic format.
| https://mathoverflow.net/users/6748 | Knaster-Kuratowski-Mazurkiewicz (KKM ) Thoerem | There is a [book](http://books.google.co.uk/books?id=h_DaTNA8mw8C&printsec=frontcover&dq=kkm+theory&hl=en&ei=zm5ITJvNN8n24ga9yZneDA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCwQ6AEwAA#v=onepage&q&f=false) devoted entirely to KKM theory and its applications. It contains
an outline of the classical KKM theorem as well as a number of generalizations. Another standard reference is ["Fixed point theory"](http://books.google.co.uk/books?id=4_iJAoLSq3cC&printsec=frontcover&dq=dugundji+fixed+points&source=bl&ots=2lOqh3YAne&sig=JdRiTZFR0yHSEoWT8N4jjOZqQAo&hl=en&ei=VGxITLn9E4vNjAeg0PnUDg&sa=X&oi=book_result&ct=result&resnum=3&ved=0CB8Q6AEwAg#v=onepage&q=dugundji%2520fixed%2520points&f=false) by Granas and Dugundji.
| 3 | https://mathoverflow.net/users/5371 | 32969 | 21,394 |
https://mathoverflow.net/questions/32954 | 8 | A recent question on the notion and notation of multiplicative integrals
( [What is the standard notation for a multiplicative integral?](https://mathoverflow.net/questions/32705/what-is-the-standard-notation-for-a-multiplicative-integral) ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:
$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \Big( \frac{x}{m} \Big)!\,\Big( \frac{x-1}{m} \Big)!\dots \Big( \frac{x-m+1}{m} \Big)!\, \,$$
For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:
$$\prod\_{k=1}^{m}\Big(\frac{k}{m} \Big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$
Take a logarithm; divide by $m$ and let $m\to\infty$: one finds
$$\int\_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$
or, as a multiplicative integral
$$\prod\_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$
Now the question: *How to evaluate the above integral
by means of standard integral
calculus?*
I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.
| https://mathoverflow.net/users/6101 | Multiplicative integral of $\Gamma(x)$ | Here a start:
We have the [reflection formula](http://en.wikipedia.org/wiki/Gamma_function#Properties)
$$z! (1-z)! = \frac{\pi z (1-z)}{\sin (\pi z)}.$$
Taking $\log$'s,
$$\log (z!) + \log ((1-z)!) = \log \pi + \log z + \log (1-z) - \log \sin (\pi z).$$
Split our integral in half and rearrange it
$$\int\_0^1 \log (z!) \ dz = \int\_0^{1/2} \left( \log (z!) + \log ((1-z)!) \right) dz.$$
So we have three elementary integrals to deal with, plus
$$\int\_0^{1/2} \log \sin(\pi z) \ dz. \quad (\*)$$
According to Mathematica, $(\*) = - \log(2)/2$. So, if we can find a clean proof of this fact, we will have evaluated the integral. This may be difficult, because the indefinite integral $\int \log \sin(\pi z) \ dz$ involves dilogarithms. To me, $(\*)$ looks like a good target for residues. Anyone want to finish it off?
Edit: Here is a way to calculate $(\*)$: Denote $I=\int\_{0}^{\pi/2}\log(\sin x) \ dx=\int\_{0}^{\pi/2}\log(\cos x) \ dx$ and so $$2I=\int\_{0}^{\pi/2}\log(\frac{\sin 2x}{2}) \ dx=\int\_{0}^{\pi/2}\log(\sin 2x) \ dx-\frac{\pi \log 2}{2}=I-\frac{\pi \log 2}{2}$$
| 11 | https://mathoverflow.net/users/297 | 32979 | 21,401 |
https://mathoverflow.net/questions/32964 | 17 | The content of this note was the topic of a lecture by Günter Harder at the School on Automorphic Forms, Trieste 2000. The actual problem comes from the article
[A little bit of number theory](http://sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/bit-ps.pdf) by Langlands.
The problem is about a connection between two quite different objects. The first object
is the following pair of positive definite quadratic forms:
$$ P(x,y,u,v) = x^2 + xy + 3y^2 + u^2 + uv + 3v^2 $$
$$ Q(x,y,u,v) = 2(x^2 + y^2 + u^2 + v^2) + 2xu + xv + yu - 2yv $$
The second object is the elliptic curve
$$ E: y^2 + y = x^3 - x^2 - 10x - 20. $$
To each of our objects we now associate a series of integers.
For each integer $k \ge 0$ define
$$ n(P,k) = | \{(a,b,c,d) \in {\mathbb Z}^4: P(a,b,c,d) = k\} |, $$
$$ n(Q,k) = | \{(a,b,c,d) \in {\mathbb Z}^4: Q(a,b,c,d) = k\} |. $$
As a matter of fact, these integers are divisible by $4$ for any
$k \ge 1$ because of the transformations $(a,b,c,d) \to (-a,-b,-c,-d)$
and $(a,b,c,d) \to (c,d,a,b)$.
For any prime $p \ne 11$ we now put
$$ a\_p = |E({\mathbb F}\_p)| - (p+1),$$
where $E({\mathbb F}\_p)$ is the elliptic curve over ${\mathbb F}\_p$ defined above.
Then Langlands claims
**For any prime $p \ne 11$, we have**
$ 4a\_p = n(P,p) - n(Q,p).$
The "classical" explanation proceeds as follows: Given the series of integers $n(P,p)$ and $n(Q,p)$, we form the generating series
$$ \Theta\_P(q) = \sum \limits\_{k=0}^\infty n(P,k) q^k
= 1 + 4q + 4q^2 + 8q^3 + \ldots, $$
$$ \Theta\_Q(q) = \sum \limits\_{k=0}^\infty n(Q,k) q^k
= 1 + 12q^2 + 12q^3 + \ldots. $$
If we put $q = e^{2\pi i z}$ for $z$ in the upper half plane, then $\Theta\_P$ and $\Theta\_Q$ become ${\mathbb Z}$-periodic holomorphic functions on the upper half plane. As a matter of fact, the classical theory of modular forms shows that the function
$$ f(z) = \frac14 (\Theta\_P(q) - \Theta\_Q(q))
= q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 - 2q^9 - 2q^{10}
+ q^{11} - 2q^{12} \ldots $$
is a modular form (in fact a cusp form since it vanishes at $\infty$) of weight $2$ for
$\Gamma\_0(11)$. More precisely, we have $ f(z) = \eta(z)^2 \eta(11z)^2,$ where $\eta(z)$ is Dedekind's eta function, a modular form of weight $\frac12$.
Now we have connected the quadratic forms to a cusp form for $\Gamma\_0(11)$. This group has two orbits on the projective line over the rationals, which means that the associated Riemann surface can be compactified by adding twocusps: the result is a compact Riemann surface $X\_0(11)$ of genus $1$. Already Fricke has given a model for this Riemann surface: he found that $X\_0(11) \simeq E$ for the elliptic curve defined above.
Now consider the space of cusp forms for $\Gamma\_0(11)$. There are Hecke operators $T\_p$ acting on it, and since it has dimension $1$, we must have $T\_p f = \lambda\_p f$ for
certain eigenvalues $\lambda\_p \in {\mathbb Z}$. A classical result due to Hecke then predicts that the eigenvalue $\lambda\_p$ is the $p$-th coefficient in the $q$-expansion of $f(z)$. Eichler-Shimura finally tells us that $\lambda\_p = a\_p$. Putting everything together gives Langlands' claim.
Way back then I asked Harder how all this follows from the general Langlands conjecture, and he replied that he did not know. Langlands himself said his examples came "from 16 of Jacquet-Langlands". So here's my question:
*Does anyone here know how to dream up concrete results like the one above from Langlands' conjectures, or from "16 of Jacquet-Langlands"?*
| https://mathoverflow.net/users/3503 | Quaternary quadratic forms and Elliptic curves via Langlands? | Chapter 16 of Jacquet--Langlands is about the Jacquet--Langlands correspondence, which concerns the transfer of automorphic forms from quaternion algebras to the group $GL\_2$.
The modularity of the theta series that you write down is a (very) special case of this
correspondence.
But probably Langlands more had in mind going the other way, in the following sense:
in Chapter 16, Jacquet and Langlands not only show the existence of transfer, they characterize its image.
In particular, their results show that the modular form $f$ is in the image of transfer from
the definite quaternion algebra $D\_{11}$ ramified at $\infty$ and 11. Thus one knows {\em a priori}
that there has to be a formula relating $f$ to the $\Theta$-series of some rank four quadratic forms associated to $D\_{11}$; it is then a simple matter to compute them
precisely (here one uses the fact that $f$ is a Hecke eigenform, and the compatibility of
transfer with the Hecke action), and hence obtain the formula $f = (\Theta\_P - \Theta\_Q)/4.$
The question of characterizing the image of transfer is an automorphic interpretation of what is classically sometimes called the Eichler basis problem: the problem of computing the span of the theta series arising from definite quaternion algebras. The name comes from the fact that the particular case considered here (but with 11 replaced by an arbitrary prime $p$), namely the fact that modular forms of weight two and prime conductor $p$ are spanned by theta series coming from $D\_p$, was I think first proved by Eichler in 1955.
| 21 | https://mathoverflow.net/users/2874 | 32981 | 21,403 |
https://mathoverflow.net/questions/32990 | 1 | We know that $A$ embeds into $A$\*\* (the double dual space of $A$ ). Is the following true? If $\Psi$ is in $A$\*\* and weak\* continuous, is there an element $a \in A$ such that $ \Psi$ is the evaluation functional at $a$? That is to say, $\Psi(f)=f(a)$ for any $ f \in A^{\*}$?
| https://mathoverflow.net/users/7360 | Double dual space of a C* algebra A | This has nothing to do with operator algebras as it is well-known that if $B$ is a linear topological space (local convexity and Hausdorff-ness hypothesis included), the topological dual of $B^{\ast}$ endowed with the wk\*-topology is $B$ itself, that is, every wk\*-continuous functional on $B^{\ast}$ is an evaluation functional. For the proof, see for example the first section of chapter V in J. B. Conway's Functional Analysis textbook.
Regards,
G. Rodrigues
| 8 | https://mathoverflow.net/users/2562 | 32993 | 21,410 |
https://mathoverflow.net/questions/32957 | 4 | Let $X$ be a circle that with one corner (i.e. think of a triangle where we smooth out two of the vertices). Now let us consider the topological torus $M \cong \mathbb{T}^n$ which is the product of $n$ copies of $X$. Note that $M$ contains $n$ distinct circles along which it is not smooth.
Finally, suppose we are given a function $f:M \rightarrow \mathbb{R}$ which is continuous on all of $M$ and is smooth wherever $M$ is smooth. Is there any way to conclude that there exists a critical point on the smooth part of $M$? What if we replace some of the non-smooth circles with smooth ones? That is take $M$ to be the product of $k$ smooth circles, and $n -k$ non-smooth ones?
I'm not concerned whether the critical point is non-degenerate.
An easy example is the case $n = 1$: The $\min$ or $\max$ of $f$ will correspond to a critical point on the smooth part of $X$. This example tells us that generically we should expect $f$ to have a critical point on the smooth part of $M$ since, generically, the $\min$ or $\max$ should not lie on the measure zero subset of $M$ that is not smooth.
I heard of something called Stratified Morse Theory, but I'm not sure if this applies, or whether there is a more elementary way to think about the problem not using Stratified Morse Theory.
| https://mathoverflow.net/users/nan | Morse Theory on Non-smooth Manifolds | Answer to your question is negative if $n\ge 1$ ($n$ is a number of smooth circles and I suppose that there is non-smooth circles). Indeed, in that case your manifold is homeomorphic to torus $T^k$ and the homeomorphism $\varphi\colon M\to T^k$ could be taken smooth on the smooth part of $M$. The image of non-smooth part is a "big" subset (more than finite), denote that subset by $A$. There exists a smooth function $f$ on $T^k$, such that all its critical points are contained in $A$. (For any function on $T^k$ with a finite number of critical points there is a diffeomorphism of $T^k$ sending the set of critical points to $A$). Function $\varphi^\*f$ is continuous, smooth on the smooth part of $M$ and has no smooth critical points.
| 5 | https://mathoverflow.net/users/2823 | 33006 | 21,416 |
https://mathoverflow.net/questions/32966 | 3 | Let me refer to Jech's "Set Theory" Chap. 33 Determinacy:
"With each subset A of $\omega^\omega$ we associate the following game $G\_A$, played by two players I and II. First I chooses a natural number $a\_0$, then II chooses a natural number $b\_0$, then I chooses $a\_1$, then II chooses $b\_1$, and so on. The game ends after $\omega$ steps; if the resulting sequence $\langle a\_0, b\_0, a\_1, b\_1, ...\rangle$ is in A, then I wins, otherwise II wins.
A strategy (for I or II) is a rule that tells the player what move to make depending on the previous moves of both players. A strategy is a winning strategy if the player who follows it always wins. The game $G\_A$ is determined if one of the players has a winning strategy.
The Axiom of Determinacy (AD) states that for every subset A of $\omega^\omega$, the game $G\_A$ is determined."
Now, there is some apparent lack of symmetry in the definition of the $G\_A$ game: the player who plays first (I) attempts for a sequence in A.
What happens if we interchange the roles of both players? I. e. if we let the player who plays first attempt for a sequence not in A? Let us call this game $G'\_A$
Is it the case that for every subset A, A is determined wrt $G\_A$ iff A is determined wrt $G'\_A$?
| https://mathoverflow.net/users/6466 | Determinacy interchanging the roles of both players | The answer is no. The game $G'(A)$ you describe is just the same as the complementary game $G(A^c)$ from I's point of view. So the question is: if one of the players has a winning strategy in $G(A)$, does one of the players have a winning strategy in $G(A^c)$? Using AC one can construct a set $A$ for which this is not the case.
We do a recursive construction, much as in the usual recursive construction of a nondetermined set. The main difference is set out in advance a strategy $\sigma$ for I to be a winning strategy for $A$. Fix any strategy $\sigma$ for I (so it takes a finite sequences of even length as input and gives natural numbers as output). For a given infinite sequence $y$ of moves for II we let $\sigma\*y$ denote the member of $\omega^\omega$ resulting when I follows $\sigma$ and II plays out $y$.
Set $Z=\{\sigma\*y:y\in\omega^\omega\}$. We will make sure $Z\subseteq A$, then certainly $\sigma$ will be a winning strategy for I in $G(A)$.
So we also define $\langle a\_\alpha\rangle\_{\alpha<2^\omega}$ and $\langle b\_\alpha\rangle\_{\alpha<2^\omega}$ so that no $a\_\alpha$ is equal to any $b\_\beta$ and both sets are disjoint from $Z$. Let $\langle\sigma\_\alpha\rangle\_{\alpha<2^\omega}$ enumerate all of I's strategies and let $\langle\tau\_\alpha\rangle\_{\alpha<2^\omega}$ all of II's strategies. Suppose $\langle a\_\beta\rangle\_{\beta<\alpha}$ and $\langle b\_\beta\rangle\_{\beta<\alpha}$ have been defined. First consider $\sigma\_\alpha$; if it equals $\sigma$ we don't do anything. Otherwise it disagrees with $\sigma$ on some $s$; there are $2^\omega$ many $x$ which are extensions of $s$, for each such $x$ we have $\sigma\_\alpha\*x\not\in Z$ so pick one which also isn't equal to an $a$ or a $b$ so far and make it $a\_\alpha$.
Next consider $\tau\_\alpha$. There are continuum many $x$ that I can play so that $\tau\_\alpha\*x$ does not land in $Z$ (just have $x$ give a first move different from that prescribed by $\sigma$). Using a cardinality argument we get some $x$ such that $\tau\_\alpha\*x$ is not in $Z$ and not equal to anything chosen so far, and set it equal to $b\_\alpha$.
At the end let $A$ equal $Z$ together with all of the $a\_\alpha$. Then $\sigma$ is a winning strategy for I in $G(A)$ because every play according to $\sigma$ lands in $Z$. But no strategy wins $A^c$. We've made sure that for every strategy for I there is a play which equals an $a\_\alpha$ and thus lands in $A$ and so that strategy cannot be winning in $G(A^c)$. And similarly we've made sure that for every strategy for II there is a play equalling one of the $b\_\beta$ and thus landing in $A^c$ and not winning for II in $G(A^c)$.
| 7 | https://mathoverflow.net/users/2436 | 33008 | 21,418 |
https://mathoverflow.net/questions/33010 | 4 | It's not difficult to find a function $f\colon \mathbb R \to \mathbb R$ such that its restriction to any (not trivial) interval is surjective. Does anyone know whether such a function is necessarily not (Lebesgue) measurable? I'm pretty sure this is the case, but I cannot prove it.
Here is an example of such an $f$. Let $\sim$ be the equivalence relation defined by $x \sim y$ if and only if $x-y \in \mathbb Q$, with $x$ and $y$ in $\mathbb R$. Let $C$ be $\mathbb R/\sim$ (here we need the axiom of choice), and denote with $\pi$ the projection. Since $\mathbb Q$ is countable, $C$ must have the cardinality of the continuum. Let $\varphi \colon C \to \mathbb R$ a bijection. Then $f:=\varphi \circ \pi$ has the required property.
| https://mathoverflow.net/users/7845 | Real-valued function "so surjective" that should be non measurable | No. [Conway's base-13 function](http://en.wikipedia.org/wiki/Conway_base_13_function) is measurable.
See this [question](https://mathoverflow.net/questions/32641/is-conways-base-13-function-measurable).
| 6 | https://mathoverflow.net/users/1335 | 33012 | 21,420 |
https://mathoverflow.net/questions/32923 | 18 | I'm currently trying to understand the concepts and theory behind some of the common proof verifiers out there, but am not quite sure on the exact nature and construction of the sort of systems/proof calculi they use. Are they essentially based on higher-order logics that use Henkin semantics, or is there something more to it? As I understand, extending Henkin semantics to higher-order logic does not render the formal system any less sound, though I am not too clear on that.
Though I'm mainly looking for a general answer with useful examples, here are a few specific questions:
* What exactly is the role of type theory in creating higher-order logics? Same goes with category theory/model theory, which I believe is an alternative.
* Is extending a) natural deduction, b) sequent calculus, or c) some other formal system the best way to go for creating higher order logics?
* Where does typed lambda calculus come into proof verification?
* Are there any other approaches than higher order logic to proof verification?
* What are the limitations/shortcomings of existing proof verification systems (see below)?
The Wikipedia pages on proof verification programs such as [HOL Light](http://en.wikipedia.org/wiki/HOL_Light) [Coq](http://en.wikipedia.org/wiki/Calculus_of_constructions), and [Metamath](http://us.metamath.org/) give some idea, but these pages contain limited/unclear information, and there are rather few specific high-level resources elsewhere. There are so many variations on formal logics/systems used in proof theory that I'm not sure quite what the base ideas of these systems are - what is required or optimal and what is open to experimentation.
Perhaps a good way of answering this, certainly one I would appreciate, would be a brief guide (albeit with some technical detail/specifics) on how one might go about generating a complete proof calculus (proof verification system) from scratch? Any other information in the form of explanations and examples would be great too, however.
| https://mathoverflow.net/users/602 | How do proof verifiers work? |
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Don't think of type theory, categorical logic, and model theory as *alternatives* to one another. Each step on the progression forgets progressively more structure, and whether that structure is essence or clutter depends on the problem you are trying to solve. Roughly speaking, the two poles are type theory and model theory, which focus on proofs and provability, respectively.
To a model theorist, two propositions are the same if they have the same provability/truth value. To a type theorist, equisatisfiability means that we have a proof of the biimplication, which is obviously not the same thing as the propositions being the same. (In fact, even the right notion of equivalence for proofs is still not settled to type theorists' satisfaction.)
Categorical logicians tend to move between these two poles; on the one hand, gadgets like Lawvere doctrines and topoi are essentially model-theoretic, since they are provability models. On the other hand, gadgets like cartesian closed categories give models of proofs, up to $\beta\eta$-equivalence.
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It depends on what you are doing. If you are building a computerized tool, then typically either natural deduction or sequent calculus is the way to go, because these calculi both line up with human practice and help constrain proof search in ways helpful to computers. It makes sense to cook up a sequent calculus or natural deduction system even if the theory you want to use (e.g., set theory) is not normally cast in these terms.
On the other hand, model theory has been spectacularly successful in applications to mathematics, and this is in part *because* it does not have a built-in notion of proof system -- so there is simply less machinery you need to reinterpret before you can apply it to a mathematical problem. (The corresponding use of type theory is much less developed; homotopy theorists are in the very earliest stages of turning dependent type theory into ordinary mathematics.)
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Every well-behaved intuitionistic logic has a corresponding typed lambda calculus. See Sorensen and Urcyczyn's [Lectures on the Curry-Howard Correspondence](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.107.3462) for (many) more details.
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Yes and no. If you're interested in actual, serious mathematics, then there is no alternative to HOL or the moral equivalent (such as dependent type theory or set theory) because mathematics deals intrinsically with higher-order concepts.
However, large portions of any development involve no logically or conceptually complex arguments: they are just symbol management, often involving decidable theories. This is often amenable to automation, if the problems in question are not stated in unnaturally higher-order language. (Sometimes, as in the case of the Kepler conjecture, there is an artificial way of stating the problem in a simple theory. This is essentially the reason why Hales' proof relies so heavily on computers: he very carefully reduced the Kepler conjecture to a collection of machine-checkable statements about real closed fields.)
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The main difficulty with these tools is finding the right balance between automation and abstraction. Basically, the more expressive the logic, the harder automated proof search becomes, and the more easily and naturally you can define abstract theories that can be used in many different contexts.
| 23 | https://mathoverflow.net/users/1610 | 33019 | 21,424 |
https://mathoverflow.net/questions/32757 | 8 | It is usual to mention theorems of the kind:
Th. Assume there is a proper class of Woodin cardinals, $\mathbb{P} $ is a partial order and $G \subseteq \mathbb{P}$ is V-generic, then $V \models \phi \iff V[G] \models \phi$
where $\phi$ is some set theoretic statement (like "the Strong Omega Conjecture holds"), as some sort of evidence that $\phi$ is a statement less intractable than other statements like CH which are not focing-invariant.
My question is, in wich sense are these statements more tractable? What kind of "empirical evidence" gives support to the hope that they can be decided by large cardinal axioms?
| https://mathoverflow.net/users/6466 | Tractability of forcing-invariant statements under large cardinals | Typically, generic absoluteness is a consequence of a stronger property, that in many cases is really the goal one is after. To explain this stronger property, let me begin by reviewing two important absoluteness results.
1) The first is **Mostowski's Absoluteness**. Suppose $\phi$ is $\Sigma^1\_1(a)$, where $a\in\omega^\omega$. This means that $\phi(x)$ for $x\in\omega^\omega$, asserts
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where $R$ is a predicate recursive in $a$. These statements are very absolute: Suppose that $M$ is a well-founded model of a decent fragment of ZF, and that $a,x\in M$. Then $\phi(x)$ holds iff $M\models \phi(x)$.
In particular, *whether or not $\phi(x)$ holds cannot be changed by forcing, or by passing to an inner or outer model*. Note that $M$ could be countable. In fact, only needs to be an $\omega$-model; well-foundedness is not necessary.
This is how the result is usually stated. What is going on is the following:
Suppose that $T$ is a tree (in the descriptive set theoretic sense) and that $T\in M$. Then $T$ is ill-founded iff $M\models T$ is ill-founded.
In particular, $T$ could be the tree associated to $\phi$. This is the tree of all $(s,t)$ such that $s,t$ are finite sequences of the same length $l$, and $\forall n\le l$ $ R(s\upharpoonright l,t\upharpoonright n,a\upharpoonright n)$. So $T$ is the tree of attempts to verify $\phi$: $\phi(x)$ holds iff (there is a $y$ such that for all $n$, $(x\upharpoonright n,y\upharpoonright n)\in T$) iff the tree $T\_x$ is ill-founded. Recall that $T\_x$ consists of all $t$ such that, if $l$ is the length of $t$, then $(x\upharpoonright n,t\upharpoonright n)\in T$ for all $n\le l$.
The point is that $T$ is a very simple object. As soon as $T,x$ are present, $T\_x$ can be built, and the result of the construction of $T\_x$ is the same whether it is performed in $V$ or in $M$. Since well-foundedness is absolute, whether or not $T\_x$ is ill-founded is the same whether we check this in $M$ or in $V$. Of course, $T\_x$ is ill-founded iff $\phi(x)$ holds.
The moral is that the truth of $\Sigma^1\_1$ statements is certified by trees. And I think that this is saying that in a strong sense, $\Sigma^1\_1$ statements are very tractable. All we need to check their validity is a very easy to build tree and, once we have it, the tree is our certificate of truth or falsehood, this cannot be altered.
Recall that proofs in first-order logic can be described by means of certain finite trees. If something is provable, the tree is a very robust certificate. This is a natural weakening of that idea.
Of course one could argue that, if a $\Sigma^1\_1$ statement is not provable, then in fact it may be very hard to establish its truth value, so tractability is not clear. Sure. But, independently of whether or not one can prove something or other, the certificate establishing this truth value is present from the beginning. One does not need to worry that this truth value may change by changing the model one is investigating.
2) The second, and best known, absoluteness result, is **Shoenfield's absoluteness**. Suppose $\phi$ is $\Sigma^1\_2(a)$. This means that $\phi(x)$ holds iff
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where $R$ is recursive in $a$. Let $M$ be any transitive model of a decent fragment of ZFC, and suppose that $\omega\_1\subset M$ and $a,x\in M$. Then $\phi(x)$ holds iff $M\models\phi(x)$.
This is again a very strong absoluteness statement. Again, if one manages to show the *consistency* of $\phi(x)$ by, for example, passing to an inner model or a forcing extension, then in fact one has proved its *truth*.
Again, one could say that if $\phi$ is not provable, then it is in fact not very tractable at all. But the point is that to investigate $\phi$, one can use *any* tools whatsoever. One only needs to worry about its consistency, for example, and one can make use of any combinatorial statements that one could add to the universe by forcing.
Just as in the previous case, $\Sigma^1\_2$ statements can be certified by trees. The tree associated to $\phi$ is more complicated than in the previous case, and it is now a tree of $\omega\times\omega\_1$. (Jech's and Kanamori's books explain carefully its construction.) Again, the tree is very absolute: As soon as we have $a$ and all the countable ordinals at our disposal, the tree can be constructed. (One comparing two models $M\subset V$, we mean all the countable ordinals of $V$, even if $M$'s version of $\omega\_1$ is smaller.)
3) Generic absoluteness of a statement $\phi$ is typically a consequence of the existence of **absolutely complemented** trees associated to $\phi$. In fact, all generic absoluteness results I'm aware of are established by proving that there are such trees ("conditional" generic absoluteness results, such as only for proper forcings, or only in the presence of additional axioms, are somewhat different). This is a direct generalization of the situations above.
To define the key concept, recall that if $A$ is a tree of $\omega\times X$, then the projection $p[A]$ of $A$ is the set of all $x\in\omega^\omega$ such that $\exists f\in X^\omega\forall n\in\omega\,(x\upharpoonright n,f\upharpoonright n)\in A$.
Two (proper class) trees $A$ and $B$ on $\omega\times ORD$ are absolutely complemented iff:
1. $p[A]\cap p[B]=\emptyset$ and $p[A]\cup p[B]=\omega^\omega$.
2. Item 1 holds in all generic extensions.
A statement $\phi$ admits such a pair iff, in addition,
1. In any forcing extension, $\phi(x)$ iff $x\in p[A]$.
The idea is that this is a precise, formal, definable approximation to the intuitive statement one would actually like, namely, that there are such trees describing $\phi$ that have this ``complementary'' behavior in all outer models. First-order logic limits ourselves to considering forcing extensions.
Let me point out that $\Sigma^1\_1$ and $\Sigma^1\_2$ statements admit absolutely complemented pairs, so the existence of such a pair is a natural (far reaching) generalization of the two cases above.
Once we accept large cardinals, we can show that much larger classes than $\Sigma^1\_2$ admit absolutely complemented trees. For example, any projective statement does. Once again, the point is that as soon as we have the large cardinals and real parameters in our universe, we have the trees, and the trees certify in unambiguous forcing-unchangeable terms, whether the statements hold at any given real. It is in this sense that we consider these statements more tractable.
Here is a rough sketch of an example I particularly like, due to Martin-Solovay (for measurables) and Woodin (in its optimal form). For details, see my paper with Ralf Schindler, ``projective well-orderings of the reals,'' Arch. Math. Logic (2006) 45:783–793:
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The right hand side means that for any $\Sigma^1\_3$ statement $\phi$ (so now we have three alternations of quantifiers) and any two step forcing ${\mathbb P}∗\dot{\mathbb Q}$, for any $G$, ${\mathbb P}$-generic over $V$, any $H$, ${\mathbb Q}$-generic over $V[G]$, and for any real $x$ in $V[G]$, we have
$$ V[G]\models\phi(x)\Longleftrightarrow V[G][H]\models\phi(x). $$
The left hand side of $(\*)$ is a weakening of "there is a proper class of measurable cardinals", which is how the statement is usually presented.
The proof of the implication from left to right in $(\*)$ goes by building a tree of attempts to find a witness to the negation of a $\Sigma^1\_2$ statement. The goal is that if such a witness can be added by forcing, then in fact we can find one in the ground model. If there is a forcing adding a witness, there is a countable transitive model where this is the case. Essentially, the tree gives the construction of such a model, bit by bit, and if we have a branch, then we have such a model.
So: If there is a witness added in a forcing extension, the tree will have there a branch. So it is illfounded. By absoluteness of well-foundedness, the tree has a branch in $V$. The sharps are used to ``stretch'' the model so that we can use Shoenfield absoluteness, and conclude that there must be a witness in $V$.
4) Projective absoluteness, a consequence of large cardinals, is established by showing the existence of absolutely complemented trees for any projective statement. The theory of universally Baire sets originates with this concept, and the closely related notion of homogeneously Suslin trees. All of this is also connected to determinacy. Once again, to drive the point home: Generic absoluteness is not the goal. The goal is the existence of the pair of trees. Once we have them, we have a strong certificate of truth or falsehood of a statement. I do not know if one is to accept the search for such trees as a more tractable problem than the original statement whose pair of trees we are now searching for. But it certainly says that consistency of the statement, using large cardinals or any combinatorial tools whatsoever, is enough to have a proof of the statement. This seems much more hopeful and generous an approach than if only proofs in the usual sense are allowed. The existence of these trees for projective statements is what I meant in a comment by ``large cardinals settle second order arithmetic.'' Put yet another way: If you show, for example, that a projective statement is not 'decidable' (in the presence of large cardinals), meaning that it is consistent and so is its negation, then you have either actually showed that certain large cardinals are inconsistent, or you have found a way of changing the truth of arithmetic statements, and both of these options are much more significant events than the proof of whatever the projective statement you were interested in was. More likely than not, the truth value of the statement will be uncovered at some point, and you know there is no ambiguity as of what it would be, since the witnessing trees are already present in the universe.
(In spite of its length, I am not completely happy with this answer, but I would need to get much more technical to expand on the many interesting points that your question raises. Hopefully there is some food for thought here. For nice references to some of the issues I mention here, Woodin's article in the Notices is a good place to start, and Steel's paper on the derived model theorem has much of the details.)
| 14 | https://mathoverflow.net/users/6085 | 33022 | 21,426 |
https://mathoverflow.net/questions/33028 | 7 | Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under at-most-countable unions?
| https://mathoverflow.net/users/nan | Can iterating countable unions give every set? (ZF) | Moti Gitik proved (assuming large cardinals) that all uncountable alephs can have cofinality ω. [*All uncountable cardinals can be singular*, Israel J. Math. 35 (1980), 61–88] I believe this may be the model you're looking for, but I don't know what happens to non-wellorderable sets in that model. According to the abstract copied below, this model is very close to what you have in mind.
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Every wellorderable set $S$ in this model belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions. The proof is by induction on the cardinality of the infinite set $S$. As in the abstract, we may write $S = \bigcup\_{i<\omega} S\_i$ where $|S\_i| < |S|$ for each $i$. By the induction hypothesis, each $S\_i$ belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions, and thus $S$ belongs to this closure too.
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Actually, every set $S$ in this model belongs to the closure of $\{\{s\} : s \in S\}$ under iterated countable unions; this is essentially what Gitik's Theorem 6.3 says.
| 9 | https://mathoverflow.net/users/2000 | 33032 | 21,434 |
https://mathoverflow.net/questions/33037 | 7 | Let $S$ = { $a^2b^3$ : $a, b \in \mathbb{Z}\_{>1}$ }.
Does there exist $n$ such that $n$, $n+1 \in S$?
Motivation: I was thinking about [Question on consecutive integers with similar prime factorizations](https://mathoverflow.net/questions/32412/question-on-consecutive-integers-with-similar-prime-factorizations), wondering whether any such pair had to have prime signatures with at least one 1. This would follow if the answer to the above question is negative. (This would also follow from weaker versions of the above question too, such as taking out perfect $n$th powers from $S$.)
Please note that $a$ and $b$ in the set definition are not allowed to be equal to 1. Otherwise, there'd be solutions like 8, 9 or 465124, 465125. (465124 = $(2\cdot 11 \cdot 31)^2$ and 465125 = $61^25^3$.)
| https://mathoverflow.net/users/7434 | Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1? | As once remarked by Mahler, $x^2 - 8 y^2 = 1$ has infinitely many solutions with $27 | x$.
| 25 | https://mathoverflow.net/users/nan | 33044 | 21,440 |
https://mathoverflow.net/questions/33043 | 10 | Suppose I have a graph $G$ with vertex set $V$, edge set $E \subseteq {V \choose 2}$, a poistive integer $d$, and a weight function $w:E \to \mathbb{R}^{+}$. Is there a nice algorithmic way to decide if there is an assignment of vertices to points in Euclidean space, i.e. a function $f: V(G) \to \mathbb{R}^d$ such that $|f(x)-f(y)| = w( \{x,y \})$ whenever $\{ x, , y\} \in E$, where $|.|$ is the Euclidean norm? There is no harm in insisting that the weight function respects the triangle inequality.
The question I am most interested in is efficiently deciding whether there exists such a function $f$, for a given graph $G$, dimension $d$, and weight function $w$, but it might also be interesting to know how to try to find a map that does the job but with "small distortion".
For example, quadratic optimization tells us something...
Cases of special interest: (1) We have a complete graph $G=K\_n$, i.e. a finite metric space. (2) The weight function is constant, i.e. we want to know: is $G$ a unit distance graph in $\mathbb{R}^d$? (Sometimes people want "unit distance graph" to also mean that $f$ is injective, but for my purposes it is fine for vertices to lie on top of each other.) Even the case of $f$ constant and $d=2$ is interesting, as this could be useful for a computational attack on the Hadwiger-Nelson unit coloring problem.
I've noticed that this question is equivalent to asking if a certain real algebraic variety of degree $2$ is nonempty, but I'm not sure if that is a helpful observation, other than it guarantees, for example, that is it algorithmically decidable.
| https://mathoverflow.net/users/4558 | Algorithm for embedding a graph with metric constraints | Let's say the graph is complete, and has weights on edges that satisfy triangle inequality. If you want an isometric embedding (which your original question indicates), then there's a necessary and sufficient characterization: the squares of the distances must be of *negative type*: specifically, given the $D\_{ij} = d^2\_{ij}$ values, then they must satisfy the inequality
$$ \sum\_{i,j} b\_i b\_j D\_{ij} \le 0, \sum\_i b\_i = 0$$ for all such $b\_i$. All of this is discussed rather well in the book by [Deza and Laurent](http://www.springer.com/mathematics/numbers/book/978-3-540-61611-5).
It gets really interesting when you allow for some distortion. The special case where G is the complete graph and the weights (while not being constant) satisfy the triangle inequality is the well known metric embedding problem, which has been studied extensively in the theoretical computer science community because of connections to multicommodity flows, and also for data mining problems. A great source is the paper by [Linial, London and Rabinovich](http://www.cs.huji.ac.il/~nati/PAPERS/eran1.ps.gz)
Let the distortion be the largest value of $w(x,y)/d(x,y)$ (wlog we can assume the embedding always shrinks distances)
1. there's always an embedding into Euclidean space that ensures a distortion $O(\log n)$. The dimension of the space can then be brought down to $O(\log n)$ by applying the Johnson-Lindenstrauss Lemma. This result is due to Bourgain, and the algorithm itself is quite simple: pick $K$ subsets of the input at random, each of size $\log n$, and for each point, let its $i^{th}$ coordinate in the embedding be its distance to the $i^{th}$ such subset. The original proof uses $K = \log^2 n$, and then we use the JL lemma to reduce the dimension back to $\log n$.
1a. this basic construction also works for all $\ell\_p$-norms for $1 \le p \le 2$, but the dimensionality reduction of course doesn't.
1b. The construction is tight: the lower bound comes from considering the shortest path metric induced by a constant-degree expander.
The case when you only care to preserve some of the distances is more interesting. While I'm not sure what has been considered for this case, there's another whole body of work that considers the induced shortest path metrics for special graphs (trees, series parallel graphs, planar graphs and so on). One of the most interesting conjectures in this area whether the metric space induced by planar graphs admits a constant-factor distortion embedding into Euclidean space.
| 12 | https://mathoverflow.net/users/972 | 33047 | 21,441 |
https://mathoverflow.net/questions/33036 | 3 | The reason I am wondering this is that all of the reductions from 3-SAT => quadratic programming (or similar NP-hard reductions) involve encoding the underlying NP-hard problem into feasibility testing. If you take out that trick, can you still find another way to encode it?
EDIT: Killed the duality stuff, don't know what I was thinking.
| https://mathoverflow.net/users/4642 | Is quadratic programming still NP-hard if you have bounds and a feasible point? | No; I think what you're observing is a side effect of reducing from deicision problems; if you tried to encode an NP optimization problem, you'd end up using more than just the feasibility machinery.
Take MAX-CUT, with variables $x\_i\in\{-1,+1\}$ indicating taking a vertex or not, and
$W\in\mathbb{R}^{n\times n}$ a matrix of edge weights. Since cut capacity can be written as $\frac 1 2\sum\_{i<j} W\_{ij}(1-x\_ix\_j)$, the optimization problem has form
$$
\min\ x^TWx \quad\quad \textrm{subject to} \quad \forall i\centerdot x\_i^2 = 1.
$$
Even the real-valued relaxation $x\_i\in [-1,+1]$ is problematic since $W$ is in general indefinite. Like I said above, since we're trying to solve an NP optimization problem, we are definitely relying on the optimization machinery.
As a final point, I'm not sure why you brought duality into the picture, since the usual guarantees of strong duality are murky once nonconvexity enters the picture ....
| 3 | https://mathoverflow.net/users/2621 | 33048 | 21,442 |
https://mathoverflow.net/questions/33049 | 0 | Let $D$ be the open unit disk, and $J$ a Jordan arc (that is, a homeomorphic copy of $[0, 1]$) that lies in $D$, except $J(0)$ lies on the boundary of $D$, say $J(0)=1$. I would like to see that $D\smallsetminus J([0, 1])$ is a path-connected topological space. Please help, if you can. Thanks!
| https://mathoverflow.net/users/7305 | A Jordan arc in the unit disk | It's certainly the case that $\mathbb{R}^2\setminus J$ is path connected.
So any two points in $D\setminus J$ are joined by a path in $\mathbb{R}^2$
missing $J$. If this path isn't in $D$ it hits the boundary of $J$
but then replace part of the path by an arc of the boundary of $D$.
Then one can replace this part of the arc by an arc just inside the boundary
(there is $\epsilon>0$ such that any closed arc on the boundary not containing
$J(0)$ has distance $>\epsilon$ from $J$).
| 4 | https://mathoverflow.net/users/4213 | 33053 | 21,446 |
https://mathoverflow.net/questions/33021 | 28 | In pondering [this](https://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities) MO question and people's efforts to answer it, and recalling also something that I learned in my youth about using Morse theory ideas to prove some results of Lefschetz in the complex case, I seem to have learned two things -- things that I suppose are absorbed in the cradle by those who study algebraic geometry as opposed to learning it by osmosis -- or maybe I haven't got them quite right. Anyway:
(1) Blowing up a surface at a smooth point does not change the fundamental group, and (therefore ?) the fundamental group of a smooth projective surface is a birational invariant.
(2) Surfaces in $\mathbb P^3$ are simply connected, and more generally for a set $X\subset\mathbb P^n$ defined by a single homogeneous equation of degree $>0$ the pair $(\mathbb P^n,X)$ is at least $(n-1)$-connected. That is, the relative homotopy groups and therefore the relative homology groups vanish up through dimension $n-1$.)
(This is all over the complex numbers.)
Is this correct? And, taking off from (1), what are some other simple statements about invariants from homotopy theory that are birational invariants? And what are the first things to know about birational invariants that do not come from topology?
EDIT I wish I could accept more than one answer.
| https://mathoverflow.net/users/6666 | Birational invariants and fundamental groups | I would like to mention one more homotopy invariant of smooth projective varieties that is also a birational invariant. If $X$ is a smooth projective variety, then the torsion subgroup $T(X)$ of ${\rm H}^3(X,\mathbb{Z})$ is a birational invariant. This is explained in the beautiful paper "Some elementary examples of unirational varieties which are not rational" by Artin-Mumford, which starts exactly outlining a "homotopical" approach to showing that there exist unirational threefolds that are not rational. Their homotopic criterion is that rational varieties $X$ have trivial group $T(X)$ and they construct unirational varieties with non-trivial two-torsion in such group, showing that these are examples of unirational, non rational varieties.
Finally, for surfaces the quantity $K^2+\rho$ is a birational invariant, where $K^2$ is the self intersection of the canonical divisor class on $X$ and $\rho$ is the rank of the N\'eron-Severi group of $X$). This is not too deep, as it is immediate knowing that a birational map of surfaces is a composition of blow ups and blow downs of smooth points (which for surfaces is quite easy!) and the above quantity is obviously invariant under a blow up. The reason for mentioning it, is that the N\'eron-Severi group can be defined in terms of Hodge theory, which, while not entirely homotopical, certainly has a homotopical feel to it. Moreover, many birational invariants are defined using Hodge structures, so it seemed useful to point it out!
| 19 | https://mathoverflow.net/users/4344 | 33061 | 21,449 |
https://mathoverflow.net/questions/33058 | 18 | Recently the elliptic curve $E:y^2+y=x^3-x^2$ of conductor $11$ (which appears in [my answer](https://mathoverflow.net/questions/11747/galoisian-sets-of-prime-numbers/32939#32939)) became my favourite elliptic over $\bf Q$ because the associated modular form
$$
F=q\prod\_{n>0}(1-q^n)^2(1-q^{11n})^2
$$
is such a nice "$\eta$-product". (This modular form is also associated to the isogenous elliptic curve
$y^2+y=x^3-x^2-10x-20$ which appears in [Franz's question](https://mathoverflow.net/questions/32964/quaternary-quadratic-forms-and-elliptic-curves-via-langlands).)
**Question.** Are there other elliptic curves over $\bf Q$ which have a simple minimal equation and whose associated modular form is a nice $\eta$-product or even a nice $\eta$-quotient?
I know two references which might have a bearing on the question
--- [Koike's article on McKay's conjecture](https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-95/issue-none/On-McKays-conjecture/nmj/1118787564.full)
and
--- [p.18](https://books.google.com/books?id=MLdRYIg6pDkC&lpg=PR1&dq=ono%25252520web%25252520of%25252520modularity&pg=PA18#v=onepage&q&f=false) of Ono's *Web of modularity* on $\eta$-quotients.
Can someone provide a partial or exhaustive list of such nice pairs $(E,F)$ ?
| https://mathoverflow.net/users/2821 | Eta-products and modular elliptic curves | There is an exhaustive list in the paper [Y. Martin and K. Ono, Eta-Quotients and Elliptic
Curves, *Proc. Amer. Math Soc.* **125** (1997), no. 11, 3169--3176].
Suppose that $E\_N$ is an elliptic curve of conductor
$N$, then the corresponding $L$-series is assigned to the eta product
$$
\eta(a\tau)\eta(ab\tau)\eta(ac\tau)\eta(abc\tau),
$$
where $a+ab+ac+abc=24$, $a,b,c\in\mathbb Z$,
for the following values of $N$ and $(b,c)$:
$$ \begin{align\*}
N &\quad (b,c)\cr
11 &\quad (1,11)\cr
14 &\quad (2,7)\cr
15 &\quad (3,5)\cr
20 &\quad (1,5)\cr
24 &\quad (2,3)\cr
27 &\quad (1,3)\cr
32 &\quad (1,2)\cr
36 &\quad (1,1)\cr
\end{align\*}
$$
It is probably more exciting that all these elliptic curves and their $L$-series at $s=2$ appear in Boyd's conjectures on Mahler's measure. For a nice review of this story see
[M.D. Rogers, Hypergeometric formulas for lattice sums and Mahler measures, [arXiv:0806.3590](http://arxiv.org/abs/0806.3590)] and the original paper [D.W. Boyd, Mahler's measure and special values of $L$-functions, *Experiment. Math.* **7** (1998) 37--82].
| 21 | https://mathoverflow.net/users/4953 | 33063 | 21,450 |
https://mathoverflow.net/questions/32716 | 6 | The Cheeger constant of a finite graph measures the "bottleneckedness" of the graph, and is defined as:
$$h(G) := \min\Bigg\lbrace\frac{|\partial A|}{|A|} \Bigg| A\subset V, 0<|A|\leq \frac{|V|}{2} \Bigg\rbrace$$
Here $V$ is the vertex set of $G$ and $\partial A$ denotes the collection of all edges going from a vertex in $A$ to a vertex in $V\setminus A$. The idea is that $h(G)$ is small if there is a bottleneck somewhere in $G$.
Now let $G$ have vertices $\lbrace 1,2,\ldots,n\rbrace^3\subset\mathbb{Z}^3$, and with an edge between two vertices if the distance between them is 1. Suppose that $n$ is even. Then it seems intuitively obvious that the minimum should be achieved with an "orthogonal half", that is $A= \lbrace 1,2,\ldots,n/2\rbrace\times\lbrace 1,2,\ldots,n\rbrace\times\lbrace 1,2,\ldots,n\rbrace$, and so $h(G)$ would be $n^2/(n^3/2) = 2/n$. Is this in fact the minimum, and how could one prove such a thing?
| https://mathoverflow.net/users/6015 | What is the Cheeger constant of a cubical subset of the cubic lattice? | The result (for 3 dimensions and I think easily generalises to any dimension) follows from Theorem 3 of the Bollobás and Leader paper. The theorem (in 3 dimensions) states that for any subset $A$ of the vertices $V$ of a cubical grid of side length $N$ with $|A|\leq\frac{N^3}{2}$ that $$|\partial A| \geq \min\_{r=1,2,3}\left\lbrace|A|^{1-1/r}rN^{(3/r)-1}\right\rbrace$$
So:
$$\min\_{r=1,2,3}\left\lbrace\left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N}\right\rbrace \leq \frac{|\partial A|}{|A|}$$
Now $|A| \leq \frac{N^3}{2}$, so $2 \leq \frac{N^3}{|A|}$, so $\frac{r2^{1/r}}{N} \leq \left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N}$.
We can check for $r=1,2,3$ that $2\leq r2^{1/r}$
so we get that
$$ \frac{2}{N} \leq \left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N} $$
for each $r$, and therefore for the minimum, and so the ``orthogonal half'' subset of the cube,
$(1,2,\ldots,N/2)\times (1,2,\ldots,N)\times (1,2,\ldots,N)$ which gives $\frac{|\partial A|}{|A|}=\frac{N^2}{N^3/2} = \frac{2}{N}$, is best possible.
| 2 | https://mathoverflow.net/users/6015 | 33073 | 21,457 |
https://mathoverflow.net/questions/33046 | 11 | From [Wikipedia](http://en.wikipedia.org/wiki/Oracle_machine) (bold emphasis at the end is mine):
>
> In complexity theory and computability theory, an oracle machine is an abstract machine used to study decision problems. It can be visualized as a Turing machine with a black box, called an oracle, which is able to decide certain decision problems in a single operation. The problem can be of any complexity class. **Even undecidable problems, like the halting problem, can be used.**
>
>
>
Isn't assuming the existence of a machine which can decide the halting problem... problematic? The way I've heard it explained is that it's only problematic if you assume it can solve its own halting problem or any of the "super-oracles" above it. However, if an oracle O can solve the halting problem for machine M, can't M just use O to solve its own halting problem? Isn't that a contradiction, from which all propositions follow?
I'm sure I'm making an elementary mistake, so please point it out :)
| https://mathoverflow.net/users/2592 | Aren't "oracle machines" unsound concepts? | Oracle machines are not "problematic". Let us consider oracle machines that have the halting problem as their oracle. Now, by "the halting problem" we mean the collection of all Turing machines (without oracle) that halt when started with an empty tape.
This can be decided by an oracle machine with the halting problem as its oracle.
However, such oracle machines with halting problem oracle cannot solve the halting problem
for Turing machines with the ordinary halting problem as their oracle.
I.e., there is no oracle machine using the usual halting problem as an oracle that decides
the set of all oracle machines with the halting problem as their oracle that halt on the empty input.
In recursion theory, there is the jump operator or Turing jump that assigns to every problem
$P$ (set of words, set of natural numbers, whatever you prefer), the set of all
(codes of) oracle Turing machines with the oracle $P$ that halt on the empty tape.
The Turing jump of a problem $P$ is strictly more complicated (more "uncomputable") than
$P$ itself, i.e., no machine with oracle $P$ decides the jump of $P$.
| 17 | https://mathoverflow.net/users/7743 | 33075 | 21,459 |
https://mathoverflow.net/questions/33084 | 7 | I have heard sometimes that the only dimensions $k$ for which there exists a "*good*" smooth product $P:S^k\times S^k\to S^k$ are $k = 0,1,3,7$ (the above products corresponding to $\mathbb Z\_2, U(1)\subset\mathbb C$, the product of unit quaternions and of unit Cayley numbers).
I would like to ask for references about such a result.
More precisely, I am interested in finding out how is it that one can define "*good*" so that the above is true (with proofs, preferably).
| https://mathoverflow.net/users/5628 | Are there good product rules on the $k$-sphere? | I guess that what you want to say is that for $k=0, 1,3, 7$ the sphere $S^k$ is a *H-space* (as in the comment).
Also, Lie groups are a word to look at for having ``good'' products (better than the above, this works for $S^k$ with $k= 0,1,3$ not $7$, see the comment below).
See Theorem 2.16 of the following [nice book](http://www.math.cornell.edu/~hatcher/VBKT/VBdoublepage.pdf).
| 7 | https://mathoverflow.net/users/5753 | 33086 | 21,468 |
https://mathoverflow.net/questions/31308 | 3 | Apologies if my question is poorly phrased. I'm a computer scientist trying to teach myself about generalized functions. (Simple explanations are preferred. -- Thanks.)
One of the references I'm studying states that the space of Schwartz test functions of rapid decrease is the set of infinitely differentiable functions: $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ such that for all natural numbers $n$ and $r$,
$\lim\_{x\rightarrow\pm\infty} |x^n \varphi^{(r)}(x)|$
What I would like to know is why is necessary or important for test functions to decay rapidly in this manner? i.e. faster than powers of polynomials. I'd appreciate an explanation of the intuition behind this statement and if possible a simple example.
Thanks.
EDIT: the OP is actually interested in a particular 1994 paper on "Spatial Statistics" by
Kent and Mardia,
1994 Link between kriging and thin plate splines (with J. T. Kent). In Probability, Statistics and Optimization (F. P. Kelly ed.). Wiley, New York, pp 325-339.
Both are in Statistics at Leeds,
<http://www.amsta.leeds.ac.uk/~sta6kvm/>
<http://www.maths.leeds.ac.uk/~john/>
<http://www.amsta.leeds.ac.uk/~sta6kvm/SpatialStatistics.html>
Scanned article:
<http://www.gigasize.com/get.php?d=90wl2lgf49c>
FROM THE OP: Here is motivation for my question: I'm trying to understand a paper that replaces an integral $$\int f(\omega) d\omega$$ with $$\int \frac{|\omega|^{2p + 2}}{ (1 + |\omega|^2)^{p+1}} \; f(\omega) \; d\omega$$ where $p \ge 0$ ($p = -1$ yields to the unintegrable expression) because $f(\omega)$ contains a singularity at the origin i.e. is of the form $\frac{1}{\omega^2}.$
LATER, ALSO FROM THE OP:
I understand some parts of the paper but not all of it. For example, I am unable to justify the equations (2.5) and (2.7). Why do they take these forms and not some other form?
| https://mathoverflow.net/users/7486 | Splines, harmonic analysis, singular integrals. | I believe I now have the answer to the question. The power of $\omega$ appear from the taylor expansion of $e^{i\omega.t\_j}$ (in section 2.3 of Kent and Mardia's paper)
Thanks.
(Apologies for the seeming bit of self promotion, but I've tagged this as the correct answer.)
| 1 | https://mathoverflow.net/users/7486 | 33087 | 21,469 |
https://mathoverflow.net/questions/33088 | 6 | How is entropy of a general probability measure defined?
| https://mathoverflow.net/users/7699 | Entropy of a general prob. measure | The Entropy of a function $f$ with respect to a measure $\mu$ is
$$ Ent\_{\mu}(f)=\int f \log f d\mu - \int f d\mu \log(\int f d\mu ) $$
The entropy of a probability distribution $P$ with respect to $\mu$ is given by $ Ent(\frac{dP }{d\mu })$. I a not aware of a general definition that would not implie a reference (here it is $\mu$) measure ...
| 9 | https://mathoverflow.net/users/6531 | 33090 | 21,471 |
https://mathoverflow.net/questions/33096 | 28 | Hi everyone, the summer break is coming and I am thinking of reading something about mathematical logic. Could anyone please give me some reading materials on this subject?
| https://mathoverflow.net/users/3849 | Reading materials for mathematical logic | Here are a few suggestions (which depending on your background may be more or less useful):
1. [Logic and Structure](https://rads.stackoverflow.com/amzn/click/3540208798) by Dirk van Dalen. I have used this as a textbook when teaching mathematical logic and for that purpose it is decent. Some people find it a bit dry, but at least it covers a large amount of material in a reasonably clear manner.
2. [Mathematical Logic](https://rads.stackoverflow.com/amzn/click/1568811357) by Joseph R. Shoenfield. This book is, I think, regarded by many logicians as being the gold standard text on the subject.
3. [A Course in Mathematical Logic](https://rads.stackoverflow.com/amzn/click/0720428440) by John Bell and Moshe Machover. This is my personal favorite textbook in mathematical logic. (Unfortunately, it's a North Holland book and so is a bit less affordable.)
4. [A Course in Mathematical Logic for Mathematicians](https://rads.stackoverflow.com/amzn/click/1441906142) by Yuri I. Manin (with contributions from Boris Zilber). I think that pretty much anything written by Manin is worth taking seriously and this book is no exception.
5. [Notes on Logic and Set Theory](https://rads.stackoverflow.com/amzn/click/0521336929) by Peter T. Johnstone. This is a delightful little (literally) book on logic which is highly recommended (perhaps in conjunction with one of the other larger books from this list).
6. [The Mathematics of Metamathematics](https://rads.stackoverflow.com/amzn/click/B002FKDA1Y) by Helena Rasiowa and Roman Sikorski. This is a nice book which gives a lattice theoretic development of mathematical logic. (Difficult to find, but worth a look if your library has a copy.)
7. [Introduction to Metamathematics](https://rads.stackoverflow.com/amzn/click/0923891579) by Stephen C. Kleene. A classic text in mathematical logic which is still a rewarding read.
I hope these (admittedly biased) suggestions are some use!
| 27 | https://mathoverflow.net/users/6485 | 33102 | 21,479 |
https://mathoverflow.net/questions/33112 | 5 | Can anyone estimate N such that Prob( 0 is in the convex hull of $N$ points ) >= .95
for points uniformly scatterered in $[-1,1]^d$, $d = 2, 3, 4, 10$ ?
The application is nearest-neghbour interpolation:
given values $z\_j$ at sample points $X\_j$, and a query point $P$,
one chooses the $N$ $X\_j$ nearest to $P$ ($N$ fixed) and averages their $z\_j$.
If $P$ is not in the convex hull of the $N$ $X\_j$,
the interpolation will be one-sided, not so good.
I'd like to be able to say
"taking 6 neighbors in 2d, 10 in 3d, is seldom one-sided".
If anyone could point me to selfcontained pseudocode for the function Inhull( $N$ points )
(without calling full LP), that would be useful too.
(Please add tags interpolation convex-geometry ?)
| https://mathoverflow.net/users/6749 | Estimate probability( 0 is in the convex hull of N random points ) ? | This is a classical and essentially geometric problem. In fact, *the answer does not depend
on the distribution of the points* (as long as the distribution is centrally symmetric).
The following result is due to Wendel ([link](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=wendel&s5=A%2520problem%2520in%2520geometric%2520probability&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq)).
>
> **Theorem.** If $X\_1$, ..., $X\_N$ are i.i.d. random points in $R^d$ whose distribution
> is symmetric with respect to $0$ and assigns measure zero to every hyperplane
> through $0$, then
> $$\mathbb P(0\notin \mbox{conv}\{X\_1,\dots,X\_N\})=\frac{1}{2^{N-1}}\sum\limits\_{k=0}^{d-1}{N-1 \choose k}.$$
>
>
>
The proof is straightforward. Let $\mu$ be the distribution of $X\_k$, and set
$$ f(x\_1,\dots,x\_N) = \begin{cases} 1, & \mbox{if } x\_1,\dots,x\_N\ \mbox{ lie in an open halfspace of $\mathbb R^d$ with $0$ in the boundary}, \newline 0, & \mbox{else.} \end{cases}$$
Then due to the invariance of $\mu$ under reflection in the origin, we have that
$$\mathbb P(0\notin \mbox{conv}\{X\_1,\dots,X\_N\})=\int\_{\mathbb R^d}\dots \int\_{\mathbb R^d} \frac{1}{2^N}\sum\limits\_{\varepsilon\_i=\pm1}f(\varepsilon\_1x\_1,\dots,\varepsilon\_Nx\_N)\ \mu(dx\_1)\dots\mu(dx\_N).$$
Now, the sum
$$C(N,d)=\sum\limits\_{\varepsilon\_i=\pm1}f(\varepsilon\_1x\_1,\dots,\varepsilon\_Nx\_N)$$
can be interpreted as the number of connected components of the set $\mathbb R^d\backslash (H\_1\cup\dots\cup H\_N)$ induced by the hyperplanes $H\_1$, ..., $H\_N$ through $0$ which are in general position. But there is a classical calculation going back to to Steiner and Schläfli, which shows that
$$C(N,d)= 2\sum\limits\_{k=0}^{d-1}{N-1 \choose k}.$$
| 17 | https://mathoverflow.net/users/5371 | 33132 | 21,500 |
https://mathoverflow.net/questions/33103 | 7 | This problem arose when trying to understand the stack of twisted stable maps into a stack (specifically BG), as introduced by Dan Abramovich, Angelo Vistoli and several co-authors (Olsson, Graber, Corti,...). However, my specific question can be formulated without mentioning such beasts and I will do so. If anyone wants background I can give some.
Suppose that we have the following set-up. One is given:
* A smooth projective curve *C*
* and a finite abelian group *G* acting on *C*,
* such that the quotient map $\pi : C \to C/G$ is an admissible cover.
Suppose moreover that $C/G \cong \mathbf{P}^1$. Over the complex numbers, one can make the following computation. For each ramification point of the covering, consider a small loop centered around that point, with orientation induced by the complex structure. Choose a lifting of that loop to a path on *C*. The path will start and end in the same fiber. Since the restriction of $\pi$ away from the ramification locus is a *G*-torsor, the difference between start- and endpoint will give us a well-defined element of *G*. (This uses that *G* is abelian -- in general, one would only get an element up to conjugation, since there are several choices of liftings of the loop.) Now consider the product of all these elements over all ramification points. By considering the fundamental group of $\mathbf{P}^1$ minus the ramification locus, it is clear that this product is the identity in *G*.
I would like to express this computation algebraically, i.e. without recourse to any monodromy or the classical fundamental group, working over an arbitrary base where the order of *G* is invertible. Unless I'm mistaken, one can still define an evaluation map associating an element of *G* to each ramification point, since the following (to me rather mysterious) construction should work. Consider the stack quotient $[C/G]$. This is a twisted curve in the sense of Abramovich et al, which basically means that the ramification points on *C/G* have been cut out and replaced by cyclotomic gerbes ("stacky points"). Now this twisted curve has an actual *G*-torsor over it, so we get a map $[C/G] \to BG$. Restricting it to one of the stacky points, we obtain a cyclotomic gerbe over the base scheme with a *G*-torsor over it. But this is by definition an object of the *rigidified inertia stack* of *BG*, and the points of the (rigidified) inertia stack correspond to the elements of *G*.
Question 1: Is there a way of formulating this without using the language of twisted curves, i.e. associating an element of *G* to each ramification point only in terms of the admissible cover? There should be, since the moduli space of maps from twisted stable curves to *BG* is isomorphic to the moduli space of stable curves with an admissible cover which is a *G*-torsor away from the branch locus, but I don't see any sensible way of expressing such a construction algebraically.
Question 2: Can one show that in the algebraic setting, the product of the elements over all ramification points is the identity?
| https://mathoverflow.net/users/1310 | Twisted curves, admissible covers, and an algebraic analogue of a specific monodromy computation | (written hurriedly, hope it is semi-helpful and -correct)
Let's say you're over some field k. If it's not separably closed, basechange until it is. The strictly completed local ring of a ramification point in your base P^1 is going to look like k[[t]]. Now remove that point to give yourself Spec k((t)). The restriction of your cover to this punctured formal neighborhood is going to be a G-cover of Spec k((t)), which is to say a map from the absolute Galois group of k((t)) to G. But since you've wisely demanded that the order of G is prime to char k, this map kills wild inertia, and factors through the tame inertia group, which is cyclic. So the image of this map at least gives you a well-defined cyclic subgroup of G. You have to be a bit more careful to get an element of G, and indeed there isn't really a canonical choice locally; there is a choice of generator of tame inertia involved. Via class field theory you can make these choices "compatibly" at different ramification points (e.g. this will make the product vanish, as you say) but I don't think there's a "correct" such compatible choice.
| 6 | https://mathoverflow.net/users/431 | 33136 | 21,504 |
https://mathoverflow.net/questions/33138 | 7 | Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:
1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.
2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).
Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?
I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum\_i \lambda\_i p\_i$, where $p\_i \in P$ and $\lambda\_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.
| https://mathoverflow.net/users/7842 | Question about von Neumann algebra generated by a complete algebra of projections | The answer "yes" follows from Theorem 2.8 of Bade's "[On Boolean algebras of projections and algebras of operators](http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0073954-0/home.html)," 1955, which is in the more general context of algebras of operators on a Banach space.
Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "[Weak and strong limits of spectral operators](http://projecteuclid.org/euclid.pjm/1103044796)," 1954.
| 6 | https://mathoverflow.net/users/1119 | 33144 | 21,511 |
https://mathoverflow.net/questions/33158 | 11 | Name a theorem T that has a proof based upon the truth of a conjecture C, and also has another proof based upon the falsehood of the same conjecture C, but for longtime has no known direct proof that is independent of C. For instance, it would be a claim that can be proved if P=NP, and can also be proved if P is different from NP.
I apologize if the question was previously asked (I also apologize for the title if it does not faithfully reflect the question).
| https://mathoverflow.net/users/7874 | Use of Conjectures to Prove a Theorem | If you are asking for an example of such a method in action, then you have the theorem of Hecke-Deuring-Heilbronn that $h(D) \rightarrow \infty$ as $D \rightarrow \infty$, where $h(D)$ is the class number of the imaginary quadratic field with discriminant $D$.
The Hecke part is that the result is true if there are no Siegel zeros of L-functions for imaginary quadratic fields. Siegel zeros are exceptional zeros occurring in the real line in the interval $(\frac{1}{2}, 1)$. The Deuring-Helbronn part is that the result is true if there are Siegel zeros. The proof uses an effect of "repulsion" of such zeros, which is called the Deuring-Heilbronn phenomenon. This is all explained by Dorian Goldfeld in a bulletin article, "[Gauss' Class number problem for Imaginary Quadratic Fields](http://www.ams.org/journals/bull/1985-13-01/S0273-0979-1985-15352-2/)".
The existence of Siegel zeros is a stronger version of the negation of the Generalized Riemann hypothesis. Hopefully in future the generalized Riemann hypothesis would be proved and thus hopefully it will be shown that the study of Sigel zeros had been just the study of the empty set.
---
Later story(added just for additional information): This method was later strengthened by Landau, Siegel and so on, and finally with more recent developments on the Birch-Swinnerton-Dyer conjecture by Gross and Zagier, an effective version of this theorem was proved by Dorian Goldfeld, and the explicit constants were computed by Joseph Oesterlé. Thus the Gauss class number problem was solved in its entirety.
Thanks to Keith Conrad for correcting ambiguities.
| 20 | https://mathoverflow.net/users/2938 | 33159 | 21,520 |
https://mathoverflow.net/questions/33162 | 14 | I don't know any number theory, so excuse me if the following notions have names that I'm not using.
For a positive natural number $n\in{\mathbb N}\_{\geq 1}$, define $Log(n)\in{\mathbb N}$ to be the ``total exponent" of $n$. That is, in the prime factorization of $n$ it is the total number of primes being multiplied together (counted with multiplicity); for example $Log(20)=3.$ I'll define $log\_2(n)\in{\mathbb R}$ to be the usual log-base-2 of $n$, so $log\_2(20)\approx 4.32$.
One can think of $log\_2(n)$ as "the most factors that $n$ could have" and think of $Log(n)$ as the number of factors it actually has. Define $D(n)$ to be the ratio of those quantities $$D(n)=\frac{Log(n)}{log\_2(n)}\in(0,1],$$ and call it the *divisibility of $n$*. Hence, powers of 2 are maximally divisible, and large primes have divisibility close to 0. Another example: $D(5040)=\frac{8}{12.3}\approx 0.65$, whereas $D(5041)\approx\frac{2}{12.3}\approx 0.16$.
**Question:** What is the expected divisibility $D(n)$ for a positive integer $n$? That is, if we define $$E(p):=\frac{\sum\_{n=1}^p D(n)}{p},$$ the *expected divisibility for integers between 1 and $p$*, I want to know the value of $$E:=lim\_{p\rightarrow\infty}E(p),$$ the *expected divisibility for positive integers*.
Hints:
1. I once wrote and ran a program to determine $E(p)$ for input $p$. My recollection is a bit faint, but I believe it calculated $E(10^9)$ to be about $0.19.$
2. A friend of mine who is a professor in number theory at a university once guessed that $E$ should be 0. I never understood why that would be.
| https://mathoverflow.net/users/2811 | How divisible is the average integer? | Hopefully I've read all your notation correctly. If so, by playing (very) fast and loose with heuristics, I think your friend is right that the answer is 0.
Your function $Log(n)$ is the additive function $\Omega(n)$. According to the mathworld entry
<http://mathworld.wolfram.com/PrimeFactor.html>,
$\Omega(n)$ has been dubbed the "multiprimality of $n$" by Conway, and satisfies
$$
\Omega(n)\sim \ln\ln(n)+\text{mess},
$$
so (very roughly),
$$
D(n)\sim \frac{\ln\ln(n)}{\ln(n)},
$$
and
$$
E(p)\sim \frac{1}{p}\int\_e^p \frac{\ln\ln n}{\ln n}dn.
$$
This goes to 0 (very very slowly) as $p\rightarrow\infty$.
| 15 | https://mathoverflow.net/users/35575 | 33164 | 21,521 |
https://mathoverflow.net/questions/28776 | 37 | A theorem of Hecke (discussed in [this question](https://mathoverflow.net/questions/12352))
shows that if $L$ is a number field, then the image of the
different $\mathcal D\_L$ in the ideal class group of $L$ is a square.
Hecke's proof, and all other proofs that I know, establish this essentially by
evaluating all quadratic ideal class characters on $\mathcal D\_L$ and showing
that the result is trivial; thus they show that the image of $\mathcal D\_L$
is trivial in the ideal class group mod squares, but don't actually exhibit a square root
of $\mathcal D\_L$ in the ideal class group.
Is there any known construction (in general, or in some interesting cases) of an ideal
whose square can be shown to be equivalent (in the ideal class group) to $\mathcal D\_L$.
Note: One can ask an analogous question when one replaces the rings of integers by
Hecke algebras acting on spaces of modular forms, and then in some situations I know
that the answer is *yes*. (See [this paper](http://www.math.northwestern.edu/~emerton/pdffiles/two.pdf).) This gives me some hope that there might be a construction in
this arithmetic context too. (The parallel between Hecke's context (i.e. the number
field setting) and the Hecke algebra setting is something I learnt from Dick Gross.)
Added: Unknown's very interesting comment below seems to show that the answer is "no",
if one interprets "canonical" in a reasonable way. In light of this, I am going to ask another question which is a tightening of this one.
On second thought: Perhaps I will ask a follow-up question at some point, but I think I need more time to reflect on it. In the meantime, I wonder if there is more that one can say about this question, if not in general, then in some interesting cases.
| https://mathoverflow.net/users/2874 | Does (the ideal class of) the different of a number field have a canonical square root? | The following example shows that, in its strongest form, the answer to Professor Emerton's question is no. This answer is essentially an elaboration on what is already in the comments.
Let $p \equiv q \equiv 5 \pmod 8$. Let $K/\mathbb{Q}$ be a cyclic
extension of degree four totally ramified at $p$ and $q$ and unramified
everywhere else (it exists). To make life easier, suppose that the $2$-part of the class
group of $K$ is cyclic.
The Galois group of $K$ is
$G = \mathbb{Z}/4 \mathbb{Z}$.
Let $C$ denote the class group of $K$. I claim that $C^G$ is cyclic
of order two. Since the $2$-part of $C$ is cyclic,
this is equivalent to showing that $C\_G$ is cyclic of order two.
By class field theory, $C\_G$ corresponds to a Galois extension
$L/\mathbb{Q}$ unramified everywhere over $K$ such that there is an exact sequence
$$1 \rightarrow C\_G \rightarrow \mathrm{Gal}(L/\mathbb{Q})
\rightarrow \mathbb{Z}/4 \mathbb{Z} \rightarrow 1.$$
If $\Gamma$ is any finite group with center $Z(\Gamma)$, then an easy
exercise shows that $\Gamma/Z(\Gamma)$ is cyclic only if it is trivial.
We deduce that $L$ is the genus field of $K$.
There is a degree four extension $M/L$ contained inside the cyclotomic field
$\mathbb{Q}(\zeta\_p,\zeta\_q)$ that is unramified over $K$ at all finite
primes. However, the congruence conditions on $p$ and $q$ force $K$ to be
(totally) real and $M$ (totally) complex. Thus $M/K$ is ramified at
the infinite primes, and $C\_G = \mathbb{Z}/2\mathbb{Z}$, and the claim is established.
We note, in passing, that $L = K(\sqrt{p}) = K(\sqrt{q})$.
Suppose there
exists a *canonical* element $\theta \in C$ such that $\theta^2 = \delta\_K$,
where $\delta\_K$ is the different of $K$.
The different $\delta\_K$ is invariant under $G$. If $\theta$ is
canonical in the strongest sense then it must also be invariant under $G$. In particular,
the element $\theta \in C^G$ must have order dividing two, and hence $\theta^2 = \delta\_K$
must be trivial in $C$. We conclude that if $\delta\_K$ is not principal,
no such $\theta$ exists.
It remains to show that there exists primes $p$ and $q$ such that
$\delta\_K$ is not principal and the $2$-part of $C$ is cyclic. A computation shows this is so for
$p = 13$ and $q = 53$. For those playing at home, $K$ can be taken to be
the splitting field
of
$$x^4 + 66 x^3 + 600 x^2 + 1088 x - 1024,$$
where $C = \mathbb{Z}/8 \mathbb{Z}$ and $\delta\_K = [4]$. $C$ is generated by (any) prime
$\mathfrak{p}$ dividing $2$, and $G$ acts on $C$ via the quotient $\mathbb{Z}/2\mathbb{Z}$, sending $\mathfrak{p}$ to $\mathfrak{p}^3$.
| 26 | https://mathoverflow.net/users/nan | 33165 | 21,522 |
https://mathoverflow.net/questions/33169 | 10 | Thm (Kronecker).- If all conjugates of an algebraic integer lie on the unit circle, then the integer is a root of unity.
Question: Can one provide a good effective version of this? That is: given that we have an algebraic integer alpha of degree <=d, can we show that alpha has a conjugate that is at least epsilon away from the unit circle, where epsilon depends only on d? It actually isn't hard to do this (from the standard proof of Kronecker, viz.: alpha, alpha^2, alpha^3... are all algebraic integers, and their minimal polynomials would eventually repeat (being bounded) if all conjugates of alpha lied on the unit circle) with
epsilon exponential on d, i.e., epsilon of the form epsilon = 1/C^d; what we actually want is an epsilon of the form 1/d^C, say.
(Question really due to B. Bukh.)
| https://mathoverflow.net/users/398 | (Good) effective version of Kronecker's theorem? | If $M(\alpha)$ is the Mahler measure of $\alpha$, then the largest conjugate of $\alpha$
has absolute value at least
$$1 + \frac{\log(M(\alpha))}{d}.$$
Lehmer's conjecture implies that this at least $O(d^{-1})$ away from one.
Dobrowolski's lower bound for $M(\alpha)$ shows that there is a conjugate at least
$O(d^{-1-\epsilon})$ away from $1$ for any $\epsilon > 0$. Better bounds are available if one has more information, for example, the signature of $\mathbf{Q}(\alpha)$, or whether $\alpha$ is conjugate to $\alpha^{-1}$ or not.
For explicit references, see
<http://www.maths.ed.ac.uk/~chris/Smyth240707.pdf>
| 14 | https://mathoverflow.net/users/nan | 33170 | 21,526 |
https://mathoverflow.net/questions/33176 | 4 | Let $G := SO\_n(R)$ be equipped with the Euclidean metric on vectors of length $n^2$. Is it true that for any $\epsilon >0$, there is a finite subgroup of $G$ which intersects every metric ball of radius $< \epsilon$ in $G$?
| https://mathoverflow.net/users/7821 | Can SO_n(R) be approximated arbitrarily well using a discrete subgroup? | Generalizing Robin's answer to arbitrary $n$:
Jordan's theorem implies that for any $n$ there is an integer $J(n)$ such that the index of a normal abelian subgroup of a finite subgroup of $GL(n,\mathbf{C})$ and hence $SO(n)$ is $\leq J(n)$. A theorem by Boris Weisfeiler (based on the classification of finite simple groups) implies that there are real $a,b$ such that $J(n)<(n+1)!n^{b+a\log n}$. See <http://www.pnas.org/content/81/16/5278.short?related-urls=yes&legid=pnas;81/16/5278>
So any finite subgroup of $SO(n)$ is included in $\leq J(n)$ copies of the maximal torus and so if one takes $n\geq 3$ and small enough $\varepsilon>0$, then for any finite subgroup $G$ of $SO(n)$ there will there elements $\varepsilon$ or further away from $G$ (with respect to any say left-invariant metric)
| 11 | https://mathoverflow.net/users/2349 | 33182 | 21,531 |
https://mathoverflow.net/questions/33181 | 3 | Let $f:C\to C'$ be a functor, and let $A$ be a locally presentable, complete, and cocomplete category. Then according to the paper I'm reading, the pullback functor, $f^\*:A^{C'}\to A^C$ (given by precomposition with $f$), admits left and right adjoints $f\_!$ and $f\_\*$. It's clear that the proof of this fact follows from the adjoint functor theorem, so it suffices to show that $f^\*$ is continuous and cocontinuous.
However, it's not clear to me how to show this fact.
Question:
Using the notation above, why is $f^\*$ continuous and cocontinuous?
Sorry if this ends up being too easy.
| https://mathoverflow.net/users/1353 | Probably easy: Why is f*:A^C'->A^C continuous and cocontinuous for any functor f:C->C'? | Whenever $A$ has all small (co)limits, small (co)limits in functor categories $A^C$ are pointwise: a cone $F : I \to A^C$ is a (co)limit iff each of its “components” or “partial evaluations” $F(c) : I \to A$ is a (co)limit.
But the property of being a *pointwise* (co)limit is obviously preserved by composition with $f:C' \to C$. So $f^\*$ is continuous and co-continuous (and you can apply the AFT).
On the other hand, as Martin says in comments, we don't really need the AFT: the adjoints are given by left and right Kan extensions, which we can write down explicitly as (co)limits, as described by Michael Warren in [this recent answer](https://mathoverflow.net/questions/32791/how-is-the-right-adjoint-f-to-the-inverse-image-functor-f-described-for-f/32808#32808).
All of this is discussed in the chapters on limits and Kan extensions in Mac Lane *Categories for the working Mathematician*. Also: we don't need the local presentability for any of this, and the limits and colimits halves each work independently of the other.
| 5 | https://mathoverflow.net/users/2273 | 33185 | 21,534 |
https://mathoverflow.net/questions/33120 | 2 | Let $n > 1$ be a square-free natural number, which is fixed. The assertion to be proved is the following:
>
> Let $p$ run through primes. Then, $$\left( \frac{n}{p} \right)$$ is equally distributed between $1$ and $-1$.
>
>
>
The precise statement of which is to be made using the appropriate asymptotic expressions.
J-P. Serre, "A Course in Arithmetic", outlines a proof of the above assertion just after the proof of Dirichlet's theorem on arithmetic progressions. That proof uses the zeta function of number fields. I am not able to shake off the feeling that it should be provable without using this. That is, it should be possible to prove this statement just using the properties of Dirichlet $L$-functions, and elementary arguments on quadratic residues. However I am not able to construct such a proof either, since I am not very skilled in this type of matters. So I ask here, is such a proof known?
| https://mathoverflow.net/users/2938 | Distribution of quadratic residues of a fixed number without using Dedekind zeta function | This answer summarizes the above discussion: Extend $p \mapsto \left( \frac{n}{p} \right)$ to a multiplicative function $\chi$ on the positive integers. By quadratic reciprocity, $\chi$ is periodic modulo $4n$, and it is multiplicative by construction, so it is a character. We know that $L(1, \chi) \neq 0$. Thus, $\lim\_{s \to 1^{+}} |\log L(s,\chi)| < \infty$.
We compute:
$$\log L(s, \chi) = - \sum \log \left( 1-\frac{\chi(p)}{p^s} \right) = \sum \frac{\chi(p)}{p^s} + O(1).$$
So $\sum \left( \frac{n}{p} \right)/p^s$ is bounded as $s \to 1^{+}$. A little more work shows that the limit as $s \to 1^{+}$ exists.
If you want to prove results like that $\sum \left( \frac{n}{p} \right)/p$ converges, or that $|\sum\_{p < N} \left( \frac{n}{p} \right)| = o(N)$, then you need Tauberian methods, as discussed in any book on analytic number theory.
| 2 | https://mathoverflow.net/users/297 | 33189 | 21,535 |
https://mathoverflow.net/questions/33191 | 6 | Let $K$ be a field and $E$ be an elliptic curve defined over $K$. It well understood the $K$-points on $E$ forms an abelian group. What is the structure of this group?(Depending on char($K$)?) Is it a direct sum of some well known abelian groups such as $\mathbb{Z}/m\mathbb{Z}$?
| https://mathoverflow.net/users/4171 | Elliptic curves — general structure of the group | First case: Complex numbers. Over $\mathbb C$ the structure as an abstract group is $\mathbb S^1 \oplus \mathbb S^1$ where $\mathbb S^1$ is the circle, i.e., $\mathbb R/\mathbb Z$. This follows as Robin Chapman notes [below](https://mathoverflow.net/questions/33191/elliptic-curves-general-structure-of-the-group#comment75333_33194), i.e., it is a complex torus in the form $\mathbb C/\Lambda$ where $\Lambda$ is a lattice in $\mathbb C$.
Let $K$ be an algebraically closed field of char $p$. If $n$ is prime to $p$, the the $n$-torsion is $\mathbb{Z}/n\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$. The $p^e$-torsion could be either trivial for all $e$, or $\mathbb{Z}/p^e\mathbb{Z}$ for all $e$.
Over a non-algebraically closed-field, this is going to be much more complicated. I will try to give just an introduction.
Over $\mathbb Q$ and number fields: Over $\mathbb Q$ or a number field, it is finitely generated by the Mordell–Weil theorem. So it has a torsion part and free part. The free part could be arbitrarily large.
The Mordell–Weil theorem is in fact true for arbitrary finitely generated fields. This is due to Néron.
Over $\mathbb Q$, the torsion part has exactly $15$ possibilities according to the [theorem of Mazur](https://planetmath.org/MazursTheoremOnTorsionOfEllipticCurves). Over number fields, this had been generalized that the torsion part is uniformly bounded.
Over finite fields, the torsion group would be $\mathbb{Z}/n\mathbb{Z} \oplus \mathbb{Z}/m\mathbb{Z}$ where $n \mid m$. There is no free part.
And it could go on like this. Please have a look at Silverman's "[Advanced topics in the Arithmetic of Elliptic Curves](https://doi.org/10.1007/978-1-4612-0851-8)" for elliptic curves over real numbers, $p$-adic numbers, function fields, etc.
| 10 | https://mathoverflow.net/users/2938 | 33194 | 21,536 |
https://mathoverflow.net/questions/33198 | 5 | There is a well-known Quillen's localization sequence for (algebraic) K-theory: $\dots\to K\_p^Y(X)\to K\_p(X)\to K\_p(X-Y)\to \dots$, where $Y\to X$ is a closed embedding of schemes.
Now suppose that $X$ is regular (and excellent of finite dimension, if needed).
Another well-known fact is that (in this case) the relative K-theory group
$K\_p^Y(X)$ is isomorphic to $K'\_p(Y)$ (some authors denote this by $G\_p(Y)$; note that $Y$ is not necessarily regular!).
Now, I tensor this long exact sequence by $\mathbb{Q}$. Can I consider the $i$-th graded piece of the $\gamma$-filtration for this long exact sequence? Certainly, $K^{}(X)\otimes \mathbb{Q}$ and $K^\*(X-Y)\otimes \mathbb{Q}$ are endowed with $\gamma$-filtration, but I am not quite sure about $K'\_p(Y)\otimes \mathbb{Q}$ (one of my problems here is that I am interested in quite a general situation). Also, could I say that the $i$-th level of the $\gamma$-filtration for $K'\_p(Y)$ is some (which one??) level of its niveau filtration?
Which references are most appropriate for these matters? I believe that for rational coefficients these things are easier than for integral ones.
| https://mathoverflow.net/users/2191 | On $\gamma$-graded pieces of the localization sequence for G-theory (i.e. for K'-theory) | The basic reference is Soule's paper "Operations en K-theorie algebrique" (Can. J. Math. 37 (1985) 488-550). Essentially, there is a grading on $K'$-theory (with rational coefficients) enabling one to interpret the localisation sequence as the long exact sequence of motivic homology. The filtration comes from the $\gamma$-filtration on $K\_\*^Y(X)$, the $K$-theory with support.
Tamme's article in the Beilinson conjectures book (<http://wwwmath.uni-muenster.de/math/u/schneider/publ/beilinson-volume/index.html>) is a good survey of this, but he sticks to the case of schemes over a field.
I am not sure I understand your final question, since even for $K$-theory of nice schemes (e.g. spectra of fields) the $\gamma$- and niveau filtrations don't agree except for $K\_0$.
| 6 | https://mathoverflow.net/users/5480 | 33203 | 21,541 |
https://mathoverflow.net/questions/33188 | 7 | Background:
-----------
The limit of a functor $F:D^{op}\to C$ is an object $\lim F$ representing the functor $$\ell F(x):=\operatorname{Psh}\_D(\ast,C(x, F(\cdot))),$$ where $\*$ denotes the terminal presheaf on $D$. (Notice that $C(x, F(\cdot)))$ is a presheaf on $D$).
We can define the limit of a functor weighted by a presheaf in much the same way (by replacing $\ast$ with a fixed presheaf on $D$ called the *weight*).
Why am I bringing this up? It is a very slick definition. Nowhere do we have anything like universal arrows popping up. Adjunctions are out of sight and out of mind. Indeed, this definition generalizes straightforwardly to S-enriched categories for S symmetric monoidal closed (and all of the other requirements you need for the S-enriched Yoneda lemma to work).
The usual definition of the Kan extension is as a functor completing a certain commutative triangle such that it is universal in a specific sense in a certain functor category (intentionally vague...). This definition is pretty annoying to work with and is avoided whenever possible by instead insisting that all Kan extensions be pointwise (for instance, Kelly does this his book on enriched categories).
Question:
---------
Does there exist a similar slick definition of the Kan extension (not necessarily pointwise)? By slick here, we mean free of adjoint functors (and their less conspicuous cousins, universal arrows) and free of commutative diagrams (translating the content of a commutative diagram into prose does not count).
| https://mathoverflow.net/users/1353 | A slick definition of the Kan extension? | Firstly, the $W$-weighted limit $\lim^W F$ is defined to be a representation of $\operatorname{Psh}\_D(W,C(-,F-))$; the definition you've given isn't even well-typed.
There is no difference at all in $\mathrm{Set}$-enriched category theory between a representation and a universal arrow -- each determines the other, and when these exist for all suitable objects then adjoints are there whether you like it or not (this is all in Mac Lane). So I'm not sure what's particularly slick about this approach.
Anyway, the non-pointwise right Kan extension is given, for $E \overset{K}{\leftarrow} C \overset{F}{\to} D $ by
$$[C,D] (G K, F) \cong [E,D] (G, \operatorname{Ran}\_K F)$$
which is exactly the same as the usual definition in terms of universal arrows. The pointwise extension is $(\operatorname{Ran}\_K F)e = \lim^{E(e,K-)} F$. The difference is discussed in Kelly chapter 4.3. Incidentally, Kelly sticks to pointwise extensions because the non-pointwise ones aren't much use, not because he doesn't like the usual definition.
Is that the kind of thing you're looking for?
| 10 | https://mathoverflow.net/users/4262 | 33206 | 21,543 |
https://mathoverflow.net/questions/33205 | 6 | Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$?
What I do know is that if $(A,\mathfrak{m})$ is Noetherian local and $M$ is finitely generated over $A$ then $\mbox{Tor }^{r+1}(M,A/\mathfrak{m})=0 $ if and only if $\mbox{proj. dim }M\leq r$.
Generally speaking, is $\mbox{Tor }$ functor as good a tool to measure projective dimension as $\mbox{Ext }$ even when the ring/module is not Noetherian or local?
I suspect we can use $\mbox{Tor }$ to measure projective dimension when ring is Neotherian local and module is finitely generated because flatness and projectivity coincide in such case.
| https://mathoverflow.net/users/5292 | Tor and projective dimension | Over the integers, the rational numbers Q are flat and so Tor^i(Q,M) = 0 for all M and all i>0. However Q is not projective so has projective dimension 1.
| 10 | https://mathoverflow.net/users/51 | 33207 | 21,544 |
https://mathoverflow.net/questions/33214 | 0 | Let $f \colon X \to Y$ be a morphism of schemes, where $X$ and $Y$ are separated integral Noetherian schemes. Does there necessarily exist a nonempty open affine $U \subset Y$ such that $f^{-1}(U)$ is affine? If $X \to Y$ is not dominant, then the answer is clearly yes (take $U$ to have empty inverse image). Even for dominant morphisms, I feel like this should be related to Chevalley's theorem that the image of a constructible set is constructible, but I can't put it together into a proof.
Note: this question is silly (see comments/answer below). The specific case I had in mind was when $f$ is unramified over the generic point of $Y$; however, if I decide I want the answer to this, I'll ask in a separate question.
| https://mathoverflow.net/users/5094 | Are morphisms of schemes generically affine | I think the answer is no. Take any field $k$, and consider the natural morphism $\mathbb P^n\_k \to spec(k)$, where $\mathbb P^n\_k$ is the $n$-dimensional projective space. Since $spec(k)$ is, as topological space, a single point and $\mathbb P^n\_k$ is not affine, your open subset cannot exist.
| 9 | https://mathoverflow.net/users/7845 | 33215 | 21,550 |
https://mathoverflow.net/questions/33160 | 1 | For the following wave equation
$
\frac{{\partial ^2 p}}{{\partial ^2 x}} + \frac{{\partial ^2 p}}{{\partial ^2 y}} = A\frac{{\partial ^2 p}}{{\partial ^2 t}} + B\frac{{\partial p}}{{\partial t}}
$
is there a way to show that there are boundary conditions at or near positive and negative infinity, for both non-zero B and B=0 conditions, and for {A,B} as rational numbers? I believe that this should follow from Sommerfeld's condition of radiation, and should perhaps be similar to conditions for the ordinary wave equation. What are these boundary conditions? Ideally, I think that the boundary conditions should involve both time and spatial derivatives.
By "positive and negative infinity" I mean that I am interested in what happens when $x \to \pm \infty
$ and $y \to \pm \infty$. I've been working on a problem where I would like to computationally solve the wave equation with boundary conditions that approximate infinity. So I suppose that this would be an imposed compatibility condition.
| https://mathoverflow.net/users/7875 | Boundary conditions of wave equation near infinity | First of all, can you be more precise in your question? You are asking about boundary conditions at infinity, and this might make sense, but... for what purpose? do you need a set of conditions that imply existence and uniqueness of a global solution? or, do you need to classify solutions of the standard Cauchy problem (with data at t=0) according to their behaviour at infinity?
Anyway, there are a couple general tools that might help you at least to clarify what you are looking for exactly:
1) If you need a tool to classify solutions according to their behaviour at infinity, then scattering theory (mentioned by Willie in his comment) might be helpful. However, its main purpose is to compare two different equations, i.e., use the solutions of a simpler equation to classify the solutions of a 'more difficult' equation. So I do not think this is what you actually need.
2) If you need to understand what might be reasonable 'data at infinity' for a Cauchy problem, then the [Kelvin transform](http://books.google.it/books?id=Xnu0o_EJrCQC&pg=PA518&lpg=PA518&dq=kelvin+transform+wave.equation&source=bl&ots=dihlwFUtjS&sig=rR7OLAKEz4uwgwooTyreGxtdXuo&hl=it&ei=VD1LTNzkJJDgOOX0hH4&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDMQ6AEwAw#v=onepage&q=kelvin%20transform%20wave.equation&f=false) might be of use. This is a space-time change of coordinates that transforms a wave equation into a wave equation, and exchanges infinity with t=0. Playing with it might give you some insight into what kind of conditions you might impose at infinity on your solution. There is also a much more sofisticated transform with a similar effect, the [Penrose transform](http://en.wikipedia.org/wiki/Penrose_transform), but this might be overkill in your case.
| 3 | https://mathoverflow.net/users/7294 | 33221 | 21,555 |
https://mathoverflow.net/questions/33228 | 7 | I recently came across a family of infinite graphs (in the context of two-dimensional convexity) that don't have induced 4-paths (paths with 4 vertices).
Note that the complement of a 4-path is again a 4-path.
Clearly, every induced $n+1$-cycle contains an induced $n$-path.
Hence, by the Strong Perfect Graph Theorem of Chudnowski, Robertson, Seymour, and Thomas,
graphs without induced 4-paths are perfect.
Can anyone provide a simple proof of that fact?
Having no induced 4-paths seems like a very strong condition.
| https://mathoverflow.net/users/7743 | Simple proof that these graphs are perfect | These P4-free graphs are also known as [cographs](http://en.wikipedia.org/wiki/Cograph). A simple proof of the perfectness of such graphs was given by Seinsche, *On a property of the class of n-colorable graphs*, J. Comb. Th. Ser. B 16 (1974), 191–193. [MR0337679](http://www.ams.org/mathscinet-getitem?mr=337679) The key to the proof is the fact that these graphs are also characterized by the property that every subset of $V(G)$ with more than one element is either not $G$-connected or not $\overline{G}$-connected. It follows that every such graph can be obtained from a single vertex by repeatedly duplicating vertices with or without an edge between the two duplicates. Since these two duplication operations preserve perfectness, all such graphs are perfect. (This quick argument is due to Lovász.)
| 11 | https://mathoverflow.net/users/2000 | 33231 | 21,559 |
https://mathoverflow.net/questions/29590 | 54 | Let $R$ be a ring. A notable theorem of **N. Jacobson** states that if the identity $x^{n}=x$ holds for every $x \in R$ and a fixed $n \geq 2$ then $R$ is a commutative ring.
The proof of the result for the cases $n=2, 3,4$ is the subject matter of several well-known exercises in **Herstein**'s *Topics in Algebra*. The corresponding proofs rely heavily on "elementary" manipulations. For instance, the proof of the case $n=3$ can be done as follows:
1) If $a, b \in R$ are such that $ab= 0$ then $ba=0$.
2) $a^{2}$ and $-a^{2}$ belong to $\mathbf{Z}(R)$ for every $a \in R$.
3) Since $(a^{2}+a)^{3} = (a^{2}+a)^{2}+(a^{2}+a)^{2}$ it follows that
$a=a+a^{2}-a^{2} = (a+a^{2})^{3}-a^{2} = (a^{2}+a)^{2}+(a^{2}+a)^{2}-a^{2}$
and whence the result. ▮
Certainly, the mind can't but boggle at the succinctness of the above solution. Actually, it is the conciseness of this argument that has prompted me to pose the present question: **is an analogous demonstration of the general theorem possible?** The one that appears in [**1**] depends on some non-trivial structure theorems for division rings.
As usual, I thank you in advance for your insightful replies, reading suggestions, web links, etc...
**References**
[**1**] I. N. Herstein, *Noncommutative rings*, The Carus Mathematical Monographs, no. 15, Mathematical Association of America, 1968.
[**2**] I. N. Herstein, Álgebra Moderna, Ed. Trillas, págs. 112, 119, and 153.
| https://mathoverflow.net/users/1593 | A condition that implies commutativity | For fixed $n \in \mathbb{N}$, Birkhoff's completeness theorem implies that such a proof must exist in the first-order equational theory of rings - as I mentioned here in a recent [post](https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30273#30273). Many years ago Stan Burris told me that John Lawrence discovered such an equational proof that works uniformly for all $n$ (possibly also for Jacobson's form $x^{n(x)} = x$). I don't know if the proof is published yet, but some clues as to how it may proceed might be gleaned from their earlier joint [work [1]](http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf)
[1](http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf) S. Burris and J. Lawrence, Term rewrite rules for finite fields.
International J. Algebra and Computation 1 (1991), 353-369.
<http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf>
| 32 | https://mathoverflow.net/users/6716 | 33234 | 21,562 |
https://mathoverflow.net/questions/33199 | 8 | In *Singular points of complex hypersurfaces*, John Milnor proves the following theorem:
Let $x \in V$ be a point on a variety $V$ in $\mathbb{R}^n$ or $\mathbb{C}^n$. Assume $x$ is either a smooth point or an isolated singularity. Let $D\_{\epsilon}$ be the closed $\epsilon$-ball about $x$, $S\_{\epsilon}$ its boundary (the sphere about $x$ of radius $\epsilon$), and $K = V \cap S\_{\epsilon}$. Then for $\epsilon$ sufficiently small, the pair $(D\_{\epsilon}, V \cap D\_{\epsilon})$ is homeomorphic to the pair $(CS\_{\epsilon}, CK)$, where $C$ denotes taking the cone. (Theorem 2.10)
In Remark 2.11, Milnor observes that this theorem "likely" holds even if $x$ is a non-isolated singularity; in particular, it is known even in this case that "a suitably chosen neighborhood of any point is homeomorphic to the cone over something."
This book was written in 1968. What is the current status of this problem?
| https://mathoverflow.net/users/5094 | Small neighborhoods of singularities on varieties | There is a good paper of Goresky, "[Triangulation of Stratified Objects](http://www.jstor.org/stable/2042563)", that I think reasonably quickly implies Milnor's result and its generalization to non-isolated singularities. The result is that any Whitney-stratified set, and in particular any algebraic variety in $\mathbb{C}^n$, is supported on a smooth triangulation. I think that you just need that and the inverse function theorem.
As I meant to explain in the comments, this theorem has sometimes been regarded as a "chore" theorem. You can look at what Goresky says: "Triangulation theorems for stratified objects have been obtained independently by Hendricks (unpublished), Johnson (unpublished), and Kato (in Japanese)". When Goresky wrote his paper, it was a messy question that did not have a well-defined status. Now the situation is a bit better and I think that this generalization of Milnor's result can be called settled. Sometimes a good author not only proves a chore theorem, but also cleans it up an elevates it to non-chore status. But a lot of chore theorems are never proven in a clean form or are never proven at all.
| 9 | https://mathoverflow.net/users/1450 | 33241 | 21,567 |
https://mathoverflow.net/questions/28125 | 13 | (1) Is the definition of [flop](http://en.wikipedia.org/wiki/Flip_%28algebraic_geometry%29) given by Wikipedia the industry standard?
(2) Regardless of the answer to (1), when is it expected that a birational transformation gives rise to a derived equivalence?
References to places where precise conjectures are recorded will be very much appreciated!
The reason I'm asking: apparently it is conjectured that different crepant resolutions are derived equivalent. On page 40 of [this](http://arxiv.org/abs/alg-geom/9506012) paper of Bondal-Orlov, they conjecture that flops induce derived equivalences. Apparently "flop" is sometimes used to mean birational transformation preserving canonical classes (without specifying the type of surgery actually being performed). So I'm interested to know whether such transformations are expected to be factorizable into (Wikipedia) flops, or produce derived equivalences for other reasons.
| https://mathoverflow.net/users/nan | What is a flop (and when are they conjectured to give derived equivalences)? | I'm not any kind of expert on this stuff and I'm not sure what the current state of this conjecture is, but Kawamata has conjectures in [this paper](http://arxiv.org/PS_cache/math/pdf/0205/0205287v3.pdf) and [this paper](http://arxiv.org/PS_cache/math/pdf/0311/0311139v2.pdf) regarding when two birational varieties have equivalent derived categories. He also discusses flops in the first paper.
He has partial results, including: if $X$ is general type and $\mathcal{D}^b(X) \cong \mathcal{D}^b(Y)$ as triangulated categories then $X$ and $Y$ are K-equivalent. This generalizes the famous theorem of Bondal-Orlov that the bounded derived category of a Fano variety determines the variety. IIRC, in the proof of his theorem he takes the kernel of the Fourier-Mukai transform that gives the equivalence, shows that the support of the kernel (meaning the union of the supports of the cohomology sheaves of the kernel) has a component $Z$ dominating both varieties and uses $Z$ for the "roof" of the K-equivalence. The assumption that $X$ is general type is used to show that the projections from $Z$ are birational.
| 5 | https://mathoverflow.net/users/7756 | 33244 | 21,569 |
https://mathoverflow.net/questions/33230 | 5 | The complex matrix exponential of a Hermitian matrix is unitary: $e^{-iH} = U$. Is there a name or a characterization for matrices Q whose real exponential is stochastic: $e^{-Q} = S$?
| https://mathoverflow.net/users/756 | Matrices whose exponential is stochastic | A matrix $A$ such that $\exp(tA)$ is (right) stochastic for all $t > 0$ should be called a "generator of a semigroup of stochastic matrices" or an "infinitesimally stochastic matrix". Clearly, since $A=\lim\_{t\to0} (\exp(tA)-I)/t$, (i) the sum of the elements in each row of $A$ has to be 0, and (ii) all non-diagonal elements must be non-negative. Conversely, a matrix $A$ satisfing (i) and (ii), for large enough $n$ produces a stochastic matrix $I+A/n$, hence $(I+A/n)^n$ and $\exp(A)=\lim\_{n\to\infty}(I+A/n)^n$ are also stochastic (and so is $\exp(tA)$). That said, I would have a look at the results of a Google search with "infinitesimally stochastic" (I can't do it now).
(Edit: as observed, the above is a stronger condition than the one you asked for; though it's a more close analog to your example.)
| 6 | https://mathoverflow.net/users/6101 | 33247 | 21,571 |
https://mathoverflow.net/questions/33250 | 9 | There's a workshop at MSRI in a couple months on the Kervaire invariant problem that I'd really like to attend. I saw Hopkins speak about it a while back without understanding much of the talk, but I'm thinking that maybe by now I'm more prepared to see what it's all about. At least, I'd like to be able to get something out of the workshop, even if I don't understand everything that's going on. I don't know much more about the problem than the [wikipedia article](http://en.wikipedia.org/wiki/Kervaire_invariant) can tell me. So, I'm looking for:
(a) recommendations for reading to acquaint myself with the necessary background information, and
(b) a sketch of the story -- I can somewhat follow the intro to the [paper](http://arxiv.org/PS_cache/arxiv/pdf/0908/0908.3724v1.pdf) itself, but seeing some of the details spelled out would help me start to actually understand what's going on.
(Or, unfortunately possibly: (c) a warning that it's probably too difficult to pick up all the material in just a few months and that it's not worth my time to try.)
By the way, wikipedia says that Kervaire used the invariant to create a 10-dimensional PL manifold with no differentiable structure, but I thought that the invariant is zero in dimension 10. What's the deal there?
| https://mathoverflow.net/users/303 | references / general idea of kervaire invariant problem | Here's a recent survey article by Victor Snaith:
<http://chucha.math.cinvestav.mx/morfismos/v13n2/arfsurveyMFMS.pdf>
(I think there's also a copy on the arxiv)
| 6 | https://mathoverflow.net/users/6646 | 33251 | 21,573 |
https://mathoverflow.net/questions/33257 | 6 | The article I have checked for Baker's theorem is [Waldschmidt's](http://www.math.jussieu.fr/~miw/articles/pdf/GDL/SemBxMSGB.pdf). But the article and the citations therein are from the time of '88. Question:
>
> What is the the strongest known lower bound for Baker's theorem on linear forms on logarithms?
>
>
>
Similarly for p-adic Baker.
| https://mathoverflow.net/users/2938 | Strongest known version of Baker's theorem | There is a big difference between linear forms in *many* logarithms and in *two* (or three) logarithms. The first case is covered in the archimedean case by the work of E. Matveev; Matveev's original works are hard even to specialists but there is a very nice survey [Yu. Nesterenko, Linear forms in logarithms of rational numbers, in *Diophantine approximation* (Cetraro, 2000), 53--106, Lecture Notes in Math., 1819, Springer, Berlin, 2003. MR2009829 (2004i:11082)]. The $p$-adic case was mostly done by Kunrui Yu. The best estimate for the case of two logarithms, which is of importance because of Tijdeman's application to Catalan's equation, is given in [[M. Laurent, M. Mignotte et Y. Nesterenko, Formes linéaires en deux logarithmes et déterminants d'interpolation, *J. Number Theory* **55** (1995), no. 2, 285--321. MR1366574 (96h:11073)]](http://dx.doi.org/10.1006/jnth.1995.1141). The latest news in the last direction (also in relation to Catalan's) are reviewed in [M. Mignotte, Linear forms in two and three logarithms and interpolation determinants. *Diophantine equations*, 151--166, Tata Inst. Fund. Res. Stud. Math., 20, Tata Inst. Fund. Res., Mumbai, 2008. MR1500224 (2010h:11119)]
| 7 | https://mathoverflow.net/users/4953 | 33259 | 21,577 |
https://mathoverflow.net/questions/33226 | 1 | Let $a\_1,a\_2 \ldots a\_n$ be a real-valued sequense with $a\_i=O(N^{-1})$.
How do I estimate Discrete Fourier Transform (DFT) of this sequence?
$$\hat{a}\_{j}=\sum\_{r=1}^{K}a\_{r}\exp\Bigl(-2\pi ij\frac{r}{K}\Bigr),\qquad K=O(N^\alpha),\quad \alpha<1$$
Can I say that DFT sequence $\operatorname{Re}[\hat{a}\_{i}]=O(N^{-1})$?
| https://mathoverflow.net/users/3589 | Estimation of DFT | The answer is no, if you mean an uniform bound in $j$. Here is the example:
Fix $j$ and define
$$
a\_r = \begin{cases} \frac{1}{N}, & Re(\exp(-2\pi i j r/ K)) \geq 0;\\\
0, & otherwise.\end{cases}
$$
It is than easy to estimate that the number of $a\_r = \frac{1}{N}$ is comparable to $K$. Even more is true, one has that the number of $a\_r = \frac{1}{N}$ such that $Re(\exp(-2\pi i j r/ K)) \geq \sigma$ is comparable to $K$ for any $\sigma > 0$.
This implies that
$$
Re(\hat{a}\_j) \geq c N^{1 - \alpha}
$$
for some different $c$.
| 4 | https://mathoverflow.net/users/3983 | 33263 | 21,580 |
https://mathoverflow.net/questions/33282 | 11 | Does there exist a set $A$ such that $A=\{A\}$ ?
Edit(Peter LL): Such sets are called Quine atoms.
[Naive set theory By Paul Richard Halmos](http://dc108.4shared.com/download/YNLXDssM/Halmos_-_Naive_set_theory.djvu?tsid=20100725-105252-871bd162) On page three, the same question is asked.
Using the usual set notation, I tried to construct such a set: First
with finite number of brackets and it turns out that after deleting
those finite number of pairs of brackets, we circle back to the original
question.
For instance, assuming $A=\{B\}$, we proceed as follows:
$\{B\}=\{A\}\Leftrightarrow
B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original
equation.
So the only remaining possibility is to have infinitely many pairs of
brackets, but I can't make sense of such set. (Literally, such a set is
both a subset and an element of itself. Further more, It can be shown
that it is singleton.)
For some time, I thought this set is unique and corresponded to $\infty$
in some set-theoretic construction of naturals.
To recap, my question is whether this set exists and if so what
"concrete" examples there are. (Maybe this set is axiomatically
prevented from existing.)
| https://mathoverflow.net/users/5627 | Can we have A={A} ? | In standard set theory (ZF) this kind of set is forbidden because of the [axiom of foundation](http://en.wikipedia.org/wiki/Axiom_of_foundation).
There are alternative axiomatisations of set theory, some of which do
not have an equivalent of the axiom of foundation. This is called
non-well-founded set theory. See e.g. [Aczel's anti-foundation\_axiom](http://en.wikipedia.org/wiki/Aczel%27s_anti-foundation_axiom), where there is a unique set such that $x = \{x\}$.
| 33 | https://mathoverflow.net/users/1310 | 33283 | 21,584 |
https://mathoverflow.net/questions/20940 | 12 | In EGA IV, Sec. 16, Grothendieck defines the sheaf of principal parts as follows: Let $f:X\rightarrow S$ be a morphism of schemes and $\Delta:X\rightarrow X\times\_S X$ the diagonal morphism associated to $f$. $\Delta$ is an immersion, so the corresponding morphism $\Delta^{-1}\mathcal{O}\_{X\times\_S X}\rightarrow\mathcal{O}\_X$ is surjective. Let $\mathcal{I}$ denote its kernel and define the sheaves of principal parts as
$$\mathcal{P}\_{X/S}^n:=\Delta^{-1}( \mathcal{O}\_{X\times\_S X}) / \mathcal{I}^{n+1}$$
In their book on Crystalline cohomology, Berthelot and Ogus define the sheaf of principal parts $\mathcal{P}^n\_{X/S}$ as
$$(\mathcal{O}\_X\otimes\_{f^{-1}\mathcal{O}\_S}\mathcal{O}\_X)/\mathbf{I}^{n+1},$$
where $\mathbf{I}$ is the kernel of the multiplication map from the tensor product to $\mathcal{O}\_X$.
My question is probably simple, but I don't know how to see it:
Why are those definitions equivalent if $X$ and $S$ are not affine and $n>0$?
I've not seen the second definition anywhere else, although it seems somewhat nicer than the first one...
| https://mathoverflow.net/users/259 | Sheaves of Principal parts | The statement holds in general if $f : X \to S$ is a morphism of locally ringed spaces. The fibred product of locally ringed spaces can be constructed explicitly without gluing constructions, and also restricts to the fibred product of schemes. See [this article](http://maddin.110mb.com/pdf/faserprodukte.pdf) (german; shall I translate it?) for details. I will make use of the explicit description given there. Also I use stalks all over the place. Probably this is not the most elegant proof, but it works.
First we construct a homomorphism $\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X \to \Delta^{-1} \mathcal{O}\_{X \times\_S X}$. For that we compute the stalks at some point $x \in X$ lying over $s \in S$:
$(\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X)\_x = \mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x},$
$(\Delta^{-1} \mathcal{O}\_{X \times\_S X})\_x = \mathcal{O}\_{X \times\_S X,\Delta(x)} = (\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x})\_{\mathfrak{q}}$,
where $\mathfrak{q}$ is the kernel of the canonical homomorphism
$\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x} \to \kappa(x), a \otimes b \mapsto \overline{ab}.$
Thus we get, at least, homomorphisms between the stalks (namely localizations). In order to get sheaf homomorphisms out of them, the following easy lemma is useful:
(\*) Let $F,G$ be sheaves on a topological space $X$ and for every $x \in X$ let $s\_x : F\_x \to G\_x$ be a homomorphism. Suppose that they fit together in the sense that for every open $U$, every section $f \in F(U)$ and every $x \in U$ there is some open neighborhood $x \in W \subseteq U$ and some section $g \in G(W)$ such that $s\_y$ maps $f\_y$ to $g\_y$ for all $y \in W$. Then there is a sheaf homomorphism $s : F \to G$ inducing $s$.
This can be applied in the above situation: Every section in a neighborhood of $x$ in $\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X$ induced by an element in $\mathcal{O}\_X(U) \otimes\_{\mathcal{O}\_S(V)} \mathcal{O}\_X(U)$ for some neighborhoods $U$ of $x$ and $V$ of $s$ such that $U \subseteq f^{-1}(V)$. This yields a section in $\mathcal{O}\_{X \times\_S X}$ on the basic-open subset $\Omega(U,U,V;1)=U \times\_V U$ and thus a section of $\Delta^{-1} \mathcal{O}\_{X \times\_S X}$ on $U$. It is easily seen, that this construction yields the natural map on the stalks.
Thus we have a homomorphism $\alpha : \mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X \to \Delta^{-1} \mathcal{O}\_{X \times\_S X}$. Now let $J$ be the kernel of the multiplication map $\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X \to \mathcal{O}\_X$ and $I$ be the kernel of the homomorphism $\Delta^\# : \Delta^{-1} \mathcal{O}\_{X \times\_S X} \to \mathcal{O}\_X$. Then for every $n \geq 1$ our $\alpha$ restricts to a homomorphism
$(\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X)/J^n \to (\Delta^{-1} \mathcal{O}\_{X \times\_S X})/I^n,$
which is given at $x \in X$ by the natural map
$(\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x}) / \mathfrak{p}^n \to ((\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x}) / \mathfrak{p}^n)\_{\mathfrak{q}}$,
where $\mathfrak{p} \subseteq \mathfrak{q}$ is the kernel of the multiplication map $\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x} \to \mathcal{O}\_{X,x}$.
We want to show that this map is an isomorphism, i.e. that the localization at $\mathfrak{q}$ is not needed. For that it is enough to show that every element in $\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x}$, whose image in $\mathcal{O}\_{X,x}$ is invertible, is invertible modulo $\mathfrak{p}^n$. Or in other words: Preimages of units are units with respect to the projection
$(\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x}) / \mathfrak{p}^n \to (\mathcal{O}\_{X,x} \otimes\_{\mathcal{O}\_{S,s}} \mathcal{O}\_{X,x}) / \mathfrak{p}^1 \cong \mathcal{O}\_{X,x}$.
However, this follows from the observation that the kernel $\mathfrak{p}^1 / \mathfrak{p}^n$ is nilpotent; cf. also [this question](https://mathoverflow.net/questions/31495/when-does-a-ring-surjection-imply-a-surjection-of-the-group-of-units/32919#32919).
I'm sure that there is also a proof which avoids stalks at all.
EDIT: So here is a direct construction of the homomorphism $\mathcal{O}\_X \otimes\_{f^{-1} \mathcal{O}\_S} \mathcal{O}\_X \to \Delta^{-1} \mathcal{O}\_{X \times\_S X}$:
Let $p\_1,p\_2$ be the projections $X \times\_S X \to X$. Then we have for $i=1,2$ the homomorphism
$\mathcal{O}\_X \to {p\_i}\_\* \mathcal{O}\_{X \times\_S X} \to {p\_i}\_\* \Delta\_\* \Delta^{-1} \mathcal{O}\_{X \times\_S X} = (p\_i \Delta)\_\* \Delta^{-1} \mathcal{O}\_{X \times\_S X} = \Delta^{-1} \mathcal{O}\_{X \times\_S X}$,
and they commute over $f^{-1} \mathcal{O}\_S$. Thus we get the desired homomorphism. But I think stalks are convenient when we want to show that this is an isomorphism when modding out the ideals.
| 8 | https://mathoverflow.net/users/2841 | 33286 | 21,586 |
https://mathoverflow.net/questions/33265 | 55 | It is widely stated that Fermat wrote his famous note on sums of powers ("Fermat's last theorem") in, or around, 1637. How do we know the date, if the note was only discovered after his death, in 1665?
My interest in this stems from the fact that if this is true, we can be absolutely certain that whatever proof Fermat in mind was wrong, and he must have noticed (or he would have mentioned it to his correspondents in later years). On the other hand, if the note had been written much later the reasoning would fail. I have used this argument previously in talks for the general public, rather acritically, and I would like to make sure it is sound.
| https://mathoverflow.net/users/4790 | How do we know that Fermat wrote his famous note in 1637? | Not only do we not know the date, we don't even know whether he wrote the remark at all.
For all we know it might have been invented by his son Samuel, who published his father's comments.
In his letters, Fermat never mentioned the general case at all, but quite often posed the problem of solving the cases $n=3$ and $n=4$. I am almost certain that Fermat discovered infinite descent around 1640, which means that in 1637 he did not have any chance of proving FLT for exponent 4 (let alone in general).
In 1637, Fermat also stated the polygonal number theorem and claimed to have a proof; this is just about as unlikely as in the case of FLT -- I guess Fermat wasn't really careful in these early days.
Let me also mention that Fermat posed FLT for $n=3$ always as a problem or as a question, and did not claim unambiguously to have a proof; my interpretation is that he did not have a proof for $n = 3$, and that he knew he did not have one.
**Edit** Let me briefly quote two letters from Fermat:
I. [Oeuvres](http://wlym.com/~animations/fermat/) II, 202--205, letter to Roberval Aug. 1640 Fermat claims that if $p = 4n-1$ be prime, then $p$ does not divide a sum of two squares $x^2 + y^2$ with $\gcd(x,y) = 1$. Then he writes
*I have to admit frankly that I have found nothing in number theory
that has pleased me as much as the demonstration of this proposition,
and I would be very pleased if you made the effort of finding it, if
only for learning whether I estimate my invention more highly than it
deserves.*
This looks as if Fermat had just discovered "his method" of descent.
Starting from $x^2 + y^2 = pr$ one has to show that there is a prime
$q \equiv 3 \bmod 4$ dividing $r$ which is strictly less than $p$.
II. In his letter to Carcavi from Aug. 1659 (Oeuvres II, 431--436), Fermat writes:
*I then considered certain questions which, although negative, do
not remain to receive a very great difficulty, for it will be easily
seen that the method of applying descent is completely different from
the preceding [questions]. Such cases include the following:*
1. There is no cube that can be divided into two cubes.
2. There is only one square number which, augmented by $2$,
makes a cube, namely $25$.
3. There are only two square numbers which, augmented by $4$,
make a cube, namely $4$ and $121$.
4. All squared powers of $2$ augmented by $1$ are prime numbers.
My interpretation of this is that Fermat lists four results which he
believes can be proved using his method of descent. In my opinion
this implies that Fermat did not have a proof of FLT for exponent $3$
in 1659.
**Edit 2**
In light of the discission at [wiki.fr](http://fr.wikipedia.org/wiki/Discussion:D%C3%A9monstrations_du_dernier_th%C3%A9or%C3%A8me_de_Fermat) let
me add a couple of additional remarks along with a promise that a nonelectronic
publication of my views on Fermat will appear within the next two years (if I can
find a publisher, that is).
A search in google books for "hanc marginis" and Fermat for the
years up to 1900 reveals several hits, none of which claims that
the remark was written around 1637; in particular there are no dates
given in Fermat's Oeuvres or in Heath's Diophantus. Starting with
Dickson's history, this changes dramatically, and nowadays the date
1637 seems to be firmly attached to this entry.
The dating of the entry seems to come from a letter written by
Fermat to J. de Sainte-Croix via Mersenne mentioned in Nurdin's
answer; this letter is not dated, but since Descartes, in a letter
to Mersenne from 1638, refers to a result he credits to Sainte-Croix,
but which Fermat claims he has discovered, it is believed that Fermat's
letter to Mersenne was written well before that date. The reasons for
dating it to September 1636 are not explained in Fermat's Oeuvres.
In this letter, Fermat poses the problem of finding two fourth
powers whose sum is a fourth power, and of finding two cubes whose
sum is a cube. The reasoning seems to be that in 1636, Fermat
had not yet found (or believed to have found) a proof of the general
theorem, so the entry must have been written at a later date.
Since he did not refer to the general theorem in any of his
existant letters, it is also believed that he soon found his
mistake, so the entry cannot have been written at a time when
Fermat was mature enough to find sufficiently difficult proofs.
Let me also add that the following dates can be deduced from
Fermat's letters:
* 1638 Numbers 4n-1 are not sums of two rational squares
* 1640 Fermat's Little Theorem
* 1640 Discovery of infinite descent; used for showing that
(1) primes 4n-1 do not divide sums of to squares.
* 1640 Statement of the Two-Squares Theorem
* 1641 - 1645 Proof of (2) FLT for exponent 4
* later: Proof of (3) the Two-Squares Theorem
It is impossible to attach any dates between 1644 and 1654 to
Fermat's discoveries since he either wrote hardly any letter
in this period, or all of them are lost.
Fermat claimed to have discovered infinite descent in connection
with results such as (1), and that he at first could apply it
only to negative statements such as (2), whereas it took him a
long time until he could use his method for proving positive
statements such as (3). Thus the proofs of (1) - (2) - (3) were
found in this order.
This means in particular that if Fermat's entry in his Diophantus
was written around 1637, then the marvellous proof must have been
a proof that does not use infinite descent.
I would also like to remark that the Fermat equation for exponents
3 and 4 had already been studied by Arab mathematicians, such as
Al-Khujandi and Al-Khazin, who both attempted proving that there
are no solutions. The cubic equation also shows up in problems
posed by Frenicle and van Schooten in response to Fermat's
challenge to the English mathematicians.
| 102 | https://mathoverflow.net/users/3503 | 33289 | 21,589 |
https://mathoverflow.net/questions/33269 | 34 | For any number field $K$, the Fontaine-Mazur conjecture predicts that any potentially semistable $p$-adic representation of the absolute Galois group $G\_K$ of $K$ that is almost everywhere unramified comes from algebraic geometry (i.e., is a subquotient of the etale cohomology of some variety over $K$, up to Tate twist). As far as I can see, the only cases where any progress has been made concerns the case that $K$ is totally real or CM.
This made me wonder: Is the Fontaine-Mazur conjecture known to be true for $1$-dimensional representations for any number field $K$? For CM fields, the theory of CM abelian varieties gives varieties whose cohomology realizes nontrivial characters (and I guess that easy variations should produce all characters). What are the geometric objects appearing for other fields?
[edit: The word 'geometric' is avoided now, see the comments.]
| https://mathoverflow.net/users/6074 | Fontaine-Mazur for GL_1 | Let $\chi$ be a one-dimensional geometric (in the sense of FM) $p$-adic Galois representation of $G\_K$ and let $\psi$ be the Hecke character of $K$ associated to $\chi$ by class field theory. The fact that $\chi$ is de Rham (=pst) at all primes above $p$ imples that $\psi$ is an *algebraic* Hecke character. Generally, the only algebraic Hecke characters of $K$ are of the form $(\text{finite order})\cdot\mathcal{N}^n$ where $\mathcal{N}$ is the norm character. Under class field theory, $\mathcal{N}$ corresponds to the cyclotomic character, so it comes from geometry; additionally, any finite order character comes from geometry (it arises as the subquotient of the $H^0$ of a zero-dimensional variety). The only time there are more algebraic Hecke characters is when $K$ contains a CM field. Denoting $L$ the maximal CM field in $K$, every algebraic Hecke character of $K$ is of the form $(\text{finite order})\cdot(\psi\_L\circ\mathcal{N}\_{K/L})$ where $\psi\_L$ is an algebraic Hecke character of $L$ and $\mathcal{N}\_{K/L}$ is the norm from $K$ to $L$. Again, finite order characters come from geometry, so this case is reduced to the CM case. As you've mentioned the CM case has been dealt with, so Fontaine–Mazur is true for $\mathrm{GL}(1)$.
| 20 | https://mathoverflow.net/users/1021 | 33296 | 21,593 |
https://mathoverflow.net/questions/33303 | 2 | Say we have $n$-gons $P$ and $Q$. Is there any necessary condition for $Q = f(P)$, for some linear transformation $f : \mathbb{R}^2 \to \mathbb{R}^2$?
Sorry if this is too elementary / general.
| https://mathoverflow.net/users/2503 | Linear transformation takes a polygon to another one. | Jesse Douglas studied linear transformations of polygons on the complex plane in 1930s. He proved, in particular, that a transformation $z\_i{}'=\sum\_{i=1}^na\_{ij}z\_j$ (all numbers are complex) will transform a polygon $\pi=(z\_1,\cdots,z\_n)$ into a polygon $\pi'=(z\_1{}',\cdots,z\_n{}')$ if, and only if, the matrix $a\_{ij}$ is cyclic, that is, if, and only if, $a\_{ij}=\alpha\_{j-i}$, $\alpha\_{j-i}=\alpha\_k$ if $k\equiv j-1\ (\text{mod}\,n)$. (See his article ["On linear polygon transformations"](http://www.ams.org/journals/bull/1940-46-06/S0002-9904-1940-07259-3/S0002-9904-1940-07259-3.pdf), Bull. Amer. Math. Soc. 46, (1940) pp. 551 - 560.)
| 5 | https://mathoverflow.net/users/5371 | 33309 | 21,600 |
https://mathoverflow.net/questions/33292 | 10 | It is apparent that the abc conjecture is deeply related to Arakelov theory. In one direction, it is shown in S. Lang, "Introduction to Arakelov Theory", that a certain height inequality in Arakelov theory implies the abc conjecture. I am wondering about the other direction and precise implications. Are there such results?
More importantly, if there exist such results, what are some Diophantine implications? That is, what are the Diophantine implications of abc conjecture, that factor through Arakelov Theory/Arithmetic Geometry?
(I once read that Joseph Oesterle came up with abc conjecture while trying to do computations towards the Taniyama-Shimura conjecture. But that story does not make clear the connection with Arakelov theory.)
| https://mathoverflow.net/users/2938 | Implications of the abc conjecture in Arakelov theory | ABC is equivalent to the conjectured height inequality that Lang (or more precisely Vojta, in an appendix to Lang's book, following the ABC appendix he wrote for his own book) uses. This is shown in several papers by van Frankenhuijsen.
<http://research.uvu.edu/machiel/papers/abcrvhi.pdf>
<http://research.uvu.edu/machiel/papers/ABCRothMord.pdf>
<http://research.uvu.edu/machiel/bibliography.html>
So ABC is equivalent to some of Vojta's conjectures in arithmetic geometry.
Also, Elkies' "ABC implies effective Mordell" is a sort of counter-application in that it shows that, given the ABC conjecture, one does not need Arakelov methods to prove the Mordell conjecture.
<http://imrn.oxfordjournals.org/cgi/pdf_extract/1991/7/99>
Lang wrote an article on Diophantine inequalities related to ABC that outlines some of the relationships between conjectures that were known 20 years ago.
<http://www.ams.org/bull/1990-23-01/S0273-0979-1990-15899-9/S0273-0979-1990-15899-9.pdf>
| 8 | https://mathoverflow.net/users/6579 | 33316 | 21,604 |
https://mathoverflow.net/questions/33294 | 16 | Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero commutative ring. I know that it's true if $A$ is Noetherian or integral domain. I thought it was not true in general but I came up with something that looks like a proof and I can't figure out where it went wrong.
| https://mathoverflow.net/users/5292 | Cardinal of maximal linearly independent subsets of a free module | I think I have a counter-example. Let $A$ be the ring of functions $f$ from $\mathbb{C}^2 \setminus (0,0) \to \mathbb{C}$ such there is a polynomial $\widetilde{f} \in \mathbb{C}[x,y]$ such that $\widetilde{f}(x,y)=f(x,y)$ for all but finitely many $(x,y)$ in $\mathbb{C}^2$.
Map $A$ into $A^2$ by $f \mapsto (fx, fy)$. We check that this is injective: If $fx=0$ then $f$ is zero off of the $x$-axis. Similarly, if $fy=0$, then $f$ is zero off of the $y$-axis. So $(fx, fy) = (0,0)$ implies that $f$ is zero everywhere on $\mathbb{C}^2 \setminus (0,0)$.
We now claim that there do not exist $(u,v)$ in $A^2$ such that $(f,g) \mapsto (fx+gu, \ fy+gv)$ is injective. Suppose such a $(u,v)$ exists. Let $\widetilde{u}$ and $\widetilde{v}$ be the polynomials in $\mathbb{C}[x,y]$ which coincide with $u$ and $v$ at all but finitely many points. Let $\Delta=\widetilde{u} y - \widetilde{v} x$. Since $\Delta$ is a polynomial which vanishes at $(0,0)$, it is not a non-zero constant. Thus, $\Delta$ vanishes on an entire infinite subset of $\mathbb{C}^2$. Let $(p,q)$ be a point in $\mathbb{C}^2 \setminus (0,0)$ such that $\Delta(p,q)=0$, $\widetilde{u}(p,q)= u(p,q)$ and $\widetilde{v}(p,q)=v(p,q)$.
So $q u(p,q) - p v(p,q) =0$. Since $(p,q) \neq (0,0)$, there is some $k \in \mathbb{C}$ such that $(u(p,q), v(p,q)) = (kp, kq)$. Take $f$ to be $-k$ at $(p,q)$ and $0$ elsewhere; let $g$ be $1$ at $(p,q)$ and $0$ elsewhere. So $(fx+gu, fy+gv)=0$, and the map $(f,g) \mapsto (fx+gu, \ fy+gv)$ is not injective.
| 19 | https://mathoverflow.net/users/297 | 33321 | 21,609 |
https://mathoverflow.net/questions/33223 | 2 | A Perfectly Matched Layer (PML) is an absorbing boundary condition (ABC) which can be used to approximate free-field conditions for the numerical solution of wave equation problems.
[PML note](http://www-math.mit.edu/~stevenj/18.369/pml.pdf)
The PML is normally applied to a PDE using the following transformation:
$
\frac{\partial }{{\partial x}} \to \frac{1}{{1 + i\frac{{\sigma (x)}}{\omega }}}\frac{\partial }{{\partial x}}
$
In the above, $i = \sqrt { - 1}$ and $\sigma(x)$ is a function of position in the ABC. Now apparently
$
\frac{{\partial ^2 }}{{\partial x^2 }} \to \frac{1}{s}\frac{\partial }{{\partial x}}\left( {\frac{1}{s}\frac{\partial }{{\partial x}}} \right)
$
$
s = 1 + i\frac{{\sigma (x)}}{\omega }
$
But is it possible to apply a coordinate stretching in the following fashion:
$
\frac{{\partial ^2 }}{{\partial x^2 }} \to \frac{1}{u}\frac{{\partial ^2 }}{{\partial x^2 }}
$
The coordinate stretching performed in this fashion would be similar in function to the stretching performed by the transformation applied to $\partial /\partial x$.
Essentially what I would like to do is to apply the coordinate stretching directly to $\partial ^2 /\partial x^2$. I've looked in the PML literature for a very long time, and it seems that most interest is in the application of the coordinate stretching directly to $\partial /\partial x$.
Moreover, I can imagine the coordinate-stretching occurring in a similar fashion to stretching a rubber sheet. If the stretching is being done to the coordinates of $\partial /\partial x$, then what is happening to $\partial ^2 /\partial x^2$?
| https://mathoverflow.net/users/7875 | Application of coordinate-stretching transformation for Perfectly Matched Layer | Now, I am not familiar with the PML method to categorically say that it is generally impossible to have an absorbing boundary layer with the second order transform that you seek. But I can say confidently that if you are looking from the point of view of the complex coordinate transformation, then you *cannot* have a transformation that substitutes
$$\frac{\partial^2}{\partial x^2}\to \frac{1}{u}\frac{\partial^2}{\partial x^2} $$
This is because the second coordinate derivative is **not** tensorial in nature.
Let $y\_i$ and $x\_i$ be two difference coordinate systems for the same domain, such that $y\_i = y\_i(x\_1,\ldots,x\_n)$ can be written as functions of $x\_j$s. The first derivative of any function is tensorial, in the sense that the coefficients
$$ \frac{\partial f}{\partial x\_i} = \sum\_{j} \frac{\partial y\_j}{\partial x\_i} \frac{\partial f}{\partial y\_j} $$
the matrix $(\partial y\_j / \partial x\_i)$ is sometimes called the *Jacobian matrix* for the coordinate transformation.
However, the second derivative of functions are, in general, not tensorial. So usually we have
$$ \frac{\partial^2 f}{\partial x\_i \partial x\_j} \neq \sum\_{kl} \frac{\partial y\_k}{\partial x\_i} \frac{\partial y\_l}{\partial x\_j} \frac{\partial^2 f}{\partial y\_k \partial y\_l}$$
The only exception is when $f$ has a critical point: if $\frac{\partial f}{\partial y\_k} = 0$ for each $k$, then the above expression is actually an equality. In general the action of a coordinate transformation on any higher order (order $>1$) partial differential operator will introduce lower order terms. The best way to explain this is in terms of *connections* on the tangent bundle in the language of differential geometry.
Another way to illustrate this is to consider the case where the underlying domain is one-dimensional, just the real line $\mathbb{R}$. Now let $v$ be any non-vanishing continuous vector field on $\mathbb{R}$, then $v = \phi(x) \partial\_x$ for some continuous, non-vanishing coefficient function $\phi$. For any such $v$ we can integrate it to obtain a coordinate function $y$ such that $\partial\_y = v$. Now,
$$ \partial\_y^2 f = v(v(f)) = \phi \partial\_x(\phi \partial\_x f) = \phi^2 \partial\_x^2 f + \phi \partial\_x\phi \partial\_x f$$
And you see that unless $\phi$ is the constant function (in which case $y$ is just a dilation of $x$), in general the second derivative relative to the $y$ coordinate system will depend on first derivatives in the $x$ coordinate system.
This all just goes to say that in general, if you are looking for a transformation that directly substitutes
$$ \frac{\partial^2}{\partial x^2} \to \frac{1}{u}\frac{\partial^2}{\partial x^2} $$
it cannot be obtained by any sort of coordinate transformation.
| 1 | https://mathoverflow.net/users/3948 | 33337 | 21,622 |
https://mathoverflow.net/questions/33335 | 3 | Recently I discussed an experiment with a friend. Assume we start a random experiment. At first there is an array with size 100 000, all set to 0. We calculate at each round a random number modulo 2 and select one random position in that array. If the number in the array is 1, nothing is changed and otherwise the pre-computed value is set. The question is: how many distinct hash values would we have added in 1%, 5%, 50%, 95%, 99% of all cases?
Example: 4 rounds with array of size 10:
```
Array Position random number
[0,...,0] 5 0
[0,...,0] 7 1
[0,...0,1,0,0,0] 6 1
[0,..0,.1,1,0,0,0] 6 0
[0,..0,.1,1,0,0,0] 2 0
```
First we considered this a somehow simple problem. But after thinking about for some hours, searching the web and asking some math students we couldn't find a solution.So do you know a probability distribution for this problem?
| https://mathoverflow.net/users/6674 | Looking for a probability distribution | This is equivalent to (among other names) the Coupon Collector problem. Your are asking about the distribution of the number of coupons collected after $t$ steps, when the total number of possible coupons is $n$.
<http://en.wikipedia.org/wiki/Coupon_collector%27s_problem>
ADDED: this and related distributions are also studied under other names such as Birthday Problem, random mappings, and random hashing. Kolchin-Sevastyanov-Chistyakov *Random Allocations*, Knuth *The Art Of Computer Programming, vol. 2*, and Flajolet & Sedgwick *Analytic Combinatorics* all discuss these problems and may contain the precise asymptotics of the distribution you are looking for.
| 7 | https://mathoverflow.net/users/6579 | 33342 | 21,625 |
https://mathoverflow.net/questions/33312 | 16 | You notice a stop-light ahead of you and it is currently red. You can't run the red light, so you will have to brake, but braking wastes energy and you want to be as fuel efficient as possible. What braking strategy maximizes efficiency?
Let's set down some notation and move slowly toward a well-defined question. Suppose you are currently a distance $d$ from the stop-light, and suppose that the stop-light is on a timer whereby it switches from red to green after $T$ seconds. You know the value of $T$, but you don't know how far in the cycle the stop-light is right now -- perhaps it will turn green in 1 second or perhaps in $T$ seconds. So if $t$ is the amount of time until it actually turns green, then $t$ is a random variable uniformly distributed on $[0,T]$.
Your initial speed is $v$, so that if you don't slow down you'll be at the light in $\frac{d}{v}$ seconds. If $t<\frac{d}{v}$ then ``you win" by not slowing down, because the light will turn green before you get to it and you will have lost no energy to heat. So if $T\leq\frac{d}{v}$ then clearly the best strategy is not to slow down. Thus we may suppose $T>\frac{d}{v}$. We assume no friction.
I'm looking for a strategy for applying the brake minimally, not knowing the status of the stop-light's cycle. Perhaps we apply the brakes uniformly to end up stopped at the light, or perhaps we do not apply the brakes at all until we are almost to the stop-light, or perhaps we apply the brakes at the very beginning and coast at that reduced speed until we get very close to the light. What strategy minimizes the expected brake usage?
More precisely, I'm looking for a non-increasing differentiable function (for the car's velocity in terms of its distance to the stop-light) $$f\colon[0,d]\to{\mathbb R}\_{\geq 0}$$ such that $f(d)=v$, $f(0)=0$, and such that if you solve the differential equation to find velocity in terms of time, then the expected value of that velocity at time $t$ is maximized.
| https://mathoverflow.net/users/2811 | What braking strategy is most fuel-efficient? | I think the problem would have been more naturally stated in the context of bicycles. In any case, the answer is as follows:
You are looking for an optimal velocity function $v: [0, T] \to \mathbb{R}\_{\geq 0}$ satisfying some conditions. Each such function represents the strategy, "if the light is still red at time $t$, travel at speed $v(t)$; when the light turns green, coast." One of the conditions on $v$ is that you may not run the red light. In terms of the function $v$, this condition may be written as $\int\_0^T v(t) \, dt \leq d$.
The quantity you wish to compute is the expected speed at which you will pass through the light after it turns green. By the givens (uniform distribution, the nature of our strategy), this expected speed is precisely the average value of $v(t)$, i.e., it is $\frac{1}{T} \int\_0^T v(t) \, dt$.
Putting the last two paragraphs together, we see that the optimal expected speed is $\frac{d}{T}$. Moreover, this expected speed is achieved for *any* choice $v(t)$ with the property $\int\_0^T v(t) \, dt = d$, i.e., for any strategy that will get you to the stoplight within time $T$.
Added in edit:
I agree with Willie Wong that maximizing the expected kinetic energy with which you pass through the light should be more physically relevant to, say, a bicyclist coasting on a shallow down-hill.
| 5 | https://mathoverflow.net/users/4658 | 33347 | 21,627 |
https://mathoverflow.net/questions/33322 | 10 | Background
----------
I am working through a particular result in a paper of Cherlin, Shelah, and Shi, and am satisfied that it follows from basic model theory material - but I'm stuck on one point in the background material.
In Hodges' "A Shorter Model Theory", Lemma 7.2.5 on page 191 seems to have an unused assumption. He
considers, for a $\forall\_2$ $L$-theory $T$, an existential formula $\phi$ and the set of all universal formulas implied (under $T$) by $\phi$. He calls this set the *resultant* of $\phi$:
$Res\_\phi(x) =$ {universal $\psi(x)$ | $T \vdash \forall x (\phi(x) \rightarrow \psi(x))$}
Then he shows that an arbitrary $L$-structure $A$
and element $a$ satisfy $Res\_\phi(a)$ iff there is some extension $B$
of $A$ with $B \models T$ and $B \models \phi(a)$.
Key Issue
---------
I can't tell where Hodges is using the assumption that $T$ is $\forall\_2$. The given proof (below) looks like it goes through without that assumption:
$\Rightarrow$ Suppose some extension $B$ of $A$ satisfies $\phi(a)$ and $T$. Then $B$ satisfies
$Res\_\phi(a)$. Any universal formula satisfied in $B$ must be satisfied in the substructure $A$, also. Since $Res\_\phi(x)$ contains only universal formulas, $B \models Res\_\phi(a) \implies A \models Res\_\phi(a)$.
$\Leftarrow$: Suppose $A$ satisfies $Res\_\phi(a)$. If there is no extension
$B$ of $A$ that satisfies $\phi(a)$ and $T$, then add to the language a constant $c$ representing $a$, and consider (in the language $L\_c$) $Diag(A) \cup \{\phi(c)\} \cup T$. By assumption, this set of sentences has no model, so by compactness
$$
T \vdash \phi(c) \rightarrow \neg \sigma(c,d^A)
$$
for some $d^A$ in $A$ and some
quantifier-free formula $\sigma$. Since $c$ and $d^A$ do not appear in $T$, we can replace
them with universally quantified variables:
$$
T \vdash \forall x [ \phi(x) \rightarrow \forall y \neg \sigma(x,y) ]
$$
Thus, $\forall y(\neg \sigma(x,y)) \in Res\_\phi(x)$ so $A \models \forall y (\neg \sigma(a,y))$. But we got $\sigma$ by considering the $L\_c$-diagram of $A$, so
$A \models \exists y \sigma(a,y)$. Contradiction.
I don't see anywhere that the $\forall\_2$ assumption on T comes in.
Perhaps this assumption is just for context in the rest of the neighboring material?
| https://mathoverflow.net/users/4594 | Extra assumption in Hodges' lemma on the resultant of a first-order formula? | The result doesn't need the assumption that $T$ is an $\forall\_2$-theory, nor does it need that the formula $\phi(x)$ is existential. The relevant general result is that a structure $A$ has an extension satisfying a theory $T$ (in the same vocabulary) if and only if $A$ satisfies all the universal sentences that are provable from $T$. To get the result quoted in the question, minus the unnecessary hypotheses, apply this general result not to the given $T$ but to $T$ plus $\phi(c)$, where $c$ is a new constant symbol interpreted as $a$. (Actually, there is an even more general version of the result; see the "Model Extension Theorem" in Section 5.2 of Shoenfield's "Mathematical Logic" or Lemma 3.5.10(i) of Hinman's "Fundamentals of Mathematical Logic.")
| 8 | https://mathoverflow.net/users/6794 | 33350 | 21,628 |
https://mathoverflow.net/questions/33339 | 7 | Sorry for the vague title, I can't think of a better one that isn't overly long.
Suppose that $S$ is a commuting set of projection operators on a Hilbert space. I'll introduce the following notation: if $p \in S$, let $p^+ \equiv p$ and $p^- \equiv 1 - p$. Let $I \equiv ${$+, -$}. The projections are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$; this makes the set of all projections into a complete lattice. Is the following identity true?
$\sup\_{f \in I^S} \inf\_{p \in S} p^{f(p)} = 1$
In the case where $S$ is finite with elements $p\_1, p\_2, \ldots p\_n$, the left hand side of this equation is simply the product over $i$ of $p\_i + (1 - p\_i)$, so I'm interested in whether this can be generalised to the infinite case. It's easy to see that the following two statements are equivalent to the above:
If $\inf\_p p^{f(p)} x = 0$ for all $f$, then $x = 0$
If $\sup\_p p^{f(p)} x = x$ for all $f$, then $x = 0$
but I have no idea how to prove either of these.
My reason for asking is that I'm trying to show that, if $\mathcal{H}\_1$ and $\mathcal{H}\_2$ are Hilbert spaces, then if a projection on $\mathcal{H}\_1 \otimes \mathcal{H}\_2$ is of the form $\sup\_i p\_i \otimes q\_i$, with $p\_i$ and $q\_i$ drawn from some complete Boolean algebras of projections on $\mathcal{H}\_1$ and $\mathcal{H}\_2$ respectively, then the $q\_i$ may be chosen to satisfy $q\_i q\_j = 0$ when ever $i \neq j$. So if anybody knows of an alternative way to prove that, or knows that it's false, then by all means say so.
| https://mathoverflow.net/users/7842 | Question about projections on a Hilbert space | Here's a counterexample. For each natural number $n$, let $D\_n$ be the set of those $x\in[0,1]$ whose binary expansion has a 1 in the $n$-th place. Then the operation of multiplication by the characteristic function of $D\_n$ is a projection operator $p\_n$ on $L^2[0,1]$. Let $S$ be the set of these operators $p\_n$; they commute. For any $f:S\to I$, the infimum of the operators $p^{f(p)}$ is zero. For example, if $f$ were identically $+$, then the range of $p\_n$ consists of functions supported on $D\_n$ (up to measure 0, of course), and so the infimum would project to functions supported on the intersection of the $D\_n$, i.e., a one-point set (up to measure 0 again); such functions are 0 in $L^2$. For the general case, where some of the values of $f$ are $-$, you can use the same argument with some of the $D\_n$ replaced by their complements.
| 5 | https://mathoverflow.net/users/6794 | 33352 | 21,629 |
https://mathoverflow.net/questions/33353 | 2 | Suppose $a,b$ are two matrices (arbitrary for now), and I have a function defined on a space of matrices, $T(x) = a x b$. This function is a linear and bounded transform on the a finite dimensional vector space of matrices, so can be represented as a matrix. Say $x$ is $m$ by $n$, then you can write $x$ as column vector of $mn$ entries (row major ordering, lets say), and work out the corresponding matrix representation for $T$, call it $M$.
Now suppose all matrices here are $n$ by $n$. So on the one hand, this is an expensive way to represent $T$, as it would take on the order of $n^6$ operations to apply in this way, vs $2 n^3$, from the definition (two $n$ by $n$ matrix-matrix mults, vs one $n^2$ by $n^2$).
So it seems that, in the space of arbitrary $n^2$ by $n^2$ matrices, there is a subset which can be represented as $x \mapsto a x b$, for some $n$ by $n$ matrices $a,b$.
So I'm wondering if there's generally a name for maps of this form, $x \mapsto a x b$, or if anyone generally has any comments. I know this looks like change of basis, but I'm thinking more generally than that. This may be a silly or ill-posed question, in which case I won't be offended if you say so :).
I'm asking because this shows up in Lagrange interpolation of functions of two (real) variables, and I'd like to know what to call the (c++) function which evaluates the transform, right now I'm calling it 'lagrange\_tensor', but I'm interested generally.
Also it may be nicer to work with the transpose of $b$, so $T = x \mapsto a x b^t$. And if either $a$ or $b$ is the identity, then the matrix $M$ above has many zeros, so that's one way to see why it's an expensive representation. As a side note, the set of operators of this form is not a vector space, as $axb + cxd \not = (a+c)x(b+d).$ Actually, this probably means it's not very interesting and I just answered my own question...
thanks
| https://mathoverflow.net/users/7895 | naming for the map $T = x \mapsto a x b$ | I'd go for the obvious *two-sided multiplication operator*. A Google search shows this has been used indeed.
| 2 | https://mathoverflow.net/users/6101 | 33361 | 21,632 |
https://mathoverflow.net/questions/33366 | 7 | You might think that the title is an overstatement of a well-known fact but it is the best title I can come up with for the wonders the intersection operator does in some fields of math.
Recently,(on summer vacation) I was studying one subject after another and after changing about three subjects, I began to notice that in all these the set-theoretic intersection operator always carried over some property of the parent sets to the one obtained after the intersection.
To summarize briefly:
>
> Let $A$ and $B$ be two sets, say with property P. Then, $A \cap B$ has property P.
>
>
>
Evidences, some trivial:
Topology
--------
* The intersection of two open sets is open.
* The intersection of two closed sets is closed.
* The nonempty intersection of two subspaces of a metric space is a metric space.
......and so forth.
Algebra
-------
* The intersection of two subspaces of a vector space is a vector.
* The intersection of two subgroups of a group is a group(w.r.t the same binary operation and clearly the intersection is between the underlying sets).
* The intersection of two sub-fields of a field is a field.
......the list continues.
The third subject was Graph Theory, but I haven't yet come across the notion of intersection.
Now I would like to ask ~~whether this trend always holds or~~ whether there is some underlying principle each discipline abides by when using the notion of intersection. ~~Is there any property deviating from this trend?~~ What are the reasons for the ubiquity of the quoted property?
| https://mathoverflow.net/users/5627 | The unprecedented success of the “intersection” operator | When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group.
However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or products), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.)
The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets.
To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it.
Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used.
| 15 | https://mathoverflow.net/users/6579 | 33373 | 21,635 |
https://mathoverflow.net/questions/33356 | 1 | Burnside's Lemma / Counting Formula says that the number of orbits of an action is equal to the average number of fixed points of the acting permutations. In my case, I'm particularly interested in the sizes of the orbits themselves for a particular action.
Is there a result as general as Burnside's Lemma but that deals with the sizes of orbits? Or, more specifically, does the following setup seem familiar to anyone?
**Additional Motivation:** The motivation for my question is something like the following.
Suppose we have a set of $m$ sharply transitive permutations $\Pi = \{\pi\_1, \ldots, \pi\_m\}$ on a set $X$. Here I mean that for $x,y \in X$, there is a unique $\pi \in \Pi$ such that $\pi(x) = y$.
The permutations cooperate "nicely" in the following sense. Let $A$ be an $m \times m$ matrix with off-diagonal entries drawn from $\{1, \ldots, m\}$ such that for $i \neq j$ whenever $\pi\_i (x) = \pi\_j (y)$, it follows that $\pi\_j (x) = \pi\_{A(i,j)} (y)$.
Let $G$ be the permutation group generated by $\Pi$. In the general case I'm working on, I really only know $A$, and I'm shooting for a statement of the form: "If $A$ has a certain property, then $|G(x)|$ is divisible by $m$."
The origins of the matrix $A$ while of course essential to proving anything like this statement remain sufficiently messy that I'd rather not get into it.
| https://mathoverflow.net/users/2971 | Does Burnside's Lemma / Counting Formula have a Cousin? | Maybe you're after the [Orbit-Stabiliser Theorem](http://en.wikipedia.org/wiki/Group_action). Let $G$ be a group that acts on a set $X$ and let $x \in X$. Then $|G|=|G\_x||G(x)|$ where $G\_x$ is the stabliser of $x$ in $G$ and $G(x)$ is the orbit of $x$.
| 2 | https://mathoverflow.net/users/2264 | 33376 | 21,637 |
https://mathoverflow.net/questions/33370 | 3 | In derived algebraic geometry there are several different setting,i.e., sometimes we use $E\_{\infty}$ring,somethings we use dg-algebra,... It is for different situations. But could someone give some examples illustrating under what problem we use relevant "derived structure" ($E\_{\infty}$ ring, dg-algebra...)? Some motivations?
| https://mathoverflow.net/users/2391 | different "derived structure" in derived algebraic geometry | Here are the two motivations I know of:
Number 1 comes from algebraic topology. The definitive reference is <http://www.math.harvard.edu/~lurie/papers/survey.pdf> , which explains it much better than I can. Very roughly, complex oriented cohomology theories are represented by E-oo rings and are classified by their group-laws. The group law is what the cohomoly theory does to the tensor product of line bundles. Some interesting group laws come from elliptic curves. So if you are very daring, you can take the moduli space of elliptic curves, and assign to each point the E-oo ring representing the cohomology theory fitting to the group law of the elliptic curve. The result should be a sheaf of E-oo rings on the moduli space of elliptic curves, or in other words a derived structure of E-oo rings.
Number 2 comes from virtual fundamental classes in algebraic geometry. Kontsevich in section 1.4 of <http://arxiv.org/pdf/hep-th/9405035> suggested that for specific types of moduli spaces (those with a perfect obstruction theory) their should exist a derived structure of dg-algebras. The dg-algebra structure should come from local presentations as intersection of submanifolds as discussed in this question:
[Serre intersection formula and derived algebraic geometry?](https://mathoverflow.net/questions/12236/serre-intersection-formula-and-derived-algebraic-geometry) . Kontsevich suggested that via a Riemann-Roch formula you should get the virtual fundamental class needed in Donaldson-Thomas theory and Gromov-Witten Theory.
| 5 | https://mathoverflow.net/users/473 | 33378 | 21,639 |
https://mathoverflow.net/questions/33348 | 7 | Is there anywhere where I can read a complete proof in English of this theorem by Borel and Tits:
>
> Suppose that $G$ is a simple algebraic group over an infinite field $k$, and that $H$ is a subgroup of $G(k)$ containing the subgroup of $G(k)$ generated by the rational points of the unipotent radicals of the $k$-parabolic subgroups, and that $\alpha \colon H \to G'(k')$ is a homomorphism, where $G'$ is a simple algebraic group over an infinite field $k'$, such that $\alpha(G'')$ is Zariski dense in $G'$. Then there exists a homomorphism $\phi\colon k \to k'$, a $k'$-isogeny $\beta\colon G^\phi\to G'$ with $d \beta \ne 0$, and a homomorphism $\gamma\colon H \to Z\_{G'(k')}$ to the centre, all three unique, such that $\alpha(h)=\gamma(h)\beta(\phi^0(h))$ for all $h \in H$.
>
>
>
There is a proof in French in Borel and Tits, *Homomorphismes 'abstraits' de groupes algebriques simples*, **Annals of Mathematics**, Second Series, Vol. 97, No. 3 (May, 1973), pp. 499-571.
| https://mathoverflow.net/users/7909 | theorem of Borel and Tits | This is too long for a comment, but like others who have commented I don't expect to find anything like a "complete proof" of the Borel-Tits theorem written in English. Borel and Tits have each written at times in French, English, German, but their
serious joint work has been in French and is not especially hard to follow. The actual mathematics is difficult, however, requiring a lengthy technical treatment in their 1973 Annals paper on abstract homomorphisms. (Even so, they stopped short of allowing anisotropic groups even while suspecting there were would be some good results in that case; perhaps later work by people like Gopal Prasad sheds light there.)
The underlying question originates with work of Dieudonne, O'Meara, and others on the automorphisms of various classical groups over fields (then more general rings), including compact groups. On the other hand, Steinberg gave in 1960 a unified treatment of the automorphisms of finite Chevalley groups, which I imitated for infinite Chevalley groups in 1967 using more from algebraic groups. What Borel and Tits did was far more comprehensive and sophisticated than any of these concrete investigations. There was a short survey given in French, but also an earlier 1968 preview in English by both authors (not indexed in MathSciNet) which gives a short overview of the method of proof:
*On "abstract" homomorphisms of simple algebraic groups*, pp. 75-82, Proc. of the Bombay Colloquium in Algebraic Geometry, 1968. (This was published in book form for Tata Institute).
As the time lag between 1968 and 1973 suggests, the full proofs took a lot of work but achieved near-definitive results in a reasonably unified framework.
| 11 | https://mathoverflow.net/users/4231 | 33381 | 21,642 |
https://mathoverflow.net/questions/33374 | 9 | I am interested in finding a lower bound of the sum:
$$\sum\_{i=0}^d \left(\genfrac{}{}{0pt}{}{n}{i}\right)
\left(\genfrac{}{}{0pt}{}{m}{k-i}\right)$$
when $d < k$ (and assuming both $n\geq k$, $m\geq k$). When $d=k$ this sum is equal to
$$\left(\genfrac{}{}{0pt}{}{n+m}{k}\right) $$ (the Chu--Vandermonde identity). What I would like to know if there is some good and standard lower bound for the first $d$ terms of this sum in terms of the relation between $d$ and $k$. Is there some standard reference book where I can hope to find such a bound?
| https://mathoverflow.net/users/7917 | Partial sums of the Chu--Vandermonde identity | Carla, I don't think it is a good idea to give a detailed solution here.
My point is that this problem is pretty standard: if you look inside
[N.G. de Bruijn's *Asymptotic Methods in Analysis*](http://books.google.com/books?isbn=0486642216),
especially in the 3rd chapter, you will find that your sum (generically) falls into the
category *c* (p. 54, a comparatively small number of terms somewhere in the middle);
the method is discussed in full details (Section 3.4) on the example of a very similar binomial sum.
| 6 | https://mathoverflow.net/users/4953 | 33382 | 21,643 |
https://mathoverflow.net/questions/33242 | 5 | I have a "continuous" linear programming problem that involves maximizing a linear function over a curved convex space. In typical LP problems, the convex space is a polytope, but in this case the convex space is piecewise curved -- that is, it has faces, edges, and vertices, but the edges aren't straight and the faces aren't flat. Instead of being specified by a finite number of linear inequalities, I have a continuously infinite number. I'm interested in estimating solutions numerically, and my current method is to approximate the surface by a polytope, which means discretizing the continuously infinite number of constraints into a very large finite number of constraints. Unfortunately, typical linear programming algorithms run in something like cubic-time in the number of constraints, so I'm getting a huge performance hit as I make the discretization finer. Firstly, I'm interested to know if this kind of problem has been studied before, and what's been done. Secondly, I'm looking for good strategies for approaching my problem numerically (good LP packages, suggested algorithms, optimizations, etc.).
For concreteness, here is a simplified version of the problem I'm trying to solve:
I have $N$ fixed functions $f\_i:[0,\infty]\to \mathbb{R}$. I want to find $x\_i$ $(i=1,\dots,N)$ that minimize $\sum\_{i=1}^N x\_i f\_i(0)$, subject to the constraints:
$\sum\_{i=1}^N x\_i f\_i(1) = 1$, and
$\sum\_{i=1}^N x\_i f\_i(y) \geq 0$ for all $y>2$
More succinctly, if we define the function $F(y)=\sum\_{i=1}^N x\_i f\_i(y)$, then I want to minimize $F(0)$ subject to the condition that $F(1)=1$, and $F(y)$ is positive on the entire interval $[2,\infty)$. Note that this latter positivity condition is really an infinite number of linear constraints on the $x\_i$'s, one for each $y$. A specific $y\_0$ restricts me to the half-space $F(y\_0) \geq 0$ in the space of $x\_i$'s. As I vary $y\_0$ between 2 and infinity, these half-spaces change continuously, carving out a curved convex shape. The geometry of this shape depends implicitly (and in a complicated way) on the functions $f\_i$.
The reason I suspect there should be an approach that's better than just discretizing the number of constraints is that continuity of the $f\_i$'s implies a kind of local structure on the space of constraints that becomes invisible under discretization. If we sit on the boundary of our convex space (so that at least N constraints are saturated, corresponding to some $y\_k$), and we want to move along the boundary, then generically only those constraints corresponding to small neighborhoods of the $y\_k$ are important. Sometimes when the function $F(y)$ develops a new zero, new $y$ can become important, but this is nongeneric.
NOTE: I asked this question first on stackoverflow.net, and was told it was a nonstandard enough CS problem that I should ask about it here.
| https://mathoverflow.net/users/6996 | Continuous Linear Programming: Estimating a Solution | Even though the $f\_i$ are not polynomials I'll give the answer in that case because it is very nice and it seems like there is some interest. I have to stress in advance though that the answer exploits the resulting algebraic structure in a fundamental way, and so is unlikely to extend to the case when the $f\_i$ are not polynomials.
First of all, a semidefinite program (SDP) is an optimization problem with matrix variables, linear objective, and positive semidefiniteness constraints on symmetric (real) matrices in addition to the standard linear (in)equalities allowed in linear programming. They are a generalization of linear programs (LP) and are vastly more expressive. LPs are the case when the matrices are constrained to be diagonal. SDP can also be viewed as a noncommutative version of LP.
The relationship with semi-infinite programming suggested by Gilead is I think the fact that one can view the constraint "A is positive semidefinite" as $x^TAx\geq 0$ for all $x$, which is an infinite number of constraints. On the other hand, one can view any convex constraint in this way, because any closed convex set can be described by (infinitely many) linear inequalities.
Theoretically SDPs are important both because many problems can be written as SDPs and because they can be solved using interior point methods in polynomial time, almost as efficiently as LPs in theory. In practice, the technology is much newer than that for LPs so one cannot solve SDPs which are nearly as big using off-the-shelf software, but those days seem to be getting closer.
To see how to turn your problem into an SDP if the $f\_i$ were polynomials, let $x\_i$ be your decision variables. Note that $f\_i(1)$ is just a constant, so your first constraint is just a linear equation on the $x\_i$ and that is no problem in an SDP. For the others, we need to do a little work.
Let $H$ denote the operator which sends a symmetric matrix to its sums along antidiagonals, so
$H: \begin{bmatrix}a & b \\\\ b & c \end{bmatrix}\mapsto\begin{bmatrix}a & 2b & c\end{bmatrix}$, and so on for bigger matrices. If we identify polynomials with their sequences of coefficients, then $p = q^2$ as polynomials if and only if $p = H(qq^T)$ as vectors. Therefore $p$ is a sum of squares (SOS) if and only if $p = H(Q)$ for a positive semidefinite $Q$ (any such $Q$ is the sum of matrices of the form $qq^T$). Now, if a polynomial $p$ is SOS then it is automatically nonnegative everywhere. Conversely, one can show than any univariate nonnegative polynomial is a sum of squares. This gives us an exact characterization of nonnegative polynomials in terms of positive semidefinite matrices. Thinking of the matrix $Q$ as a new decision variable, we can write the constraint "$p$ is nonnegative" in a semidefinite program; the solver will find a $Q$ which certifies this.
Similarly, a polynomial $p$ is nonnegative on an interval $[w,\infty)$ if and only if it is of the form $p(x) = SOS\_1(x) + (x-w)\cdot SOS\_2(x)$ for some SOS polynomials $SOS\_i$. Again, one direction is obvious and the other requires a little effort (write $p$ in factored form and group the factors cleverly). Therefore we can also write nonnegativity on an interval in terms of positive semidefinite matrices, and hence use it as a constraint in an SDP.
Now note that when we use $H$ to define the constraint for a polynomial $p$ to be SOS, we are writing linear equality constraints between the coefficients of $p$ and some linear functionals of the matrix inside $H$. Similarly for when we write the constraint that $p$ is nonnegative on an interval: it is a linear equality between some decision variables, plus the constraint that certain matrices (the ones defining the SOS polynomials) are positive semidefinite.
Until now we've been thinking of $p$ as a constant polynomial. But because arbitrary linear equalities between decision variables are allowed in an SDP, we can just as easily write the constraint "$\sum\_{i=1}^N x\_i f\_i(y)$ is nonnegative for $y\in[2,\infty)$" using this method, because our polynomial in question has coefficients which are linear in the decision variables.
Putting this all together gives a semidefinite program which would express exactly what you want in the case that the $f\_i$ are polynomials. One could then either find a feasible $x\_i$, prove that none exists, or optimize a linear functional of the $x\_i$ all in polynomial time. Unfortunately because of the way we have used the algebraic structure of the problem, this is unlikely to extend to non-polynomial $f\_i$.
Finally, I should note that if you're interested in this sort of thing I highly recommend checking out my advisor Pablo Parrilo's course notes on MIT OpenCourseWare. You can find the link on his website.
| 6 | https://mathoverflow.net/users/5963 | 33391 | 21,649 |
https://mathoverflow.net/questions/33389 | 2 | Consider Schrödinger's *time-independent* equation
$$
-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi.
$$
In typical examples, the potential $V(x)$ has discontinuities, called *potential jumps*.
Outside these discontinuities of the potential, the wave function is required to be twice differentiable in order to solve Schrödinger's equation.
In order to control what happens at the discontinuities of $V$ the following assumption seems to be standard (see, for instance, Keith Hannabus' *An Introduction to Quantum Theory*):
>
> **Assumption**: The wave function and its derivative are continuous at a potential jump.
>
>
>
**Questions**:
1) Why is it necessary for a (physically meaningful) solution to fulfill this condition?
2) Why is it, on the other hand, okay to abandon twofold differentiability?
Edit: One thing that just became clear to me is that the above assumption garanties for a well-defined probability/particle current.
| https://mathoverflow.net/users/1291 | Justification for the matching condition for the wave function at potential jumps. Why is it both restrictive enough and sufficiently general? | To answer your first question:
Actually the assumption is *not* that the wave function and its derivative are continuous. That follows from the Schrödinger equation once you make the assumption that the probability amplitude $\langle \psi|\psi\rangle$ remains finite. That is the physical assumption. This is discussed in Chapter 1 of the first volume of *Quantum mechanics* by Cohen-Tannoudji, Diu and Laloe, for example. (Google books only has the second volume in English, it seems.)
More generally, you may have potentials which are distributional, in which case the wave function may still be continuous, but not even once-differentiable.
To answer your second question:
Once you deduce that the wave function is continuous, the equation itself tells you that the wave function cannot be twice differentiable, since the second derivative is given in terms of the potential, and this is not continuous.
| 8 | https://mathoverflow.net/users/394 | 33398 | 21,654 |
https://mathoverflow.net/questions/33035 | 3 | Let V be a vector space over Z/2, and let X be a subset of V. Is there an algorithm to find the largest possible subspace of V which doesn't intersect X? Is it NP complete?
| https://mathoverflow.net/users/5463 | Largest subspace that doesn't intersect a given set | The problem is NP hard. Here's a reduction to it from 4-colorability. Given a graph with vertex set $G$ [not $V$ because that's supposed to be a vector space] and edge set $E$, form a vector space $V$ over $\mathbb{Z}/2$ having $G$ as a basis. Identify each edge $e$ with the vector that is the difference of the two endpoints of $e$, and let $S$ be the set of these edges-qua-vectors. [I know that "difference" is the same as "sum" here, but it helps to think of the edges as differences.] Then each of the following statements is easily equivalent to the next, for any fixed natural number $t$. [In the end, I'll only need the case $t=2$.]
(1) $V$ has a subspace of codimension at most $t$ that misses $S$.
(2) There are $t$ linear functionals $f:V\to\mathbb{Z}/2$ such that each edge $e$ is sent to 1 by at least one of these functionals.
(3) There are $t$ linear functionals $f:V\to\mathbb{Z}/2$ such that, for each edge $e$, at least one of these functionals takes different values at the two endpoints of $e$.
(4) There are $t$ functions $g:G\to\mathbb{Z}/2$ such that, for each edge $e$, at least one of these functions takes different values at the two endpoints of $e$.
(5) There is a function $h$ from $G$ to $(\mathbb{Z}/2)^t$ taking different values at the two ends of each edge.
(6) The graph $(G,E)$ is $2^t$-colorable.
In particular, $V$ has a codimension-2 subspace missing $S$ if and only if $G$ is 4-colorable. Since 4-colorability is known to be an NP-complete problem, the vector subspace problem is also NP-hard.
I believe the "correct" generality for this idea is what's called the critical problem for matroids. What I've presented is the special case of graphic matroids.
| 6 | https://mathoverflow.net/users/6794 | 33407 | 21,657 |
https://mathoverflow.net/questions/33360 | 1 | Let $T$ be a torus defined over a field $K$ of characteristic $p>0$. Suppose that $T$ acts (algebraically) on some vector space $V$ (over the same field $K$). Let $W$ be a subspace of $V$. Now consider $N\_T(W)$, the normalizer in $T$ of $W$. I would like to be sure that this is a closed connected subgroup of $T$; i.e. I'd like it to be a torus. Is this true in general?
Here's a more specific question (supposing that the first question is too hard, or is not true in general). Consider $G=UT$ a solvable linear algebraic subgroup of $GL\_r$. Suppose that $U\_1$ is some subgroup of $U$. I'd like to be sure that $N\_T(U\_1)$ is a subtorus of $T$. I'm most interested in the situation when $p$ is large (I can imagine that this statement may only be true for $p>r$, say, but this is fine for the application I have in mind.)
| https://mathoverflow.net/users/801 | Tori acting on vector spaces | Your more specific question is not actually more specific. $\mathbb{G}\_m^n \ltimes \mathbb{G}\_a^n$ embeds in $GL\_{2n}$ as $\left( \begin{smallmatrix} T & U \\ 0 & 1 \end{smallmatrix} \right)$, where $U$ is embedded along the diagonal. If $U$ is any $T$-rep, we can embed $T \ltimes U$ into $\mathbb{G}\_m^n \ltimes \mathbb{G}\_a^n$ for $n$ large enough (after a possible extension of the base field, see discussion below). And, of course, every vector subspace of $U$ is a subgroup. So all of the examples BConrad and I bring up will occur in examples of the form you discuss.
You can ensure that $N\_T(U)$ is smooth for $p$ large: the equations defining $N\_T(U)$ will all be of the form $\chi\_i(t)=1$ for various characters $\chi\_i$ of $T$. Once $p$ gets up above the torsion in $\mathrm{Char}(T)/\langle \chi\_1, \chi\_2, \ldots, \chi\_N \rangle$, this will be smooth. But, as my example points out, you can't force this to be connected.
| 3 | https://mathoverflow.net/users/297 | 33412 | 21,659 |
https://mathoverflow.net/questions/33385 | 7 | Is there a functor that preserves all small limits but not a large one?
| https://mathoverflow.net/users/6414 | Preservation of limits | Let's try this. I'll use colimits, so take the opposite.
The class of all ordinals is ordered. Add one more element $\infty$ at the end, bigger than all of them. View this "large ordered set" as a large category $\cal C$. A small diagram in $\cal C$ has colimit $\infty$ if $\infty$ occurs in the diagram, and otherwise it has a colimit less than $\infty$. But the large diagram consisting of everything except $\infty$ has colimit $\infty$. The functor to the ordered set $\lbrace 0<1\rbrace$ that sends $\infty$ to $1$ and everything else to $0$ preserves small colimits but not all colimits.
| 18 | https://mathoverflow.net/users/6666 | 33414 | 21,660 |
https://mathoverflow.net/questions/32894 | 10 | Does every 4-regular graph contain a cycle of length (number of edges in the cycle) $1 (\mod3)$? Are there only finitely many exceptions?
I suspect such cycles exist for most 3-regular graphs but 4-regularity is enough for what I'm investigating.
| https://mathoverflow.net/users/2384 | Cycles of length 1(mod 3) in regular graphs | According to [this paper](http://dx.doi.org/10.1016/S0166-218X%2801%2900190-1), N. Dean et al. have shown that if a simple graph *G*
* is 2-connected,
* has minimum degree at least 3, and
* is not isomorphic to the Petersen graph,
then *G* contains a cycle of length 1 mod 3.
Now consider any 4-regular graph *H*. If I'm not mistaken, it's fairly easy to show that *H* contains a subgraph *G* such that two nodes of *G* have degree 3, all other nodes of *G* have degree 4, and *G* is 2-connected. Clearly *G* can't be the Petersen graph, and thus the above theorem implies that *G* (and therefore also *H*) contains a cycle of length 1 mod 3.
**Edit:** Here is how would construct *G* given *H*. Recall that *H* has a 2-factorisation, i.e., it consists of cycles such that each node of *H* is "covered" by exactly 2 cycles. If there are no articulation points in *H*, the claim is clear. Otherwise consider the [block tree](http://en.wikipedia.org/wiki/Biconnected_component) of *H* and pick a biconnected component *K* that contains only one articulation point; let *x* be the articulation point. Consider the subgraph *K*' = *K* − *x*; this graph consists of one path *P* and a set of cycles, and each node of *K*' is covered by either *P* and one cycle or exactly 2 cycles. It may be the case that *K*' contains articulation points. However, each articulation point is an endpoint of a bridge, and each bridge is an edge of *P*. Remove all bridges and let *G* be one of the connected components in the resulting bridgeless graph.
| 7 | https://mathoverflow.net/users/7170 | 33423 | 21,665 |
https://mathoverflow.net/questions/33427 | 15 | Lately my studies have been focusing on learning the machinery of K-Theory, and I thought that learning the Atiyah-Singer Index Theorem would be a good way to see K-Theory in action a bit and to learn a deep result on the way. From what I have read, there are a few methods of proof of Atiyah-Singer, one of which uses K-Theory. Also from what I have read, it seems that I have most of the background knowledge to approach the proof of Atiyah-Singer.
However, it doesn't seem that there is a standard reference or sequence of references to go to in order to learn the proof of this theorem. In particular, I am not sure which book(s) would be best to look at to see a proof of Atiyah-Singer which utilizes K-Theory. I have found a few that seem to take the K-Theory approach to the theorem, but I have no way of telling how good or useful they are.
So what I am asking for is a roadmap or a reference to a proof of the Atiyah-Singer Index Theorem that uses K-Theory, and of course other advice concerning learning this theorem is welcome as well.
| https://mathoverflow.net/users/3664 | Roadmap to a proof of the Atiyah-Singer Index Theorem which uses K-Theory | The original paper by Atiyah and SInger at <http://www.jstor.org/stable/1970715> is as good as anything
| 16 | https://mathoverflow.net/users/51 | 33428 | 21,667 |
https://mathoverflow.net/questions/30877 | 6 | If I take $\mathcal{A}$ = coherent sheaves on $X$\* up to isomorphism, then there are two things I could do which come to mind.
The first is noticing that $(\mathcal{A},\oplus)$ is a monoid and subsequently applying Grothendieck's $K$-functor to it obtaining a group $K(\mathcal{A})$.
The second is to take the free abelian group generated by $\mathcal{A}$ quotiented out by relations
$B = A + C$ for any exact sequence
$$0 \to A \to B \to C \to 0$$
obtaining a second group $L(\mathcal{A})$.
Are these two groups $K(\mathcal{A})$ and $L(\mathcal{A})$ isomorphic?
If not, which one is the 'correct' one?
\*I'm not sure what $X$ should stand for. I'd be happy to assume smooth and projective variety over $\mathbb{C}$, or maybe just locally noetherian scheme, or maybe just locally ringed space or etc.
| https://mathoverflow.net/users/3701 | How is K-theory defined for coherent sheaves? | The Grothendieck group on $X$ is defined (in your notation) as $K(\mathcal{B})$, where $\mathcal{B}$ is the monoid of f.g. locally free sheaves on $X$, (i.e. vector bundles on $X$.) The group $L(\mathcal{A})$ is usually denoted as $G(X)$, or $K'(X)$. If $X$ is smooth, the embedding $\mathcal{B} \subset \mathcal{A}$ induces an isomorphism from $K(\mathcal{B})$ to $L(\mathcal{A})$ (i.e. from $K\_0(X)$ to $G(X)$.)
Chuck Weibel's notes for his K-theory book (which can be found on his website) is a good place to find some of this information.
| 6 | https://mathoverflow.net/users/339 | 33444 | 21,676 |
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