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https://mathoverflow.net/questions/25642 | 7 | It's a big, famous, hard problem in operator algebras to determine if the von Neumann algebras $L(F\_2)$ and $L(F\_3)$ are isomorphic, or not. Here $F\_n$ is the free group on n generators and $L(F\_n)$ is the weak-operator-topology closure of the group algebra $\mathbb C[F\_n]$ acting naturally on the Hilbert space $\ell^2(F\_n)$.
I presume it must be known if the algebras $\mathbb C[F\_2]$ and $\mathbb C[F\_3]$ are isomorphic or not. But from casually asking a few algebraists, I've never had any luck in finding this out (I admit to not working very hard on this!) I'm guessing some (co)homology theories must help...? What about for replacing $\mathbb C$ by a more general ring?
| https://mathoverflow.net/users/406 | Telling group algebras apart | Well, yes. Imagine that you have an algebra $A$ over $\mathbb{C}$ and you want to find out whether it is $\mathbb{C}[F\_2]$ or $\mathbb{C}[F\_3]$. Pick any one-dimensional $A$-module $M$ and compute $\operatorname{Ext}^1\_A(M,M)$. If $A=\mathbb{C}[F\_2]$, you'll get a $2$-dimensional vector space over $\mathbb{C}$, while if $A=\mathbb{C}[F\_3]$, you'll get a $3$-dimensional vector space.
| 10 | https://mathoverflow.net/users/2106 | 25653 | 16,811 |
https://mathoverflow.net/questions/25592 | 49 | While trying to get some perspective on the extensive literature about highest weight modules for affine Lie algebras relative to "level" (work by Feigin, E. Frenkel, Gaitsgory, Kac, ....), I run into the notion of *dual Coxeter number* but am uncertain about the extent of its influence in Lie theory. The term was probably introduced by Victor Kac and is often denoted by $h^\vee$ (sometimes by $g$ or another symbol). It occurs for example in the 1990 third edition of his book *Infinite Dimensional Lie Algebras* in Section 6.1. (The first edition goes back to 1983.) It also occurs a lot in the mathematical physics literature related to representations of affine Lie algebras. And it occurs in a 2009 paper by D. Panyushev in *Advances* which studies the structure of complex simple Lie algebras.
>
> Where in Lie theory does the dual Coxeter number play a natural role (and why)?
>
>
>
A further question is whether it would be more accurate historically to refer instead to the *Kac number of a root system*, since the definition of $h^\vee$ is not directly related to the work of Coxeter in group theory.
BACKGROUND: To recall briefly where the *Coxeter number* $h$ comes from, it was introduced by Coxeter and later given its current name (by Bourbaki?). Coxeter was studying a finite reflection group $W$ acting irreducibly on a real Euclidean space of dimension $n$: Weyl groups of root systems belonging to simple complex Lie algebras (types $A--G$), these being crystallographic, together with the remaining dihedral groups and two others. The product of the $n$ canonical generators of $W$ has order $h$, well-defined because the Coxeter graph is a tree. Its eigenvalues are powers of a primitive $h$th root of 1 (the "exponents"): $1=m\_1 \leq \dots \leq m\_n = h-1$. Moreover, the $d\_i = m\_i+1$ are the degrees of fundamental polynomial invariants of $W$ and have product $|W|$.
In the Weyl group case, where there is an irreducible root system (but types $B\_n, C\_n$ yield the same $W$), work of several people including Kostant led to the fact that $h$ is 1 plus the sum of coefficients of the highest root relative to a basis of simple roots. On the other hand, the dual Coxeter number is 1 plus the sum of coefficients of the highest *short* root of the dual root system. For respective types $B\_n, C\_n, F\_4, G\_2$, the resulting values of $h, h^\vee$ are then $2n, 2n, 12, 6$ and $2n-1, n+1, 9,4$. This gets pretty far from Coxeter's framework.
One place where $h^\vee$ clearly plays an essential role is in the study of a highest weight module for an affine Lie algebra, where the canonical central element $c$ acts by a scalar (the *level* or *central charge*). The "critical" level $-h^\vee$ has been especially challenging, since here the theory seems to resemble the characteristic $p$ situation rather than the classical one.
| https://mathoverflow.net/users/4231 | What role does the "dual Coxeter number" play in Lie theory (and should it be called the "Kac number")? | The dual Coxeter number comes up naturally as a normalization factor for invariant bilinear forms on the Lie algebra: according to Kac's book which you quote, $2h^{\vee}$ is the ratio between the Killing form and the "minimal" bilnear form (the trace form for $sl\_n$), which has the property that the square of the length of the maximal root is 2.
This minimal form corresponds to the minimal affine Kac-Moody group corresponding to the Lie algebra, or equivalently to the minimal line bundle on the affine Grassmannian or the moduli spaces of G-bundles on curves (the generator of the Picard group). As a result, the $-2h^\vee$-th power of the basic ample line bundle on the Grassmannian or moduli space of bundles (which is associated to the level given by the Killing form) ends up being identified with the canonical line bundle, and in particular the $h^\vee$th power is a square-root of the canonical bundle, or spin structure. (This is analogous to the role of $\rho$ for the finite flag variety.)
Thus the critical level arises naturally geometrically -- it corresponds to half-forms on the Grassmannian/moduli spaces. The basic yoga of quantization (or of unitary/normalized induction of representations) tells us that classical symmetries are "shifted" by half-forms - cf $\rho$-shifts in representation theory. Likewise the critical shift for affine algebras.. for example the Feigin-Frenkel theorem is the analogue of the Harish-Chandra isomorphism: the center of the enveloping algebra at critical level (rather than level 0 as one might naively guess, ignoring half-form twists) is isomorphic to the algebra of invariant polynomials on the (dual of the) Lie algebra. (This can be said more canonically keeping track of symmetries of change of variable, magic word being "opers", but let's ignore that).
One can say all this very naturally algebraically (without resorting to geometry) -- $\rho$ can be described as the square root of the modular character of the Borel subalgebra (up to sign or something, not being very careful here). The critical level has a similar description in terms of the positive half (Taylor series part) of the Kac-Moody algebra - if you try to define the modular character of this half you are quickly led to semiinfinite determinants etc, ie to the previous geometric story, and so one can assert that the critical level "is" half the modular character of the positive loop subalgebra.
| 26 | https://mathoverflow.net/users/582 | 25680 | 16,827 |
https://mathoverflow.net/questions/25655 | 15 | Let $M$ be a module over a ring $R$. In nice situations (though I don't know what exactly nice means...) the following two numbers are equal:
1.) The codimension of the support of $M$
2.) The biggest $k$ such that $\text{Ext}^k(M,.)$ doesn't vanish
Why do we expect this intuitively? Why should lengths of injective/projective/flat resolutions have anything to do with the support?
| https://mathoverflow.net/users/2837 | Why do modules with small support have high Exts? | To understand what "nice" is in your sense has been a very interesting question in commutative algebra.
In the following discussion I will assume, unless otherwise notice, that $(R,m,k)$ is Noetherian local, and $M$ is finitely generated. Let $(1)$ be the codimension of support of
$M$ and $(2)$ be the biggest non-vanishing index of $\text{Ext}(M,-)$.
First, the number (2) is finite forces $M$ to have finite projective dimension by taking $N=k$ the residue field. So we will assume $\text{pd}\ M <\infty$. Then, as BCrd pointed out:
$$ (2) = \text{pd} \ M = \text{depth} \ R - \text{depth} \ M$$
The first inequality is easy by computing Ext via a projective res. of M + Nakayama lemma. The second is the Auslander-Buchsbaum theorem.
On the other hand:
$$(1) = \text{dim} \ R - \text{dim} \ M $$
So
$$(1) - (2) = (\text{dim} \ R - \text{depth} \ R) - (\text{dim} \ M-\text{depth} \ M) $$
Thus, if both $R,M$ are Cohen-Macaulay (which by def. means dim=depth) and $\text{pd}\ M <\infty$ then $(1) = (2)$. If $R$ is "more Cohen-Macaulay" then $M$, we will have $(1)<(2)$.
The situation described in Emerton's answer is also very interesting. In general, the smallest index for which $\text{Ext}^i(M,N) \neq 0$ is the biggest length of an $N$-regular sequence inside the annihilator of $M$. When $N=R$ this number is called the *grade* of $M$, which I will call (3).
It is easy to see that $(3) \leq (1)$ in general. One can prove that $(1) = (3)$ if $R$ is Gorenstein as follows: By Local Duality, $\text{Ext}^i(M,R)$ is Matlis dual to the local cohomology module $\text{H}\_{m}^{d-i}(M)$, here $d= \text{dim}\ R$. It is not hard to show that local cohomology vanish beyond $\text{dim}\ M$, QED.
Amazingly, it has been an open conjecture for 50 years that $(1)=(3)$ whenever $M$ has finite projective dimension!
For "intuitive" understanding, I would offer the following: often when study modules of finite projective dimension one draw inspirations from those of the form $R$ modulo a regular sequence (so the resolution is a Koszul complex). In such case one can easily see that $(1) = (2) =(3)$.
EDIT: I got too caught up in the results and forgot your main question: why bigger codimension implies bigger projective resolution? A very low-tech way to see it is: bigger codimension means bigger annihilator of $M$. Now each element of the annihilator of $M$ gives a non-trivial relation on elements of $M$, namely $ax=0$, so it is not surprising that the modules with bigger annihilators have more complicated resolutions.
| 14 | https://mathoverflow.net/users/2083 | 25681 | 16,828 |
https://mathoverflow.net/questions/25674 | 3 | I would like to find those integers $x,y$ that satisfies $y^2=x^3+1$. Is there some elementary way to find those?
| https://mathoverflow.net/users/6266 | Integer points of an elliptic curve | I'm not sure the following counts as "elementary" but it's certainly not too difficult.
First show your curve has rank 0. To do this, I asked SAGE, but it's not so hard to do by hand. If you rewrite your curve $C$ as $y^2 = x^3-3x^2+3x$ by renaming $x+1$ as $x$,
then you can write down the 2-isogeneous curve $\overline{C}:y^2=x^3+6x^2-3x$. You are now in the setup to apply the algorithm discussed in Ch. III.6 of the book "Rational Points on Elliptic Curves" by Silverman-Tate. (Which book I highly recommend for learning how to solve this sort of problem.)
The upshot is just that $C$ has rank 0, which implies by general theory (Nagell-Lutz) that all its rational points are integral. Moreover, the Nagell-Lutz theorem even says what possible y-coordinates can occur: either $y$ is zero (for points of order 2) or else $y$ divides the discriminant. In your case this says that for integral points, $y=0$ or $y$
divides $-27$. The rest is easy.
| 8 | https://mathoverflow.net/users/412 | 25690 | 16,834 |
https://mathoverflow.net/questions/25664 | 6 | What would be the consequence of requiring that any choice function be computable; i.e. using as the foundational basis ZF + ACC? Does it make a difference if we admit definable functions?
I guess I am sometimes bothered by the thought that any random choice over an uncountable set by definition would seem to almost certainly return a non-computable member. This seems impractical and perhaps even problematic, considering that major branches of mathematics such as for example analysis, with only few notable exceptions, mainly operate within the computable or definable realm.
Presumably an immediate consequence would be that the Banach–Tarski paradox and similar theorems related to unmeasurable sets would fail. But would there be more fundamental consequences?
| https://mathoverflow.net/users/1320 | Axiom of Computable Choice versus Axiom of Choice | The Axiom of Choice is a principle that applies to arbitrary families of arbitrary sets, and this is a realm where the concept of recusive functions or Turing Turing computability simply does not apply. For example, mathematicians may use AC to select elements of subsets of a (possibly uncountable dimension) vector space in order to form a basis---you iteratively pick a vector outside the span of what you have so far---and it simply doesn't make sense in this generality for such a function to be recursive or equivalent to a recursive function. This is why people were objecting to your question in the comments.
But let me try to make some sense of the question. Suppose we restrict our choice principle to families of sets that each have at least one computable member. Then, I claim that there is a definable choice function, for we may select from each set in the family the computable member that is computed by the smallest program. Thus, this formulation of what you might mean by ACC is simply provable in ZF. But this function will not in general be itself computable (it merely selects computable members, and this is not the same), and indeed, since the domain of the function is not necessarily $N$, the concept of this function being computable isn't always sensible.
If you intend to have a version of ACC that only applies to countably-indexed families $\langle A\_n| n\in N\rangle$, with each $A\_n$ a set containing subsets of $N$, at least one of which is computable, then it is provable in ZF that we cannot insist there is always a computable function $f$ such that $f(n)$ is a program computing a member of $A\_n$. The reason is that there are only countably many such functions $f$, and we may easily diagonalize against them to produce a bad sequence $\langle A\_n | n\in N\rangle$.
If you replace *computable* with *definable*, then your principle has a much better chance. One subtle issue, however, is that "being definable" is not first order expressible in set theory, so you cannot state your ACC principle that way. Rather, one can use "ordinal-definable", which is expressible. The principle that every set contains an ordinal-definable element is equivalent (by a previous MO question) to the set-theoretic principle $V=HOD$. And this principle implies AC, in the form that under V=HOD, every family of nonempty sets admits an ordinal definable choice function.
One can also impose more restrictive versisons of definability on the choice functions or on the families. For example, perhaps one wants principles that only apply to sets of reals, and one wants to know when there are, say, projective choice functions. This phenomenon is called *uniformization*, and is extensively studied in descriptive set theory.
So there are a variety of ways to make sense of your question, and these different interpretations give different answers. So it all depends on what you mean.
| 14 | https://mathoverflow.net/users/1946 | 25694 | 16,837 |
https://mathoverflow.net/questions/25534 | 20 | Consider the following optimization problem:
**Problem:** find a monic polynomial $p(x)$ of degree $n$ which minimizes $\max\_{x \in [-1,1]} |p(x)|$.
The solution is given by Chebyshev polynomials:
**Theorem:** Let $T\_n(x) = cos (n \cdot cos^{-1} x)$. Then $(1/2^{n-1}) T\_n$ is a monic polynomial of degree $n$ which achieves the above minimum.
The proof of this fact is short and surprisingly free of messy calculations. From the definition of $T\_n$ you derive a recurrence relation expressing $T\_n$ in terms of $T\_{n-1}, T\_{n-2}$, which shows that $T\_n$ are indeed polynomials. Then
you argue that $(1/2^{n-1}) T\_n$ is monic and achieves its extrema $\pm 1/2^{n-1}$ at least $n+1$ times in $[-1,1]$, from which the above theorem easily follows. If you'd like a nice exposition of this argument which does not skip any steps, [this](http://www.johndcook.com/ChebyshevPolynomials.pdf) is short and clear.
However, I don't get much enlightenment from this proof: it feels pulled out of a
hat. For example, it gives me no clue about which other polynomial optimization problems
have similar solutions.
**My question:** Is there a natural and motivated sequence of steps which, starting from the above optimization problem, leads to Chebyshev polynomials?
**Update:** I changed the title to better reflect the question I am asking.
| https://mathoverflow.net/users/1407 | Is there an intuitive explanation for an extremal property of Chebyshev polynomials? | Well, let's try to avoid the hat.
Consider the dual (and obviously equivalent) problem: find the polynomial $p(x):[-1,1]\rightarrow [-1,1]$ of degree $n$ with the greatest possible leading coefficient. We have some information on values of $p$, and need something about its coefficient. Let's try Lagrange's interpolation. Take some $n+1$ values $t\_1 < t\_2 < \dots < t\_{n+1}$ from $[-1,1]$ and write down (for $u(x)=(x-t\_1)\dots(x-t\_{n+1})$) the formula
$$
p(x)=\sum p(t\_i) \frac{u(x)/(x-t\_i)}{u'(t\_i)}.
$$
Then take a look on coefficient of $x^n$. It equals
$$
\sum \frac{p(t\_i)}{u'(t\_i)}.
$$
We know that $|p(t\_i)|\leq 1$, so the leading coefficient does not exceed
$
\sum 1/|u'(t\_i)|.
$
Ok, when does equality occur? The answer is: $p$ should take values $(-1)^{n-i+1}$ in $t\_i$. That is, we have to find a polynomial of degree $n$ with $n+1$ extremal values $\pm 1$ on $[-1,1]$. This may hold only if $t\_1=-1$, $t\_{n+1}=1$, and $t\_2$, $\dots$, $t\_n$ are roots of $p'$. So, $1-p^2(x)$ should be divisible by $(1-x^2)p'(x)$. Hereon the trigonometic substitution $x=\cos t$, $p=\cos f$ is very natural, as we know that $1-f^2$ is divisible by $f'$ for $f=\cos$. So we invent Chebyshev's polynomials.
Also, it is seen from Lagrange formula that they are extremal in many other problems with restrictions $|p(x)|\leq 1$ on $[-1,1]$. For example, the value in each specific point $x\_0>1$ is maximized also for Chebyshev polynomial, it is proved by exactly the same way.
| 9 | https://mathoverflow.net/users/4312 | 25697 | 16,839 |
https://mathoverflow.net/questions/25637 | 2 | Does anybody know a translation of [Föllmer: Calcul d'Ito sans probabilités](http://archive.numdam.org/ARCHIVE/SPS/SPS_1981__15_/SPS_1981__15__143_0/SPS_1981__15__143_0.pdf) in English or German?
It seems to be a very interesting text - Abstract: "It is shown that if a deterministic continuous curve has a 'quadratic variation' in a suitable sense (which however depends explicitly on a nested sequence of time subdivisions, for example the standard dyadic one), then it satisfies a deterministic 'Ito formula' when composed with a twice differentiable function. Thus the only place where probability really appears in the derivation of Ito's formula is in the fact that, given any sequence of subdivisions, almost every path of a semimartingale admits a quadratic variation relative to this sequence (though no path may exist which has a quadratic variation relative to all sequences)"
See also: [Convergence and non-convergence of left-point and mid-point Riemann sums](https://mathoverflow.net/questions/16664/convergence-and-non-convergence-of-left-point-and-mid-point-riemann-sums) and first answer.
Other references along the same lines would also be appreciated - Thank you!
| https://mathoverflow.net/users/1047 | Föllmer: "Calcul d'Ito sans probabilités" in English or German? | Föllmer's approach was mainly adopted by specialists in Mathematical Finance.
Have a look at [*Introduction to Stochastic Calculus for Finance*](http://books.google.co.uk/books?id=KKLa1j-diXwC&printsec=frontcover&dq=Introduction+to+Stochastic+Calculus+for+Finance%3A+A+New+Didactic+Approach&source=bl&ots=-L3eKT8OjQ&sig=3Md3B6gg4LLy2Uzzzy8fs37Rwqo&hl=en&ei=4HT5S6P-OI6Q4gbxh9AI&sa=X&oi=book_result&ct=result&resnum=5&ved=0CC8Q6AEwBA#v=onepage&q&f=false) by D. Sondermann. This is an intro lecture course based on the work of Föllmer. And the book contains an English translation
of the original article by Föllmer in the Appendix.
| 6 | https://mathoverflow.net/users/5371 | 25706 | 16,847 |
https://mathoverflow.net/questions/25707 | 8 | If $p$ is a prime of the form $4n+3$, the class number $h$ of $Q[\sqrt{-p}]$ can be expressed using the number $V$ of quadratic residues and $N$ nonresidues in the interval $[1,\frac{p-1}{2}]$:
* If $p=8n+7$ then $h=V-N$
* If $p=8n+3$ then $h=\frac{1}{3}(V-N)$
This result seems so simple and elegant, but its proof (which I saw in Number Theory by Borevich & Shafarevich - p. 346), while definitely beautiful, is not very short and is based on the analytic class number formula. And so I didn't feel that the proof really helped "demystify" the result.
This leads me to ask: Can someone see intuition for this result? Any short heuristic argument that would lead to it? Any obvious meaning of $h=V-N$? In short, "why" is it true?
| https://mathoverflow.net/users/5860 | Intuition for a formula that expresses the class number of an imaginary quadratic field by counting quadratic residues | I'll have to be brief; I can think of two reasons "why":
1. Cauchy and Jacobi proved that for a prime ideal ${\mathfrak p}$ in a complex quadratic number field with prime discriminant, the h-th power of ${\mathfrak p}$ (with $h$ as in your question) is principal. Their technique was what we nowadays know as the Stickelberger ideal.
You should find references in Ireland-Rosen.
2. Venkov, in the 1920s, gave an arithmetic proof of a large part of Dirichlet's class number formulas based on Gauss's work on ternary quadratic forms and the arithmetic of quaternions. There are modern accounts floating around, and I remember the names Rehm and Shemanske in this connection (if google doesn't help, I'll provide you with references when I'm back from the holidays -). This approach is probably more involved than Dirichlet's, so I don't know whether it explains anything.
EDIT: Let me first give you the references for 2.
* B.A. Venkov, *On the arithmetic of quaternion algebras* (Russ.),
Izv. Akad. Nauk (1922), 205--220, 221--246; ibid. (1929), 489--509, 532--562, 607--622
* H.P. Rehm, *On a theorem of Gauss concerning the number of solutions
of the equation $x^2 + y^2 + z^2 = m$*, in "Selected
topics on ternary forms and norms" (O. Taussky, ed.), Dekker 1976
* Th.R. Shemanske, *Representations of ternary quadratic forms and the
class number of imaginary quadratic fields*,
Pac. J. Math. {\bf 122} (1986), 223--250
As for 1, let $K/{\mathbb Q}$ be a finite abelian extension with Galois group $G$
and conductor $m$. Let $\sigma\_a$ denote the restriction of the automorphism
$\zeta\_m \to \zeta\_m^a$ of ${\mathbb Q}(\zeta\_m)$ to $K$. Then
$$ \theta(K) = \frac{1}{m} \sum a\sigma\_a^{-1} \in {\mathbb Q}[G], $$
where the sum is over all $0 < a < m$ with $(a,m) = 1$, is called the *Stickelberger element* corresponding to $K$. The fact that $(b-\sigma\_b)\theta \in {\mathbb Z}[G]$ for integers $b$ coprime to $m$ allows us to define the *Stickelberger ideal*
$I\_0(K)$ as the ideal in ${\mathbb Z}[G]$ generated by elements of the form
$(b-\sigma\_b)\theta$. Set $I(K) = {\mathbb Z}[G] \cap \theta {\mathbb Z}[G]$.
If $K = {\mathbb Q}(\zeta\_m)$ is a full cyclotomic field, then $I(K) = I\_0(K)$.
Stickelberger's Theorem says that if $K/{\mathbb Q}$ is an abelian extension, then the Stickelberger ideal $I(K)$ annihilates the class group $Cl(K)$.
Applied to quadratic extensions, you get the theorem first proved by Cauchy and Jacobi mentioned above. See my Reciprocity Laws, Chapter 11, for details.
| 4 | https://mathoverflow.net/users/3503 | 25709 | 16,849 |
https://mathoverflow.net/questions/25208 | 6 | Let $\omega^\omega$ be Baire space. If $A,B\subseteq\omega^\omega$ we say that $A$ is Wadge reducible to $B$ (written $A\leq\_w B$) if there is a continuous function $f:\omega^\omega\rightarrow\omega^\omega$ with $x\in A$ if and only if $f(x)\in B$. (In other words $A$ is a continuous preimage of $B$). By identifying sets with $A\leq\_w B$ and $B\leq\_w A$ we induce a partial ordering on the set of corresponding equivalence classes, or Wadge degrees. Under the axiom of determinacy AD, it turns out by the so-called Wadge lemma that this hierarchy is almost linearly ordered, meaning we get a linear order if we identify a degree $a$ and the degree consisting of complements of members of $a$. Clearly the Borel sets will form an initial segment of this hierarchy. What I am curious about is if this is still the case if we drop the AD assumption.
Even without AD, determinacy holds for the Borel sets and so the Borel Wadge degrees will still be almost linear ordered - indeed almost well-ordered. For non-Borel degrees using AC it is possible to get a lot of bad behavior; for example many incomparable Wadge degrees. But all the ways I can figure to do such things is to enumerate all continuous functions and build simultaneously the incomparable sets by diagonalizing against the continuous functions. An argument like this is (I think) rather unlikely to produce a Borel set.
So specifically my question is: is it possible for there to be a non-Borel set which is Wadge-incomparable to some Borel set? This is the same as asking if it is possible for there to be a Borel set $B$ and a non-Borel set $A$ with $B\not\leq\_w A$.
| https://mathoverflow.net/users/2436 | A subset of Baire space Wadge incomparable to a Borel set? | The same type of diagonalization should allow you to do this.
Suppose $B$ is a Borel set that is not $F\_\sigma$.
Let $(f\_\alpha:\alpha<2^{\aleph\_0})$ list all continuous functions.
At any stage $\alpha<\aleph\_0$ we will have $A\_\alpha$ and $C\_\alpha$ disjoint
subsets of $\omega^\omega$ of size less than $2^{\aleph\_0}$. Elements in $A\_\alpha$ will
end up in our set $A$ while elements of $C\_\alpha$ will not. That is $A=\bigcup A\_\alpha$
and $A\cap C\_\alpha=\emptyset$ for all $\alpha$.
At stage $\alpha$ we make sure $f\_\alpha$ will not be a reduction of $B$ to $A$.
case 1: there is $y$ in the image of $f\_\alpha$ but not in $A\_\alpha\cup C\_\alpha$.
Say $y=f(x)$. If $x\in B$, put $y\in C\_{\alpha+1}$, otherwise put $y\in A\_{\alpha+1}$.
In either case, we have insured $f\_\alpha$ is not a reduction.
case 2: there is $x$ in B with $f\_\alpha(x)\in C\_\alpha$ or $x\not\in B$ with
$f\_\alpha(x)\in A\_\alpha$.
In this case we already see $f\_\alpha$ is not a reduction and do nothing.
case 3: otherwise $f\_\alpha$ maps $B$ into $A\_\alpha$ and the complement into $C\_\alpha$.
But the continuous image of a Borel is either countable or has size continuum. Thus the images of $B$ and its complement both must be countable. But then $B$ is $F\_\sigma$
and its complement is $F\_\sigma$, a contradiction.
| 6 | https://mathoverflow.net/users/5849 | 25712 | 16,852 |
https://mathoverflow.net/questions/25714 | 6 | I am looking for a reference to the following result.
>
> Let $f:\mathbb R^m\to\mathbb R$ be a convex function.
> Then $f$ is differentiable at all points of outside of a countable union of $(m-1)$-rectifiable sets.
>
>
>
**Comments:**
* $n$-rectifiable set is an image of Lipschitz map from bounded domain in $\mathbb R^n$
* I checked Federer's *"Geometric Measure Theory"*, but I might miss the right place.
* **Extract from the Greg's answer** *(for those who are lazy to read the paper):* In [this paper](http://dml.cz/dmlcz/101616), it is given a complete characterization of subsets of nondifferatiable points of a convex function. Namely, it is proved that $A$ is a such a set if and only if it can be covered by countably many graphs of DC-functions. ("DC-function" = "Difference of Convex functions".)
| https://mathoverflow.net/users/1441 | The set of non-smooth points of a convex function is (m - 1)-rectifiable | The paper [On the differentiation of convex functions in finite and infinite dimensional spaces](http://dml.cz/dmlcz/101616) by Zajíček primarily deals with the general question in Banach space, but it looks like it has a summary of the situation (as of 1979) that tells you everything that you might want to know. It has references to closer papers by Anderson-Klee and Besicovitch.
I just Googled around and eventually got to this paper.
| 6 | https://mathoverflow.net/users/1450 | 25716 | 16,854 |
https://mathoverflow.net/questions/25620 | 10 | Given any two points on a hyperbolic paraboloid ($xy = z$ or $z = (x^2 - y^2)/2$) how does one find the geodesic between them?
I know that since the hyperbolic paraboloid is doubly ruled, some of the geodesics are lines. However, I have very little idea of how to find the geodesics between arbitrary points.
If an exact answer cannot be given, then I would be interested in a technique of approximating the geodesics.
Finally, I'd like to add that this is outside my area of expertise (I'm an algebraist!) so you'll have to explain it to me.
Thanks!
| https://mathoverflow.net/users/6270 | Geodesics on a hyperbolic paraboloid | It is standard differential geometry to find the differential equation for the geodesics on this surface. (But I could easily have made a mistake in the calculation anyway.) Since it is a complete negatively curved surface, there is exactly one geodesic connecting any two points. You have a curve
$\vec{p}(t) = (x(t),y(t),z(t))$ on the surface $z = xy$. The geodesic equation is
$$\vec{p}''(t) \propto \vec{\nabla}z \oplus -1$$
(the acceleration is perpendicular to the surface), which expands to
$$(x'',y'',x''y+2x'y' + xy'') \propto (y,x,-1).$$
Thus
$$\frac{x''}{y} = \frac{y''}{x} = -(x''y+2x'y'+xy'').$$
That is an equation that Mathematica can solve. The only tricky part is to solve it with boundary conditions at both ends, where you may as well assume that the endpoints are at $t=0$ and $t=1$. For that purpose it could be better to minimize the curve's energy, by definition
$$E[\vec{p}] = \int\_0^1 |\vec{p}'(t)|^2 dt = \int\_0^1 (x'^2+y'^2 + (x'y+xy')^2) dt.$$
I don't know the most convenient way to do this numerically, but somehow it should be possible. Again, since the surface is negatively curved, this energy functional is well-behaved.
I don't know whether there is a closed form solution in elementary functions. There is a closed form solution for a [round paraboloid](https://mathworld.wolfram.com/ParaboloidGeodesic.html), but it's messy.
| 8 | https://mathoverflow.net/users/1450 | 25721 | 16,858 |
https://mathoverflow.net/questions/25726 | 4 | I have an array of points with their coordinates X and Y. Each point represents a bus stop.
I need to sort the points in a sequence by giving them sequence numbers, so that the path from the first to the last is the shortest.
For a convention, let's call the array "points", so we access and write by calling
```
point(i) for the point
point(i)(x) and point(i)(y) for the coordinates
point(n) is the sequence number.
```
It's also possible to use the "for each" loop:
```
for each point in points
point(x)
point(y)
point(n)
```
The sequence needs to go through all the points of the array.
I searched a lot on the internet, and the only thing I could find is Dijkstra's algorithm, which is not exactly what I am looking for, since I need to go through all the points.
It would be great if one of you guys knew something about that...
Thanks a lot.
Julien
| https://mathoverflow.net/users/6275 | Algorithm for the shortest path through all the points of a 2D cloud | If you only care about the length of the path between the first and last bus stops, then it looks like you are trying to solve the shortest Hamiltonian path problem (HPP). This is related to the more widely studied traveling salesman problem, see [TSP](http://en.wikipedia.org/wiki/Travelling_salesman_problem). Since your points actually lie in the plane, you are considering the special case of the Euclidean Hamiltonian path problem. For Euclidean TSP, there is a [polynomial-time approximation scheme](http://en.wikipedia.org/wiki/Polynomial-time_approximation_scheme), so I am guessing that the same is true for Euclidean HPP. Also, there are some heuristics based on neural networks, that appear to work well. See this [paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6VC5-3YJYG1B-5&_user=10&_coverDate=04%2F30%2F2000&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1345451288&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=fc6dbe054dae78d3e1c8dcacd48c03df).
| 7 | https://mathoverflow.net/users/2233 | 25730 | 16,865 |
https://mathoverflow.net/questions/25687 | 9 | Is it true that if $\operatorname{Ext}^{1}\_{A}(P,A/I)=0$ forall $I$ then $P$ is projective?
Similar statements are true for flat and injective modules, but I'm beginning to suspect that projective modules cannot be characterized solely by ideals.
| https://mathoverflow.net/users/5292 | Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective? | Answer is yes if $A$ is Noetherian and $P$ is finitely generated. Indeed, your
condition
implies that $Ext^1(P, N)=0$ for any finitely generated $A$-module $N$, which implies that
$P$ is projective.
| 10 | https://mathoverflow.net/users/6277 | 25733 | 16,868 |
https://mathoverflow.net/questions/25740 | 0 | My question is local and coordinate-full: I have an open neighborhood $0 \in U \subseteq \mathbb R^n$, and I'm allowed to make it smaller around $0$. On this neighborhood, I have a constant-rank-$k$ smooth integrable distribution. "Smoothness" means that I have $k\leq n$ many vector fields $v\_1,\dots,v\_k$ — I will work in coordinates $x^1,\dots,x^n$, so $v\_a = \sum\_{i=1}^n v\_a^i(x) \frac{\partial}{\partial x^i}$ — and the distribution at $x$ is the span in ${\rm T}\_xU$ of $\{v\_1(x),\dots,v\_k(x)\}$. "Constant-rank" means that at every $x\in U$, the vectors $\{v\_a(x)\}$ are linearly independent. And "integrable" means that the commutator of any two vector fields on the distribution is in the distribution, i.e. there is a tensor $f^c\_{a,b}(x)$ such that
$$\sum\_{i=1}^k \left( v\_a^i(x)\, \frac{\partial v\_b(x)^j}{\partial x^i} - v\_b^i(x)\, \frac{\partial v\_a(x)^j}{\partial x^i} \right) = \sum\_{c=1}^k f^c\_{a,b}(x)\,v^j\_c(x) $$
So $\{v\_1,\dots,v\_k\}$ is a basis for the distribution. A different basis is given by $\{w\_1,\dots,w\_k\}$ if $w\_a^i(x) = \sum\_b A\_a^b(x)\,v\_b^i(x)$ for some pointwise-invertible matrix-valued function $A$.
>
> My question is whether there is a basis, which in abuse of notation I will call $\{v\}$, such that the structure constants $f^c\_{a,b}$ satisfy $\sum\_a f^a\_{a,b}(x) = 0$ for every $x\in U$?
>
>
>
| https://mathoverflow.net/users/78 | Does every smooth integrable constant-rank distribution have a basis in which the structure constants are traceless? | Am I missing something? By the Frobenius theorem, there exist co-ordinates $y^1, \dots, y^n$ such that $\partial/\partial y^1, \dots, \partial/\partial y^k$ span the same distribution. So these $k$ co-ordinate vector fields form a basis where the structure constants vanish identically.
| 3 | https://mathoverflow.net/users/613 | 25741 | 16,871 |
https://mathoverflow.net/questions/25734 | 7 | By the fundamental theorem of algebra, the algebraic closure $\mathbb{K}$ of $\mathbb{Q}$ decomposes as $\mathbb{K} = F \oplus i F$ where $F = \mathbb{R} \cap \mathbb{K}$ (the intersection is in $\mathbb{C}$). I want to know if there is a purely algebraic way to characterize $F$, i.e. without invoking any analysis, topology, or transcendental number theory. I am asking this because I noticed that it is often convenient when working with examples in characteristic 0 algebraic number theory to give preference to the real roots of a polynomial, and I am wondering if there is a canonical algebraic way to formulate this preference. It doesn't seem like an object built out of lots and lots of transcendental extensions should be so fundamental to purely algebraic examples.
Here are some specific questions that I have been playing with.
Is there a purely algebraic way to distinguish between the splitting fields of $x^2 + 2$ and $x^2 - 2$?
Is there a purely algebraic way to distinguish the real root among the three roots of $x^3 - 2$ in a splitting field?
Of course, the relevant algebraic structures can't be invariant under $\mathbb{Q}$ automorphisms. But I don't see why one can't just be a little bit imaginative.
(Examples edited, changing 1 to 2)
| https://mathoverflow.net/users/4362 | Is there a purely algebraic criterion which characterizes the real algebraic numbers? | Paul: you ask if there is a way to algebraically characterize the field of real algebraic numbers. As a specific field in $\mathbf C$, no there's not a good algebraic characterization, but as an abstract field *yes there is a characterization*. This field is one particular example (and the only concrete one at that) of a *real closure* of $\mathbf Q$. Any two real closures of $\mathbf Q$ are isomorphic to each other.
If you pick a number field $K$ other than the rationals, you can contemplate its real closures: maximal algebraic extensions of $K$ which admit an ordering. Assuming this is possible at least once (e.g., ${\mathbf Q}(i)$ has no real closure), then you can ask if the real closures of $K$ are all isomorphic to each other respecting the embedding of $K$ into them. Nope.
Consider $K = {\mathbf Q}(a)$ where $a^2 = 2$. If we stuff $K$ into the real algebraic numbers by sending $a$ to $\sqrt{2}$ then $a$ is a square in the real algebraic numbers (it's the square of $\sqrt[4]{2}$, which is a real algebraic number: we're talking about concrete real numbers that are algebraic). But if we stuff $K$ into the real algebraic numbers by sending $a$ to $-\sqrt{2}$ then $a$ is not a square in the real algebraic numbers (all squares in the real numbers are positive). Therefore these two embeddings of $K$ into the real alg. numbers are not compatible with each other *as extensions of $K$*. That is, there is no automorphism of the real algebraic numbers which commutes with these two embeddings of $K$ into it. In other words, real closures of $K$ are not all isomorphic as extensions of $K$.
Theorem: Let $K$ be a number field. Every real closure of $K$, up to isomorphism as an extension of $K$, looks like the real algebraic numbers using some real embedding of $K$, and different real embeddings lead to non-isomorphic real closures as extensions of $K$.
I think I got that right. If I screwed up I'm sure BCnrd will let me know. :)
| 15 | https://mathoverflow.net/users/3272 | 25742 | 16,872 |
https://mathoverflow.net/questions/10282 | 27 | Other than the standard baby Rudin, Spivak, and Stein-Shakarchi, are there other alternative and comprehensive analysis texts at the undergraduate level? For example something that has general results that would serve as a very good reference book for specialist analysts in any field, whether functional, complex and measure theorists. Like change of limits, convergence of series etc.
I notice [the question on undergraduate textbooks](https://mathoverflow.net/questions/761/undergraduate-level-math-books) has few responses regarding analysis books of this sort.
| https://mathoverflow.net/users/nan | Alternative undergraduate analysis texts | Nobody has mentioned Folland's "Real Analysis with Applications"?? This was the textbook for my undergraduate real analysis course (measure theory, Banach spaces, Hilbert spaces), and I still go back to it all the time. I am not yet all that experienced (I just finished my third year of graduate school), but overall I have gotten more use out of this book than any other that I own.
It has the most comprehensive swath of applications of analysis of any introductory text I have ever encountered: basic functional analysis, Fourier theory, probability theory, distributions, Hausdorff measures, Haar measure, smooth measures, and more. The early material is covered with all the appropriate detail, while the later material quickly provides the essential definitions and results needed to come to grips with an unfamiliar idea in the literature. Also, the exercises are abundant and uniformly fantastic. My only complaints are that some of the later proofs are hard to read, and there is sadly no discussion of the spectral theorem.
| 14 | https://mathoverflow.net/users/4362 | 25746 | 16,876 |
https://mathoverflow.net/questions/25747 | 13 | We have no topologists on our faculty, and from time to time I get to teach our topology course. I know that there are examples of inequivalent knots with the same Homfly polynomial, and I know that there are non-trivial knots with trivial Alexander polynomial, but I don't know whether the question has been settled as to whether there is a non-trivial knot (or link?) with a trivial Homfly polynomial. I'd like to give my students up-to-date information on this, and being outside the area I don't know quite where to look.
| https://mathoverflow.net/users/3684 | Is there a non-trivial knot with trivial Homfly polynomial? | All I know is that in their 2003 paper Eliahou, Kauffman and Thistlethwaite
write that they did not find any links with trivial HOMFLY-PT. Although they do find links
with both trivial Jones and Alexander.
<http://www.math.uic.edu/~kauffman/ekt.pdf>
My guess would be there exist links with trivial HOMFLY-PT but no such knots.
Although as Qiaochu mentions it is still open whether there are knots with trivial Jones, Joergen Andersen claims on his website that there are no knots with trivial
Colored Jones. (a.k.a Jones of all the cables of the knot)
<http://home.imf.au.dk/andersen/>
| 6 | https://mathoverflow.net/users/692 | 25760 | 16,882 |
https://mathoverflow.net/questions/25758 | 15 | Once again, the question says it all.
My motivation is the article on factorization I am writing. I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater than one, the obstruction to factoriality is the nonvanishing of the (Weil) divisor class group $\operatorname{Cl}(R)$, not the Picard group $\operatorname{Pic}(R)$ (equivalently, the Cartier divisor class group).
My understanding is that it is equivalent to find a normal Noetherian domain with vanishing Picard group which is not locally factorial. I would be especially happy to see an example among affine domains, i.e., in which the domain is finitely generated as an algebra over some field.
| https://mathoverflow.net/users/1149 | Seeking Noetherian normal domain with vanishing Picard group but not a UFD | If the group $C\_2$ acts on $\mathbb{C}[x,y]$ by sending $x$ to $-x$ and $y$ to $-y$, then the fixed subalgebra is the domain $\mathbb{C}[x^2,y^2,xy]$. This is Noetherian and normal. A theorem of Nakajima gives the isomorphism type of the divisor class group (in this case, $C\_2$.) A theorem of Kang asserts for such examples (polynomial rings over $\mathbb{C}$, fixed subalgebra of finite group action), the Picard group is *always* trivial, so one can find many examples.
The reference for all of this is Benson's [Polynomial Invariants of Finite Groups](http://books.google.com/books?id=jxIXivzKFdgC&printsec=frontcover&dq=benson+polynomial+invariants+of+finite+groups&source=bl&ots=O7gr6Ujm7n&sig=XyB4HTSvlMJkxkJ8m7EhmGCF1lA&hl=en&ei=VVX6S-XbJoT58AbWx4H-Cg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q&f=false).
Kang's theorem is [Theorem 3.6.1](http://books.google.com/books?id=jxIXivzKFdgC&lpg=PP1&ots=O7gr6Ujm7n&dq=benson%2520polynomial%2520invariants%2520of%2520finite%2520groups&pg=PA33#v=onepage&q&f=false); Nakajima's theorem is [Theorem 3.9.2](http://books.google.com/books?id=jxIXivzKFdgC&lpg=PP1&ots=O7gr6Ujm7n&dq=benson%2520polynomial%2520invariants%2520of%2520finite%2520groups&pg=PA38#v=onepage&q&f=false) (and Corollary 3.9.3).
Steve
| 19 | https://mathoverflow.net/users/1446 | 25763 | 16,885 |
https://mathoverflow.net/questions/25782 | 0 | I am reading something where this is used extensively, but it is not defined anywhere and no references are given, and I can't find any.
| https://mathoverflow.net/users/2300 | In a k-linear category, what is the tensor product between a hom space and an object? | If $\mathcal{C}$ is a $k$-linear category, $X \in \mathcal{C}$, and $V$ is any $k$-vector space (in particular, it could be a hom of two objects in $\mathcal{C}$), then $V \otimes X$ (sometimes written $V \odot Y$ to avoid confusion with a monoidal structure) is the object representing the functor $\mathcal{C} \to \operatorname{Vect}$ sending $Y \in \mathcal{C}$ to $\operatorname{Hom}\_{\operatorname{Vect}}(V, \mathcal{C}(X, Y))$, if such an object exists. This is a special case of the notion of [copower](http://ncatlab.org/nlab/show/copower).
In the case where $\mathcal{C}$ is a tensor category with internal homs, the construction agrees with the tensoring by the internal hom.
| 6 | https://mathoverflow.net/users/396 | 25783 | 16,897 |
https://mathoverflow.net/questions/25035 | 4 | I'm responsible for a charity donation site. We're about the change the site design, and we want to know the best way of detecting if the distribution of donations changes after the design. The problem is the data is quite clumpy, particularly around \$5, \$10, \$20, \$25 and \$50 values, with \$15 being relatively rare. There are nonetheless other real values, particularly between \$20 and \$100.
The consequence is that the stdev is three times the mean, so my first approach, a T-Test with some correction for the skew, doesn't seem feasible. If our re-design only has a moderate impact it's unlikely a T-test will be able to detect it with certainty.
Binning into \$5 bins and running a G-Test individually means I can make statements about individual categories, but I'm worried about drawing overall conclusions from such a series of measurements. I've read briefly about using Fisher's method for combining p-values but I'm not sure how to explain the result, particularly as a positive result could mean *fewer* \$10 donations but more \$15 donations.
I'm certain that clumpy data is a known phenomenon, but Googling hasn't helped, and my background is CS, not stats. Would anyone know the best way to handle this?
| https://mathoverflow.net/users/6141 | Detecting a shift in distribution where the distribution is clumpy | The [Kolmogorv-Smirnov](http://en.wikipedia.org/wiki/Kolmogorov%25E2%2580%2593Smirnov_test) test suggested by Steve Huntsman seems to have done the job.
| 1 | https://mathoverflow.net/users/6141 | 25786 | 16,900 |
https://mathoverflow.net/questions/25774 | 6 | Where can we find Deligne's paper " Theorie de Hodge I"?
| https://mathoverflow.net/users/3945 | Where can we find Deligne's paper " Theorie de Hodge I"? | [Voici](http://math.harvard.edu/~tdp/Deligne-Theorie.de.Hodge-1-single-page.pdf).
| 19 | https://mathoverflow.net/users/307 | 25787 | 16,901 |
https://mathoverflow.net/questions/25794 | 66 | Let $\chi$ be a Dirichlet character and $L(1,\chi)$ the associated L-function evaluated at $s=1$. What would be the 'shortest' proof of the non-vanishing of $L(1,\chi)$?
Background: The non-vanishing of $L(1,\chi)$ plays an essential role in the proof of Dirichlet´s theorem on primes in arithmetic progressions. In his "Introduction to analytic number theory", T. M. Apostol gives an elementary proof of the above fact estimating various sums in a few lemmas in the context of a proof of the aforementioned Dirichlet theorem. While his approach has the advantage of being self-contained and not requiring much of a background, it is quite lenghty. In their "Analytic number theory", H. Iwaniec and E. Kowalski remark that in Dirichlet´s original proof the non-vanishing of $L(1,\chi)$ for real Dirichlet characters is a simple consequence of Dirichlet´s class number formula. However, in both approaches it is necessary to distinguish between real and complex Dirichlet characters. Hence my two "sub"-questions:
1) Is there a proof that avoids the distinction between the complex and real case?
2) Are there in general other proof strategies for $L(1,\chi)\neq 0$ that can be considered shorter and/or more elegant than the two mentioned above?
| https://mathoverflow.net/users/1849 | Shortest/Most elegant proof for $L(1,\chi)\neq 0$ | I like the proof by Paul Monsky:
'Simplifying the Proof of Dirichlet's Theorem'
American Mathematical Monthly, Vol. 100 (1993), pp. 861-862.
Naturally this does maintain the distinction between real and complex
as whatever you do, the complex case always seems to be easier
as one would have two vanishing L-functions for the price of one.
I incorporated this argument into my note on a "real-variable" proof
of Dirichlet's theorem at
<http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/dirichlet.pdf> .
There are proofs, notably in Serre's *Course in Arithmetic*
which claim to treat the real and complex case on the same footing.
But this is an illusion; it pretends the complex case is as hard
as the real case. Serre considers the product $\zeta\_m(s)=\prod L(s,\chi)$
where $\chi$ ranges over the modulo $m$ Dirichlet characters.
If one of the $L(1,\chi)$ vanishes then $\zeta\_m(s)$ is bounded as $s\to 1$
and Serre obtains a contradiction by using Landau's theorem on
the abscissa of convergence of a positive Dirichlet series. But all
this subtlety is only needed for the case of real $\chi$. In the non-real
case, at least two of the $L(1,\chi)$ vanish so that $\zeta\_m(s)\to0$
as $s\to1$. But it's elementary that $\zeta\_m(s)>1$ for real $s>1$
and the contradiction is immediate, without the need of Landau's subtle result.
**Added** (25/5/2010) I like the Ingham/Bateman method. It is superficially
elegant, but as I said in the comments, it makes the complex case as hard
as the real. Again it reduces to using Landau's result or a choice of other
trickery.
What one should look at is not $\zeta(s)^2L(s,\chi)L(s,\overline\chi)$
but
$$G(s)=\zeta(s)^6 L(s,\chi)^4 L(s,\overline\chi)^4 L(s,\chi^2)L(s,\overline\chi^2)$$
(cf the famous proof of nonvanishing of $\zeta$ on $s=1+it$ by
Mertens).
Unless $\chi$ is real-valued this function will vanish at $s=1$ if
$L(1,\chi)=0$. But one shows that $\log G(s)$ is a Dirichlet
series with nonnegative coefficients and we get an immediate contradiction
without any subtle lemmas. Again it shows that the real case is the hard one.
For real $\chi$ then $G(s)=[\zeta(s)L(s,\chi)]^8$ while Ingham/Bateman
would have us consider $[\zeta(s)L(s,\chi)]^2$. This leads us to
the realization that for real $\chi$ we should look at $\zeta(s)L(s,\chi)$
which is the Dedekind zeta function of a quadratic field. (So if one
is minded to prove the nonvanishing by showing that a Dedekind zeta function
has a pole, quadratic fields suffice, and one needn't bother with cyclotomic
fields).
But we can do more. Let $t$ be real and consider
$$G\_t(s)=
\zeta(s)^6 L(s+it,\chi)^4 L(s-it,\overline\chi)^4
L(s+2it,\chi^2)L(s-2it,\overline\chi^2).$$
Unless both $t=0$ and $\chi$ is real, if $L(1+it,\chi)=0$ one gets
a contradiction just as before. So the nonvanishing of any $L(s,\chi)$
on the line $1+it$ is easy **except** at $1$ for real $\chi$.
This special case really does seem to be deeper!
**Added** (26/5/2010)
The argument I outlined with the function $G\_t(s)$ is well-known to extend
to a proof for a zero-free region of the L-function to the left of the line
$1+it$. At least it does when unless $t=0$ and $\chi$ is real-valued.
In that case it breaks down and we get the phenomenon of the Siegel zero;
the possible zero of $L(s,\chi)$ for $\chi$ real-valued, just to the left of
$1$ on the real line. So the extra difficulty of proving $L(1,\chi)\ne0$
for $\chi$ real-valued is liked to the persistent intractability of
showing that Siegel zeroes never exist.
| 34 | https://mathoverflow.net/users/4213 | 25797 | 16,907 |
https://mathoverflow.net/questions/25756 | 3 | In [[Föllmer 81]](http://archive.numdam.org/ARCHIVE/SPS/SPS_1981__15_/SPS_1981__15__143_0/SPS_1981__15__143_0.pdf) (English translation to be found [here](http://books.google.co.uk/books?id=KKLa1j-diXwC&printsec=frontcover&dq=Introduction+to+Stochastic+Calculus+for+Finance%3A+A+New+Didactic+Approach&source=bl&ots=-L3eKT8OjQ&sig=3Md3B6gg4LLy2Uzzzy8fs37Rwqo&hl=en&ei=4HT5S6P-OI6Q4gbxh9AI&sa=X&oi=book_result&ct=result&resnum=5&ved=0CC8Q6AEwBA#v=onepage&q&f=false)) writes: "The class of processes of quadratic variation is clearly larger than the class of semimartingales: Just consider a deterministic process of quadratic variation which is of unbounded variation."
Could anyone please give me examples (with references) of *deterministic processes of quadratic variation which are of unbounded variation*? Thank you!
(P.S.: What seems to make these deterministic processes interesting is that you also have to use [Ito integrals](http://en.wikipedia.org/wiki/Ito_calculus) to integrate them)
| https://mathoverflow.net/users/1047 | Examples of deterministic processes of quadratic variation which are of unbounded variation | Take $f:[0,1]\to\mathbb{R}$ such that $f(0)=0$ and it interpolates linearly between $f(1/n)=\frac{(-1)^n}{n}$ for any natural $n$.
| 2 | https://mathoverflow.net/users/2968 | 25806 | 16,914 |
https://mathoverflow.net/questions/25800 | 6 | I'd be grateful for a reference for the following result, which I believe to be true, and
should be well-known.
Let the continuous functions $f\_0,f\_1,\cdots,f\_n: [0,1]\rightarrow [0,\infty)$ be given
and consider the problem of maximizing the integral
$$\int\_0^1 f\_0(x)d\mu(x)$$
over all *positive* Borel measures $\mu$ on [0,1], which satisfy the constraints
$$\int\_0^1 f\_k(x)d\mu(x)=1,\;\;1\leq k\leq n.$$
Then, if a solution exists, the maximum is attained by a linear combination of (at most) $n$
shifted $\delta$-functions:
$$\mu=\sum\_{j=1}^n \alpha\_j\delta(x-x\_j),\;\;\;x\_j\in [0,1]$$
| https://mathoverflow.net/users/5365 | A simple infinite dimensional optimization problem | This is a particular case of the Generalized Moment Problem.
The result you are looking for can be found in the [first chapter](http://www.worldscibooks.com/etextbook/p665/p665_chap01.pdf) of *Moments, Positive Polynomials and Their Applications* by Jean-Bernard Lasserre (Theorem 1.3).
The proof follows from a general result from measure theory.
>
> **Theorem**. Let $f\_1, \dots , f\_m : X\to\mathbb R$ be Borel measurable on a measurable
> space $X$ and let $\mu$ be a probability measure on $X$ such that $f\_i$ is
> integrable with respect to $\mu$ for each $i = 1, \dots, m$. Then there exists a
> probability measure $\nu$ with finite support on $X$, such that:
> $$\int\_X f\_id\mu=\int\_Xf\_i d\nu,\quad i = 1,\dots,m.$$
> Moreover, the support of $\nu$ may consist of at most $m+1$ points.
>
>
>
| 6 | https://mathoverflow.net/users/5371 | 25808 | 16,915 |
https://mathoverflow.net/questions/24693 | 24 | I heard this puzzle from Bob Koca. Suppose we play misere tic-tac-toe (a.k.a. noughts and crosses) where both players are X. Who wins?
That particular puzzle is easy to solve, but more generally, has $n \times n$ impartial tic tac toe, in both normal and misere forms, been studied before?
---
EDIT: Thane Plambeck's paper, mentioned at the end of his answer below, coined the term **Notakto** for this game. That name seems to have caught on; for example, there is now a [Wikipedia article on Notakto](https://en.wikipedia.org/wiki/Notakto).
| https://mathoverflow.net/users/3106 | Neutral tic tac toe | It's possible to give a complete theory of 3x3 misere "X-only" tic-tac-toe disjunctive sums by introducing the 18-element commutative monoid $Q$ given by the presentation
$Q = \langle\ a,b,c,d\ | \ a^2=1,\ b^3=b,\ b^2c=c,\ c^3=ac^2,\ b^2d=d,\ cd=ad,\ d^2=c^2 \rangle\ $.
Such a "disjunctive" game of 3x3 neutral tic-tac-toe is played not just with *one* tic-tac-toe board (as has been previously discussed in this thread), but more generally with an arbitrary (finite) number of such boards forming the start position. On a player's move, he or she selects a single one of the boards, and makes an X on it (a board that already has a three-in-a-row configuration of X's is considered out-of-play). Play ends when every board has a three-in-a-row configuration, and the player who completes the last three-in-a-row on the last available board is the loser.
The game analyzed already in this thread corresponds to play on a 3x3 single board.
The monoid Q arises as the misere quotient of the impartial game
G = 4 + {2+,0}
{2+,0} is the canonical form of the 3x3 single board start position, and "4" is the nim-heap of size 4, which also happens to occur as a position in this game. I'm using the notation of John Conway's On Numbers and Games, on page 141, Figure 32.
One way to think of Q is that it captures the misere analogue of the "nimbers" and "nim addition" that are used in normal play disjunctive impartial game analyses, localized to the play of this particular impartial game, neutral 3x3 tic-tac-toe.
I performed these calculations partly using Mathematica, and partly using Aaron N. Siegel's "MisereSolver" program.
See also
<http://arxiv.org/abs/math/0501315>
<http://arxiv.org/abs/math/0609825>
<http://arxiv.org/abs/0705.2404>
<http://www.miseregames.org>
It's possible to build a dictionary that assigns an element of Q to each of the conceivable 102 non-isomorphic positions in 3x3 single-board neutral tic-tac-toe. (I mean "non-isomorphic" under a reflection or rotation of the board. In making this count, I'm including positions that couldn't be reached in actuality because they have too many completed rows of X's, but that doesn't matter since all those elements are assigned the identity element of Q). To determine the outcome of a multi-board position (ie, whether the position is an N-position -- a Next player to move wins in best play, or alternatively, a P-position-- second player to move wins), what a person does is multiply the corresponding elements of Q from the dictionary together, and reduce them via the relations in the presentation Q that I started with above, arriving at a word in the alphabet a,b,c,d.
If that word ends up being one of the four words {a, b^2, bc, c^2 }, the position is P-position; otherwise, it's an N-position.
I'm guessing the the 4x4 game does not have a finite misere quotient, but I don't know for sure.
If people want more details, I'm happy to send them. Google my name for my email address.
Best wishes
Thane Plambeck
Postscript (added 8 Jan 2013) Here's a paper <http://arxiv.org/abs/1301.1672> I just put up in the arXiv that has more details.
| 10 | https://mathoverflow.net/users/6295 | 25811 | 16,917 |
https://mathoverflow.net/questions/25825 | 9 | Recall that the (first) Weyl algebra over $\mathbb{C}$ is the algebra generated by $x,y$ with the relation $yx-xy=1$. It can be realized as the algebra of polynomial differential operators in 1 variable, i.e. $\mathbb{C}[x]$ is a faithful representation, where $x$ acts by multiplication by $x$ and $y$ acts by $\frac{\partial}{\partial x}$.
In many places I've seen the q-Weyl algebra defined as the algebra generated by $x,y$ with relations yx-qxy=1, where q is some fixed nonzero scalar. This seems like a natural way to do it, and seems to be quite common.
Now in the notes [Introduction to representation theory](http://arxiv.org/abs/0901.0827) Etingof et al. define an algebra which they call the q-Weyl algebra. This is the $\mathbb{C}$-algebra generated by $x,x^{-1},y,y^{-1}$ with the relations $xx^{-1} = x^{-1}x = 1$, $yy^{-1}= y^{-1}y = 1$ and $xy=qyx$ where q is some fixed nonzero scalar.
My question then is: What is the reason for the name 'q-Weyl algebra' for the algebra defined by Etingof et al.?
| https://mathoverflow.net/users/135 | Why is this algebra called the q-Weyl algebra? | Suppose that $x$ and $y$ obey $xy-yx=h$, where $h$ is a central element. Set $X$ and $Y$ to be $e^x$ and $e^y$. For now, don't worry too much about what this exponentiation means. Then $XY=q YX$, where $q=e^h$.
If we interpret $X$ and $Y$ as operators on functions then $(Xf)(x)=e^x f(x)$ and $(Yf)(x) = f(x+h)$. You can check that $XY$ does indeed equal $q YX$. So the $q$-commutation relation can be seen as an exponentiation of the standard Weyl relation.
| 18 | https://mathoverflow.net/users/297 | 25828 | 16,930 |
https://mathoverflow.net/questions/25829 | 2 | I would like to know if there is a equation for the maximum number of shortest paths that pass through *r* where *r* is a node contained in any path from node *s* (a fixed node, i mean, *s* is the only source of paths) to any node *t* in an unweighted undirected acyclic graph. I've searched and found [this work](http://www.cse.fau.edu/~jie/research/publications/Publication_files/msp.pdf) that show this number for grid graphs, but I'm interested in this number for general topology graph.
I would be grateful for any reference for a work in this subject, or a suggestion how to start to solve this problem. Thanks in advance.
**Edit:** Sorry, the graph I'm interested is loop-free as Hans Stricker pointed, but it is cyclic.
| https://mathoverflow.net/users/6299 | Maximum number of shortest-paths | Judging from the link you provide, you have three distinct vertices s,t,d and want to compute the number of shortest walks P(s,d,t) from s to d that contain t. The reason I use "walks" instead of "paths" is because of graphs like:
```
s----d----t
```
where we must reuse edges.
If you really mean acyclic graph, then P(s,d,t)=1 if there is a path containing vertices s,t and d and P(s,d,t)=0 otherwise.
Let P(s,t) be the number of shortest walks between s and t and P(s,s)=1. If s, d and t are all distinct then P(s,d,t)=P(s,t)P(d,t).
In Section 2.4 of the paper you link to, they describe the algorithm for finding P(s,t). That is $P(s,t)=\sum\_{v \in V} P(s,v)$ where V is the set of neighbours of t for which P(s,v) is minimal. The difficulty is identifying which neighbours of t minimise P(s,v).
One way is to construct a set Sn of paths of length n=1,2,... (this can be done recursively), from t until you find some neighbour of s. Then count 1 for each path that ends in a neighbour of s and 0 otherwise. Another way would be to compute P(s,t) and then use a backtracking algorithm.
There's not going to be an exciting formula for P(s,d,t) in general, since it depends on the graphs' structure. This is much like how there's not going to be an exciting formula for the number of edges in a graph (unless you restrict to some specific class of graphs).
| 1 | https://mathoverflow.net/users/2264 | 25838 | 16,937 |
https://mathoverflow.net/questions/25836 | 4 | By the Pati-Salam group I refer to SU(2) x SU(2) x SU(4). It can be obtained as the group of isometries of the 8 dimensional manifold $S^3 \times S^5$, but I wonder if this is the only 8 dimensional manifold having this group of isometries.
This particular manifold is interesting because a quotient by any U(1) will produce a 7 dimensional manifold whose isometry group is the unbroken standard model group, as pointed out by Witten time ago. But my particular curiosity comes because Non Commutative Geometry gets the Pati Salam group from a different setup: the finite algebra $M\_2(H) \oplus M\_4(C)$, related perhaps to deformations of *even* spheres.
| https://mathoverflow.net/users/4037 | Manifolds whose isometry group is Pati-Salam? | $S^3 \times S^5$ has isometry group $SO\_4(\mathbb{R}) \times SO\_6(\mathbb{R})$, which has $SU(2) \times SU(2) \times SU(4)$ as a four-fold cover. Since it appears that you aren't worrying too much about central terms, we can replace $S^3$ with $\mathbb{R}P^3$, $S^5$ with $\mathbb{R}P^5$, or take a quotient by a diagonal group of order 2.
I'm pretty sure these are the only connected choices, because we can characterize homogeneous orbits by the stabilizers of points. In this case, you need a closed subgroup of Pati-Salam of dimension at least 13 whose intersection with each factor group is not the whole factor. There just aren't that many subgroups of suitably large dimension: we need a diagonally embedded $SU(2)$ (possibly with a central translate) to get dimension at least 3 in the first factors, and we need Spin(5) in the last factor to get dimension at least 10. This forces the orbits to be connected components.
| 5 | https://mathoverflow.net/users/121 | 25839 | 16,938 |
https://mathoverflow.net/questions/25803 | 1 | Hi,
Can anyone familiar with the book 'A Probabilistic Theory of Pattern
Recognition' or the theory described help me out?
See quote from chapter 12, 'Vapnik-Chervonenkis Theory', of 'A
Probabilistic Theory of Pattern Recognition' below. I'm not following
the geometric setup outlined there.
Some specific questions:
The hyperrectangle referred to is as in
<http://en.wikipedia.org/wiki/Hyperrectangle>, right?
Does $\phi \in C$ correspond to hyperrectangles of arbitrary dimension?
When the text says "assign the smallest hyperrectangle containing
these points", I assume it means both in the sense of smallest
dimension as well as size. I'm not sure whether this means the
$\phi\_i$ are of fixed dimension. I think so.
However, my main source of perplexity is the sentence beginning
"Clearly, for each $\phi$..." and ending "on the boundary of the
hyperrectangle". I've no idea why this would be true.
There is a similar setup earlier in the book, Section 4.5, but this
talks about hyperplanes, and is easier to follow.
No doubt I'm misreading or something. Clarifications
appreciated.
```
Regards, Faheem.
```
(Some background to the extract that follows.)
In what follows, d is just a fixed integer. I guess it represents the dimension.
Suppose you have a classifier
$$ \phi: \mathbb{R}^d \longrightarrow \{0, 1\}$$
and $n$ ordered pairs $(X\_i, Y\_i)$, where $X\_i \in \mathbb{R}^d$ are the data values,
and $Y\_i\in \{0,1\}$ are the correct classifications of the $X\_i$.
Then the empirical error (or risk) of $\phi$ is
$$ \hat{L}\_n(\phi) = \frac{1}{n} \sum\_{i=1}^n I\_{\{\phi(X\_i)\neq Y\_i\}} $$
that is, the number of errors made by the classifier is counted and normalized.
From pg 191 of 'A Probabilistic Theory of Pattern Recognition' by
Devroye et al, chapter 12, 'Vapnik-Chervonenkis Theory'.
Let $C$ be the class of classifiers assigning 1 to those $x$ contained
in a closed hyperrectangle and 0 to all other points, Then a
classifier minimizing the empirical error $\hat{L}\_n(\phi)$ over all
$\phi \in C$ may be obtained by the following algorithm: to each
2d-tuple $(X\_{i\_1}, X\_{i\_{2d}})$ of points from $X\_1,\dots, X\_n$,
assign the smallest hyperrectangle containing these points. If we
assume that $X$ has a density, then the points $X\_1,\dots, X\_n$ are in
general position with probability one. This way we obtain at most ${n
\choose 2d}$ sets. Let $\phi\_i$ correspond to the $i$-th such
hyperrectangle, that is, the one assigning 1 to those $x$ contained in
the hyperrectangle, and 0 to other points. Clearly, for each $\phi \in
C$, there exists a $\phi\_i$, i = 1, ${n \choose 2d}$, such that
$$\phi(X\_j) = \phi\_i(X\_j)$$
for all $X\_k$, except possibly for those on the boundary of the
hyperrectangle. Since the points are in general position, there are at
most $2d$ such points. Therefore, if we select a classifier
$\hat{\phi}$ among $\phi\_1, \phi\_{{n \choose 2d}}$ to minimize the
empirical error, then it approximately minimized the empirical error
over the whole class $C$ as well.
| https://mathoverflow.net/users/6293 | A question about Chapter 12 (Vapnik-Chervonenkis Theory) of 'A Probabilistic Theory of Pattern Recognition' | Yes, a hyperrectangle is a generalisation of rectangle to higher dimensions. Here the data is given by points in $\mathbb{R}^d$, so the hyperrectangles are all of that dimension. As with all such algorithms you need to find a way to get a handle on the set of classifiers, so rather than the infinite class of all hyperrectangles of dimension $d$, the choice will be from the $n \choose 2 d$ hyperrectangles of dimension $d$, each of which is the smallest for some choice of $2 d$ of the data points.
For example, if the data consisted of 1000 points in $\mathbb{R}^2$, rather then the infinite class of all rectangles, we confine ourselves to the $1000 \choose 4$ rectangles which minimally contain a subset of 4 of the data points.
One task then is to show that the best of this finite set is almost as good as the best of all the hyperrectangles -- good in the sense that were the data points each labelled $+$ or $-$, the $+$s would be best separated from the $-$s. The argument claims that for each hyperrectangle there will be one from the finite set agreeing with it except for a small number of points on the boundary equal to the number of faces of the hyperrectangles. In the example above, it says that for any rectangle, there is one in the set of $1000 \choose 4$ which encloses exactly the same points, except possibly for 4 on the boundary. Not completely obvious, I agree.
Edit: If hyperrectangles are restricted to have lines parallel to the axes, then it is obvious. Judging from problem 11.6 on page 183, this may well be the case.
| 2 | https://mathoverflow.net/users/447 | 25856 | 16,949 |
https://mathoverflow.net/questions/25846 | 9 | Problem: Consider a random walk on the lattice $\mathbb{Z}^2$ where on each iteration a particle either stays at its current location or moves to a neighboring vertex with probability 1/5. We start the random walk with one particle at the origin. For each $n \geq 1$ and $x \in \mathbb{Z}^2$ let $p\_n(x)$ be the probability of finding the particle at $x$ after $n$ iterations. For two points $x,y \in \mathbb{Z}^2$ let $|\cdot|$ denote the Euclidean distance of $x$ and $y$ via the standard embedding $\mathbb{Z}^2 \subset \mathbb{R}^2$.
For what $n$ is it true that $|x| \leq |y| \Rightarrow p\_n(x) \geq p\_n(y)$? What kind of techniques are available to prove statements like this? Barring arithmetic mistakes I have verified this up to n=6 via explicit computation.
Please forgive me if this is actually a trivial question (I know very little about random walks). I would also be very happy with suggested approaches or references.
A Little Motivation/Another Problem: Suppose we list the elements of $\mathbb{Z}^2$ is ascending order by Euclidean distance from the origin, $z\_1 \leq z\_2 \leq \cdots$, and then set $D\_n = \cup\_{i=1}^n z\_i$. For various reasons I have been dealing with these $D\_n$ and would like to consider analogues in other groups. Hence I would very much like to have a "$\mathbb{Z}^2$-intrinsic" characterization of these $\{D\_n\}$, i.e. it would be nice to have a characterization of $D\_n$ that only used group or graph theoretic statements about $\mathbb{Z}^2$. Most importantly I do not want to mention the specific embedding of $\mathbb{Z}^2$ into $\mathbb{R}^2$.
Note: The $D\_n$ are not exactly well defined since there are choices involved in the list $z\_1 \leq z\_2 \leq \cdots $. So I am actually interested in characterizing them up to the forced ambiguity.
| https://mathoverflow.net/users/4345 | Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance | As written the statement is false for $n=3$: note that $p\_3(2,2) = 0$ but $p\_3(3,0) > 0$, while $|(2,2)| < |(3,0)|$. Similar counterexamples exist for all $n\geq 5$. So for larger $n$ you would at least need some extra condition about $L^1$ norms to guarantee that you can't have $|x|<|y|$ with $p\_n(x)=0$ and $p\_n(y)>0$. I would guess that this would still be too weak, however.
| 3 | https://mathoverflow.net/users/5963 | 25859 | 16,951 |
https://mathoverflow.net/questions/25862 | 15 | This is a very basic question and might be way too easy for MO. I am learning analysis in a very backwards way. This is a question about complex Hilbert spaces but here's how I came to it: I have in the past written a paper about (amongst other things) compact endomorphisms of $p$-adic Banach spaces (and indeed of Banach modules over a $p$-adic Banach algebra), and in this paper I continually used the notion of a "matrix" of an endomorphism as an essential crutch when doing calculations and proofs. I wondered at the time where more "conceptual" proofs existed, and probably they do, but I was too lazy to find them.
Now I find myself learning the basic theory of certain endomorphisms of complex separable Hilbert spaces (continuous, compact, Hilbert-Schmidt and trace class operators) and my instinct, probably wrong, is to learn the theory in precisely the same way. So this is the sort of question I find myself asking.
Say $H$ is a separable Hilbert space with orthonomal basis $(e\_i)\_{i\in\mathbf{Z}\_{\geq1}}$. Say $T$ is a continuous linear map $H\to H$. Then $T$ is completely determined by its "matrix" $(a\_{ij})$ with $Te\_i=\sum\_ja\_{ji}e\_j$. But are there completely "elementary" conditions which completely classify which collections of complex numbers $(a\_{ij})$ arise as "matrices" of continuous operators?
I will ask a more precise question at the end, but let me, for the sake of exposition, tell you what the the answer is in the $p$-adic world.
In the $p$-adic world, $\sum\_na\_n$ converges iff $a\_n\to 0$, and life is easy: the answer to the question in the $p$-adic world is that $(a\_{ij})$ represents a continuous operator iff
(1) For all $i$, $\sum\_j|a\_{ji}|^2<\infty$ (equivalently, $a\_{ji}\to 0$ as $j\to\infty$), and
(2) there's a universal bound $B$ such that $|a\_{ij}|\leq B$ for all $i,j$.
[there is no inner product in the $p$-adic case, so no adjoint, and the conditions come out being asymmetric in $i$ and $j$].
See for example pages 8--9 of [this paper of mine](http://www2.imperial.ac.uk/~buzzard/maths/research/papers/eigenvarieties.pdf), although of course this isn't due to me---it's in Serre's paper on compact operators on $p$-adic Banach spaces from the 60s---see Proposition 3 of Serre's paper. In particular, in the $p$-adic world, one can identify the continuous maps $H\to H$ (here $H$ is a $p$-adic Banach space with countable ON basis $(e\_i)$) with the collection of bounded sequences in $H$, the identification sending $T$ to $(Te\_i)$.
In the real/complex world though, the analogue of this result fails: the sequence $(e\_1,e\_1,e\_1,\ldots)$ is a perfectly good bounded sequence, but there is no continuous linear map $H\to H$ sending $e\_i$ to $e\_1$ for all $i$ (where would $\sum\_n(1/n)e\_n$ go?).
Let's consider the finite rank case, so $T$ is a continuous linear map $H\to H$ with image landing in $\mathbf{C}e\_1$. Then by Riesz's theorem, $T$ is just "inner product with an element of $H$ and then multiply by $e\_1$". Hence we have an *additional* condition on the $a\_{ij}$, namely that $\sum\_j|a\_{ij}|^2<\infty$. Furthermore a continuous linear map is bounded, as is its adjoint.
This makes me wonder whether the following is true, or whether this is still too naive:
Q) Say $(a\_{ij})$ $(i,j\in\mathbf{Z}\_{\geq1})$ is a collection of complex numbers satisfying the following:
There is a real number $B$ such that
1) For all $i$, $\sum\_j|a\_{ij}|^2\leq B$
2) For all $j$, $\sum\_j|a\_{ij}|^2\leq B$
Then is there a unique continuous linear map $T:H\to H$ with $Te\_i=\sum\_ja\_{ji}e\_i$?
My guess is that this is still too naive. Can someone give me an explicit counterexample? Or, even better, a correct "elementary" list of conditions characterising the continuous endomorphisms of a Hilbert space?
On the other hand, it clearly isn't a complete waste of time to think about matrix coefficients. For example there's a bijection between Hilbert-Schmidt operators $T:H\to H$ and collections $(a\_{ij})$ of complexes with $\sum\_{i,j}|a\_{ij}|^2<\infty$, something which perhaps the experts don't use but which I find incredibly psychologically useful.
| https://mathoverflow.net/users/1384 | Naive questions about "matrices" representing endomorphisms of Hilbert spaces. | Chapter V of Halmos' "A Hilbert space problem book" is called "Infinite matrices".
It contains lots of nice results and problems, and also the statement that "there are no elegant and usable necessary and sufficient conditions [for a matrix to be the matrix of an operator]".
| 19 | https://mathoverflow.net/users/5743 | 25864 | 16,954 |
https://mathoverflow.net/questions/25870 | 1 | I know coefficients of some function in basis $p\_j,j=1...K$ where
$p\_{j}(x)=\sum\_{s\in Z}a\_{s,j}\exp\left(-2\pi i(j+sK)x\right)$
With respect to inner product $(f,g)=\int\_{0}^{1}f(x)\overline{g(x)}dx$ this basis is orthogonal.
What kind of basis is it? Can I somehow use Fourier analysis framework for this basis?
One more question about this basis. How to solve this integral equation?
$$a\_{j}=\int\_{-\infty}^{\infty}g(x)p\_{j}(x)dx$$
$g-$unknown function. Could you give me some hint, or some links?
| https://mathoverflow.net/users/3589 | some strange orthogonal basis and an integral equation with it | If $a\_{s,j}$ is in $\ell\_2$ for any fixed $j$, $p\_j$ forms an orthogonal basis for some finite dimensional sub-space of $L^2(\mathbb{T})$. If $f(x) = \sum\_{j=0}^K b\_j p\_j(x)$, then its fourier coefficients are just given by $\hat{f}(\xi) = a\_{s,j} b\_j$ where $j = \xi \mod K$ and $s = \lfloor \xi / K \rfloor$, so surely normal Fourier analytic techniques apply.
I don't quite understand your question about "What kind of basis is it?" Can you clarify?
| 1 | https://mathoverflow.net/users/3948 | 25872 | 16,958 |
https://mathoverflow.net/questions/25866 | 15 | A maybe trivial question about fiber bundles (I'm not an expert, and I didn't find quickly an answer looking here and there). Suppose you are given fiber bundle $p\colon E\to M$, where
$E,M$ are smooth manifolds and $p$ is smoothly locally trivial. Also suppose that the fiber $F$ of the bundle is smoothly contractible. Is that true that $p$ admits a smooth section?
| https://mathoverflow.net/users/6206 | smooth sections of smooth fiber bundles | The answer is: Yes (at least for finite dimensional manifolds).
In fact you only need that the fiber is contractible not smoothly contractible. Take any continuous section $s\_0 \colon B \to E$. cover $B$ by open sets $U\_i$ such that the bundle is trivializable over each $U\_i$, also make sure that the closure of each $U\_i$ is compact and that the cover is locally finite.
Furthermore, give $E$ any complete Riemannian structure. This provides each fiber $E\_x$ with an induced Riemannian structure, which is also complete. We use this to define the obvious supremum distance between any two sections of $E$ over any sub-space of $B$. We may also construct a continuous map $r \colon E \to \mathbb{R}\_+$ such that the ball of radius $r(x)$ and center $x$ in each fiber is geodesically convex.
The construction now goes in 2 steps:
1) local construction: for each $i$ we may find a smooth section $s \colon U\_i \to E$ such that the supremum distance defined above is smaller than $r$ on all the points of $S\_0$ restricted to $U\_i$. This is easy and follows from smooth approximation of any function from $U\_i \to F$ defining a section $U\_i \to U\_i \times F$.
2) global construction: use a partion of unity to get a global construction. This is now possible because we were carefull enough to create the smooth local sections such that they lie in a geodesically convex neighborhood.
While finishing this Andrew Stacey posted a similar answer, but it seemed a waste not to poste this also. Especially since we are focussing on different details.
| 9 | https://mathoverflow.net/users/4500 | 25885 | 16,967 |
https://mathoverflow.net/questions/25863 | 7 | [Torsors](http://math.ucr.edu/home/baez/torsors.html) are defined as a special kind of group action. I am wondering whether the analogous notion exists for monoid actions. Some references would be helpful.
In general I'm interesting in the notion of 'subtraction/division' induced by having a torsor. My application is in computer science. A monoid is used to capture modifications to a computer program and the monoid action corresponds to performing the modification on the program. If I have a torsor-like entity, I can take two software entities and produce the modification required to convert one into the other.
The answer is that any such monoid will automatically be a group. In my application, it seems that I will only get close to the notion of torsor if my modifications have inverses, which they do not.
| https://mathoverflow.net/users/2620 | Torsors for monoids | One common definition of torsor under a group $G$ is a (**Edit:** nonempty) set $X$ together with an action $act: G \times X \to X$, such that the map $(act,id): G \times X \to X \times X$ is a bijection of sets. The definition is still meaningful if you replace "group" with "monoid".
**Edit:** As Torsten has pointed out, replacing "group" with "monoid" doesn't give you any additional generality, since the bijection endows $G$ with inverses. We can instead ask that the map $(act,id)$ be injective. No, that doesn't specialize correctly to the setting of groups. I think I'll quit while I'm behind, and let this stand as a cautionary tale for others.
Allowing $X$ to be empty yields a notion of pseudotorsor, which seems kind of useless in a bare set-theoretic context. It seems to come up in algebraic geometry, though.
| 2 | https://mathoverflow.net/users/121 | 25886 | 16,968 |
https://mathoverflow.net/questions/25873 | 16 | Let $A$ be a commutative algebra, say over $\mathbb{C}$.
Giving a grading on $A$ corresponds at least morally to giving a $\mathbb{C}^\*$ action on spec(A): $A\_i$ can be thought of as those functions on which $t$ acts by multiplication with $t^i$. Similary a graded $A$ module is just a $\mathbb{C}^\*$ equivariant sheaf.
Now I want to know, if there is also a geometric interpretation of filtered rings/modules.
| https://mathoverflow.net/users/2837 | Geometric interpretation of filtered rings and modules | To a filtered algebra $(A,F)$ one can assign its Rees algebra $R=\bigoplus\_i F\_iA$. It is a graded algebra containing the algebra of polynomials in one variable $\mathbb{C}[t]$ naturally embedded as the subalgebra generated by the element $t\in R\_1$ corresponding to the element $1\in F\_1A$. So the algebra $R$ defines a $\mathbb{C}^\*$-equivariant quasi-coherent sheaf of algebras $\mathcal{R}$ over the affine line $\operatorname{Spec}\mathbb{C}[t]$. The algebra $A$ can be recovered as the fiber of $\mathcal{R}$ at the point $t=1$, and the associated graded algebra $\operatorname{gr}\_FA$ is the fiber of $\mathcal{R}$ at $t=0$. Filtered $A$-modules correspond to $\mathbb{C}^\*$-equivariant quasi-coherent sheaves of modules over $\mathcal{R}$.
The algebra $R$ is a torsion-free $\mathbb{C}[t]$-module, so the quasi-coherent sheaf $\mathcal{R}$ over $\operatorname{Spec}\mathbb{C}[t]$ has to be torsion-free. This description does not take into accout the issue of completeness of the filtration $F$ (in case it extends also in the decreasing direction), which requires a separate consideration.
| 26 | https://mathoverflow.net/users/2106 | 25888 | 16,970 |
https://mathoverflow.net/questions/25821 | 0 | In [a previous question](https://mathoverflow.net/questions/25740/does-every-smooth-integrable-constant-rank-distribution-have-a-basis-in-which-the/25741), I asked an utterly trivial question, which Deane Yang correctly pointed out was utterly trivial. I will now ask a similar question, which is the one I meant to ask last time; I hope it's not similarly trivial.
I am working on $\mathbb R^n$, although in fact any manifold with volume form is good enough. And my question is local: I have an open neighborhood $U \ni 0$, and you are allowed to make it smaller if you want.
Recall that a **constant-rank distribution** $D$ on $U$ is a vector subbundle of the tangent bundle ${\rm T}U$. Let's fix the rank to be $k\leq n$, and suppose that everything is smooth: $\Gamma(D)$ is a $C^\infty$-submodule of $\Gamma({\rm T}U)$. The distribution is **involutive** if $\Gamma(D)$ is a Lie subalgebra of $\Gamma({\rm T}U)$ (over $\mathbb R$ — the Lie bracket on $\Gamma({\rm T}U)$ is not $C^\infty$-linear). The distribution $D$ is **smooth** if $\Gamma(D)$ is a free rank-$k$ module over $C^\infty$, i.e. if $\Gamma(D)$ has a **basis** $\{v\_1,\dots,v\_k\}$ so that $\Gamma(D) = \operatorname{Span}\_{C^\infty}\{v\_1,\dots,v\_k\}$.
So, suppose that on $U \subseteq \mathbb R^n$ I have a smooth involutive rank-$k$ distribution. Given a basis $v\_1,\dots,v\_k \in \Gamma(D)$ (and I will use coordinates $x^1,\dots,x^n$ on $\mathbb R^n$, so I will write $v\_a = \sum\_i v\_a^i(x) \frac{\partial}{\partial x^i}$), then I can define the **structure coefficients** $f\_{a,b}^c(x)$, $a,b,c = 1,\dots,k$, via $[v\_a,v\_b] = \sum\_c f\_{a,b}^c v\_c$, or, in coordinates:
$$ \sum\_i \left( v\_a^i(x)\,\frac{\partial v\_b^j}{\partial x^i} - v\_b^i(x)\,\frac{\partial v\_a^j}{\partial x^i}\right) = \sum\_c f\_{a,b}^c(x)\,v\_c^j(x) $$
>
> *Question:* Does there exist a basis $\{v\_a\}$ for a given smooth involutive constant-rank distribution so that for each $a=1,\dots,k$ and each $x\in U$, we have $\displaystyle \sum\_b f^b\_{a,b}(x) = \sum\_i \frac{\partial v^i\_a(x)}{\partial x^i}$?
>
>
>
For example, by Frobenius theorem (my utterly trivial question), I can find a basis so that the LHS vanishes for each $a$. Or, by [another of my trivial questions](https://mathoverflow.net/questions/22694/does-a-smooth-constant-rank-integrable-distribution-have-a-local-basis-of-d), I could make the basis entirely divergence-free. But I don't think I can simultaneously make the basis consist of divergence-free vector fields.
>
> *Question:* If so, how many choices of such a basis do I have? Clearly ${\rm GL}(k,\mathbb R)$ acts on the set of choices; are there more?
>
>
>
| https://mathoverflow.net/users/78 | Does a smooth, constant-rank, integrable distribution have a basis in which the traces of the structure constants are the divergences of the corresponding basis elements? | The answer to the question in the title is yes. More precisely I claim one may always find a divergence free basis where also all structure constants vanish. The following is a hopefully valid proof by induction on the rank k of the distribution: if k=1 then by an answer to your previous question one may always find a divergence free vector field spanning the distribution. In the case k>1 start by choosing a nowhere zero vector field $v\_1$ which is in the distribution and divergence free. Also pick a n-1 dimensional submanifold $A\_0\subset M$ transversal to $v\_1$ (everything is local near a (fixed) point $p\in M$). On $A\_0$ we have a rank $k-1$ involutive distribution (obtained by restricting the big one) and a volume form obtained by restricting $i\_{v\_1}\omega$ (where $\omega$ denotes the original volume form on $M$). By induction there is a base $\tilde{v}\_2,\ldots,\tilde{v}\_k$ of this distribution on $A\_0$ which is in involution and divergence free for $i\_{v\_1}\omega$. With the flow of $v\_1$ we extend these vector fields to obtain a full basis $v\_1,\ldots,v\_k$ of the rank $k$ distribution on $M$. The obtained basis will be in involution by construction. It remains to check that the $v\_2,\ldots,v\_k$ are divergence free for $\omega$. This follows from the fact that $\omega=dt\wedge i\_{v\_1}\omega$ where $t\in C^\infty(M)$ denotes the function which has value $0$ on $A\_0$ and is otherwise determined by the flow of $v\_1$. (Apply the usual rules for calculus with vector fields and forms.)
But you should really make sure I made no mistake.
If the proof is right it should also allow you to conclude that all manifolds with an integrable distribution and volume form look locally the same. (i.e. like the standard coordinate model). Hence the symmetry group is quite big (infinite dimensional I think).
| 2 | https://mathoverflow.net/users/745 | 25896 | 16,977 |
https://mathoverflow.net/questions/25875 | 2 | Given $k$ an algebraically closed field, I know that that a maximal ideal $\mathfrak{m}$ of $A = k[X\_1,\cdots,X\_n]$ is just a $\langle X\_1-a\_1,\cdots,X\_n-a\_n \rangle $ (Nullstellensatz).
Knowing that, it seems intuitive that $\mathfrak{m}$ can not be generated by less than $n$ elements. Is that true? In that case, how can I show that ?
(Actually, I need that, just after using Nakayama's lemma, to show that $\dim\_{A/\mathfrak{m}=k} \mathfrak{m}/\mathfrak{m}^2$ is greater than $n$.)
(I've seen things that might be relevant like "Krull height theorem" but I think such things take place in a more general context and I have some difficulties, first to understand them, and second to adapt them...)
Thank you.
| https://mathoverflow.net/users/6187 | Generators of a maximal ideal of $k[X_1,\cdots,X_n]$ | Since you mentioned Krull's height theorem (= the generalized principal ideal theorem) and having difficulty applying it, I thought you or someone else might appreciate seeing how this works: it is quite straightforward.
The generalized principal ideal theorem is as follows: let $R$ be a Noetherian ring and $I$ a proper ideal of $R$ which can be generated by $n$ elements. Let $\mathfrak{p}$ be a prime ideal which is minimal among all primes containing $I$. Then $\mathfrak{p}$ has height at most $n$, that is, there *do not* exist prime ideals $\mathfrak{p}\_0,\ldots,\mathfrak{p}\_n$
such that
$\mathfrak{p}\_0 \subsetneq \mathfrak{p\_1} \subsetneq \ldots \subsetneq \mathfrak{p\_n} \subsetneq \mathfrak{p}$.
[For a deduction of this from Krull's Principal Ideal Theorem, see e.g. Theorem 96 on p. 70 of [Commutative algebra](http://alpha.math.uga.edu/%7Epete/integral.pdf).]
Let us apply this with $R = k[x\_1,\ldots,x\_n]$ and the ideal $I = \langle x\_1 - a\_1,\ldots,x\_n - a\_n \rangle$. $I$ is itself a maximal -- hence prime -- ideal, since $R/I \cong k$. Thus the generalized principal ideal theorem simply says that $I$ cannot be generated by fewer elements than its height. But its height is certainly *at least* $n$. No geometry is needed here: just define $\mathfrak{p}\_0 = 0$ and for $1 \leq i \leq n$, $\mathfrak{p}\_i = \langle x\_1 - a\_1,\ldots,x\_i - a\_i \rangle$.
Finally, a comment: I did not use that $k$ was algebraically closed *per se* but only worked with maximal ideals of this particular form. On the other hand, it is still true over an arbitrary field $k$ that every maximal ideal of $k[x\_1,\ldots,x\_n]$ has height $n$ and can be generated by $n$ elements (and no fewer, by Krull's theorem): see Corollary 130 on p. 83 of the document linked to above.
| 7 | https://mathoverflow.net/users/1149 | 25898 | 16,978 |
https://mathoverflow.net/questions/20704 | 8 | In many cases, the recurrence equations that people are solving involves index of only non-negative values. Here I have a recurrence equation which arises from transport of light in an infinite 1D chain:
$a\_m=\sum \_{j=1}^{\infty } \left(T\_ja\_{m+j}+T\_ja\_{m-j}\right) + \delta \_{m,0}$
where $\delta\_{m,0}$ is the Kronecker delta function. i.e.:
$\delta\_{i,j} = \begin{cases} & 1 \text{ if } i=j \\ & 0 \text{ if } i \neq j \end{cases}$
Here I would like to solve $a\_m$, where the index of m is from negative infinity to positive infinity, while $T\_j$ is a given sequence, and p is just a given constant.
Defining the generating function $G(z)=\sum \_{k=-\infty }^{\infty } a\_kz^k$, I found that:
$G(z)=\frac{1}{1-\sum \_{k=1}^{\infty } t\_k\left(z^{-k}+z^k\right)}$
The problem is, how am I going to do series expansion on G? Doing a simple expansion of $\frac{1}{1-\sum \_{k=1}^{\infty } t\_k\left(z^{-k}+z^k\right)}=\sum \_{j=0}^{\infty } \left(\sum \_{k=1}^{\infty } t\_k\left(z^{-k}+z^k\right)\right){}^j$ won't help. Since the power is too difficult to expand out.
And contour integration isn't helping as well, since it is too difficult to compute analytically or numerically too.
Here I would like to ask about direction in obtaining analytical solution, or approximated one.
And in my case, my function G is given by:
$G(z)=\left(1+\frac{3i}{2r^3}\left(r^2\left(\ln \left(1-\frac{e^{i r}}{z}\right)+\ln \left(1-e^{i r}z\right)\right)\right)-i r\left(\text{Li}\_2\left(\frac{e^{i r}}{z}\right)+\text{Li}\_2\left(e^{i r}z\right)\right)+\text{Li}\_3\left(\frac{e^{i r}}{z}\right)+\text{Li}\_3\left(e^{i r}z\right)\right){}^{-1}$
p.s.:I have posted the same problem in [Voofie](http://www.voofie.com/content/44/solving-recurrence-equation-with-indexes-from-negative-infinity-to-positive-infinity/).
| https://mathoverflow.net/users/5217 | Solving recurrence equation with indexes from negative infinity to positive infinity | I see your problem more like a linear equation on an infinite dimensional space of doubly infinite sequences than as a recurrence equation, since there are no initial values from which start to build up the solution. In the following I will assume that $\sum\_{j=1}^\infty|t\_j|<\infty$ and that $\mathbf{a}=(a\_m)$ is bounded. The linear operator $T$ defined on $\ell^\infty(\mathbb{Z})$ by
$$
(T\mathbf{a})\_m=\sum\_{j=1}^\infty t\_j(a\_{m+j}+a\_{m-j})
$$
is bounded with operator norm $\|T\|\le2\sum\_{j=1}^\infty|t\_j|$. Your equation can be written as
$$
(I-T)\mathbf{a}= \mathbf{b},
$$
where $I$ is the identity operator and $\mathbf{b}\in\ell^\infty(\mathbb{Z})$. If $\sum\_{j=1}^\infty|t\_j|<\dfrac{1}{2}$, then $I-T$ is invertible and the above equation has a unique solution for all $\mathbf{b}\in\ell^\infty(\mathbb{Z})$, given by
$$
\mathbf{a}=(I-T)^{-1}\mathbf{b}=(I+T+T^2+T^3+\dots)\mathbf{b}.
$$
This is an explicit formula that in practice may be useless, although one can get an approximation by computing a few terms of the sum.
Just to check that this can really give a solution, let's study the particular case in which $t\_1=t$ and $t\_j=0$ for all $j>1$, and $(\mathbf{b})\_m=\delta\_{m,0}$. Then we find that
$$
a\_m=a\_{-m}=\sum\_{k=1}^\infty\binom{2k+m}{k+m}t^{2k+m}=t^m{}\_2F\_1\left(\frac{m+1}{2},\frac{m+2}{2},m+1,4t^2\right),
$$
where ${}\_2F\_1$ is the hypergeometric function. We see that indeed one must have $|t|<\dfrac{1}{2}$ for this to make sense.
This analysis doesn't mean that there are no other solutions under different conditions, but I think that it will be difficult to avoid the requirement that $\sum\_{j=1}^\infty|t\_j|<\infty$.
| 4 | https://mathoverflow.net/users/1168 | 25900 | 16,980 |
https://mathoverflow.net/questions/25897 | 0 | **Problem**
Consider the following data set:
```
YEAR;AMOUNT;MEASUREMENTS
1985;9.53013698630137;365
1986;11.086301369863;365
1987;13.0712328767123;365
1988;11.9248633879781;366
1989;10.2191780821918;365
1990;7.41933085501859;269
1991;12.1751396648045;358
1992;9.7037037037037;108
1993;13.1452261306533;199
1994;8.70697674418605;215
1995;10.5224615384615;325
1996;7.59776536312849;358
1997;10.5065753424658;365
1998;10.3983561643836;365
1999;12.971381031614;601
2000;10.3513661202186;732
```
The years 1990 and 1992 to 1994 have a low measurement count.
**Update: Context**
I am creating a system that allows the general public to create charts on climate. The purpose of the chart is to show general climate trends (through linear and non-linear regression analysis). My concern is that insufficient measurements taken throughout the year will skew the data in misleading ways.
I am using statistical analysis software for the PostgreSQL database (i.e., PL/R) to perform the calculations. I do not know if I can tell PL/R to "give less weight" to annual averages with fewer than 365 measurements. Also, if all the measurements were made in winter, then it seriously does not reflect the average maximum temperature for the year -- not even close. I also do not know if I can tell PL/R to assign weight based on when during the year the majority of measurements were made.
I have created two charts illustrating the difference between removing and keeping the sub-200 measurement counts:
* [Chart 1](https://i.imgur.com/Gnpdn.png)
* [Chart 2](https://i.stack.imgur.com/6hgeq.png)
**Questions**
1. Should the short years be excluded from the list?
2. If 365 measurements is the norm, what are the minimum number of measurements needed to be statistically significant (i.e., not skew the correlation)?
Thank you!
| https://mathoverflow.net/users/5908 | When to throw away data | I cannot comment due to rep requirements but Alekk's comment reg weight- If you use hierarchical bayesian (also known as multi-level models) ideas to process the data then the observations coming from years 1990 and 1992 to 1994 will receive lower weight when estimating the parameters of interest.
In general, throwing out data is not a good idea as there is some information (perhaps it is weak) in those years.
**Update/Edit**
Disclaimer: I do not know anything about climate science. FWIW, my thoughts are below:
In response to your updated context, you need a way to model the climate data to account for seasonality and time trends in your data. The typical way to model time series data is to use [autoregressive models](http://en.wikipedia.org/wiki/Autoregressive_model). Typically, these models assume stationarity which means that the data is assumed to fluctuate around a long term average. In the context of climate science, you will have to relax this somewhat to see if the process in fact has a long term average that is increasing/decreasing. I do not think there is a canned routine that will do this for you but I could be wrong. I would encourage you to discuss the issue with a climate researcher.
| 4 | https://mathoverflow.net/users/4660 | 25908 | 16,985 |
https://mathoverflow.net/questions/25901 | 9 | This might be standard, but I have not seen it before:
Let $K$ be an algebraically closed field (of characteristic 0 if necessary). Let $G$ be the orthogonal group ${\bf O}(m)$ or the symplectic group ${\bf Sp}(2n)$, and let $\mathfrak{g}$ be its Lie algebra. Is there a classification of the adjoint orbits of $G$ acting on $\mathfrak{g}$? An answer for the special orthogonal group ${\bf SO}(m)$ is also welcome. I am aware of the classification of nilpotent and semisimple orbits.
I am looking for an answer kind of like Jordan normal form in the case that $G$ is the general linear group. My guess is that the classification is Jordan normal form plus some other invariant.
| https://mathoverflow.net/users/321 | Classification of adjoint orbits for orthogonal and symplectic Lie algebras? | "Classification" can mean more than one thing, but it's useful to be aware of the extensive development of *adjoint quotients* by Kostant, Steinberg, Springer, Slodowy, and others. This makes sense for all semisimple (or reductive) groups and their Lie algebras over any algebraically closed field, perhaps avoiding a few small prime characteristics. Older sources include Steinberg's 1965 IHES paper on regular elements ([MSN](http://www.ams.org/mathscinet-getitem?mr=180554) and [article](http://www.numdam.org/item?id=PMIHES_1965__25__49_0)) and the Springer–Steinberg portion of the
1970 Lecture Notes in Math. vol. 131 ([MSN](http://www.ams.org/mathscinet-getitem?mr=268192) and [chapter](https://link.springer.com/chapter/10.1007/BFb0081546)). (For an overview with references, based partly on Steinberg's Tata lectures, see Chapter 3 of my 1995 AMS book *Conjugacy Classes in Semisimple Algebraic Groups* ([MSN](http://www.ams.org/mathscinet-getitem?mr=1343976), [book](http://dx.doi.org/10.1090/surv/043), and [chapter](http://www.ams.org/books/surv/043/04)).) While the general focus has been on developing a picture of the collection of all classes or orbits as some kind of "quotient", quite a few special features of the classical types are also brought out in the Springer-Steinberg notes. As suggested by Victor, Roger Carter's book *Finite Groups of Lie Type* ([MSN](http://www.ams.org/mathscinet-getitem?mr=794307)) also has a lot of related material but with special emphasis on nilpotent orbits. The Jordan decomposition does reduce many classification questions to the nilpotent case, at least in principle, if you are willing to deal with various centralizers along the way.
[ADDED] The paper in J. Math. Physics linked below gives a nice concrete answer to the original question, building on some of the older theory but using mainly tools from linear algebra and basic group theory. This is the traditional approach of most physicists, though papers in this mixed journal are sometimes unreliable and contain mathematics of the sort probably not usable in physics but also not publishable in math journals. Djokovic and his collaborators are more reliable than most, fortunately, and he has written many papers using parts of Lie theory as well. One downside is the narrower perspective than found in the notes of Springer–Steinberg, for instance. But it all depends on whether you want to work over other fields or want to organize the classes/orbits more conceptually.
Here is a MathSciNet reference:
>
> MR708648 (85g:15018) 15A21 (17B99 17C99)
> Djokovic´,D. Zˇ . [¯Dokovic´, Dragomir Zˇ .] (3-WTRL); Patera, J. [Patera, Jirˇ´ı] (3-MTRL-R);
> Winternitz, P. (3-MTRL-R); Zassenhaus, H. (1-OHS)
> Normal forms of elements of classical real and complex Lie and Jordan algebras.
> J. Math. Phys. 24 (1983), no. 6, 1363–1374 (review by R.C. King).
>
>
>
([MSN](http://www.ams.org/mathscinet-getitem?mr=708648) and [article](http://dx.doi.org/10.1063/1.525868).) Their references include the work by Burgoyne–Cushman, Milnor, Springer-Steinberg mentioned by Bruce and me.
| 6 | https://mathoverflow.net/users/4231 | 25914 | 16,990 |
https://mathoverflow.net/questions/25894 | 9 | Apologies for not knowing exactly what I'm looking for, but I'd appreciate any general pointers to get me started.
I'm interested in efficient representations (graphical and otherwise) of finite graphs with repeated structure --- something like coding theory for graph structures. For example, an $N$ by $N$ grid graph where each non-boundary vertex has four neighbors (up, right, left, down) could be represented with far fewer than O($N^2$) parameters. A $N$ by $N$ grid where each non-boundary vertex has 8 neighbors (up left, up, up right, ...) should take more parameters to represent, but not many more. I'm imagining a notation or diagram that uses a base template, then defines a local connectivity pattern and a repetition structure.
I'd like to be able to ask (and start to answer) questions like:
1. What is the minimum number of parameters needed to describe a particular graph structure?
2. Are there some problems that become fixed-parameter tractable for graphs that can be described with $k$ parameters?
Is there a branch of graph theory that studies these types of questions? I'd be grateful for pointers to starting points or related work.
| https://mathoverflow.net/users/2785 | Representing repeated structure in graphs | I am not sure what exactly is meant by a *repeated structure* but at least some covering graphs should qualify. Covering graphs may be quite large, however they admit concise description via *voltage graphs*. The edges of a small base graph or voltage graph are equipped with group elements, called *voltages*. For instance, an n x n tessellation of a torus can be described by a single vertex with two loops and group elements (0,1) and (1,0) taken from the group $Z\_n \times Z\_n$ assigned to the loops.
Of course most graphs admit no compression in this way.
| 3 | https://mathoverflow.net/users/4400 | 25930 | 16,998 |
https://mathoverflow.net/questions/25922 | 14 | Suppose that $X$ and $Y$ are smooth projective varieties which are birationally equivalent. I would like to have that $$\textrm{deg} \ \textrm{td}(X) = \textrm{deg} \ \textrm{td}(Y).$$ Invoking the Hirzebruch-Riemann-Roch theorem, this boils down to showing that $$ \chi(X,\mathcal{O}\_X) = \chi(Y,\mathcal{O}\_Y).$$
This is probably a basic fact. A stronger statement is apparently shown in *Birationale Transformation von linearen Scharen auf algebraischen Mannigfaltigkeiten* by van der Waerden. The only problem is that I can't seem to find it in that article (probably because I don't read that well German).
For $\dim X =2$ one can prove this as Hartshorne does as follows.
Any birational transformation of nonsingular projective curves can be factored into a sequence of monoidal transformations and their inverses. For such a monoidal transformation, the result follows from Proposition 3.4 in Chapter V of Hartshorne.
Does this work in the general case?
| https://mathoverflow.net/users/4333 | Is the Euler characteristic a birational invariant | If you are willing to stick to characteristic zero, then you can assume that there is actually a morphism $f\colon X\longrightarrow Y$ realizing the birational equivalence (reason: look at the graph $\Gamma\subset X\times Y$ realizing the birational equivalence and take its closure, use resolution of singularities to resolve $\Gamma$, and then replace $X$ by $\Gamma$). In this case, $f\_{\*}\mathcal{O}\_{X}=\mathcal{O}\_{Y}$, and all higher direct images are zero, the Leray spectral sequence then implies that the Euler characteristics are equal.
More generally, if $Y$ has *rational singularities* and $f\colon X\longrightarrow Y$ is a proper birational map, with $X$ smooth, then $f\_{\*}\mathcal{O}\_{X}=\mathcal{O}\_Y$ and all higher direct images are zero (this is the definition of rational singularities) and so the same conclusion follows. Smooth varieties have rational singularities! (The computation for smooth varieties is necessary to show that the definition makes sense, i.e., that checking that this property holds for one resolution $X$ implies that it holds for all resolutions).
| 14 | https://mathoverflow.net/users/1055 | 25934 | 17,000 |
https://mathoverflow.net/questions/25878 | 12 | This question is the outcome of a few naive thoughts, without reading the proof of Gelfand-Neumark theorem.
Given a compact Hausdorff space $X$, the algebra of complex continuous functions on it is enough to capture everything on its space. In fact, by the Gelfand-Neumark theorem, it is enough to consider the commutative C\*-algebras instead of considering compact Hausdorff spaces.
The important thing here is that $C\*$-algebras have a *complex structure*. The real structure is not enough. Given the algebra $C(X, \mathbb R) \oplus C(X, \mathbb R)$ of real continuous functions on $X$, the algebra $C(X, \mathbb{C})$ is simply the direct sum $C(X, \mathbb R) \oplus C(X, \mathbb R)$, as a Banach algebra(and this can be given a complex structure, (seeing it as the complexification...)). But to obtain a C\*-algebra, we need an additional C\*-algebra, and the obvious way, ie, defining $(f + ig)$\* $= (f - ig)$ does not work out. More precisely, the C\* identity does not hold.
So one cannot weaken (as it stands) the condition in the Gelfand-Neumark theorem that we need the algebra of *complex* continuous functions on the space $X$, since we do need the C\* structure. Of course, this is without an explicit counterexample. Which brings us to:
>
> Qn 1. Please given an example of two non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that the function algebras $C(X, \mathbb R)$ and $C(Y, \mathbb R)$ are isomorphic(as real Banach algebras)?
>
>
>
(Here I am hoping that such an example exists).
Then again,
>
> Qn 2. From the above it appears that the structure of complex numbers is involved when the algebra of complex functions captures the topology on the space. So how exactly is this happening?
>
>
>
(The vague notions concerning this are something like: the complex plane minus a point contains nontrivial $1$-cycles, so perhaps the continuous maps to the complex plane might perhaps capture all the information in the first homology, etc..)..
**Note** : Edited in response to the answers. Fixed the concerns of Andrew Stacey, and changed Gelfand-Naimark to Gelfand-Neumark, as suggested by Dmitri Pavlov.
| https://mathoverflow.net/users/6031 | Relevance of the complex structure of a function algebra for capturing the topology on a space. | Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff. For each closed subset $K$ of $X$, define $\mathcal{I}\_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$. The map $K\mapsto\mathcal{I}\_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$. Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required. I remember doing this as an [exercise](http://books.google.com/books?id=-OdfXeNmrT0C&lpg=PR3&dq=douglas%2520banach&pg=PA54#v=onepage&q&f=false) in Douglas's *Banach algebra techniques in operator theory* in the complex setting, but the same proof works in the real setting.
---
Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof. Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)
I will take it for granted that each $\mathcal{I}\_K$ is a closed ideal. This doesn't require that the space is Hausdorff (nor that $K$ is closed). Suppose that $K\_1$ and $K\_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K\_2\setminus K\_1$. Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K\_1$ but $f(x)=1.$ Thus, $f$ is in $\mathcal{I}\_{K\_1}\setminus\mathcal{I}\_{K\_2}$, and this shows that $K\mapsto \mathcal{I}\_K$ is injective. The work is in showing that it is surjective.
Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K\_\mathcal{I}=\cap\_{f\in\mathcal{I}}f^{-1}(0)$, so that $K\_\mathcal{I}$ is a closed subset of $X$. *Claim:* $\mathcal{I}=\mathcal{I}\_{K\_\mathcal{I}}$.
It is immediate from the definition of $K\_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K\_\mathcal{I}$, so that $\mathcal{I}\subseteq\mathcal{I}\_{K\_\mathcal{I}}.$ Let $f$ be an element of $\mathcal{I}\_{K\_\mathcal{I}}$. Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|\_\infty<3\epsilon$. Define $U\_0=f^{-1}(-\epsilon,\epsilon)$, so $U\_0$ is an open set containing $K\_\mathcal{I}$. For each $y\in X\setminus U\_0$, because $y\notin K\_\mathcal{I}$ there is an $f\_y\in \mathcal{I}$ such that $f\_y(y)\neq0$. Define $$g\_y=\frac{f(y)}{f\_y(y)}f\_y$$ and $U\_y=\{x\in X:|g\_y(x)-f(x)|<\epsilon\}$. Then $U\_y$ is an open set containing $y$. The closed set $X\setminus U\_0$ is compact, so there are finitely many points $y\_1,\dots,y\_n\in X\setminus U\_0$ such that $U\_{y\_1},\ldots,U\_{y\_n}$ cover $X\setminus U\_0$. Relabel: $U\_k = U\_{y\_k}$ and $g\_k=g\_{y\_k}$. Let $\varphi\_0,\varphi\_1,\ldots,\varphi\_n$ be a partition of unity subordinate to the open cover $U\_0,U\_1,\ldots,U\_n$. Finally, define $g=\varphi\_1 g\_1+\cdots+\varphi\_n g\_n$. That should do it.
In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points. (Maximal ideals are actually always closed in a Banach algebra, real or complex.)
| 7 | https://mathoverflow.net/users/1119 | 25942 | 17,007 |
https://mathoverflow.net/questions/25944 | 8 | Suppose $F\_1$ and $F\_2$ are free groups, and suppose $\alpha:F\_1 \to F\_2$ is a surjective homomorphism. Then, because $F\_2$ is free, the homomorphism splits, and we get a subgroup $H$ of $F\_1$ isomorphic to $F\_2$ and a retraction of $F\_1$ onto $H$, i.e., a surjective map to $H$ that restricts to the identity on $H$ (with kernel a normal complement to $H$).
Question: Can we find a freely generating set $A$ for $F\_1$ and a freely generating set $B$ for $H$ such that $B$ is a subset of $A$ and the retraction sends all elements of $B$ to themselves and sends all elements of $A \setminus B$ to the identity element?
The corresponding statement for free abelian groups is true: simply pick a (free abelian) generating set for the retraction image and the kernel and take their union to get a freely generating set for the whole group. [Note: Any subgroup of a free abelian group is free abelian.] But this technique of taking a freely generating set for the kernel fails in the non-abelian case, because the kernel is too big.
| https://mathoverflow.net/users/3040 | Does every retraction of free groups arise from projection to a subset of a freely generating set? | No. This is explicitly stated in the paragraph above Theorem 1 of:
[Turner, Edward C, Test words for automorphisms of free groups.
Bull. London Math. Soc. 28 (1996), no. 3, 255--263.](http://blms.oxfordjournals.org/cgi/reprint/28/3/255.pdf)
The author refers to Proposition 1.
**EDIT:**
Let's give an explicit example. Let $F=\langle a,b\rangle$ and let $g=a[b,a]=ab^{-1}a^{-1}ba$. Clearly $\langle g\rangle$ is a retract of $F$. But the Whitehead graph of $g$ is two triangles glued along an edge. This is Whitehead-reduced, so $g$ is not part of a free basis for $F$.
| 13 | https://mathoverflow.net/users/1463 | 25945 | 17,008 |
https://mathoverflow.net/questions/25943 | 16 | The following questions occurred to me.
This is not research mathematics, just idle curiosity.
Apologies if it is inappropriate.
1. Suppose you have a fixed volume *V* of maleable material,
perhaps clay.
The goal is to form it into a shape *S* (convex or nonconvex) that would roll down an inclined
plane as fast as possible.
The plane is tilted at θ with respect to the horizontal.
The race track that quantifies "as fast as possible"
is of length *L*. The shape *S* must be
entirely behind a starting plane orthogonal to the inclined plane,
and its race is finished when it is entirely ahead of a finishing plane,
*L* distant, again orthogonal to the inclined plane.
So if *S* is a disk of radius *r*, its center of
gravity will have to travel a total distance of *L* + 2*r*
to complete the race. *S* is released at rest, and rolls
under gravity. Assume that the materials of *S* and of the
inclined plane have a sufficient coefficient of friction μ
such that there is no slippage, just pure rolling.
The stipulation that *S* start behind and end
ahead of the orthogonal planes suggests that the optimal shape
is an arbitrarily thin (and therefore arbitrarily long) cylinder, to minimize *r*.
But this does not accord with (my) physical intuition.
What am I missing here?
2. A variation on the same problem replaces
the inclined plane with an inclined half-cylinder of radius *R*, like a rain gutter.
It may be necessary to assume a relation between *R* and the volume *V* of material
to make this problem reasonable (so that *S* fits in the gutter). But certainly this version's
answer could not be an arbitrarily long cylinder!
| https://mathoverflow.net/users/6094 | Fastest Rolling Shape? | There is one thing that's way more important than the precise definition of how the race starts and ends: the moment of intertia of $S$, which determines how much energy is wasted on the rotation. Let us assume that $S$ is cylindrically symmetrical with mass $m$, radius $R$, moment of intertia $I$, and let $k=I/mR^2$. Then from the conservation of energy,
$$\frac{I\omega^2}{2}+\frac{mv^2}{2}=mgx\sin\theta,$$
and using that $v=\omega R,$ we get that
$$\frac{dx}{dt}=v=\sqrt{\frac{2g\sin\theta}{k+1}}x^{1/2}.$$
This already makes it clear that in order to minimize the rolling time, $k$ must be as small as possible, but this can also be quantified by solving the separable ODE and finding that
$$T=\sqrt{\frac{2(k+1)d}{g\sin\theta}}\geq \sqrt{\frac{2L}{g\sin\theta}},$$
where $d=L+2R$ is the distance traveled. The conclusion, in the cylindrically symmetrical case of radius $R$, is that $S$ should be as close as possible to the infinitely thin heavy rod with a disk-shaped thin "flap" of radius $R$ attached, so that $k$ is just slightly over $1$. In a practical sense, imagine a wheel pair with very light rims and a very heavy axle. [I like to think of this situation as "anti-flywheel problem": we want to *minimize* rotation energy accumulated by $S$, so contrary to the usual flywheel, the wheel is light and the axle is heavy.] If your "malleable material" has uniform density and $R$ may vary, one sequence of mathematical solutions approaching the time bound consists of solid cylinders with radii $r\_i$ shrinking to 0 and growing widths $\ell\_i$, with two $w$-thin disks of radius $R\_i$ attached, where
$$\pi(\ell\_i r\_{i}^2+2w R\_i^2)=V, \quad r\_i=o(R\_i), \quad R\_i\to 0.$$
[I've only put in an explicit sequence because of a clarification request in the comments; I actually think that this level of detail hinders understanding]. It's also clear that the cylindrical symmetry restriction is irrelevant.
The second problem is more difficult: I don't have much to say on it other than that, by the same token, a solid ball is better than a disk/cylinder even if it could be fitted in the "gutter".
P.S.: "Physical" intuition may be deceptive: many people assume that a ball will roll according to the same law of motion as a cylinder or an ideal point mass. This actually happened to some of my mathematically oriented classmates in a physics lab assignment, who tried to make up data in order to avoid doing tedious experiments.
| 11 | https://mathoverflow.net/users/5740 | 25950 | 17,012 |
https://mathoverflow.net/questions/25956 | 10 | Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.
In [my
answer](https://mathoverflow.net/questions/25794/shortest-most-elegant-proof-for-l1-chi-neq-0/25822#25822) to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples
of the real characters modulo 3 and 4. The proof goes as follows: one takes
$$
f(x)=\sum\_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum\_{j=1}^{m-1}\chi(j)x^j
$$
and uses Abel's theorem to write
$$
L(1,\chi)=\int\_0^1f(x)dx;
$$
since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.
Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of
the polynomial
$$
g\_\chi(x)=\sum\_{n=1}^{m-1}\chi(n)x^n
$$
on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works
perfectly further; for example,
$$
g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5
$$
or
$$
g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.
$$
Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is
always holds for small values $m\le30$.
**Question.** Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$,
is it true that $g\_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for
which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed
in the literature?
| https://mathoverflow.net/users/4953 | Positivity of $L(1,\chi)$ for real Dirichlet's character | These are called Fekete polynomials, and you can find out a great deal about them [here](http://www.dms.umontreal.ca/~andrew/PDF/fekete.pdf). Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.
| 5 | https://mathoverflow.net/users/1464 | 25958 | 17,017 |
https://mathoverflow.net/questions/25911 | 7 | For some context see [Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance](https://mathoverflow.net/questions/25846/random-walks-in-z2-z2-intrinsic-characterization-of-euclidean-distance)
As per Noah's answer and JBL's comment this was false as stated. However, I think the following reformulation is interesting.
As before we consider a random walk on $\mathbb{Z}^2$ where a particle either stays at its vertex or moves to a neighbor with probability 1/5. We start the process with a particle at the origin. For $x \in \mathbb{Z}^2$ we let $p\_n(x)$ denote the probability that we find the particle at $x$ after $n$ iterations. Let $\left|\cdot\right|$ denote the Euclidean distance of two points in $\mathbb{Z}^2$ via the standard embedding of $\mathbb{Z}^2 \subset \mathbb{R}^2$.
Now for the reformulated question: For each $n$, let $C\_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| \leq \left|y\right| \Rightarrow p\_n(x) \geq p\_n(y)$. }
$$
Does $\lim\_{n\to\infty} C\_n = \infty$? If so, how fast does this diverge?
EDIT: As per George Lowther's comment, I now find it quite probable that $\lim\inf\_{n\to\infty} C\_n \leq 5$ if not $C\_n = 5$ for all large $n$.
A natural attempt to salvage the question is the following: For each $n$, let $\tilde{C}\_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| < \left|y\right| \Rightarrow p\_n(x) > p\_n(y)$. }
$$
Again we ask if $\lim\_{n\to\infty} \tilde{C}\_n = \infty$ and if so, how fast this diverges.
| https://mathoverflow.net/users/4345 | Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance Part II | As I mentioned in my comment - what you are suggesting implies that the probability of being at (3,4) is the same as being at (5,0) for all large $n$. That seems unlikely, and would guess that $C\_n=5$ for $n$ large.
---
The answer to your modified question is yes! $\tilde C\_n$ tends to infinity as $n$ goes to infinity. (Phew! It took me a couple of revisions to prove this, but hopefully the calculations below are now correct).
In fact, $\tilde C\_n\ge c\sqrt{n}$ for some positive constants c. I think that you can also show that $\tilde C\_n\le C\sqrt{n}$ for some other constant $C$ but I'm not completely sure yet, although it should follow from a closer examination of my expression below for $p\_n(x)$.
You can derive an asymptotic expansion for $p\_n(x)$ in 1/n. Evaluating this to second order is enough to answer your question. After $n$ steps the distribution of the particle will be approximately normal with variance 2n/5 in both dimensions, so we expect to get $p\_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}$ to leading order.
The idea is to note that you are repeatedly applying a linear operator,
$$
p\_{n+1}=Lp\_n,\ Lp(x) \equiv (p(x)+p(x-e\_1)+p(x+e\_1)+p(x-e\_2)+p(x+e\_2))/5
$$
where $e\_1=(1,0)$, $e\_2=(0,1)$. In finite dimensional spaces, you would solve this by decomposing $p\_0$ into a sum of eigenvectors and for large n, the dominant term of $L^np\_0$ will be that corresponding to the largest eigenvalue. In this case, the infinite dimensional operator $L$ has a continuous spectrum, and is diagonalized by a Fourier transform.
$$
p\_0(x)=1\_{\lbrace x=0\rbrace}=\int\_{-[\frac12,\frac12]^2}e^{2\pi ix\cdot u}\,du.
$$
Noting that $e^{2\pi ix\cdot u}$ (as a function of $x$) is an eigenvector of $L$,
$$
Le^{2\pi ix\cdot u}=\left(\frac15+\frac25\cos(2\pi u\_1)+\frac25\cos(2\pi u\_2)\right)e^{2\pi ix\cdot u}
$$
gives the following for $p\_n$,
$$
p\_n(x)=L^np\_0(x)=\int\_{[-\frac12,\frac12]^2}\left(\frac15+\frac25\cos(2\pi u\_1)+\frac25\cos(2\pi u\_2)\right)^ne^{2\pi ix\cdot u}\,du.
$$
The term inside the parentheses is less than 1 in absolute value everywhere away from the origin, so looks like a Dirac delta when raised to a high power n. Using a Taylor expansion to second order,
$$
\left(\frac15+\frac25\cos(2\pi u\_1)+\frac25\cos(2\pi u\_2)\right)^n
=e^{-\frac45\pi^2n\vert u\vert^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u\_1^2u\_2^2)+O(n\vert u\vert^6)\right).
$$
This expansion is valid over any domain on which $n\vert u\vert^6$ is bounded. Say, $\vert u\vert\le n^{-1/6}$. Outside of this domain, the integrand above is bounded by $e^{-cn(n^{-1/6})^2}=e^{-cn^{2/3}}$ for a constant c, which is much smaller than O(1/n^3) and can be neglected. Then,
$$
p\_n(x)=\int\_{\mathbb{R}^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u\_1^2u\_2^2)+O(n\vert u\vert^6)\right)e^{-\frac45\pi^2n\vert u\vert^2+2\pi ix\cdot u}\,du.
$$
Here I not only substituted in the second order approximation to the integrand, but also extended the range of integration out to infinity. This is fine, because it can be shown that the value of this integral over $\vert u\vert\ge n^{-1/6}$ has size of the order of no more than $e^{-cn^{2/3}}$, so vanishes much faster than $O(1/n^3)$. Substituting in $v=\sqrt{\frac{8n}{5}}\pi u$ also shows that the $O(nu^6)$ term in the integrand vanishes at rate $1/n^3$, giving the following.
$$
p\_n(x)=\frac{5}{8\pi^2n}\int\_{\mathbb{R}^2}\left(1+\frac{1}{24n}(7\vert v\vert^4-20v\_1^2v\_2^2)\right)e^{-\frac12\vert v\vert^2+i\sqrt{\frac{5}{2n}}x\cdot v}\,dv+O(n^{-3}).
$$
This integral can be computed,
$$
p\_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{1}{24n}\left(36-\frac{90}{n}\vert x\vert^2+\frac{175}{4n^2}\vert x\vert^4-\frac{125}{n^2}x\_1^2x\_2^2\right)\right)+O(n^{-3}).
$$
This is a bit messy, but the exact coefficients are not too important. What matters is the general form of the expression.
The leading order term also agrees with the guess above based on it being approximately normal.
Also, for any fixed $\vert x\vert \lt\vert y\vert$, the leading order term in $p\_n(x)-p\_n(y)$ will dominate for large n, giving $p\_n(x)\gt p\_n(y)$. So, $\tilde C\_n\to\infty$.
Consider $\vert x\vert\le c\sqrt{n}$ for some $c\le1$. Then,
$$
p\_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{3}{2n}\right)+O(c^2n^{-2}).
$$
If $\vert x\vert\lt\vert y\vert\le c\sqrt{n}$ then $\vert y\vert^2-\vert x\vert^2\ge 1$ (as it is a nonzero integer)
$$
\begin{align}
p\_n(x)-p\_n(y)&=\frac{5}{4\pi n}\left(1+\frac{3}{2n}\right)e^{-\frac{5}{4n}\vert x\vert^2}\left(1-e^{-\frac{5}{4n}(\vert y\vert^2-\vert x\vert^2)}\right)+O(c^2n^{-2})\\
&\ge\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}(1-e^{-\frac{5}{4n}})+O(c^2n^{-2})\\
&=\frac{25}{16\pi n^2}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+O(c^2)\right).
\end{align}
$$
As long as $c$ is chosen small enough that the $O(c^2)$ term is always greater than -1, this expression will be positive. So $p\_n(x)\gt p\_n(y)$ for all $\vert x\vert\lt\vert y\vert\le c\sqrt{n}$, giving $\tilde C\_n\ge c\sqrt{n}$.
| 8 | https://mathoverflow.net/users/1004 | 25959 | 17,018 |
https://mathoverflow.net/questions/25826 | 2 | The Nakai-Moishezon criterion states that a line bundle $L$ over a surface $X$ is ample iff $L \cdot L > 0$ and $L \cdot C > 0$ for every curve $C$.
We can use this criterion to check that if $X$ is the product of two elliptic curves, then lots of divisors of $X$ are not ample. The fibers of the projection maps of $X$ to its factors have zero self intersection and hence cannot be ample.
Question: is there an Abelian surface such that everyone of its curves is ample?
This is what I attempted. I don't believe it leads anywhere, tough...
Suppose $X$ is an Abelian surface that is not the product of two elliptic curves. Suppose that $C\_1$ and $C\_2$ are two curves in $X$ representing different homology classes. Then, they must intersect [fix an element $\theta \in X$ such that $\theta$ sends $C\_1$ to a curve that intersects $C\_2$...]. So, all that matters is to check that $C\_1 \cdot C\_1 > 0$.
We do it by contradiction. Assume that $C\_1 \cdot C\_1 = 0$. By acting with the inverse of a point of $C\_1$ on $C\_1$, we can assume that the identity element of $X$ is in $C\_1$. Since $C\_1 \cdot C\_1 = 0$, $C\_1$ is a subgroup of $X$, furthermore, it is smooth [just act on $C\_1$ with $C\_1$ itself]. So, we have a mapt $X \rightarrow X/C\_1$, a elliptic fibration of $X$ with elliptic fibers. If this was a trivial HOLOMORPHIC bundle then we would get the contradiction we sought. But that is very unlikely to be the case.
| https://mathoverflow.net/users/3314 | Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory) | This answer exists simply to record that BCnrd and Bjorn Poonen gave excellent answers in the comments above. If someone votes up my answer, this will be removed from this list of unanswered questions. (And, as I have made this answer CW, I will not gain any reputation.)
| 5 | https://mathoverflow.net/users/297 | 25962 | 17,021 |
https://mathoverflow.net/questions/25919 | 0 | Suppose $f$ is a continuous function of infinitely many real variables, and that 0 is an "identity element" for $f$ in the sense that
$$ f(0,\alpha,\beta,\gamma,\dots) = f(\alpha,\beta,\gamma,\dots). $$
Has anyone thought about the following limit (in particular, is there anything in the literature on it)?
$$ \lim\_{\Delta\alpha\to0}\frac{f(\Delta\alpha,\ \alpha,\ \beta,\ \gamma,\dots) - f(\alpha+\Delta\alpha,\ \beta,\ \gamma,\ \dots) }{\Delta\alpha} $$
In case it makes anyone feel any better, for my purposes it may suffice to assume all but finitely many of the variable are zero (but of course with no prior finite bound on how many nonzero ones there are).
| https://mathoverflow.net/users/6316 | A derivative of sorts? | I think Willie Wong's answer here suffices. This is just a case of looking at something from a suddenly different point of view and missing the obvious since it's not the way I'd been looking at it before.
I had an occasion to think about the difference
$$
f(\alpha,\beta,\gamma,\delta,\dots) - f(\alpha+\beta,\gamma,\delta,\dots)
$$
and was just playing around with that.
Continuity in the first variable is enough for the present purpose; I mentioned it only to rule out an obvious reason why the limit might not exist.
The functions I was thinking about were somewhat similar to the sum of products I mentioned in one of my comments above.
| 0 | https://mathoverflow.net/users/6316 | 25969 | 17,025 |
https://mathoverflow.net/questions/25977 | 16 | the definition of compact space is: A subset K of a metric space X is said to be compact if every open cover of K contains finite subcovers. What is the meaning of defining a space is "compact". I found the explanation on wikipedia : "In mathematics, more specifically general topology and metric topology, a compact space is an abstract mathematical space in which, intuitively, whenever one takes an infinite number of "steps" in the space, eventually one must get arbitrarily close to some other point of the space. " I can not understand it, or at least I can not get this by its definition. Can anybody help me?
| https://mathoverflow.net/users/6324 | How to understand the concept of compact space | Some heuristic remarks are helpful only to a subset of readers. (Maybe that's true of all heuristics, as a meta-heuristic - if everyone accepts a rough explanation, it's something rather more than that.)
Non-compactness is about being able to "move off to infinity" in some way in a space. On the real line you can do that to the left, or right: but bend the line round to fill all but one point on a circle (which is compact) and you see the difference having the "other point" near which you end up. This example of real line versus circle is too simple, really. Another way you can "go off to infinity" in a space is by having paths branching out infinitely (as in König's lemma, which supplies another kind of intuition).
Compactness is a major topological concept because the various ways you might try to "trap" movement within a space to prevent "escape" to infinity can be summed up in a single idea (for metric spaces, let's say). The definition by open sets is cleaner, but the definition by sequences having to accumulate on themselves (not necessarily to converge, but to have at least one convergent subsequence) is somewhat quicker to say. If you restrict attention to spaces that are manifolds, you can think of continuous paths and whether they have to wind back close to themselves or not.
| 20 | https://mathoverflow.net/users/6153 | 25981 | 17,032 |
https://mathoverflow.net/questions/25978 | 0 | I heard that there is a result which is proved that RL\subseteq L^{4/3}, but I don't which paper have proved it.
Can someone tell me this paper?
| https://mathoverflow.net/users/6326 | A result about LSpace and RLSpace | I think that the currently best known bound is L^{3/2} in
Michael E. Saks, Shiyu Zhou: RSPACE(S) \subseteq DSPACE(S3/2). FOCS 1995 344-353
There was a paper showing Symmetric Log space in L^{4/3}
R. Armoni, A. Ta-Shma, A. Wigderson, S. Zhou.
A (log n )^{4/3} space algorithm for (s,t) connectivity in undirected graphs
Preliminary version in Proceedings of the 29th STOC, pp. 230-239, 1997.
J. ACM, vol. 47, no. 2, 294-311, 2000.
(by now it is known that symmetric log spaces is in log space)
| 6 | https://mathoverflow.net/users/6327 | 25985 | 17,035 |
https://mathoverflow.net/questions/25948 | 9 | In commutative Iwasawa theory, the main conjecture states that the p-adic L-function generates the characteristic ideal of an algebraic object. Non-commutative Iwasawa theory seems to mimik this - except that the existence of the object on the analytic side (to my knowledge) is still conjectural. My question is: why is it so hard to define a "non-commutative" p-adic L-function? And on a more technical note: In Coates-Fukaya-Kato-Sujatha-Venjakob, this conjectural function only exists for primes of good ordinary reduction. This seems to imply that in the excluded cases, things go horribly wrong. Does anybody know what or why?
| https://mathoverflow.net/users/5730 | non-commutative iwasawa theory | First a short answer. I don't think one can say that the commutative analytic side is known, as you do. It is fully known only in the cyclotomic $\mathbb{Z}\_{p}$ situation, assuming the ETNC and in the crystalline case (see below). The answer to your "why is it so hard" question is: because $p$-adic $L$-functions are linked with $B\_dR$ and so require subtle knowledge of $p$-adic Hodge theory which is lacking in the non-commutative case (and also mostly in the general commutative case). The answer to your technical question is: because in the ordinary case, there is a concrete incarnation of $D\_{dR}$ which allows for a definition of the required trivialization. Outside the ordinary case, no such concrete incarnation is known and so things are two orders of magnitude harder, already in the commutative case (again, see below for some more details).
In both the commutative and non-commutative situation, if you want to construct a $p$-adic $L$-function in the style of Kato, Perrin-Riou, Colmez (basically in the cyclotomic $\mathbb{Z}\_{p}$-extension case) and Fukaya-Kato, CFKSV (in the non-commutative but still number field tower case) you need two things: one is an equivariant basis of the fundamental line (more or less the determinant of the motivic cohomology of your motive, or more concretely some sort of Euler system), the other is what is usually called a reciprocity law or Coleman map or $\epsilon$-isomorphism to transform this into a "function" or "measure" or "element in a localized $K\_{1}$" (depending what you mean by $p$-adic $L$-function).
Slightly more precisely and in a more technical language, in order to formulate a conjectural setting allowing for the existence of a $p$-adic $L$-function, you need a trivialization of some complexes of Galois cohomology which is compatible with "evaluation at characters" (without this compatibility, there can be no interpolation property worth its name). This trivialization involves very subtle properties of the ring $B\_{dR}$.
Now, in the cyclotomic $\mathbb{Z}\_{p}$-extension case, such a trivialization is known to exist for any crystalline motive thanks to very deep results of many people, but most notably Perrin-Riou, Kato-Kurihara-Tsuji, Colmez and Benois-Berger. So in that case, if you $assume$ the Equivariant Tamagawa Number Conjecture, then the $p$-adic $L$-function is well-defined. This is only in this (very limited) sense that the analytic object you refer to in your question is well-defined.
In the general commutative case, very little is known, because one has to consider the $D\_{dR}$ of Galois representations with coefficients in rings of large dimension and this is extremely hard though spectacular progresses are made each day in that respect. In the good ordinary tower of number fields case, be it commutative or non-commutative, the required trivialization is known because in that case there is a "concrete incarnation" of $D\_{dR}$.
Finally, in the general non-commutative case, almost nothing is known because, as far as I know, the collective knowledge we have on the behaviour of $D\_{dR}$ for non-commutative rings of large dimension is very far from what would be required only to formulate a question.
| 5 | https://mathoverflow.net/users/2284 | 25989 | 17,039 |
https://mathoverflow.net/questions/25993 | 23 | Consider a compact subset $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?
I guess it's wrong, because the topological assumption is invariant w.r.t homeomorphism, in contrast to being of zero Lebesgue measure. But I don't see any simple counterexample.
| https://mathoverflow.net/users/6129 | Sets with positive Lebesgue measure boundary | Construct a Cantor set of positive measure in much the same way as you make the `standard' Cantor set but make sure the lengths of the deleted intervals add up to 1/2, say.
Let $U$ be the union of the intervals that are deleted at the even-numbered steps and let $V$ be the union of the intervals deleted at the odd-numbered steps. The Cantor set is the common boundary of $U$ and $V$; their closures are as required.
| 30 | https://mathoverflow.net/users/5903 | 26000 | 17,044 |
https://mathoverflow.net/questions/26001 | 53 | I asked myself, which spaces have the property that $X^2$ is homeomorphic to $X$. I started to look at some examples like $\mathbb{N}^2 \cong \mathbb{N}$, $\mathbb{R}^2\ncong \mathbb{R}, C^2\cong C$ (for the cantor set $C$). And then I got stuck, when I considered the rationals. So the question is:
Is $\mathbb{Q}^2$ homeomorphic to $\mathbb{Q}$ ?
| https://mathoverflow.net/users/3969 | Are the rationals homeomorphic to any power of the rationals? | Yes, Sierpinski proved that every countable metric
space without isolated points is homeomorphic to the rationals:
<http://at.yorku.ca/p/a/c/a/25.htm> .
An amusing consequence of Sierpinski's theorem is that
$\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$. Of course here one
$\mathbb{Q}$ has the order topology, and the other has the $p$-adic
topology (for your favourite prime $p$) :-)
| 91 | https://mathoverflow.net/users/4213 | 26009 | 17,050 |
https://mathoverflow.net/questions/25971 | -4 | What is this video trying to tell us?
<http://www.youtube.com/watch?v=JX3VmDgiFnY>
The statement that fractional linear transformations correspond to rotations of the sphere under the stereographic projection is wrong (since for example some fractional linear transformations have only one fixed point, which is impossible for the rotation).
| https://mathoverflow.net/users/2260 | Meaning of the Mobius transformations video | Any Möbius transformations is a composite of a rotation of *S*2 (3 degrees of freedom), along with a translation and dilation of ℝ2 (3 degrees of freedom), adding up to the six dimensions of the Lie group PSL(2,ℂ) = group of Möbius transformations.
In the video, the translations are depicted by letting the sphere move left and right on the surface of the plane, while the dilations are depicted by lifting the sphere in the third dimension.
| 4 | https://mathoverflow.net/users/5690 | 26014 | 17,053 |
https://mathoverflow.net/questions/26013 | 4 | Let $R$ be a local artinian Gorenstein ring and $M$ a finitely generated $R$-module, then
$\mathrm{Ext}\_R^1(M,M) = 0$ if only if $M$ is projective?
| https://mathoverflow.net/users/5775 | Selforthogonal modules over Artinian Gorenstein rings | This would be a very strong version of the Auslander-Reiten Conjecture ([see here](https://www.leuschke.org/research/AR-Conjecture/), for example) in the Gorenstein case. The Conjecture is still open, though many partial results are known.
By the way, [this paper](https://doi.org/10.1016/j.jalgebra.2008.04.027) (ScienceDirect link, may not be visible to everyone) claims a proof of the Conjecture in the Gorenstein case, which would give an affirmative answer to your question, but I --- and several other people I've talked to --- believe there is a gap in the proof. Their assumption is that $\mathrm{Ext}\_R^{i}(M,M)=0$ for $i =1,2$. The questionable step is at the top of page 2163, the second line, where they say "therefore ... is exact".
Edit 2023-05-15: The paper I referred to above is: "When are Torsionless modules projective?", by Rong Luo and Zhaoyong Huang,
Journal of Algebra 320 (2008) 2156-2164, MR2437647.
| 7 | https://mathoverflow.net/users/460 | 26017 | 17,055 |
https://mathoverflow.net/questions/18902 | 5 | Are there well-known or interesting applied problems (especially of the real-time signal processing sort) where arbitrarily long time series of small (say $d \equiv \dim \le 30$ for a nominal bound, and preferably sparse) matrices arise naturally?
I am especially interested in problems that can be mapped onto a setup in which for each event of a reasonably nice [point process](http://en.wikipedia.org/wiki/Point_process) on $\mathbb{R}$ (the simplest two such processes would be a Poisson or discrete-time process) there is an associated pair $(j,k) \in \{1,\dots,d\}^2$. In this case time-windowed sums $N\_{jk}(t)$ of the various pairs can be formed in an obvious way (although there may be plenty of subtlety or freedom in the windowing itself): these supply such a matrix time series.
Each such pair $(j,k)$ could be regarded as a transition from server $j$ to another (possibly identical) server $k$ in a closed queue with $d$ servers and infinitely many clients. It is not hard to see that in the setting of communication networks, this framework amounts to a very general form of [traffic analysis](http://en.wikipedia.org/wiki/Traffic_analysis). Such an application should not be considered for an answer: it's already been covered.
A slightly more restrictive but simpler example is where the pairs $(j,k)$ are inherited from a [cadlag](http://en.wikipedia.org/wiki/Cadlag) random walk on the root lattice
$A\_{d-1} :=\left \{x \in \mathbb{Z}^d : \sum\_{j=1}^d x\_j = 0\right \}$.
Examples of this sort would also be of considerable interest to me.
| https://mathoverflow.net/users/1847 | Are there interesting problems involving arbitrarily long time series of small matrices? | A very significant application in the context of communications engineering is the modelling of multiple-input-multiple-output (MIMO) communications channels.
These channels are typically modeled by complex $n \times m$ matrices where $n$ is the number of receive antennas and $m$ is the number transmit antennas. The $(i,j)$ entry in the matrix describes the channel between the $i$th transmit antenna and the $j$th receive antenna. In most applications $n$ and $m$ are reasonably small, less than 16. Also, in most real world applications the channel (and hence the matrix) changes over time. This gives you your *time series of matrices*. In some situations the matrix will even be sparse because some transmit antennas might not *see* some receive antennas.
There is a seriously huge amount of literature on the MIMO channel and a large amount of it deals with the static case, i.e. for the sake of simplicity it is assumed that the channel doesn't change with time. However there are also many papers that deal with the time varying case. For example:
[Chen and Su, "MIMO Channel Estimation in Correlated Fading Environments"](http://dl.comsoc.org/twc/Public/2010/mar/chen3.html)
I unfortunately am not an expert in MIMO, but I do know some people who are and could ask them for more details if you were interested.
| 2 | https://mathoverflow.net/users/5378 | 26022 | 17,056 |
https://mathoverflow.net/questions/25991 | 3 | Suppose we have a full exceptional collection (F1,...,Fn) of coherent sheaves on a smooth projective variety X. The number n of sheaves in this collection is equal to the rank of the Grothendieck group K0(X).
Is there any relation between n and the rank of the Picard group Pic(X) or the dimension of X?
| https://mathoverflow.net/users/6330 | Number of sheaves in a full exceptional collection | As a far as I know, there is no obvious relation between the number of objects in a full exceptional collection and the dimension of $X$, unless you impose extra conditions on the full exceptional collection, as in Bondal and Polishchuk's 'Homological properties of associative algebras', where they prove that when the collection is 'geometric', then ${\rm rank}(K\_{0})(X)={\rm dim} X + 1$. Their hypotheses apply for example to $\mathbb{P}^{n}$, where the standard full exceptional collection is $\mathcal{O},\mathcal{O}(1), ..., \mathcal{O}(n)$.
But for examples like $\mathbb{P}^{1} \times \mathbb{P}^{1}$ these hypotheses do not apply, since the standard full exceptional collection is $\mathcal{O}, \mathcal{O}(1,0), \mathcal{O}(0,1), \mathcal{O}(1,1)$, so ${\rm rank}(K\_{0}(X))={\rm dim} X + 2$.
In general, a variety $X$ will not admit a full exceptional collection of objects in its derived category of coherent sheaves, for instance because $K\_{0}(X)$ is usually not free of finite rank. And some varieties cannot carry any exceptional objects at all. For instance, if $X$ is smooth projective with trivial canonical bundle, then if you checked that $Hom(E,E)$ were one dimensional as needed for $E$ to be exceptional, Serre duality would imply that $Ext^{n}(E,E)$ is also one dimensional.
To me it seems more natural to ask for a nice 'compact generator' of the derived category of a variety. This is a perfect complex $E$ on $X$ (one locally quasi-isomorphic to a finite length complex of vector bundles) that generates the category $Perf(X)$ of perfect complexes in finitely many steps for each object or $D\_{QCoh}(X)$, the unbounded derived category of quasi-coherent sheaves, in infinitely many steps. This is a generalization
of the sum $E$ of the objects in a full exceptional collection, with no extra conditions imposed on the $Ext$ algebra of $E$.
Such compact generators exist in great generality, due to work of Bondal and van den Bergh. Given such a compact generator and letting $A=RHom(E,E)$, the derived endomorphism dg algebra of $E$, we get equivalences $RHom(E,?): D\_{QCoh}(X) \rightarrow D(A)$ and $RHom(E,?): Perf(X) \rightarrow Perf(A)$, just as we would if $E$ were the sum of objects in a full exceptional collection.
One could ask if you can find concrete examples of such compact generators, and indeed you can. If $X$ is projective of dimension $d$, let $L$ be an ample, globally generated line bundle. Then it is easy to show using a standard criterion for compact generation that
$E=\mathcal{O}\oplus L \oplus \cdots \oplus L^{d}$ is a compact generator of the derived category of $X$.
So one can always say that if $X$ is projective, then the minimal number of line bundles (not necessarily exceptional) needed to generate the derived category of $X$ is bounded by ${\rm dim} X + 1$.
I would also guess that fewer line bundles will not suffice, but I'm not sure.
| 12 | https://mathoverflow.net/users/4659 | 26030 | 17,058 |
https://mathoverflow.net/questions/26020 | 2 | Hi,
Please recommend a good book on wave equations and fourier series / transforms at 3rd year undergraduate level.
Our course text is a bit dense and can be hard to follow - see the course text at <http://www.ouw.co.uk/bin/ouwsdll.dll?COURSEMS324_Mathematics_-_Pure_and_Applied#> - Block 1 - Waves.
As mentioned below normally the OU texts are very readable but I'm having a bit of trouble with this one.
Thanks!
| https://mathoverflow.net/users/6332 | Good books on wave equations and fourier analysis | Hum, unfortunately I am not familiar with the Open University course, so I am just making a guess based on the course description you linked to.
Insofar as Fourier Analysis is concerned, a decent text is Stein and Shakarchi's Fourier Analysis: an introduction. ( <http://press.princeton.edu/titles/7562.html> ) You will most likely only need chapters 1, 2, 4, and 5, with a bit of knowledge of 3. One thing good about the book is that it was written as a first course in an analysis sequence, so doesn't assume too much knowledge about real and complex analysis.
Once you have a bit of Fourier analysis under your belt, reading Korner's Fourier Analysis ( <http://books.google.com/books?id=OcZ5iKsGrmoC&lpg=PP1&dq=korner%20fourier%20analysis&pg=PR8#v=onepage&q=korner%20fourier%20analysis&f=false> ) can be enlightening and give you some feel about what one can do using the machinery.
For the applications to wave equations as mentioned in the course description, somehow I feel that a textbook in electromagnetism (Jackson or Griffiths) may contain more practical material (look at the sections on standing waves and wave-guides).
| 4 | https://mathoverflow.net/users/3948 | 26036 | 17,060 |
https://mathoverflow.net/questions/26025 | 4 | Hi people!
This my first question, here. I don't sure if it has a trivial answer, or not.
Let G a group, N normal subgroup in G. In which cases there is a subgroup in G isomorphic to G/N?
TIA
| https://mathoverflow.net/users/6334 | Given a normal subgroup N⊆G, when does G contain a subgroup isomorphic to G/N? | Assuming you're looking at the case where the isomorphism is induced by the quotient $G \to G/N$ (as per George McNinch's comment), then this should be if and only if the sequence
$$ 0 \to N \to G \to G/N \to 0$$
splits. i.e. there is a section $\sigma : G/N \to G$. This is then seen to be equivalent to $G$ being isomorphic to the semidirect product $N \rtimes G/N$.
| 8 | https://mathoverflow.net/users/1703 | 26038 | 17,061 |
https://mathoverflow.net/questions/26032 | 4 | This is related to an [older question](https://mathoverflow.net/questions/17597/prime-numbers-with-given-difference) about [prime k-tuples and constellations](http://en.wikipedia.org/wiki/Prime_k-tuple), but takes a slightly different direction.
Given an integer k, we want to find n such that the interval {n+1, ..., n+k} contains as many primes as possible. (We consider only n ≥ k to eliminate certain exceptional cases, such as {3,5,7}, which is irregular since for k=5 there can be at most 2 primes in the interval if n>2.)
There is an obvious upper bound ak on the number of primes in this interval, given by considering the numbers modulo p for all primes p ≤ k. More precisely, ak is the largest possible cardinality of a set A ⊂ {1, ..., k} such that for some n, the set n+A does not contain any numbers divisible by any prime p ≤ k.
For example, a3=2 since out of 3 consecutive numbers at least one is even, and a7=3 since out of seven consecutive numbers at least three are even, and at least one of the odd numbers is divisible by 3. Similarly, a9=4 and a13=5.
The four bounds listed so far can be achieved by taking n=4, n=10, n=10, n=36, respectively. The natural question to ask is whether for every k there is a value of n such that the interval {n+1, ..., n+k} contains ak primes, but the comments on the [older question](https://mathoverflow.net/questions/17597/prime-numbers-with-given-difference) make me suspect that this may be open. (Another natural question is whether there are infinitely many such n, but since for k=3 this is the twin prime conjecture, that's definitely out of reach at present.)
Since the natural question to ask seems very hard, my question instead is this: Is anything about the asymptotics of this problem? More precisely, I'd like to know if something like the following statement is true: "For every ε>0, there exist infinitely many pairs (k,n) such that the interval {n+1, ..., n+k} contains at least (1-ε)ak prime numbers."
There are a few different ways to tweak that statement -- for example, we could ask for infinitely many n for a fixed k, or we could let both k and n become arbitrarily large. (Of course k will need to become arbitrarily large as ε becomes small.) I'd be happy with any of them -- I'm asking this question out of curiosity rather than out of a need for a specific result.
| https://mathoverflow.net/users/5701 | Intervals with large numbers of primes | For fixed $k$ this is definitely hopeless, since it would imply that for some $b$ there are infinitely many primes $p$ such that $p + b$ is prime, and this is a well-known open problem that seems out of reach of the latest techniques for finding small gaps between primes (see this survey article of Soundararajan for example: <http://www.ams.org/journals/bull/2007-44-01/S0273-0979-06-01142-6/S0273-0979-06-01142-6.pdf>)
For similar reasons this will also be hopeless if $k$ is too small depending on $n$.
If $k$ is huge compared with $n$ then the existence of many such pairs would follow immediately from the prime number theorem if only one knew that $a\_k \leq (1 + o(1))\pi(k)$. However, I do not believe this is known and in fact I'm nigh-on certain that nothing better than $a\_k \leq 2\pi(k)$ is known. This is a result of Montgomery and Vaughan; the slightly weaker bound of $a\_k \leq (2 + o(1))\pi(k)$ follows rather easily from the Selberg upper bound sieve. Incidentally, the presence of the factor $2$ here reflects something called the parity problem in sieve theory: breaking it, even by a tiny amount, is generally very problematic.
In the previous discussion referenced above, the result of Hensley and Richards was mentioned. This is an example to show that it is \emph{not} true that $a\_k \leq \pi(k)$. As you hint in the question, one might then conjecture that there is $n$ such that $\{n+1,\dots, n+k\}$ contains $a\_k > \pi(k)$ primes, in which case one would have a violation of the triangle inequality $\pi(x+y) \leq \pi(x) + \pi(y)$. Such a conjecture would follow from the Hardy-Littlewood $k$-tuple conjecture which is, of course, hopelessly out of reach.
| 6 | https://mathoverflow.net/users/5575 | 26042 | 17,064 |
https://mathoverflow.net/questions/26043 | 10 | Consider a compact differentiable manifold $M$. We say that $f:M\to M$ and $g: M \to M$ are topologically conjugated if there exists $h:M\to M$ a homeomorphism such that $f\circ h= h \circ g$. The conjugacy class of a homeomorphism $f$ is the set of all $g$ such that $g$ is topologically conjugated to $f$.
If a homeomorphism $f: M\to M$ has infinite topological entropy (which is an invariant under topological conjugacy), then the conjugacy class of $f$ has no diffeomorphisms.
Is there any other known obstruction for a homeomorphism not to have diffeomorphisms in its conjugacy class? I would guess that yes, but I could not find one.
Is there a restriction on the dimension of the manifold?
(Remark: In dimension one, that is, in the circle, every homeomorphism has a diffeomorphism-of class $C^1$ in its conjugacy class. However, Denjoy counterexamples have no $C^2$ diffeomorphisms in theirs).
| https://mathoverflow.net/users/5753 | Topological conjugacy between homeomorphisms and diffeomorphisms | Here is an example. Consider a map $f:\mathbb R^2\to\mathbb R^2$ given by $(r,\varphi)\mapsto (r,\varphi+\sin(1/r))$ in polar coordinates $(r,\varphi)$, $0<r\le 1/\pi$. For $r>1/\pi$, let $f$ be identity, then close up the plane to make a compact manifold.
This map has zero topological entropy but has no conjugate $C^1$ diffeomorphism. Indeed, we may assume that 0 is mapped to itself by the conjugation. Then the circles centered at the origin are mapped to Jordan curves winding around the origin. There are arbitrarily small circles made of fixed points, hence the derivative of the diffeomorphism at the origin is the identity. On the other hand, there are arbitrary small circles on which our map is periodic with rotation number, say, 1/10. It is easy to see that this cannot happen on a Jordan curve winding around the origin, unless some point changes is angular coordinate by $2\pi/10$. This contradicts the fact that the derivative at 0 is the identity.
| 13 | https://mathoverflow.net/users/4354 | 26057 | 17,074 |
https://mathoverflow.net/questions/26018 | 33 | I just caught sight on arXiv a paper by Holst and Stern titled [Geometric Variational Crimes](http://arxiv.org/abs/1005.4455). Apparently a Variational Crime is an approach to solve problems using a finite element method (e.g. Galerkin) where certain assumptions about approximate solutions are violated. As you can see, even after doing a brief search in the literature my understanding of the "problem" is still rather superficial and limited (and very possibly wrong).
So, my questions are
1. Can someone give me a quick and dirty explanation of what Variational Crimes is about (and how it is useful)?
2. Does anyone know where the name comes from?
This is just out of curiosity into what other people are doing; I am by no means a numerical analyst.
| https://mathoverflow.net/users/3948 | What are "variational crimes" and who coined the term? | Thanks for your interest in the paper! (It's also nice to see something on Math Overflow that I know something about.) Your summary of variational crimes is actually pretty close to the mark: it refers to certain "abuses" of the Galerkin method, where some of the assumptions are violated, and thus the standard error estimates (e.g., [Céa's lemma](http://en.wikipedia.org/wiki/C%C3%A9a's_lemma)) are no longer valid.
Since you have the basic idea right, let me try to provide some context and motivation. As a simple example, consider Poisson's equation on some domain $ U \subset \mathbb{R}^n $ with Dirichlet boundary conditions,
$$ - \Delta u = f \text{ on } U, \quad u \rvert \_{\partial U} = 0 .$$
This can be written as a variational problem on $ V = \{ v \in H^1(U) : v \rvert\_{\partial U} = 0 \} $: Find $ u \in V $ such that
$$ \int \_U \nabla u \cdot \nabla v \; dx = \int \_U f \thinspace v \; dx , \quad \forall v \in V .$$
(You can see, using integration by parts, that any classical solution solves this variational problem.) If we define the bilinear form $ B(u,v) = \int \_U \nabla u \cdot \nabla v \; dx $ and functional $ F(v) = \int \_U \thinspace f \thinspace v \; dx $, then this problem can be written in the usual abstract form: Find $ u \in V $ such that
$$ B(u,v) = F(v) ,\quad \forall v \in V .$$
To apply the Galerkin method, we need to take a subspace $ V \_h \subset V $ (e.g., the span of some finite element basis) and solve the Galerkin variational problem: Find $ u \_h \in V \_h $ such that
$$ B(u\_h, v ) = F(v), \quad \forall v \in V\_h .$$
The problem is that, for many practical purposes, this is impossible to compute. First, the bilinear form $ B(\cdot, \cdot) $ requires us to calculate an integral exactly. In practice, this is usually not possible, so people instead approximate the integral using *numerical quadrature*. However, this is a variational crime, since using numerical quadrature replaces $ B (\cdot, \cdot) $ by some $ B\_h (\cdot, \cdot) \approx B(\cdot, \cdot) $ in the Galerkin variational principle; likewise, numerical quadrature also replaces $ F(\cdot) $ by some $ F\_h(\cdot) \approx F(\cdot) $. (This is aside from the fact that computers only use finite-precision arithmetic, so even if we had an closed formula for these integrals, there would always be some floating-point error involved.)
Moreover, if $U \subset \mathbb{R}^n $ is polyhedral, then it can be triangulated exactly, so we can get $ V\_h \subset V $ to be some finite element space supported on this piecewise-linear mesh. However, if $U$ has a curved boundary, then a piecewise-linear (or piecewise-polynomial, in the case of isoparametric elements) mesh only approximates the actual domain. Since the functions in $ V\_h$ are defined on a slightly different domain than those in $V$, in this case $ V\_h \not\subset V $.
Now, in the real world of numerical computation (engineering, etc.), people didn't worry too much about using these approximations instead of the exact Galerkin variational problem; it was a practical necessity, and the approximations seemed to converge just fine. However, these "variational crimes" meant that the abstract Galerkin error analysis was no longer valid for the modified methods. Strang pointed this out, and his lemmas quantify the additional errors introduced by these "crimes."
As far as the history/terminology: to the best of my knowledge, Strang himself coined the term "variational crime." The earliest reference I know is
[Strang, G. (1972), Variational crimes in the finite element method. In The mathematical foundations of the finite element method with applications to partial differential equations (Proc. Sympos., Univ. Maryland, Baltimore, Md., 1972), pages 689–710. Academic Press, New York.], although I haven't been able to find an electronic copy. He followed this up with a more easily-located article for a wider audience: [Strang, G. (1973), Piecewise polynomials and the finite element method. Bull. Amer. Math. Soc., 79, 1128–1137.](http://projecteuclid.org/euclid.bams/1183535131) Finally, an excellent account is given in the book [Brenner, S. C., and L. R. Scott (2008), The mathematical theory of finite element methods, volume 15 of Texts in Applied Mathematics. Springer, New York, third edition.](http://dx.doi.org/10.1007/978-0-387-75934-0); Chapter 10 is entirely about variational crimes.
| 44 | https://mathoverflow.net/users/673 | 26066 | 17,080 |
https://mathoverflow.net/questions/26031 | 16 | There are two simple and classic enumerations that still I'm puzzled about. Let's start with a simple counting problem from a well-known dynamical system.
**fact 1** Consider the "tent map" f:[0,1]→[0,1] with parameter 2, that is
>
> f(x):=2min(x,1-x).
>
>
>
Clearly, it has 2 fixed points, and more generally, for any positive integer *n*, there are 2*n* periodic points of period *n* (it's easy to count them as they are fixed points of the *n*-fold iteration of *f*, which is a piecewise linear function oscillating up and down between 0 and 1 the proper number of times). To count the number of periodic orbits of *minimal* period *n*, a plain and standard application of the Moebius inversion formula gives
>
> Number of n-orbits of (I,f) =
> $\frac{1}{n}\sum\_{d|n} \mu(d)2^{n/d}.$
>
>
>
(**rmk**: any function with similar behaviour would give the same result, e.g. *f(x)=4x(1-x)*,...&c.)
Now let's leave for a moment dynamical systems and consider the following enumeration in the theory of finite fields.
**fact 2** Clearly, there are 2n polynomials of degree *n* in $\mathbb{F}\_2[x]$. With a bit of field algebra it is not hard to compute the number *I(n)* of the irreducible ones. One can even make a completely combinatorial computation, just exploiting the unique factorization, expressed in the form:
$\frac{1}{1-2x}=\prod\_{n=1}^\infty (1-x^n)^{-I(n)}.$
One finds:
>
> Number of irreducible polynomials of
> degree *n* in $\mathbb{F} \_ 2[x]$ =
> $\frac{1}{n}\sum\_{d|n} \mu(d)2^{n/d}.$
>
>
>
**Question**: it's obvious by now: is there a natural and structured bijection between periodic orbits of *f* and irreducible polinomials in $\mathbb{F}\_2[x]$? How is interpreted the structure of one context when transported ni the other?
(**rmk**: of course, analogous identities hold for any p > 2)
| https://mathoverflow.net/users/6101 | Periodic orbits and polynomials | How I see it is every fixed point comes from an equation $T^nx=x$. Denoting $f(x)=2x$ and $g(x)=2(1-x)$ we see that this corresponds to solving equations
$$h\_1\circ h\_2\circ\cdots \circ h\_n (x)=x$$ where $h\_i\in \{f,g\}$. This is geometrically the intersection of two lines and thus gives a unique solution, and therefore a correspondence between periodic points of period $n$ and binary strings of length $n$. Now since concatenating a binary string $L$ with itself clearly gives the same $x$ with $x=L(x)=L\circ L(x)$ and you have a shift operator $L\circ h(x)=x \implies h\circ L(y)=y$ where $y=h(x)$ you get a bijection between periodic orbits and aperiodic cyclic sequences of zeros and ones, which are well known to be in bijection with the irreducible polynomials of $\mathbb F\_2 [x]$.
| 11 | https://mathoverflow.net/users/2384 | 26068 | 17,081 |
https://mathoverflow.net/questions/26059 | 14 | A Noetherian group (also sometimes called slender groups) is a group for which every subgroup is finitely generated. (Equivalently, it satisfies the ascending chain condition on subgroups).
A finitely presented group is a group with a presentation that has finitely many generators and finitely many relations.
Flipping through some search results and references, I get the impression that there should be examples of Noetherian groups that are not finitely presented (because I can locate references to "finitely presented Noetherian group", a name that shouldn't exist if being Noetherian implies being finitely presented). However, I'm not able to get an explicit reference or example. I would be grateful if somebody could point out a reference or example.
For a solvable group, being Noetherian is equivalent to being polycyclic (i.e., having a subnormal series where all the successive quotients are cyclic groups), and polycyclic groups are finitely presented. Hence, any counterexample must be a non-solvable group.
[Note: My standard example of a finitely generated group that is not finitely presented is a wreath product of the group of integers with itself. But this is far from Noetherian.]
| https://mathoverflow.net/users/3040 | Example of Noetherian group (every subgroup is finitely generated) that is not finitely presented |
>
> [Tarski monsters](http://en.wikipedia.org/wiki/Tarski_monster_group%20) provide examples of 2-generator noetherian groups that is not finitely presented.
>
>
>
Edit (YCor): Tarski monsters, as defined in the link (infinite groups of prime exponent $p$ in which every nontrivial proper is cyclic) exist for large $p$ and all currently known constructions of Tarski monsters are known to yield groups that are not finitely presented. However, it is unknown whether there exists a finitely presented Tarski monster.
| 11 | https://mathoverflow.net/users/6339 | 26073 | 17,084 |
https://mathoverflow.net/questions/26062 | 2 | Suppose that $K/\mathbb{Q}\_l$ is a finite extension, with ring of integers $\mathcal{O}\_K$. Suppose $\mathcal{G}/K$ is a (linear) algebraic group (connected+reductive), and $\Gamma\subset \mathcal{G}(K)$ is a hyperspecial maximal compact subgroup. This just means (I believe) that we can find $\tilde{\mathcal{G}}/\mathcal{O}\_K$ with $\tilde{\mathcal{G}}\_K=\mathcal{G}$ and $\tilde{\mathcal{G}}(\mathcal{O}\_K)=\Gamma\subset\tilde{\mathcal{G}}(K)$.
Suppose we have an automorphism $\alpha$ of $\mathcal{G}/K$ preserving $\Gamma=\tilde{\mathcal{G}}(\mathcal{O}\_K)$. Can we necessarily extend $\alpha$ to an automorphism of $\tilde{\mathcal{G}}$? If not, are there any nice conditions under which we can? For instance, what if $\mathcal{G}$ is simple or even simple and simply connected?
Edit: added connected, reductive hypotheses on $\mathcal{G}$, which were there in my head, but I forgot to write them...
| https://mathoverflow.net/users/3513 | Automorphism of algebraic group preserving a hyperspecial maximal compact | As noted, there seem to be some "reductive"s missing from the question. Here's
what is known: let $R$ be a Henselian discrete valuation ring with field of
fractions $K$, and let $R^{\prime}$ be the integral closure of $R$ in the
maximal unramified extension $K^{\prime}$ of $K$; a smooth affine scheme $X$
over $R$ defines a scheme $X\_{K}$ over $K$ and a subset $X(R^{\prime})$ of
$X\_{K}(K^{\prime})$; the functor $X\rightsquigarrow(X\_{K},X(R^{\prime}))$ is
fully faithful.
The proof of this is fairly easy (cf. 1.7.3 of .Bruhat, F., and Tits, J.,
Groupes reductifs sur un corps local II, Publ. Math. IHES, 60, 1984).
As stated this fails without "affine and smooth".
In general you can't replace $X(R^{\prime})$ with $X(R)$ (because $X(R)$ may
be empty). Perhaps if $X\_{k}(k)$ is Zariski
dense in $X\_{k}$ it's OK ($k$=residue field).
Added: As blt points out, Lemma 6.2 of Snowden and Wiles, arXiv:0908.1991v3,
states that, when $K$ is a finite extension of $\mathbb{Q}\_{\ell}$, $X$ is a
simply connected semisimple group, and the map on the generic fibre is an
automorphism, if $X(R)$ maps into $X(R)$ then $X(R^{\prime})$ maps into
$X(R^{\prime})$. Thus, for simply connected semisimple groups, the answer is
YES, and as BCnrd points out, that implies that the answer is YES for all
reductive groups.
| 3 | https://mathoverflow.net/users/930 | 26078 | 17,088 |
https://mathoverflow.net/questions/26084 | 8 | Given any two points x and y on a circle O, one can form four different lenses (regions between two circles, one of which is O) that have corners at x and y and make angles of 2π/3 at their corners. For three points x, y, and z, with O the circumcircle of triangle xyz, two of these lenses are outside O and two of them have z on the boundary, so there is a unique lens L(x,y) that is inside O and disjoint from z, and lenses L(y,z) and L(x,z) defined symmetrically.
These three lenses L(x,y), L(y,z) and L(x,z) intersect in a unique point c contained in O, defining c as a triangle center for the triangle xyz. There's also a closely related center c' that you can get by inversion of c through O (the intersection point of three lenses outside O). For an equilateral triangle xyz, c is (as for any triangle center) at the centroid of xyz, while c' is at the point at infinity in the one-point completion of the plane. {c,c'} is equivariant under Möbius transformations, an unusual property for a triangle center.
Another equivalent way of defining c, I think, is that it's the point for which angles xcy = π/3 + xzy, xcz = π/3 + xyz, and ycz = π/3 + yxz. This is similar to the fact that the circumcenter C is the point for which xCy = 2xzy etc.
Does anyone recognize these centers c and c'? Do they have entries in the [Encyclopedia of Triangle Centers](http://faculty.evansville.edu/ck6/encyclopedia/ETC.html)? The definitive way to answer this would be to compute trilinear coordinates and look them up using ETC's search feature, but I'm hoping that this looks familiar enough to someone here that I can skip that step. I tried searching for Möbius and inversion in the text of ETC but found nothing that way.
| https://mathoverflow.net/users/440 | Möbius-invariant triangle center? | Unless I'm mistaken you seem to describe the [isogonal conjugate](http://en.wikipedia.org/wiki/Isogonal_conjugate) of the Fermat point X(13). This is the [first isodynamic point](http://en.wikipedia.org/wiki/Isodynamic_point), or X(15) in ETC.
| 7 | https://mathoverflow.net/users/2384 | 26085 | 17,091 |
https://mathoverflow.net/questions/25757 | 6 | Are there any bounds on residues of $1/\zeta$ in roots of $\zeta$ in critical strip, which may use RH, but do not use the conjecture on simplicity of roots or something similar? I did not find such resuts in Titchmarsh, but I could miss something. Thanks!
| https://mathoverflow.net/users/4312 | Residues of $1/\zeta$ | This is actually a very difficult problem, and currently most results are highly conjectural. It essentially comes down to finding useful bounds on discrete moments of the Riemann zeta function of the form
$$J\_k(T) = \sum\_{0 < \Im(\rho) < T}{|\zeta'(\rho)|^{2k}},$$
as one can then choose the correct value of $k$ and apply partial summation.
For positive $k$, recent results of [Milinovich](http://arxiv.org/abs/0806.0786) and [Milinovich and Ng](http://arxiv.org/abs/0706.2321) show under the Riemann Hypothesis that
$$T(\log T)^{(k+1)^2} \ll J\_k(T) \ll T(\log T)^{(k+1)^2 + O(1/\log \log \log T)};$$
and slightly stronger results are known for $k = 0,1,2$ (with the latter two being under the Riemann Hypothesis).
For negative $k$, the situation is much more difficult. A conjecture of Gonek-Hejhal suggests that for all $k > -3/2$,
$$J\_k(T) \asymp T(\log T)^{(k+1)^2}$$
and this has been refined significantly by [Hughes, Keating, and O'Connell](http://www.jstor.org/stable/2665448). But it seems out of reach to prove anything significantly useful in this area; the best result so far has been by [Gonek](http://journals.cambridge.org/production/action/cjoGetFulltext?fulltextid=7010328), who proved that $J\_{-1}(T) \gg T$ assuming the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$. But this isn't useful for most applications, where an upper bound is needed. I believe it is possible to show $J\_{-1} \ll T^{2+\varepsilon}$ under the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$, though I don't have a reference for this. Also it is quite possible that this result also holds if we replace $|\zeta'(\rho)|^{-2}$ by $\left|\mathrm{Res}\_{s = \rho} \zeta(\rho)^{-1}\right|^2$, though again I don't know of a reference for this.
| 7 | https://mathoverflow.net/users/3803 | 26092 | 17,096 |
https://mathoverflow.net/questions/26098 | 4 | I want to pick a random direction in n-dimensional space. How can I do this?
The reason I want to do this is to pick a neighbor for [hill climbing optimization](http://en.wikipedia.org/wiki/Hill_climbing).
| https://mathoverflow.net/users/6343 | How to pick a random direction in n-dimensional space | You can proceed as explained at <http://mathworld.wolfram.com/HyperspherePointPicking.html>
| 11 | https://mathoverflow.net/users/1409 | 26100 | 17,101 |
https://mathoverflow.net/questions/26086 | 12 | Analytic/meromorphic continuation is a difficult problem in general. For "motivic L-functions", the idea of proving their analytic continuations by first proving their modularity goes back, I guess, to Riemann.
Here I just want to ask a purely complex-analytic question. Let's restrict ourselves to the case of one variable functions. Let $U$ be a region in the complex plane, and let $f$ be a holomorphic function on $U.$ Is there any criterion for $f$ to have analytic continuation to a larger region? And what is this "maximal domain of regularity"?
Feel free to assume $U$ and $f$ to have the shape you like, e.g. a power series on an open disk or a Dirichlet series on some half plane. I guess even if $U$ is an annulus or a punctured disk, where one can compute (theoretically or numerically) the value of the extended function (if exists) at the points inside the inner loop by Cauchy's formula, it is still difficult to decide if this extension is continuous or analytic.
| https://mathoverflow.net/users/370 | Analytic continuation of holomorphic functions | Well, in case of power series some criterions do exist. Roughly speaking, one can take the element
$$
\sum\limits\_{n=1}^{\infty}c\_nz^n,\quad z < 1,\label{1}\tag{$\ast$}$$
and consider an analytic function $\phi$, such that $\phi(n)=c\_n$ for every $n$. The element \eqref{1} can be analytically extended onto some angular domain iff $\phi(z)$ has finite exponential growth.
Let $E\subset \mathbb C$ be a closed unbounded domain and let $H(E)$ denote the set of functions such that each of them is analytic in a neighborhood of $E$.
For a function $\phi\in H(E)$, *the exponential type of $\phi$ on $E$* is defined as
$$\sigma\_\phi(E)=\limsup\limits\_{z\to\infty,\\ z\in E}\frac{\log^+|\phi(z)|}{|z|}.$$
The following result is due to LeRoy and Lindelöf.
>
> **Theorem 1**. Let $\Pi=\{z\in\mathbb C|\ \Re z \geq 0\}$. Assume that $\phi\in H(\Pi)$ is of finite exponential type $\sigma<\pi$. Then the series
> $$f(z)=\sum\limits\_{n=1}^{\infty}\phi(n)z^n$$
> can be analytically extended onto the angular domain $\{z\in\mathbb C|
> \ |\arg z|>\sigma\}$.
>
>
>
The LeRoy-Lindelöf theorem gives only a sufficient condition. A criterion can be obtained if we relax a bit the condition that $\phi$ is of finite exponential type.
Let $\Omega=\{z\in\mathbb C|\ \Re z > 0\}$ be the interior of $\Pi$. An analytic function $\phi\in H(\Omega)$
is said to be of (finite) *interior exponential type* iff
$$\sigma\_\phi^\Omega=\sup\limits\_{\{\Delta\}}\sigma\_\phi(\Delta)< \infty,$$
where $\{\Delta\}$ is the set of all closed angular domains such that $\Delta\subset \Omega\cup \{0\}.$
>
> **Theorem 2**. The element
> $$\sum\limits\_{n=1}^{\infty}c\_nz^n,\quad z < 1,$$
> can be analytically extended onto the angular domain ${{{{}}{}}}\{z\in\mathbb C|
> \ |\arg z|>\sigma\}$ for some $\sigma\in[0,\pi)$ iff there is a function $\phi\in H(\Pi)$ of interior exponential type less or equal to $\sigma$ such that $$c\_n=\phi(n),\quad n=0,1,2,\dots.$$
>
>
>
You might be interested in [this article](http://iopscience.iop.org/0025-5734/52/1/A02).
| 15 | https://mathoverflow.net/users/5371 | 26103 | 17,103 |
https://mathoverflow.net/questions/26091 | 3 | Is it true that if $\mbox{Ext}^{1}(P,M)=0$ for every *finitely generated* module $M$ then $P$ is projective? Or that if $\mbox{Ext}^{1}(M,Q)=0$ for every *finitely generated* module $M$ then $Q$ is injective?
| https://mathoverflow.net/users/5292 | Does Ext commute with direct limit? | For the first question you already have had an answer in [Is it true that if $\operatorname{Ext}^{1}\_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective?](https://mathoverflow.net/questions/25687/projective-module/25698) if $\mathrm{Ext}^1\_{\mathbb Z}(P,M)=0$, then it depends on the axioms of set-theory whether the conclusion is true or not. The answer to the second question is yes, it is one of the basic characterisation of injective modules that $\mathrm{Ext}^1(A/I,Q)=0$ for all ideals $I$ iff $Q$ is injective. As for the question in your title, the answer should be no for the second variable (irrespective of the axioms of set theory, but I am too lazy to try to come up with an example). For the first variable things are a little bit more interesting: If $M$ is the direct limit of ${M\_\alpha}$, then we have a spectral sequence with $E\_2$-term $lim^i\mathrm{Ext}^j(M\_{\alpha},Q)$ (*"lim" means inverse limit, there is some strange problem with using "varprojlim" which sometimes works and sometimes doesn't*) and converging to $\mathrm{Ext}^{i+j}(M,Q)$. Somewhat strangely this spectral sequence does not seem to formally give the above characterisation of injective modules as there is a potential $lim^1\mathrm{Hom}(M\_{\alpha},Q)$ contribution.
| 8 | https://mathoverflow.net/users/4008 | 26105 | 17,104 |
https://mathoverflow.net/questions/26075 | 14 | There's a well known theorem due to Beck that characterizes when an adjunction is monadic, that is, if $F$ is left adjoint to $G$, $G:D \to C$, $GF:=T$ is always a monad on $C$, and the adjunction is called monadic, essentially, when $D$ is the Eilenberg–Moore category $C^T$ of $T$-algebras and $G$ is the forgetful functor. (For the precise definition see <http://ncatlab.org/nlab/show/monadic+adjunction>). I was wondering if there was a similar characterization to determine when $D$ is the Kleisli category of FREE $T$-algebras?
| https://mathoverflow.net/users/4528 | Characterization of Kleisli adjunctions | There is a unique functor $\mathbf{Kl}(GF) \rightarrow \mathbf{D}$ commuting with the adjunctions from $\mathbf{C}$, since the Kleisli category is initial among adjunctions inducing the given monad; and this functor is always full and faithful, since $\mathbf{Kl}(GF)(A,B) \cong \mathbf{C}(A,GFB) \cong \mathbf{D}(FA,FB)$.
So this functor will be an equivalence iff it is essentially surjective, and an isomorphism iff it is bijective on objects. But its object map is just the object map of $F$.
So $\mathbf{Kl}(FG)$ is equivalent to $\mathbf{D}$ compatibly with the adjunctions from $\mathbf{C}$ precisely when $F$ is essentially surjective, and isomorphic just when $F$ is bijective on objects.
| 18 | https://mathoverflow.net/users/2273 | 26106 | 17,105 |
https://mathoverflow.net/questions/26104 | 1 | A space $X$ is called *locally contractible* it it has a basis of neighbourhoods which are themselves contractible spaces. CW complexes and manifolds are locally contractible. On the other hand, the path fibration $PX \to X$ space of based paths with evaluation at the endpoint as projection) admits local sections iff $X$ is *$\infty$-well-connected* (or *locally relatively contractible*, or *semi-locally contractible*), that is, has a basis of neighbourhoods $N$ such that the inclusion maps $N\hookrightarrow X$ are null homotopic. Another use of this concept is by Dold, when he proves a Dold fibration (a map with the Weak Covering Homotopy Property) over an $\infty$-well-connected space is locally homotopy trivial.
What, then, is an example of a space which is $\infty$-well-connected but not locally contractible?
---
Edit:
Note that the 1-dimensional version of this is a space that is semilocally 1-connected (or 1-well-connected, in my revisionist terminology), but not locally 1-connected.
| https://mathoverflow.net/users/4177 | An example of a space which is locally relatively contractible but not contractible? | The same counterexample as for semilocally 1-connected works: namely, you can take the cone on the Hawaiian earring space. The space itself is contractible, but no sufficiently small neighborhoods of the "bad" point at the base of the cone are 1-connected (hence not contractible).
| 7 | https://mathoverflow.net/users/360 | 26107 | 17,106 |
https://mathoverflow.net/questions/26112 | 17 | *What is a good example of a fact about the moduli space of some object telling us something useful about a specific one of the objects?*
I am currently learning about moduli spaces (in the context of the moduli space of elliptic curves). While moduli spaces do seem to be fascinating objects in themselves, I am after examples in which facts about a moduli space tell us something interesting about the specific objects that they parametrise. For example, does the study of $\mathbb RP^n$ tell us anything we don't already know about some given line through the origin (say, the $x\_1$-axis) in $\mathbb R^{n+1}$?
| https://mathoverflow.net/users/6345 | Examples of the moduli space of X giving facts about a certain X | The easiest example I can think of is the natural incidence correspondence between $\mathbb{P}^3$ and the parameter space of cubic surfaces. This can be used to show that every cubic surface contains a line; from this [it follows easily](https://mathoverflow.net/questions/20112/interesting-results-in-algebraic-geometry-accessible-to-3rd-year-undergraduates/20261#20261) that every smooth cubic surface contains exactly $27$ lines.
Another example is the moduli space of stable maps constructed by Kontsevich; this parametrizes certain maps from curves to (to stick with a simple case) $\mathbb{P}^2$. It can be used to answer the following question: given $3d-1$ points in $\mathbb{P}^2$ in general position, compute the number $N\_d$ of *rational* curves of degree $d$ passing through these points. It turns out that the values $N\_d$ satisfy a certain recursive relation which allows you to compute all these numbers starting from the obvious $N\_1 = 1$ (through $2$ points passes exactly one line). You can find the formula at the entry [Kontsevich's formula](http://rigtriv.wordpress.com/2008/12/23/kontsevichs-formula/) on [Rigorous trivialities](https://rigtriv.wordpress.com); it yields for instance $N\_2 = 1$ and $N\_3 = 12$.
Yet another example, again more elementary is the following. The Grassmannian $G = \operatorname{Gr}(1, \mathbb{P}^3)$ parametrizes lines in $\mathbb{P}^3$. The computation of the cohomology of $G$ allows you to compute the number of lines which are incident to $4$ fixed lines in general position (it turns out this number is $2$).
| 25 | https://mathoverflow.net/users/828 | 26118 | 17,112 |
https://mathoverflow.net/questions/26119 | 15 | Let $(X,d)$ be a metric space. Banach's fixed point theorem states that if $X$ is complete, then every contraction map $f:X\to X$ has a unique fixed point. A contraction map is a continuous map for which there is an real number $0\leq r < 1$ such that $d(f(x),f(y))\leq rd(x,y)$ holds for all $x,y\in X$.
>
> Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?
>
>
>
| https://mathoverflow.net/users/5952 | Converse to Banach's fixed point theorem? | The answer is no, for example look at the graph of $\sin(1/x)$ on $(0,1]$. But for more information and related questions check out "[On a converse to Banach's Fixed Point Theorem](http://www.ams.org/journals/proc/2009-137-09/S0002-9939-09-09904-3/)" by Márton Elekes.
| 11 | https://mathoverflow.net/users/2384 | 26122 | 17,115 |
https://mathoverflow.net/questions/26072 | 6 | Hello!
I'd like to understand the relation between the following two theorems:
* The "global" duality for projective schemes, as explained in [Hartshorne]: If $X$ is an equidimensional projective Cohen-Macaulay scheme of dimension $n$ over an algebraically closed field with dualizing sheaf $\omega\_X$, then for all $i$ there is a natural isomorphism $Ext^i({\mathcal F},\omega\_X)\cong H^{n-i}(X,{\mathcal F})^{\ast}$.
* The "local" duality theorem Cohen-Macaulay rings, as explained in [Bruns, Herzog]: If $(R,{\mathfrak m},k)$ is a complete local ring of dimension $d$, then for all finite $R$-modules $M$ and all $i$ there is a natural isomorphism $\text{Ext}\_R^i(M,\omega\_R)\cong\text{Hom}\_R(H\_{\mathfrak m}^{d-i}(M),E(k))$, where $\omega\_R$ is the canonical module of $R$ and $E(k)$ the injective hull of the residue field $k$.
The isomorpisms are strikingly similar, but I don't know if there is a rigorous way to deduce, say, the global duality from the local one. Can somebody explain this to me or give references?
Thank you!
| https://mathoverflow.net/users/3108 | Comparison of Local and Global Duality | In the case of varieties over a perfect field this question is explained with great detail in the book by J. Lipman:
Dualizing sheaves, differentials and residues on algebraic varieties. (French summary)
Astérisque No. 117 (1984)
known by some people as "*Lipman's blue book*". If you want to get rid of the base field then you should look at Greenlees May duality for schemes and Grothedieck duality for formal schemes, see joint work by Alonso, Jeremías & Lipman.
| 5 | https://mathoverflow.net/users/6348 | 26130 | 17,122 |
https://mathoverflow.net/questions/26083 | 18 | For example, Wikipedia states that etale cohomology was "introduced by Grothendieck in order to prove the Weil conjectures". Why are cohomologies and other topological ideas so helpful in understanding arithmetic questions?
| https://mathoverflow.net/users/4692 | Why are topological ideas so important in arithmetic? | Why are topological ideas so important in arithmetic? In some sense KConrad is of course spot on, but let me offer a completely different kind of answer.
Why are complex functions of one variable so important in arithmetic? (Zeta function, L-functions, Riemann hypothesis, Birch--Swinnerton-Dyer, modular forms, theta series, Eisenstein series...).
Why is geometry so important in arithmetic? (Faltings' theorem, applications of algebraic geometry, low-dimensional arithmetic of varieties (elliptic curves etc))
Why is K-theory so important in arithmetic? (Bloch-Kato, Voevodsky...)
Why is logic so important in arithmetic? (Julia Robinson, Matiyasevich, Ax-Kochen and then Hrusovski proving that "if it's true in char p for suff large p then it's true in char 0" in the context of some very deep statements)
Why is functional analysis so important in arithmetic? (L^2 functions on $\Gamma\backslash G$ with $G$ a semisimple Lie group being related to automorphic forms and hence to number theory via Langlands, with crucial analytic tools like the trace formula).
Why are dynamical systems so important in arithmetic? (3x+1 problem, work of Deninger, or of Lind/Ward and their school).
Here's the answer: it's because arithmetic is a very mature subject---it has been around literally thousands of years, and because it has been around so long, there is far more of a chance that someone will come along with an insight relating [insert arbitrary area of pure mathematics here] with arithmetic. So in some sense it's a historical fluke. If we were all born with continuum-many fingers which we could move only in real-analytic ways, and we didn't discover the positive integers until much later on, then arithmetic would be all new and we'd be waiting for Gauss, and real analysis would be as old as the hills, and people would be asking "why is [insert arbitrary thing] so important in real analysis"?
[PS (1) yeah I know, I was being facetious at the end, and (2) yeah I know, my list at the top is woefully incomplete]
| 29 | https://mathoverflow.net/users/1384 | 26135 | 17,126 |
https://mathoverflow.net/questions/26081 | 14 | The metric of a Riemannian manifold determines the shortest
distance between any two points.
1. I assume the reverse holds? That is, if you are given the
shortest distance *d(x,y)* between every pair of points
of a manifold *M*, the metric for *M* is determined?
I am mainly interested in compact, connected, closed 2-manifolds,
but the most general answer would be appreciated.
(Apologies in advance for my naivete.)
2. Assuming I am correct above, is there some natural subset
*S* of *M* such that knowing *d(x,y)* between every
pair of points of *S* uniquely determines the metric (up to isometry)?
For example, suppose *S* is a simple closed geodesic on
*M*? Perhaps some assumptions on *M* are necessary:
genus zero, convex, ... ?
| https://mathoverflow.net/users/6094 | Shortest-path Distances Determining the Metric? | Concerning the second question. A single closed geodesic is not enough. For example, let $S$ be the equator of the standard sphere. All distances between points of $S$ are realized by paths in $S$, so you don't have any information about the metric outside $S$, except that it is sufficiently large (so that the paths outside $S$ are not shorter).
On the positive side, for every 2D Riemannian manifold, you can construct an embedded graph $S\subset M$ such that the distances between its points determine the metric uniquely (up to an isometry). It suffices to take a union of small geodesic circles such that their encircled regions cover $M$ and lie within convexity radii of their centers. Indeed, a metric on a 2-disc whose boundary is convex and whose geodesics are all strictly minimal, is uniquely determined by the distances between boundary points.
This used to be a long-standing conjecture and was proved in 2005 by Pestov and Uhlmann:
"Two dimensional compact simple Riemannian manifolds are boundary distance rigid", Ann. of Math. 161 (2005), no. 2, 1093--1110; MR2153407 (2006c:53038).
In higher dimensions, the full conjecture is still open. Until very recently, it was known only for very special metrics (some locally symmetric ones and some splitting ones). Dima Burago and I proved it for sufficiently small regions in any Riemannian manifold, where "sufficiently small" actually means that the diameter is small relative to the maximum modulus of sectional curvature (and of course to the injectivity radius). This is in our paper "Boundary rigidity and filling volume minimality of metrics close to a flat one", Ann. of Math. 171 (2010), no. 2, 1183--1211.
So, as Will Jagy suggested in his answer, you can take the $(n-1)$-skeleton of a sufficiently fine triangulation for $S$.
I am not aware of any numerical algorithms for this problem. It is known that the solution is stable (depends continuously on the input data) but you have to take into account the derivatives of the boundary distance function. A $C^0$ approximation of the distances is not enough to find an approximation of the metric, see Paul Siegel's answer.
| 9 | https://mathoverflow.net/users/4354 | 26136 | 17,127 |
https://mathoverflow.net/questions/26134 | 9 | Hello,
in their book [Cohomology of Finite Groups](http://books.google.com/books?id=sKmshEctdw0C&dq=cohomology+of+finite+groups&printsec=frontcover&source=bn&hl=de&ei=vV3-S_3OFcSrsAb79c39CQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDIQ6AEwAw#v=onepage&q&f=false) Adem and Milgram investigate the cohomology of the finite orthogonal and symplectic groups only in case $\mathbb{F}\_2$.
Let $p$ be a prime dividing the order of $\text{O}\_n(q)$, $\text{Sp}\_n(q)$ and $q$ a prime power.
I am wondering if anything is known about $\text{H}^\ast(\text{O}\_n(q),\mathbb{F}\_p)$ or $\text{H}^\ast(\text{Sp}\_n(q),\mathbb{F}\_p)$.
I am also interested in the maximal elementary abelian $p$-subgroups of these groups.
In light of Quillen's stratification theorem these two questions are, of course, related to each other.
I would be grateful for any kind of information.
| https://mathoverflow.net/users/3102 | Cohomology of orthogonal and symplectic groups | Direct finite group computations of cohomology of the finite groups of Lie type tend to be very sparse. The case $p=2$ has special interest for topologists and does provide some explicit results. More generally, some indirect results of interest have been found in recent decades by systematically comparing cohomology of the finite groups (in the defining characteristic $p$) with rational cohomology of the ambient algebraic groups. The main work in this direction is found in papers by some combination of Cline, Parshall, Scott, van der Kallen (see their joint 1977 paper in Invent. Math.), Friedlander. But a full description of the cohomology rings is elusive. Benson and Carlson have done a lot of work in the traditional setting as well. There are lots of papers, but not many definitive results.
For maximal elementary abelian $p$-subgroups, some of the work by Avrunin, Parshall, Scott would be relevant, as well as the older LMS Lecture Notes by Kleidman and Liebeck. Perhaps the most helpful source is the third volume of an ongoing series of books on the classification of finite simple groups. See especially section 3.3 and the summary table there for groups of Lie type:MR1490581 (98j:20011) 20D05 (20-02)
Gorenstein, Daniel; Lyons, Richard (1-RTG); Solomon, Ronald (1-OHS)
The classification of the finite simple groups. Number 3. Part I. Chapter A.
Almost simple K-groups.
Mathematical Surveys and Monographs, 40.3.
American Mathematical Society, Providence, RI, 1998. xvi+419 pp.
ISBN 0-8218-0391-3
| 8 | https://mathoverflow.net/users/4231 | 26138 | 17,128 |
https://mathoverflow.net/questions/26137 | 17 | One of the recent questions, in fact
[the answer](https://mathoverflow.net/questions/25630/major-mathematical-advances-past-age-fifty/25631#25631)
to it, reminded me about the binomial sequence
$$
a\_n=\sum\_{k=0}^n{\binom{n}{k}}^2{\binom{n+k}{k}}^2,
\qquad n=0,1,2,\dots,
$$
of the Apéry numbers. The numbers $a\_n$ come as
denominators of rational approximations to $\zeta(3)$ in Apéry's
famous proof of the irrationality of the number.
There are many nice properties of the sequence, one of these
is the observation that for primes $p\ge5$,
$$
a\_{pn}\equiv a\_n\pmod{p^3},
\qquad n=0,1,2,\dots.
$$
The congruence was conjectured for $n=1$ by S. Chowla, J. Cowles and M. Cowles
and proved in the full generality by I. Gessel (1982).
There is probably nothing strange in the congruence
(which belongs, by the way, to the class of *supercongruences* as it happens
to hold for a power of prime higher than one). But already the classical
binomials behave very similarly: for primes $p\ge5$,
$$
\binom{pm}{pn}\equiv\binom mn\pmod{p^3},
$$
the result due to G.S. Kazandzidis (1968). There are many other examples
of modulo $p^3$ congruences, most of them explicitly or implicitly related
to some modular objects, but that is a different story. My question is:
what are the grounds for the above (very simple) congruence for binomial
coefficients to hold modulo $p^3$? Not modulo $p$ or $p^2$ but $p^3$.
I do not ask you to prove the supercongruence but to indicate a general
mechanism which provides some kind of evidence for it and can be used
in other similar problems.
My motivation rests upon my own research on supercongruences; most of them
are just miracles coming from nowhere...
| https://mathoverflow.net/users/4953 | Binomial supercongruences: is there any reason for them? | I learned the second congruence as a version of [Wolsteholme's theorem](http://en.wikipedia.org/wiki/Wolstenholme%2527s_theorem), and I would be a bit surprised if Kazandzidis was the first person to observe the equivalence between this form and any other form of Wolsteholme's result. As for the "reason" that this result is true, I wrote a proof for the Wikipedia page which is mostly, but not entirely, a direct counting argument and which you could call "the grounds" for the congruence. The conceptual content of the argument is as follows:
The result holds modulo $p^2$ because you can divide the $pm$-set into $m$ cycles of length $p$ and rotate them separately. You obtain $\binom{m}{n}$ equivalence classes of subsets of size 1, and the other equivalence classes have order $p^2$ or higher.
Then you can examine $\binom{-p}{p}$, interpreted as $\binom{p^4-p}{p}$. This binomial coefficient is algebraically equivalent to $\binom{2p}{p}/2$ up to sign. The orbit decomposition in the previous paragraph establishes a second relation with $\binom{2p}{p}$. The conclusion is that the mod $p^2$ contribution vanishes for one non-trivial pair $(m,n)$, and if it vanishes once, it vanishes always. This vanishing principle extends mod $p^4$ — your second binomial congruence holds mod $p^4$ — provided that it vanishes once "by accident". A prime for which this happens is called a Wolstenholme prime. Two such primes are known, 16843 and 2124679.
Another remark: There are two pieces of evidence that the $p^2$ congruence (Babbage's theorem) is entirely combinatorial, but the extension to $p^3$ (Wolstenholme) is essentially algebraic. First, that Wolstenholme's theorem doesn't hold for the primes $p=2,3$. Second, that Babbage's theorem has a $q$-analogue for Gaussian binomial coefficients, with the same orbit proof. But Wolsteholme's extension does not have a $q$-analogue as far as I know. In fact, you can tell that the known $q$-Babbage's theorem doesn't extend, because the difference polynomial doesn't have any more cyclotomic factors.
---
I stand corrected on a couple of points. First, on the math, there is a paper by George Andrews, "q-Analogs of the binomial coefficient congruences of Babbage, Wolstenholme, and Glaisher", that does give some version of a q-analogue of Wolstenholme's theorem. The proof given there is algebraic, so it does not shed light on possible combinatorial "explanations". Second, according to Andrews, the full binomial interpretation of Wolstenholme's result is due to Glaisher.
| 19 | https://mathoverflow.net/users/1450 | 26147 | 17,133 |
https://mathoverflow.net/questions/26094 | 5 | Let Σp={1,...,p}ℤ be the full shift on p symbols, and let X ⊂ Σp be a subshift -- that is, a closed σ-invariant subset, where $\sigma\colon \Sigma\_p\to \Sigma\_p$ is the left shift. Then σ is expansive, and hence there exists a measure of maximal entropy (an mme) for (X,σ).
It is well known that if X is a subshift of finite type on which σ is topologically mixing, then there is a *unique* mme. (See, for example, Rufus Bowen, *Equilibrium states and the ergodic theory of Anosov diffeomorphisms*, 1975. In fact, Bowen proves uniqueness of equilibrium states for any Hölder continuous potential φ, but let's stick with the case φ≡0 for now.)
If X is not a subshift of finite type, then less is known. For example, if X is a β-shift, then it has a unique mme, but it is not known if this holds for subshifts that are factors of a β-shift.
*Question*: Does anybody know of a subshift that is topologically mixing but does not have a unique mme? (That is, a subshift that has multiple measures of maximal entropy despite being topologically mixing.)
| https://mathoverflow.net/users/5701 | A topologically mixing subshift with multiple measures of maximal entropy | Yes. See Haydn's paper [here](http://docs.google.com/viewer?a=v&q=cache%3ANF8M7-tT3w8J%3Aciteseerx.ist.psu.edu/viewdoc/download%253Fdoi%253D10.1.1.139.9001%2526rep%253Drep1%2526type%253Dpdf+Multiple+measures+of+maximal+entropy+and+equilibrium+states+for+one-dimensional+subshifts&hl=en&gl=us&pid=bl&srcid=ADGEESjeew53q334z7SrBbXQQIqOXoo1JnuKUmfHUbD9LOEa2dM8XnO6NQn_x7T8TLlF1meRY3gJxyxWMGd6G3eMYPD28D7EwfVZLYedHnwCTH_5gUX5Z3nCSiLp2JDU1jnYZW0tahWA&sig=AHIEtbRdqFxp2blJGJKe0-FjOCWRPJHjRg).
| 5 | https://mathoverflow.net/users/1847 | 26151 | 17,135 |
https://mathoverflow.net/questions/26117 | 17 | Prime matrices as defined in the following paper [Prime matrices P. F. RIVETT AND N. I. P. MACKINNON](https://www.jstor.org/stable/3616179) carry over many properties of factorization as in natural numbers to matrices over the field of naturals.
I quote the following:
>
> A matrix in a set M of matrices is prime (naturally enough) if it is not the
> product of any other matrices in the set. We thought we would look for the
> prime matrices in the set M of all 2 x 2 matrices with entries in the non negative
> integers and with determinant 1. To our great surprise we discovered
> that: there are only two primes, and any member of M (except I) can be uniquely factorized into a product of those two.
>
>
>
[EDIT (PLC): For some reason, there seems to be some confusion on which two matrices are in question. They are as follows:]
$$ P = \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right),\ \ Q =
\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) $$
***My question is whether there are any other resources on prime matrices and if there has been any generalization beyond 2x2 matrices. What mathematics will I need to pursue the subject further?***
I have only three journal articles:
1. *Marenich, V. E.*, [**On prime matrices over distributive lattices**](https://doi.org/10.1007/s10958-009-9727-1), J. Math. Sci., New York 164, No. 2, 260-271 (2010); translation from Fundam. Prikl. Mat. 14, No. 7, 157–173 (2008). [Zbl 1288.15041](https://zbmath.org/1288.15041).
2. *Wang, Qianhua; Zhang, Zhongjun*, [**Algorithm for obtaining the proper relatively prime matrices of polynomial matrices**](https://doi.org/10.1080/00207178708547392), Int. J. Control 46, 769–784 (1987). [Zbl 0634.65030](https://zbmath.org/0634.65030).
3. *Rivett, P. F.; Mackinnon, N. I. P.*, [**Prime matrices**](https://doi.org/10.2307/3616179). Math. Gaz. 70, No. 454, 257–259 (1986).
Please specify whether I can get the article/journal or book for free.
Thanks.
| https://mathoverflow.net/users/5627 | Prime/undecomposable matrices | For a different point of view, you might like to take a look at Section 12.5 and Appendix A in the [free on-line version](http://iml.univ-mrs.fr/editions/preprint00/book/prebookdac.html) of the following book (in which you'll find some interesting open questions related to "prime matrices" of the type you described):
N. Pytheas Fogg, Substitutions in dynamics, arithmetics and combinatorics, Lecture Notes in Mathematics, vol. 1794, Springer-Verlag, Berlin, 2002.
| 6 | https://mathoverflow.net/users/3029 | 26157 | 17,137 |
https://mathoverflow.net/questions/26156 | 3 | Hello!
I'm in search of some (possibly statistical) measure for matrices. I want to classify a square matrix as having the largest numbers running along the main diagonal or along the anitdiagonal. It isn't necessarily on the diagonals, the numbers can be some rows to the left or right, so a simple trace wouldn't work. I want to incorporate the entire matrix in the calculation.
To see how I came to this problem: the matrix A is a transition matrix, for example in a Markov process (I have a CS background). The probability of going from state i to state j is given by $A\_{i,j}$. Each row sums up to 1. The states are ordered though, and I would like to know if a certain matrix makes small or no jumps (largest probabilities along main diagonal) or if huge jumps are present (largest probabilities along antidiagonal).
I don't think this problem is very hard, but I'm at loss for the terminology. Thanks!
| https://mathoverflow.net/users/6351 | "Main" diagonal of a matrix | The following very coarse invariant is perhaps useful:
The argument of the sum $\sum\_{s,t}a\_{s,t}e^{i(s-t)\pi/n}$ (where $n$ is the size of the matrix and where the sum is over all entries) should be related to a typical jump. It is close
to $0$ if the matrix is "concentrated" near the diagonal and close to $\pi$ if the matrix is "concentrated" near the antidiagonal. The modulus of the above sum indicates somehow how much the matrix is concentrated.
| 3 | https://mathoverflow.net/users/4556 | 26163 | 17,142 |
https://mathoverflow.net/questions/26175 | 19 | Deligne, in his 1987 paper on the fundamental group of $\mathbb{P}^1 \setminus \{0,1,\infty\}$ (in "Galois Groups over $\mathbb{Q}$"), defines a system of realisations for a motivic fundamental group. Namely, Tannakian categories are defined of unipotent lisse $\ell$-adic sheaves, unipotent vector bundles with a flat connection and the crystalline equivalent given by unipotent overconvergent isocrystals, whereas the automorphism groups of their fibre functors define the $\ell$-adic, de Rham and crystalline fundamental groups respectively. Now, I do not claim to have a deep understanding of these constructions, nor of the paper as a whole (or to even have read it fully in detail, given its size!), but I was left with a question: why does the word unipotent appear in all the constructions?
Having read parts of Minhyong Kim's work it is understandable that in this context, these subcategories generated by the unipotent objects are not only enough, but seemingly exactly what is needed. In other words, one still acquires the Tannakian structure (which if I'm not mistaken exists on the full categories as well) and thus has fundamental groups which are affine group schemes and can start playing the game of using unipotency and quotients in the lower central series to construct moduli spaces of path torsors, moving gradually away from abelianness and deeper into Diophantine information. This however doesn't answer the original motivation.
I apologise if the answer is hidden in Deligne's original paper or if the context makes it obvious to everyone but me.
| https://mathoverflow.net/users/386 | Unipotency in realisations of the motivic fundamental group | Essentially because the Tannakian theory gives in the unipotent case (and only in
that case) a reasonably sized answer with an easy motivic interpretation.
For the size you should be aware that already in the topological situation the
group scheme associated by Tannaka theory to the fundamental group of $\mathbb
P^1$ minus three points (i.e., the free group on two generators) is
*huge*. For one thing each irreducible representation gives rise to a
reductive quotient and there are continuous families (i.e., positive dimensional
varieties) of such representations. From this one can see that the group scheme
maps onto a product of reductive groups where the index set are the points of
some algebraic variety (see
[What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?](https://mathoverflow.net/questions/21415/what-algebraic-group-does-tannaka-krein-reconstruct-when-fed-the-category-of-modu/21459)
for the case of $1$-dimensional representations).
Added to this is the fact that most of the topological representations do not
have geometric origin and hence have no motivic interpretation. If one looks at
$\ell$-adic representations of the fundamental group over $\mathbb Q$ of $\mathbb
P^1$ minus three points (or of suitable germs if one wants a theory over
$\mathbb C$) and adds mixedness assumptions, then the Tannakian category should
have a motivic interpretation which also should be independent (in some suitable
sense) of $\ell$ and should be comparable to its cristalline equivalent. This
however all depends on the Langlands program and hence is currently beyond our
reach.
If one sticks to unipotent representations then essentially all these problems
disappear. A unipotent representation (over some field $k$ of characteristic
$0$) of the free group $F$ on two elements factors through a nilpotent quotient $\Gamma$ of
$F$ and such a nilpotent quotient has a Malcev completion, a unipotent algebraic
group $G$ over $\mathbb Q$ of dimension the rank of $\Gamma$, such that the
Tannakian category of unipotent representations of $\Gamma$ over $k$ is
equivalent to the category of $k$-representations of $G$. Passing to the limit
gives us a pro-unipotent algebraic group $G\_\infty$ over $\mathbb Q$ whose category of
$k$-representations is equivalent to the category of unipotent
$k$-representations of $F$. Furthermore, the Lie algebra of $G\_\infty$ has a
nice cohomological description; it is the free Lie algebra generated by
$H\_1(X,\mathbb Q)$, where $X$ is $\mathbb P^1$ minus three points.
The motivic side of things now comes along very gracefully: For one of several
natural categories that has an appropriate $H\_1(X)$ in it there is a
corresponding relative Tannakian description of unipotent families of objects
over $X$. As examples we have unipotent variations of rational Hodge structures,
geometrically unipotent $\mathbb Q\_{\ell}$-adic sheaves over $\mathbb Q$ and
successive extensions of constant $F$-iso-crystals. In all these cases these
categories are described by representations of a pro-unipotent algebraic group
object in the appropriate base category (rational Hodge structure, $\mathbb
Q\_{\ell}$-adic sheaves over $\mathbb Q$ and $F$-isocrystals) and in all the
cases its Lie algebra is the free Lie algebra on $H\_1(X)$.
| 16 | https://mathoverflow.net/users/4008 | 26185 | 17,156 |
https://mathoverflow.net/questions/26184 | 5 | Given a set of multivariate, quadratic, non-homogeneous equations, is there a way to decide how many non-negative roots it have?
Some explanations:
1. All the coefficients are real numbers.
2. The number of variables is the same as the number of equations, and all the equations are independent.
3. Some of the equations might actually be linear equations.
4. By "non-negative" solution, I mean a solution in which all the variables take non-negative values.
5. By some "physical" reasoning, we know it must have at least one non-negative solution.
Further explanations:
1. I know generally a set of n quadratic equations with n variables has at most $2^n$ distinct roots.
2. The background of this problem is: the set of quadratic equations is the right-hand side of the chemical rate equation. By equating is to zero, the steady-state case is being considered. As we only take one-body or two-body reactions into account, the degree of the equations are at most two. As the abundance of the molecules cannot be negative, we only care about the non-negative solutions.
3. The number of variables can be up to 1000, so simple numerical test is not practical.
I am not a math student, and I am not sure whether this kind of question is allowed here.
| https://mathoverflow.net/users/6347 | Decide how many non-negative solutions a set of multivariate quadratic equations have | Not efficiently, at least not unless the problem has some additional structure which can be exploited. The set of mixed Nash equilibria of a two-player game can be written as the nonnegative solutions of such a polynomial system. In general it is #P-hard to count the equilibria of such a game ([Conitzer and Sandholm](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WFW-4SDGR4B-1&_user=501045&_coverDate=07%2F31%2F2008&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1350784086&_rerunOrigin=scholar.google&_acct=C000022659&_version=1&_urlVersion=0&_userid=501045&md5=bbb5979cc80b1cd76609cfa40386604a)). It is even PPAD-hard (still thought to be polynomial time unsolvable) to compute a single equilibrium, although one is guaranteed to exist by Nash's Theorem.
You may want to investigate connections with Nash equilibria further -- if your problem is somehow equivalent to this one then you will have a lot of established results and techniques to build on. Also people would probably find a connection between game theory and chemistry surprising. Or if your problem is easier, looking at Nash equilibria might help you figure out what extra features your problem has that makes it solvable efficiently. If it is harder or incomparable it might still be interesting to know that.
| 4 | https://mathoverflow.net/users/5963 | 26196 | 17,164 |
https://mathoverflow.net/questions/24508 | 9 | I am looking for classes of sequence, that converge iff they contain a converging sub-sequence.
* The basic example of such sequences are monotone sequences of real numbers.
* A more interesting examples comes from metric fixed point theory:
Let $B$ be a Banach space and $f\colon B \to B$ be a continuous mapping that is non-expansive (i.e. $\lVert f(x) - f(y)\rVert \le \lVert x -y\rVert$).
Define $x\_{n+1} := \frac{1}{2} x\_n + \frac{1}{2} f(x\_n)$ for any startingpoint $x\_0\in B$.
This is the so called *Krasnoselski iteration*.
One can show that any accumulation point $\tilde{x}$ of $(x\_n)$ is a fixed point of $f$.
Since $f$ is non-expansive, it follows that
$\lVert x\_{n+1}-\tilde{x}\rVert = \frac{1}{2}\lVert (x\_{n}-\tilde{x}) + (f(x\_{n}) - f(\tilde{x}))\rVert\le \lVert x\_n -\tilde{x}\rVert$.
Hence $(x\_n)$ converges iff it contains a converging sub-sequence.
This is a special case of Ishikawa's fixed point theorem. (The Krasnoselski-Mann iteration - a generalization of the Krasnoselski iteration - also has this property.)
I am interested in this sequence because they provide very nice applications of the Bolzano-Weierstrass principle.
Do you know of any other examples of sequences with this property?
Do you know other proofs that uses this property together with the Bolzano-Weierstrass principle to prove the convergence of a sequence?
| https://mathoverflow.net/users/3365 | Sequence that converge if they have an accumulation point | The following version of the mean ergodic theorem is taken from the book of Krengel, "ergodic theorems".
Let T be a bounded linear operator in a Banach space X. The Birkhoff averages are denoted by $A\_n = {1\over n} \ \Sigma\_{k=0}^{n-1} \ T^k$.
Assume that the sequence of operator norms $||A\_n||$ is bounded independently of $n$. Then for any x and y in B, the following is equivalent :
-- y is a weak cluster point of the sequence $(A\_nx)$,
-- y is the weak limit of the sequence $(A\_nx)$,
-- y is the strong limit of the sequence $(A\_nx)$.
(note that we talk about cluster points instead of converging subsequences because we didn't assume B separable. Hence the weak topology is not necessarily metrizable.)
This theorem implies e.g. the ergodic theorem for Markov operators on $C(K)$ (sequential compactness follows from Azrela-Ascoli), or the ergodic theorem for power bounded operators defined on reflexive Banach spaces (sequential compactness follows from Eberlein-Smulian).
There is a whole set of theorems in ergodic theory along these lines. Let me mention the convergence of the one sided ergodic [Hilbert transform](http://www.ee.bgu.ac.il/~guycohen/CCL.pdf), discussed in Cohen and Cuny (see Th 3.2) as another example.
| 2 | https://mathoverflow.net/users/6129 | 26200 | 17,168 |
https://mathoverflow.net/questions/26203 | 13 | The Arnold conjecture on a closed symplectic manifold $(M,\omega)$ says in the weakest version that for a non-degenerate Hamiltonian there are at least $k$ 1-periodic orbits where $k$ is the sum of the Betti numbers of $M$. It is easy to show that one can assume that $\omega$ is integral, so I do so in the following.
On wikipedia it says that the [Arnold conjecture](https://en.wikipedia.org/wiki/Arnold_conjecture) is solved in many cases using Floer homology.
However, I was given the impression that this version of the Arnold conjecture has been solved in all cases, but are scattered around in several papers - due to several different complications.
**Question**: What is the current status of this weak Arnold conjecture precisely, and what are the refferences for these results?
Added: I know most of the details of the monotone case, so I am mostly interested in the more exotic cases.
| https://mathoverflow.net/users/4500 | Floer homology and status of the Arnold conjecture |
---
*V. I. Arnol'd, June 12, 1937 - June 3, 2010.*
The very sad news of his death is reported today [here](http://www.france-info.com/ressources-afp-2010-06-03-l-eminent-mathematicien-russe-vladimir-arnold-mort-subitement-en-450299-69-69.html).
---
After Floer, the main difficulty in solving the weak Arnol'd conjecture on a compact symplectic manifold $M$ lies in defining a Floer chain complex generated by 1-periodic orbits of an arbitrary non-degenerate Hamiltonian $H: S^1\times M \to \mathbb{R}$, in such a way that the homology is independent of $H$. Once one has that, the remaining step (proving an isomorphism with Morse homology) can be done either by a computation with small autonomous Hamiltonians, or by a "PSS" isomorphism.
When $M$ is monotone, the crucial compactness theorems for solutions to Floer's equation (used to define the candidate-differential on the Floer complex, to prove that it squares to zero, and, in a variant, to prove the invariance of the theory) can be proved using index considerations. When $M$ is Calabi-Yau, compactness needs an additional idea, that holomorphic spheres generically don't hit cylinders solving Floer's equation. This is beautifully worked out in
>
> Hofer, H.; Salamon, D. A. "Floer homology and Novikov rings." The Floer memorial volume,
> 483--524, Progr. Math., 133, Birkhäuser, Basel, 1995; MR1362838.
>
In general, where there may be holomorphic spheres with small negative Chern number, one has little choice but to allow "stable trajectories" consisting of broken Floer trajectories with holomorphic bubble-trees attached. Transversality is proved by introducing multi-valued perturbations to the equations, and this forces one to use rational coefficients. References:
>
> Fukaya, Kenji; Ono, Kaoru. "Arnold conjecture and Gromov-Witten invariant". Topology 38 (1999), no. 5, 933-1048. MR1688434
>
>
> Liu, Gang; Tian, Gang, "Floer homology and Arnold conjecture", J. Differential Geom. 49 (1998), no. 1, 1-74. MR1642105
>
[Edit: both these references offer proofs of the weak Arnol'd conjecture with rational coefficients.] For a detailed introduction to these "virtual transversality" methods, see
>
> Salamon, Dietmar, "Lectures on Floer homology". MR1702944
>
The technical complications of virtual transversality theory are notorious, and one could wish for a fully detailed textbook account.
What's left?
So far as I know, there is no proof for general manifolds that the number $h$ of 1-periodic orbits of a non-degenerate Hamiltonian is at least the sum of the mod $p$ Betti numbers. The strong Arnol'd conjecture for non-degenerate Hamiltonians, that $h$ is at least the minimum number of critical points of a Morse function, is wide open.
| 22 | https://mathoverflow.net/users/2356 | 26208 | 17,173 |
https://mathoverflow.net/questions/26213 | 1 | What properties of a canonical bundle are preserved under birational isomorphism between
smooth projective varieties. In particular, is triviality of the canonical bundle preserved?
| https://mathoverflow.net/users/6254 | Birational properties of canonical bundle | Generally speaking, to your specific question, the answer is no.
See for example Hartshorne, Chapter V, Section 3. On the other hand, numerous other properties are preserved (some of these are also mentioned in Hartshorne, for example, the geometric genus). There are more subtle things as well, see any book on the *minimal model program* for example.
With regards to your original question, you might want to try looking up *crepant resolutions*.
| 2 | https://mathoverflow.net/users/3521 | 26215 | 17,176 |
https://mathoverflow.net/questions/25817 | 16 | It is well known that every construction that can be performed with compass and straightedge alone can also be performed using origami, see:
R. Geretschlager. Euclidean Constructions and the Geometry of Origami.
Mathematics Magazine 68 (1995), no. 5, 357–371.
If one checks the paper by Geretschlager above, one sees that the construction for intersecting two circles is quite complex. Is there an easier construction?
The axioms for origami can be seen at:
<http://origami.ousaan.com/library/conste.html>
Background: I teach a course in geometry for future teachers. With this construction (and other easy constructions), it would be fairly clear that every construction that can be performed with compass and straightedge alone can also be performed using origami.
I only have found two origami constructions for intersecting two circles - one in the above reference and another in a forgotten (!!!) reference. It was a strange book written in the 1950s about various geometric constructions.
| https://mathoverflow.net/users/6270 | Origami Constructions: Intersecting two Circles | Have you found the papers by Roger Alperin? He has some very nice articles, especially in regards to relating various construction systems -- in addition to origami and compass-straightedge constructions, there are a variety of others, like the Vieten constructions, Pythagorean constructions, and even some variants on the origami constructions. Basically, each additional axiom potentially provides an entirely new class of constructions (which may or may not actually enlarge the set of valid constructions). Some of these questions end up being very geometric in nature, and others have some fantastic ties to the structure of algebraic numbers.
In any case, you can find several papers of his via Google. I believe your specific question about intersecting two circles is answered in Section 6 of Alperin's "Mathematical Origami: Another View of Alhazen's Optical Problem."
Hope that helps.
| 8 | https://mathoverflow.net/users/35575 | 26219 | 17,180 |
https://mathoverflow.net/questions/26209 | 2 | A question inspired by [Is the Euler characteristic a birational invariant](https://mathoverflow.net/questions/25922/is-the-euler-characteristic-a-birational-invariant):
As remarked in Mike Roth's answer to the above linked question, if $X$ and $Y$ are smooth projective varieties in characteristic zero that are birational, then there is a smooth $Z$
with morphisms $p: Z \rightarrow X$ and $q: Z \rightarrow Y$ that are both birational isomorphisms.
Question: Is it possible to arrange for at least one of $p$ and $q$ that the locus in $Z$ where the morphism fails to be an isomorphism is of codimension at least $2$?
Here it might be useful to assume that ${\rm dim}X={\dim Y} \geq 3$.
| https://mathoverflow.net/users/6254 | Birational correspondences and codimension where not an isomorphism | No it is not possible (see for instance Thm II:2.4 of Shafarevich: Basic Algebraic Geometry) which says that the exceptional locus is always of codimension $1$ provided the target is smooth. The crucial property is that the target variety be $\mathbb Q$-factorial. The example of Dmitri shows that this condition is necessary as the quadric cone is non-$\mathbb Q$-factorial at its apex. Note, however that in his example a $Z$ mapping to both $X$ and $Y$ will contract a divisor. An example is the blowup of the singularity whose exceptional locus is $\mathbb P^1\times \mathbb P^1$ and the mappings to $X$ and $Y$ restrict to projections on the two factors (and indeed any common map factors through that blowing up).
| 4 | https://mathoverflow.net/users/4008 | 26227 | 17,185 |
https://mathoverflow.net/questions/24859 | 3 | Please consider a two-dimensional surface populated with a set of Cartesian coordinates $(x\_i, y\_i)$ for $N$ circles with individual radii $r\_i$, where $r\_{\min} < r\_i < r\_{\max}$.
Here, the number of circles, $N$, may be large - ranging from hundreds to tens of thousands. The circles may sparsely populate the plane in some places, and in others, be 'conspiratorially' packed together. Furthermore, $r\_{\min}$ / $r\_{\max}$ are not necessary defined in such a way to allow for accurate convex hull, spline interpolation, etc.
While we can always perform a monte carlo sampling of coordinates on the plane (or over some defined lattice), is there an efficient deterministic method of calculating the exact area given by the union of all $N$ circles?
---
Update - After a more extensive literature search (and thanks to "jc" for mentioning Edelsbrunner!), I was able to find a few relevant papers in the literature. First, the problem was of finding the union of 'N' discs was first proposed by M. I. Shamos in his 1978 thesis:
Shamos, M. I. “Computational Geometry” Ph.D. thesis, Yale Univ., New Haven, CT 1978.
In 1985 Micha Sharir presented an $O(n \log^2n)$ time and $O(n)$ space deterministic algorithm for the disc intersection/union problem (based on modified Voronoi diagrams):
Sharir, M. Intersection and closest-pair problems for a set of planar discs. SIAM J. Comput. 14 (1985), pp. 448-468.
In 1988, Franz Aurenhammer presented another, more efficient O(n log n) time and O(n) space algorithm for circle (disc) intersection/union using power diagrams (generalizations of Voronoi diagrams):
Aurenhammer, F. Improved algorithms for discs and balls using power diagrams. Journal of Algorithms 9 (1985), pp. 151-161.
It would be really neat if anyone could be point me to an implementation of one of the two determistic algorithms above, perhaps in a computational geometry package (neither appear trivial to put into practice)...
| https://mathoverflow.net/users/3248 | Finding the union of N random circles arbitrarily (or conspiratorially) placed on a two-dimensional surface | CGAL has a general Voronoi diagram module that's quite customizable. While I've never used it to build power diagrams, it should not be hard to add in the right kind of distance function to generate the diagrams you need:
<https://doc.cgal.org/Manual/3.5/doc_html/cgal_manual/Voronoi_diagram_2/Chapter_main.html>
| 3 | https://mathoverflow.net/users/972 | 26232 | 17,188 |
https://mathoverflow.net/questions/26220 | 22 | I hope this question is not so elementary that it'll get me banned...
In mathematics we see a lot of impredicativity. Example of definitions involving impredicativity include: subgroup/ideal generated by a set, closure/interior of a set (in topology), topology generated by a family of sets, connected/path connected component of a point, sigma algebra generated by a family... And of course, the least upper bound property of real numbers. Impredicativity flood mathematics but there are people who don't like it. I think type theory was developed due the paradoxical and impredicative nature of "set of all sets that don't contain themselves".
I'm very ignorant on this and from what I read type theory is hopeless complicated to work this, so I ask the people who favor predicativism: all those people have fiddled with mathematics for all those years and no one ever found a contradiction (I mean in ZFC), so is it worth all that effort?
| https://mathoverflow.net/users/6361 | Impredicativity | Yes, it is worth the effort. A predicative version of an impredicative construction is typically more explicit and informative than the impredicative one. For example, consider the construction of a subgroup $\langle S \rangle$ of a group $G$ generated by the set $S$:
* **impredicative**: $\langle S \rangle$ is the intersection of all subgroups of $G$ which contain $S$.
* **predicative**: $\langle S \rangle$ consists of all finite combinations of elements of $S$ and their inverses, i.e., a typical elements is $x\_1 x\_2 \cdots x\_n$ where $x\_i \in S \cup S^{-1}$.
This can be quite useful if you want to compute with groups (i.e., with a computer), as you will definitely prefer the second description, which tells you how exactly the elements of the subgroup can be represented.
Many examples of impredicative constructions are special cases of the following theorem.
>
> **Theorem** (Knaster and Tarski): *A monotone map on a complete lattice has a least fixed-point above every point.*
>
>
>
To take two of your examples:
* *Subgroup generated by a set*: the complete lattice is the powerset $P(G)$ of the group $G$ in question, and the map $f : P(G) \to P(G)$ takes $S \subseteq G$ to $f(S) = S \cup S^{-1} \cup S \cdot S$.
* *$\sigma$-algebra generated by a family of subsets*: exercise.
There are two standard ways of proving the Knaster-Tarski theorem, one impredicative and one predicative. These exemplify the two general approaches of getting to desired objects "impredicatively from above" and "predicatively from below".
The impredicative proof goes as follows: call a point $x$ a *prefixed* point if $f(x) \leq x$. Consider the set $S$ of all prefixed points above a given point $y$ (which is not empty as it contains the top of the lattice). The least fixed-point above $y$ is the infimum $x = \inf S$ (exercise).
The predicative proof goes as follows: iterate $f$ starting with a given point $y$ to construct an increasing sequence $$y, f(y), f^2(y), \ldots, f^\omega(y), \ldots, f^\alpha(y), \ldots$$ where we have to iterate through ordinals until we're blue in the face. The iteration stops eventually, and that's the least fixed-point above $y$.
Of course, in such generality the predicative proof is hardly better than the impredicative one because we replaced one non-description with another. But in particular cases we might know something about $f$. For example, we might know that it preserves suprema of countable chains, as we do in the example of a subgroup generated by a set, in which case the iteration stops at $\omega$.
Your third example, namely the connected component of a point, can be dealt with also, but I am not sure it's any better than the impredicative construction:
* *Connected components*: the connected component of a point is a maximal connected subset containing it. You are probably thinking of the construction that says "just take the union of all connected subsets that contain the point". We could instead try the following: define $x \sim y$ to mean that for all continuous $f : X \to 2$, $f(x) = f(y)$. The connected components of $X$ are the equivalence classes of $\sim$. Therefore the connected component of a point is just its equivalence class. This is not entirely satisfactory as it replaces one bad description with another. Can we be more explicit? What if we have a nice basis for the space?
Sometime you have to reformulate the whole subject to get away from builtin impredicativity (and it is still worth doing because it gives computational meaning to theorems which are quite non-computational in the impredicative setting):
* *Closure/interior of a set*: under classical formulation of topology you can sometimes get away with predicative construction, for example if you can reduce your construction to manipulation of a countable topological basis, e.g. the interior of $S \subseteq \mathbb{R}$ is the union of all open intervals with rational endpoints that are contained in $S$. There are general formulations of topology, such as [formal topology](http://www.cs.chalmers.se/~coquand/formal.html) and [Abstract Stone Duality](http://www.paultaylor.eu/ASD/), which avoid impredicative constructions altogether.
Lastly, you mention Dedekind completeness of reals. I am not sure this is impredicative. The supremum of a non-empty bounded family of left-sided Dedekind cuts is simply their union. What is impredicative about taking the union of a family of cuts?
**Adendum**: Note that in the typical case it is the **construction**, i.e., an existence proof, which is predicative or impredicative, not the definition. For example, the group generated by a set is *defined* as the least subgroup containing the generators, which has nothing to to with predicativity/impredicativity.
P.S. You need a better MO username.
| 30 | https://mathoverflow.net/users/1176 | 26233 | 17,189 |
https://mathoverflow.net/questions/26169 | 6 | Let $G$ be a (connected) reductive group over a field $k$. Then there is a natural functor from the category of representations of $G$ to the category of representations of $G(k)$.
Under which circumstances can one say that this functor is a) full and b) faithful?
Edited for clarity:
Here we think of algebraic representations of $G$ over the field $k$ (so, for example, morphisms of $G$ to $GL(V)$ for $k$-vector spaces $V$, defined over $k$).
Given such a morphism $\phi : G \rightarrow GL(V)$, we take $k$ points to obtain a $k$-linear representation of the group $G(k)$ on the $k$-vector space $V$ (thus a homomorphism $\phi(k) : G(k) \rightarrow GL(V)(k)$).
| https://mathoverflow.net/users/1594 | Points of reductive groups | A bit of Tannakian formalism clarifies the situation. Recall that for every abstract group $\Gamma$ there is a notion of "algebraic hull" $\Gamma^{alg}$ constructed as follows: Consider pairs $(\varphi,H)$ where $H$ is an algebraic group over $k$ and $\varphi:\Gamma\to H(k)$ a homomorphism of groups with Zariski-dense image. Maps between such pairs are defined in the obvious was, and one defines
$$\Gamma^{alg} := \lim\_{(\varphi,H)}H$$
So this is an affine group scheme over $k$. The group $\Gamma^{alg}$ has a universal property, namely that for any algebraic group $H$ over $k$, any map $\Gamma \to H(k)$ factors over $\Gamma^{alg}$. From Tannakian formalism we know:
>
> The canonical map $\Gamma\to\Gamma^{alg}$ induces an equivalnence of monoidal categories between the category of finite dimensional algebraic $\Gamma^{alg}$-representations and finite dimensional $k$-representations of $\Gamma$.
>
>
>
Now, let's apply this to $\Gamma = G(k)$ for a fixed algebraic, not necessarily reductive group $G$ over $k$. The universal property of the algebraic hull yields a canonical morphism $\Gamma^{alg}\to G$ whose image is the Zariski-closure of $\Gamma$ in $G$. We have then results like this:
>
> Proposition (Deligne, LNM900, p.139): Let $f: H\to G$ be a morphism of affine group schemes over $k$ and let $\omega$ be the induced functor $Rep\_k(G) \to Rep\_k(H)$ between categories of finite dimensional representations. The morphism $f$ is faithfully flat if and only if $\omega$ is fully faithful and if for every representation $V$ of $G$ every subrepresentation of $\omega(V)$ is isomorphic to the image of a subrepresentation of $V$.
>
>
>
For instance, faithful flatness of $f$ implies surjectivity of $f$, which is therefore a necessary condition for the induced functor of representations to be fully faithful - this is what's behind Brian's comment. If you assume $G$ to be reductive then the category $Rep(G)$ is semisimple (in characteristic zero) and the last condition in Deligne's proposition can, maybe, be rephrased in a simpler way.
| 1 | https://mathoverflow.net/users/5952 | 26235 | 17,191 |
https://mathoverflow.net/questions/26243 | 15 | Given a non-integral real $\alpha$, is there an entire (see <http://en.wikipedia.org/wiki/Entire_function>) function $h(x)$ such that $x^{-\alpha}h(x)\longrightarrow 1$
for $x\rightarrow+\infty$ (with $x$ real non-negative)?
Clearly, such a function if it exists is not unique since $h(x)+e^{-x}$ and similar functions
work also.
| https://mathoverflow.net/users/4556 | Asymptotic approximation of $x^\alpha$ by entire functions | Start with an entire function $f$ such that $f(x)=1/x + O(1/x^2)$ for $x>0$, $x\rightarrow\infty$. For example $f(z)= (1-e^{-z})/z$.
Let F be some primitive for $f$: $F(z)=\int\_1^z f(s)ds$.
We have $F(x)= ln(x)+C+O(1/x)$, with C some constant ($ \ C=\int\_1^\infty \ (f(x)-{1\over x})\ dx$ ).
Then consider $h(x)=exp(\alpha F(x)-\alpha C)$.
We get ${h(x)\over x^\alpha} = exp(O(1/x))\rightarrow 1$.
| 15 | https://mathoverflow.net/users/6129 | 26245 | 17,197 |
https://mathoverflow.net/questions/26248 | 4 | Various sources claim that a maximum norm $||A||\_{max}=\max\_{i,j}|a\_{ij}|$ is not submultiplicative, i.e. $||AB||\_{max}\not\leq||A||\_{max}||B||\_{max}$.
Where can I find what norm a,b satisfy $||AB||\_{max}\leq||A||\_{a}||B||\_{b}$?
| https://mathoverflow.net/users/6358 | Submultiplicative matrix norm: Max Norm | The inequality $\|AB\|\_{\max} \leq \|A\|\_{a}\|B\|\_{b}$ for all $A$, $B$ can be achieved or destroyed just by rescaling the norms $\|\cdot\|\_a$ and $\|\cdot\|\_b$. Let's suppose that we're considering $d \times d$ matrices. If we just make sure that the two norms $\|\cdot\|\_a$ and $\|\cdot\|\_b$ are scaled so that both of them have the property $\|C\|\_i \geq \sqrt{d}\|C\|\_{max}$ for all $d \times d$ matrices $C$, then the desired inequality follows from the elementary inequality $\|AB\|\_{\max} \leq d.\|A\|\_{\max}\|B\|\_{\max}$. Conversely, if the norms are rescaled so that both of them give norm $\frac{1}{2}$ to the identity matrix, then the inequality clearly cannot hold since $\|Id\|\_{max}=1$. The fact that such rescalings exist follows from the fact that norms on a finite-dimensional space are pairwise equivalent.
The point of this is that there are a lot of norms on the space of matrices if we don't make any additional requirements on them. Is this the kind of answer you were looking for? Or do you want the two norms to have additional properties?
edited: there was a typo on the main inequality
| 7 | https://mathoverflow.net/users/1840 | 26252 | 17,202 |
https://mathoverflow.net/questions/26269 | 9 | Apologies in advance if this is too elementary.
The following is well known when $A$ is an algebraically closed field:
>
> Let $X$ be an integral closed subscheme of $P^n\_A$. Then $\Gamma(X, \mathcal{O}\_X) = A$.
>
>
>
My question is: For what other rings does the above statement hold?
There are two proofs of this (for $A$ algebraically closed) in Hartshorne. Both seem to use the fact that the integral closure of $A$ in its quotient field is just $A$ itself in a key way.
So I suspect that having $A$ integrally closed will be crucial, but I do not know. In particular, does the proof in Hartshorne still work, and if so, does it apply to nonnormal domains?
| https://mathoverflow.net/users/344 | When is it true that the ring of global regular functions on a projective variety is just the base ring? | The statement is true if and only if $A$ is an algebraically closed field.
Assume the statement is true.
Let $\mathfrak{m}$ be a maximal ideal of $A$. Then $A \to A / \mathfrak{m}$ is certainly finite. Thus, as I explain in my comment above, $\mathbb{P}^n\_{A / \mathfrak{m}}$ is an integral subscheme of some $\mathbb{P}^N\_A$. Thus, $A = A / \mathfrak{m}$; since $\mathfrak{m}$ was maximal, $A$ is a field.
Suppose $A$ is not algebraically closed. Then $A$ has a finite extension $B$; and again, as remarked in the comment, this implies that $B$ is the ring of global regular functions of some integral closed subscheme of $\mathbb{P}^N\_A$, a contradiction.
| 5 | https://mathoverflow.net/users/5094 | 26279 | 17,217 |
https://mathoverflow.net/questions/26268 | 8 | Do the fusion categories $Rep(S\_4)$ and $Rep(A\_5)$ admit non-symmetric braidings? All the other rep. cats. of finite subgroups of $SU(2)$ do (in the McKay correspondence). My guess is no.
| https://mathoverflow.net/users/6355 | Non-symmetric Braiding on finite group Representation Categories | Eric,
The answer is no for $Rep(A\_5)$ and yes for $Rep(S\_4)$,
thanks to Victor Ostrik's observation.
For a braided category $C$ let $C'$ denote its Mueger center,
i.e., the subcategory of objects $Y$ in $C$ such
that the square of braiding of $Y$ with any $X$ in $C$ is identity.
So $C$ is symmetric if $C=C'$ and $C$ is non-degenerate (or modular)
if $C'$ is trivial.
Note that $C:=Rep(A\_5)$ is simple, i.e., it has no non-trivial
proper fusion subcategories. Now if $C$ has a non-symmetric braiding
then $C' \neq C$ is a proper subcategory. So $C'$ is trivial, i.e.,
$C$ with the above braiding is non-degenerate (modular). This cannot
happen (e.g., $C$ has a simple object of dimension 5, but in
a modular category the square of dimension of any object divides
dimension of the category, thanks to the result of Etingof-Gelaki).
For $D:= Rep(S\_4)$ there is a non-symmetric braiding with $D'=Rep(S\_3)$,
namely the equivariantization of a pointed category $Vec\_{Z/2Z\oplus Z/2Z}$
with respect to an action of $S\_3$.
| 8 | https://mathoverflow.net/users/3011 | 26280 | 17,218 |
https://mathoverflow.net/questions/26266 | 1 | What is your favorite examples of spectral sequences arising naturally in arithmetic geometry?
Please explain it in some detail
| https://mathoverflow.net/users/4245 | spectral sequences in number theory | I posit the following example, in response to your ambiguous question:
The coniveau spectral sequence seems to play an important role in 'arithmetic geometry'. One instance is in class field theory for schemes:
From W. Raskind's nice survery article "Abelian class field theory of arithmetic schemes" [AMS, 1992, pgs. 100-101]:
Let $X$ be an arithmetic scheme, $n>0$ invertible on $X$. Then there is a coniveau spectral sequence (in the etale site):
$$E^{p,q}\_1 = \bigoplus \_{ x\in X^{p} } H^{q-p} (k(x), \;\mathbb{Z}/n \; (j-p)) \Rightarrow H^{p+q} (X, \mathbb{Z}/n \;(j)) $$
Without going into more details, this sequence plays an important role in defining a reciprocity map from a class group of $X$ to abelian fundamental groups.
That's all I will say for now in hopes that the above provides for motivation to delve further into studying coniveau, etc.
Finally, one of the best articles I have seen on coniveau is by Colliot-Thélène, Hoobler, and Kahn, "The Bloch-Ogus-Gabber theorem" which can be found at:
<http://www.math.jussieu.fr/~kahn/preprints/prep.html>
It might be nice to have others' remarks/comments on coniveau, but I don't have any precise questions yet.
| 9 | https://mathoverflow.net/users/4235 | 26282 | 17,220 |
https://mathoverflow.net/questions/26272 | 15 | It's well-known that Hadamard and de la Vallée-Poussin independently proved the Prime Number Theorem in 1896: that $\pi(n)=n/\log n+o(n/\log n)$. I'm curious as to a weaker result: that for any $\varepsilon>0$, $\pi(n)\gg n^{1-\varepsilon}$.
Chebyshev famously proved that if $\lim \pi(n)\log n$ exists it must be equal to 1, but I seem to remember that he also proved bounds on that value, pushing the date back to 1850 or so in that case. But were there earlier results in this direction?
| https://mathoverflow.net/users/6043 | Who first proved that there are at least n^(1-ε) primes up to n? | In the 1850's, Chebyshev gave the following explicit bound, for sufficiently large n.
$$0.92129 \, \frac{n}{\log n} < \pi(n) < 1.0556 \, \frac{n}{\log n}.$$
This is mentioned in "An introduction to the theory of the Riemann zeta function" by S. J. Patterson. There is an exercise in the first chapter that gives the lower bound following Chebyshev's method (that is, using Stirling's formula and the series $\log \ n! = \Sigma\_{p^k\leq n} \lfloor n/p^k \rfloor\log p$). S. J. Patterson cites Chebyshev as the first to obtain significant results toward the prime number theorem.
| 17 | https://mathoverflow.net/users/6129 | 26286 | 17,224 |
https://mathoverflow.net/questions/26287 | 3 | I've read in many places, including the [n-Lab page](http://ncatlab.org/nlab/show/Lie+algebroid), that a Lie algebroid (which I think of as in the first definition on the n-Lab page) is the same as a vector bundle $A \to X$ and a (properties?) derivation that makes $\Gamma(\wedge^\bullet A^\*)$ into a cochain complex (here $A^\* \to X$ is the dual vector bundle). One motivation for considering this cochain complex is that in the two extremes $A = {\rm T}X$ and $X = \{\rm pt\}$, the complex becomes respectively the deRham complex and the Chevellay-Eilenberg complex for a Lie algebra. My motivation has something to do with understanding BRST-type arguments.
I only partially understand this relationship. Given a trivialized vector bundle $(V\times X)\to X$ with a differential $d$ on $\mathcal C^\infty (X,\wedge^\bullet V)$, then I can define a Lie algebroid structure on $V\times X \to X$ — the Lie algebroid axioms are equivalent to $d^2 = 0$. Conversely, given a Lie algebroid $A\to X$ with $A$ trivialized as a bundle, I know how to interpret the standard formulas for the differential on $\Gamma(\wedge^\bullet A^\*)$ — n-Lab, for example, reproduces the standard formula.
But I do not know how to interpret the formula defining the differential without trivializing the bundle, and my interpretation, as far as I can tell, depends on the trivialization (the fiber coordinates). At least, I've tried to check that my interpretation of the formula (which does get the above statements about $d^2 = 0$, etc., correct) is invariant under changing the trivialization, and I have failed. Hence:
>
> What is a totally invariant description of the Lie algebroid CE chain complex?
>
>
>
Better: is there such a description that is simultaneously useful for explicit calculations?
| https://mathoverflow.net/users/78 | What is an obviously coordinate-independent description of the Chevellay-Eilenberg complex for a Lie algebroid? | I think that the relation is that if $\psi\colon A \to TM$ is the action of the algebroid, then we have that $df(a)=\psi(a)(f)$ for $f$ in degree zero of the CE complex and $a\in\Gamma(A)$ and $d\omega(a,b)=\omega([a,b])+\psi(a)(\omega(b))-\psi(b)(\omega(a))$, where $\omega\in\Gamma(A^\ast)$ and $a,b\in\Gamma(A)$. One then extends $d$ to the full CE complex by Leibniz' rule just as for the de Rham complex (and $d^2=0$ comes from the Jacobi identity).
| 2 | https://mathoverflow.net/users/4008 | 26293 | 17,230 |
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