ag.algebraic geometry - Schemes as a model category

I'm just learning some basics of model categories, so please forgive me if my question turns out to be trivial. I hope it does at least make sense.

A natural temptation is to relate this machinery to birational geometry; in particular one would like to find a model category structure having the birational morphisms as weak equivalences. More precisely it would be nice to have such a model structure on the category $Sch_k$ of schemes of finite type over a field $k$.

A natural problem arises: a model category is required by definition to have all small limits and colimits, and $Sch_k$ does not satisfy this. For limits the situation is not that bad. I believe the original work of Quillen required only the existence of finite limits and colimits. Since $Sch_k$ has finite products and fiber products, it has all finite limits.

On the other hand finite colimits need not exist. A simple way to see this is to realize that categorical quotients by equivalence relations do not always exist in $Sch_k$, and these are just some coequalizers. So my questions are:

1. Is there a canonical way to enlarge a category to add finite limits?




2. If this is the case, what do we obtain when applying this to $Sch_k$? The resulting category would have to contain algebraic spaces, as these arise as quotients of schemes by étale equivalence relations. How much bigger would it be?




3. Assuming one has a decent notion of birational morphism for these objects: is there a model structure on the enlarged category such that birational morphisms are the weak equivalences?

nt.number theory - Where stands functoriality in 2009?

Here are few remarks which might be relevant, although I understand almost nothing of the global Langlands program.

Lafforgue is currently working on problems relating to functoriality. There are a number of recent preprints and notes on his webpage, see for example "Quelques remarques sur le principe de fonctorialité". If you don't read French, maybe the lectures of Lafforgue in Cambridge a few months ago would be useful. They are available in various video formats at the Newton Institute webpage. To find them, see this list, and scroll down to May - there is a total of 5 talks by Lafforgue, the first one on May 5th.

My impression of Lafforgue's work is that he aims for a proof of functoriality in a fairly general setting, and (amazingly!) he hopes that the method would work also in the number field case and not only for function fields (although I might have misunderstood this). The method has at least some vague similarity with Tate's thesis, I think.

For more general background on functoriality and related things, see maybe Knapp's survey on the Langlands program, the Clay Summer School Proceedings from 2003 (here is the Google Books page), and this short note of Rapoport on Lafforgue's earlier work.

Edit: Thanks to "unknown" and David for pointing out the work of Ngo! I should have added that Laumon also gave a talk in May at the Newton Institute, on Ngo's proof, this is available here (both video and slides). See also the discussion at SBS. On the functoriality principle in general, there is also this 15-page expository presentation of Arthur.

graph theory - Spanning polytopes

Unfortunately, there is no known good characterization of which graphs are the 1-skeleton of some polytope in some dimension d. There are several interesting properties of polytope graphs that may be of interest to you:

$bullet$ In $mathbb{R}^3$, Steinitz's theorem characterizes all polytopal graphs as the planar 3-connected graphs.

$bullet$ In d > 3, the cyclic polytopes have $K_n$ as a 1-skeleton. A result of Perles implies that every graph $G$, $G$+$K_n$ is a d-polytopal graph for some n and some d, although it is unknown what the minimum n is for a given $G$.

$bullet$ It is a theorem of Balinski that the graph of a polytope in $mathbb{R}^d$ is d-connected, so graphs with polytopal spanning subgraphs must have a highly connected subgraph.

There are also results about how to determine whether a specific graph $G$ is a polytopal graph, but they do not extend into a classification of all polytopal graphs.

There have been studies into classes of abstract polytope graphs, classes which contain all the 1-skeletons of polytopes but likely contain graphs with no polytopal realization, for example in this paper of Eisenbrand, Haehnle, and Rothboss Diameter of Polyhedra: Limits of Abstraction

For more information, check out Gil Kalai's chapter 19 in the Handbook of Discrete and Computational Geometry by Jacob E. Goodman and Joseph O'Rourke (or Gil Kalai's excellent blog gilkalai.wordpress.com/ ) or Grunbaum's Convex Polyopes.

sheaf theory - Describing global sections of sheafifications

The following is a counterexample for $mathcal{F}$ a presheaf of abelian groups (or sets, if you like).

Let $X=lbrace a,b,c,drbrace$ with nontrival opens given by $lbrace a rbrace,lbrace b rbrace,U=lbrace a,b,c rbrace,V=lbrace a,b,d rbrace, Ucap V$.

Define the presheaf $mathcal{F}$ by

$mathcal{F}(lbrace a rbrace)=mathcal{F}(lbrace b rbrace)=mathbb{Z}/2mathbb{Z}$,

$mathcal{F}(U)=mathcal{F}(V)=mathcal{F}(Ucap V)=mathcal{F}(X)=mathbb{Z}$,

with the obvious restriction maps.

Then $mathcal{F}^+(X)=lbrace (x,y)inmathbb{Z}oplusmathbb{Z}| xequiv ytext{ (mod 2)}rbrace$, since the germs at $a$ and $b$ are determined by those at $c$ and $d$, and the only restrictions on $c$ and $d$ are that they give the same germs at $a$ and $b$.

Consider $(0,2)inmathcal{F}^+(X)$. The germs at $c$ and $d$ cannot come from a common section of $mathcal{F}(X)$. Any system of sections which does not include $X$ in the cover must include both $U$ and $V$, having sections 0 and 2, respectively. Of course these do not agree when restricted to $Ucap V$. QED

This construction relies crucially on the fact that the presheaf is not separated (i.e. gluing is not unique). If the presheaf $it{is}$ separated, the condition described in the question is clearly satisfied.

This construction was shown to me by Paul Balmer.

gn.general topology - The "miracle" of Heegard Floer.

From Szabo's delightfully understated response (pdf) to receiving the Veblen prize:

The joint work with Peter Ozsváth
which is noted here grew out of our
attempts to understand Seiberg-Witten
moduli spaces over three-manifolds
where the metric degenerates along a
surface. This led to the construction of Heegaard Floer homology
that involved both
topological tools, such as Heegaard diagrams, and
tools from symplectic geometry, such as holomorphic
disks with Lagrangian boundary constraints.
The time spent on investigating Heegaard Floer
homology and its relationship with problems in
low-dimensional topology was rather interesting.

Of course, if one believes that Heegaard Floer homology is somehow the limit of monopole Floer homology as one degenerates the metric in some way that depends on the Heegaard diagram, then the independence of Heegaard Floer homology from the Heegaard diagram would fall out from the metric-independence of monopole Floer homology. Unfortunately, I can't seem to find references that give any sort of precise picture of how Ozsvath and Szabo came to think that this should be the case (though it might have been a baby analogue of the picture in this paper (pdf) by Yi-Jen Lee, written a few years later).

It perhaps bears mentioning that Heegaard Floer homology wasn't the first invariant that Ozsvath and Szabo constructed based on thinking about the interaction of the Seiberg-Witten equations with a Heegaard diagram--these papers, which extract an invariant from the theta-divisor of the Heegaard surface, appear to have been based on thinking about what happens to the Seiberg-Witten equations when one has a neck Sx[-T,T] (S is the Heegaard surface) with the metric on S at t=-T itself having long cylinders over the compressing circles for one handlebody, while the metric on S at t=T has long cylinders over the compressing circles for the other handlebody.

nt.number theory - Modular forms reference

Have a look at Section 6.6 of Diamond and Shurman, A First Course in Modular Forms:

As an aside, the theorem states a bit more than you have said: for instance, when the field of Fourier coefficients is $mathbb{Q}$, you are just asserting the existence of an elliptic curve $E_{/mathbb{Q}}$ with $operatorname{End}_{mathbb{Q}}(E) otimes_{mathbb{Z}} mathbb{Q} = mathbb{Q}$: every elliptic curve over $mathbb{Q}$ has this property. You want to require an equality of L-series between the abelian variety and the modular form.

gr.group theory - Lifting units from modulus n to modulus mn.

Background

In his beautifully short answer to a previous question of mine, Robin Chapman asserted the following.

Let $m,n,r$ be natural numbers with $r$ coprime to $n$. Then there is $r' equiv r mod n$ which is coprime to
$mn$.

Letting $C_n$ denote the cyclic group of order $n$, the above statement is equivalent to this:

Every automorphism of $C_n$ lifts to an automorphism of $C_{nm}$ for all $m$.

Since that is the context of the question I asked, I thought that this fact ought to have an elementary group-theoretical derivation, but alas I have been unable to find one. I asked a number theorist colleague of mine and he gave me this "sledgehammer proof" (his words):

Since $r$ is coprime to $n$, the arithmetic progression $r + kn$ for $k=1,2,ldots$ contains an infinite number of primes (by a theorem of Dirichlet's). Since only a finite number of those primes can divide $m$, there is some $k$ for which $r'= r+kn$ is a prime which does not divide $m$, and hence neither does it divide $nm$.

Question

Is there an elementary (and preferably group-theoretical) proof of this result?

ct.category theory - Localization(s) of Categories

I've been trying to read a paper by Krause and came across a strange (to me, of course) notion of localization. After looking around for a long time, and then finding this on his site, I see that there are two notions for localization, both with significant usage online. These are namely Verdier localization and Bousfield localization. Is there a strong motivation to use one over the other? A little bit of context:

I see that Bousfield localization is defined for model categories, and this includes the notion of modules over a ring, among many many others. I don't see a similar restriction for the Verdier localization.

Verdier localization uses the (standard for ''localization'') idea of a multiplicative set S of maps which are formally inverted by a functor Q from a category T to a new category denoted T/S. Hartshorne's Residues and Duality is a reference for this. (BTW, where does the assumption that the pullback of a multiplicative map is multiplicative come from?)

Bousfield localization is stated in several places (such as the Krause reference above) as a Verdier localization composed with a right adjoint for Q, which I understand to mean a functorial way of choosing objects in the isomorphism classes, and maps in the multiplicative subsets of each Hom(A,B). It is also stated in the generality of model categories as needing three distinguished collections of morphisms: namely quasi-isomorphisms and (co)fibrations. What bothers me more is the definition as given in Krause: an exact functor L from a triangulated category T to itself for which there exists a natural transformation η:Id-->L which commutes with L (ηL=Lη) and for which ηL is invertible. As a second, smaller, question, what is encoded by the commutative condition (what would be lost without it?)? I can come up with contrived examples (using the automorphisms of the objects LX) of course, but in what precise way does η really just encode L as a natural transformation?

ca.analysis and odes - Uniqueness of the logarithm function

In my Analysis class, we started to prove a theorem that said:

Let a > 1. So there is a unique increasing function $f:(0,infty)tomathbb{R}$ so that:

1. $f(a) = 1$


2. $f(xy) = f(x) + f(y)quadforall x, y > 0$

First we supposed its existence. Then through 2 it follows that f is a group homomorphism between the multiplicative group $(0,infty)$ and the additive group $mathbb{R}$.

$f(x) = f(xcdot1) = f(x)+f(1)$, so $f(1)=0=f(xcdot x^{-1})=f(x)+f(x^{-1})$

Then $f(x^{-1}) = -f(x)$.

We affirm that $f(x^n) = n f(x)quadforall x>0, n in mathbb{N}$ (and then we proved by induction).

Let x > 0 and $nin mathbb{N}^*$. So exists $m in mathbb{Z}$ so that $a^m le x^n le a^{m+1}$

So $f(a^m) le f(x^n) le a^{m=1}$, i.e. $mle nf(x) le m+1$

And finally $m/n le f(x) le (m+1)/n$

Let $A_x = { frac{m}{n} : m in mathbb{Z},: n in mathbb{N},: a^m le x^n}$

So $f(x) = sup A_x$

He said that this means f is unique, but I can see no reason why. I've omitted some lemmas and parts of the proof, but kept all the results. It's quite clear f is $log_ax$, but why it is unique?

Edit: I can't get LaTeX to work. It all seems fine while editing, but wrong in the question page.

co.combinatorics - longest consecutive subsequence of a random permutation

The purpose of this answer is to use the second moment method to make rigorous the heuristic argument of Michael Lugo. (Here is why his argument is only heuristic: If $N$ is a nonnegative integer random variable, such as the number of length-$r$ increasing consecutive sequences in a random permutation of ${1,2,ldots,n}$, knowing that $E[N] gg 1$ does not imply that $N$ is positive with high probability, because the large expectation could be caused by a rare event in which $N$ is very large.)

Theorem: The expected length of the longest increasing block in a random permutation of ${1,2,ldots,n}$ is $r_0(n) + O(1)$ as $n to infty$, where $r_0(n)$ is the smallest positive integer such that $r_0(n)!>n$. ("Block" means consecutive subsequence $a_{i+1},a_{i+2},ldots,a_{i+r}$ for some $i$ and $r$, with no conditions on the relative sizes of the $a_i$.)

Note: As Michael pointed out, $r_0(n)$ is of order $(log n)/(log log n)$.

Proof of theorem: Let $P_r$ be the probability that there exists an increasing block of length at least $r$. The expected length of the longest increasing block is then $sum_{r ge 0} r(P_r-P_{r+1}) = sum_{r ge 1} P_r$. We will bound the latter sum from above and below.

Upper bound: The probability $P_r$ is at most the expected number of increasing blocks of length $r$, which is exactly $(n-r+1)/r!$, since for each of the $n-r+1$ values of $i$ in ${0,ldots,n-r}$ the probability that $a_{i+1},ldots,a_{i+r}$ are in increasing order is $1/r!$. Thus $P_r le n/r!$. By comparison with a geometric series with ratio $2$, we have $sum_{r > r_0(n)} P_r le P_{r_0(n)} le 1$. On the other hand $sum_{1 le r le r_0(n)} P_r le sum_{1 le r le r_0(n)} 1 le r_0(n)$, so $sum_{r ge 1} P_r le r_0(n) + 1$.

Lower bound: Here we need the second moment method. For $i in {1,ldots,n-r}$, let $Z_i$ be $1$ if $a_{i+1}<a_{i+2}<ldots<a_{i+r}$ and $a_i>a_{i+1}$, and $0$ otherwise. (The added condition $a_i>a_{i+1}$ is a trick to reduce the positive correlation between nearby $Z_i$.) The probability that $a_{i+1}<a_{i+2}<ldots<a_{i+r}$ is $1/r!$, and the probability that this holds while $a_i>a_{i+1}$ fails is $1/(r+1)!$, so $E[Z_i]=1/r! - 1/(r+1)!$. Let $Z=sum_{i=1}^{n-r} Z_i$, so
$$E[Z]=(n-r)(1/r! - 1/(r+1)!).$$
Next we compute the second moment $E[Z^2]$ by expanding $Z^2$. If $i=j$, then $E[Z_i Z_j] = E[Z_i] = 1/r! - 1/(r+1)!$; summing this over $i$ gives less than or equal to $n/r!$. If $0<|i-j|<r$, then $E[Z_i Z_j]=0$ since the inequalities are incompatible. If $|i-j|=r$, then $E[Z_i Z_j] le 1/r!^2$ (the latter is the probability if we drop the added condition in the definition of $Z_i$ and $Z_j$). If $|i-j|>r$, then $Z_i$ and $Z_j$ are independent, so $E[Z_i Z_j] = (1/r! - 1/(r+1)!)^2$. Summing over all $i$ and $j$ shows that
$$E[Z^2] le frac{n}{r!} + E[Z]^2 le left(1 + O(r!/n) right) E[Z]^2.$$
The second moment method gives the second inequality in
$$P_r ge operatorname{Prob}(Z ge 1) ge frac{E[Z]^2}{E[Z^2]} ge 1 - O(r!/n).$$
If $r le r_0(n)-2$, then $r! le (r_0(n)-1)!/(r_0(n)-1)$, so $r!/n le 1/(r_0(n)-1)$, so $P_r ge 1 - O(1/r_0(n))$. Thus
$$sum_{r ge 1} P_r ge sum_{r=1}^{r_0(n)-2} left( 1 - O(1/r_0(n)) right) = r_0(n) - O(1).$$

at.algebraic topology - CW structure on spaces obtained by attaching cells wildly

The space you get through such a process of cell attachments is at least homotopy equivalent to a CW complex, although maybe not homeomorphic (as Tyler's example shows).

This comes up in Milnor's Morse Theory (see pp. 21-24), where you build a CW complex inductively by attaching cells according to the critical points of a Morse function on a manifold. There's no reason that the indices have to be increasing, and this causes one to attach cells "in the wrong order," i.e. a cell of dimension n may be attached to existing cells of dimension n or greater. In his book on Morse Theory, Milnor shows that up to homotopy the resulting space is a CW complex. The basic point is that each attaching map can be deformed to the appropriate skeleton, and attaching a cell along two homotopic maps produces two homotopy equivalent spaces (this is a bit tricky, and Milnor attributes it to Hilton). Finally, an argument with homotopy colimits allows one to treat the case of countably many cell attachments.

I suppose that this applies to Tyler's example, and shows that the space he builds is homotopy equivalent to an infinite wedge of circles.

ag.algebraic geometry - Unstable Vector Bundles

Well, I don't know about horrible. There's a lot you can say that's good! I'll start rambling and see where I end up.

I'm going to pretend you said principal GL(n)-bundle instead of rank n vector bundle. Same thing, really, since we have the standard representation.

The collection Bun(n,C) of all principal GL(n) bundles P on a smooth curve C is a very nice geometric object: it's an Artin stack. It's not connected; the different components are labelled by topological data, like the Chern class. The tangent "space" (complex, really) to Bun(n,C) at a point P is naturally the derived global sections RGamma(C,ad(P)), where ad(P) is the associated bundle with fiber the adjoint representation of GL(n). The zero-th cohomology gives the infinitesimal automorphisms and the 1st cohomology gives the deformations. So the stabilizer group of any point V in Bun(n,C) is finite-dimensional, and the dimension of the stack is n(g-1) (by Riemann-Roch). Bun(n,C) is smooth, and unobstructed, thanks to the vanishing of H^2(C,ad(P)).

Bun(n,C) has a very nice stratification, too. It's an increasing union of quotient stacks [A/G] of projective varieties by finite-dimensional groups. Roughly, A is the stack of pairs (P,t), where t is a trivialization of P in an infinitesimal neighborhood of some point in C. Make the neighborhood large enough, i.e., r-th order, and you can kill off all the automorphisms of P. Unfortunately, except for n=1, there is no uniform bound on r that works for all bundles. So, Bun(n,C) isn't a finite type quotient stack.

You can also realize Bun(n,C) (homotopically) as the infinite type quotient stack of U(n)- connections modulo complexified gauge transformations. That's what Atiyah & Bott do in their paper "The Yang-Mills Equations on Riemann Surfaces". (They also have a nice discussion of slope-stability and the stratification.)

The top component of the stratification (those bundles where the stabilizer group is as small as possible) is the stack of (semi-)stable vector bundles. If you take the coarse moduli space of this substack, you get the usual moduli space of stable bundles.

In summary: If you drop the stability conditions, you get a lot more geometry with a similar flavor, and without the random bits of weirdness that crop up in the theory of moduli spaces. (e.g., the stack always carries a universal bundle, you don't need the rank and the chern class to be coprime.)

OK, I'll stop evangelizing now.

binomial distribution - Probability of system failure in a distributed network

I am trying to build a mathematical model of the availability of a file in a distributed file-system. The system works like this: a node $x$ stores a file $f$ (encoed using erasure codes) at $rb$ remotes nodes, $r$ is the replication-factor where $b$ is a constant. Erasure-coded files have the property that the file can be restored $iff$ at least $b$ of the remote nodes are available and return its part of the file.

The simplest approach to this is to assume that all remote nodes are independent of each other and have the same availability $p$. With these assumptions the availability of a file becomes $$pf = sum_{i=b}^{rb} binom{rb}{i} p^i(1 - p)^{rb - i}$$

Unfortunately these assumptions can introduce a non-neligible error, as shown by this paper: http://deim.urv.cat/~lluis.pamies/uploads/Main/icpp09-paper.pdf.

The other extreme is to calculate the probability of each possible combination of availaible/non-available node and take the sum of all these outcomes (which is sort of what they suggest in the paper above, just more formally than what I just described). This approach is more correct but has a computational cost of $O(2^{rb})$.

Do you guys have any ideas of a good approximation which introduces less error than the binomial distribution-aproximation but with better computational cost than $O(2^{rb})$?

You can assume that the availability-data of each node is a set of tuples consisting of $(measurement-date, node measuring, node being measured, succes/failure-bit)$

career - How does an academic mathematician educate him/herself about job opportunities outside academia?

As an undergrad who has recently been looking for jobs I would make the following suggestions...

1) Learn to program! MOST of the jobs I've run into that require mathematicians/statisticians will eventually require learning one or more of Perl/Python/R/C++/SAS. I'm in the process of writing a book on the subject (programming for mathematicians, specifically in python)

2) Learn probability and statistics! I didn't do that directly and wish I had because I run into problems all the time for which probability and statistics knowledge would be useful.

3) Kind of coupled with 1, but learn Algorithms and Data Structures. If you never program than this isn't worth doing, but algorithms is math heavy and most companies that want to hire people to do algorithms expect a base knowledge, and basic knowledge of data structures is a MUST in that scenario)

4) Try applied classes! Other departments will not eat you! I promise... unless it's the department of cannibalism...

5) FINISH projects! Make sure whenever you get involved with a project that you push as hard as you can to get something finished and a "product" to show off.

6) Talk to professors outside of the department if they do something you think looks interesting. I don't mean to stress coding so much, but coding is something that everyone outside of CS departments (since coders are a dime a dozen there) needs, at least in the hard sciences and engineering.

7) For someone who doesn't plan on going into academia... internship >> summer classes. Search far and wide for these, they're extremely valuable and show that you can handle a work environment in addition to being able to handle an academic environment.

These are most relevant to undergrads, I haven't a clue what a grad student should do b/c I haven't gotten there yet!!

at.algebraic topology - Folk Functorial Figuring

In the CRM Proceedings & Lecture Notes Volume 50 "A Celebration of the Mathematical Legacy of Raoul Bott" Herbert Shulman writes (p. 48):

"[Bott] taught many of us to think functorially, like thinking of a group as a category with one object and a morphism for each element, a manifold as a category of pairs (open set, point in the open set), and a bundle as an equivalence class of functors. When someone asked him who invented functors, he said 'functors are prehistoric!'. He talked about 'folk' theorems... theorems everyone knew, but were never written down."

As I've highlighted in bold the latter two everyday examples of categories seem less well-known to me.

The manifold-as-category seems clear enough, the open set is meant to be a coordinate neighborhood of the point. A morphism between two objects is just the transition maps between coordinate neighborhoods. What seems more natural to me would be to associate a category to a given atlas on a manifold. This makes me wonder what the classifying space of these categories looks like. How do they behave as you pass to the maximal atlas? Does anyone know? Since the transition maps (the morphisms) are homeomorphisms/ diffeomorphisms/ biholomorphisms (depending on what type manifold we have) the morphisms in this category are all invertible and so our category is a groupoid. For a connected manifold the classifying space should be a $BG$ or $K(G,1)$. What is $G$?

Finally, Can someone explain why a bundle is an equivalence class of functors? The functor part seems sort of clear, because the projection map is an open map. More explanation would be nice.

geometric rep theory - Hirzebruch-Riemann-Roch for quiver varieties?

These days, I attended a workshop at North Carolina State University. The key lecturer is Professor Nakajima. He introduced two types of quiver variety. One of them is affine, another one is quasi-projective variety.

The reference is the following:
Nakajima's quiver variety

I wonder whether is there anyone developing the Riemann-Roch theorem for this quasi projective scheme. It is known that the H-R-R for flag variety of finite dimensional Lie algebra is Weyl character formula

Is there any representation theoretical interpretation of H-R-R for quiver variety? Of course, one can say that some quiver variety correspondence to flag variety of some Lie algebra. However, I think quiver variety is more general.

The further question: Is there BGG category on quiver representation settings. I know one can construct Hall algebra associated to some quiver. It seems that one can define analogue of Verma module in this setting comparing to the Verma module for Lie algebra. Therefore, is there a Character formula here, if there exists, can it be re-interpretated as H-R-R for quiver variety?

ag.algebraic geometry - Equivalent Statements of Riemann Hypothesis in the Weil Conjectures

The reason why that inequality is equivalent to RH (for curves) is the functional equation.

I wrote out the details of the argument some years ago (in an unpublished book on zeta and $L$-functions), so it's easy to cut and paste it here. The polynomial $L(z)$ I speak about below would arise in practice as the numerator of the zeta-function of the curve, with $z = q^{-s}$ and the usual version of the Riemann hypothesis for $L(q^{-s})$ is equivalent to the statement that the reciprocal roots of $L(z)$ all have absolute value $sqrt{q}$. That's the form of the Riemann hypothesis I will be referring to in what follows.

Suppose we have a polynomial $L(z)$ over the complex numbers with constant term 1 and degree $d$, factored over its reciprocal roots:
$$
L(z) = (1 - alpha_1z)cdots(1-alpha_dz), alpha_j not= 0.
$$

Let $L*(z)$ be the polynomial with complex-conjugate coefficients to those of $L(z)$, so
$$
L*(z) = (1 - overline{alpha_1}z)cdots(1-overline{alpha_d}z).
$$
(Sorry, I want to make the asterisk into an exponent in this particular .tex situation, but it isn't working and I don't have time to figure it out.)
Assume $L(z)$ and $L*(z)$ are connected by the functional equation
$$
L(1/qz) = frac{W}{z^d}L*(z)
$$
for some constant $W$. If you compare coefficients of the same powers of $z$ on both sides, this functional equation implies the mapping $alpha mapsto q/alpha$ sends reciprocal roots of $L(z)$ to reciprocal roots of $L*(z)$ (and $W$ has absolute value $q^{d/2}$).

Lemma 1. Granting the functional equation above, the following conditions are equivalent:

i$)$ the reciprocal roots of $L(z)$ have absolute value $sqrt{q}$ (RH for $L(z)$),

ii$)$ the reciprocal roots of $L(z)$ have absolute value $leq sqrt{q}$.

Proof. We only need to show ii implies i.
Assuming ii, let $alpha$ be any reciprocal root of $L(z)$,
so $|alpha| leq sqrt{q}$. By the functional
equation, $q/alpha$ is a reciprocal
root of $L*(z)$, so $q/alpha = overline{beta}$
for some reciprocal root $beta$ of $L(z)$.
Then $|q/alpha| = |overline{beta}| = |beta| leq sqrt{q}$ and thus $sqrt{q} leq |alpha|$.
Therefore $|alpha| = sqrt{q}$ and i follows. QED

This lemma reduces the proof of the Riemann hypothesis for $L(z)$ from the equality $|alpha_j| = sqrt{q}$ for all $j$ to the upper bound $|alpha_j| leq sqrt{q}$ for all $j$. Of course the functional equation was crucial in explaining why the superficially weaker inequality implies the equality.

Next we want to show the upper bound on the $|alpha_j|$'s in part ii of Lemma 1 is equivalent to a $O$-estimate on sums of powers of the $alpha_j$'s which superficially seems weaker.

We will be interested in the sums
$$
alpha_1^n + cdots + alpha_d^n,
$$
which arise from the theory of zeta-functions as coefficients in an exponential generating function: since $L(z)$ has constant term 1, we can write (as formal power series over the complex numbers) $$L(z) = expleft(sum_{n geq 1}N_n z^n/nright)$$
and then logarithmic differentiation shows
$$
N_n = -(alpha_1^n + dots + alpha_d^n)
$$
for all $n geq 1$.

Lemma 2.
For nonzero complex numbers $alpha_1,dots,alpha_d$ and a
constant $B > 0$, the following are equivalent:

i$)$
For some $A > 0$, $|alpha_1^n + dots + alpha_d^n| leq AB^n$ for
all $n geq 1$.

ii$)$
For some $A > 0$ and positive integer $m$,
$|alpha_1^n + dots + alpha_d^n| leq AB^n$ for
all $n geq 1$ with $n equiv 0 bmod m$.

iii$)$ $|alpha_j| leq B$ for all $j$.

Part ii is saying you only need to show part i when $n$ runs through the (positive) multiples of any particular positive integer to know it is true for all positive integers $n$. It is a convenient technicality in the proof of the Riemann hypothesis for curves, but the heart of things is the connection between parts i and iii. (We'd be interested in part iii with $B = sqrt{q}$.) You could set $m = 1$ to make the proof below that ii implies iii into a proof that i implies iii. The passage from i to iii is what Dave is referring to in his answer when he cites the book by Iwaniec and Kowalski.

Proof.
Easily i implies ii and (since $|alpha_j| = |overline{alpha_j}|$)
iii implies i.
To show ii implies iii, we use a cute analytic trick. Assuming ii, the series
$$
sum_{n equiv 0 bmod m} (alpha_1^n + dots + alpha_d^n)z^n
$$
is absolutely convergent for $|z| < 1/B$, so the series defines a holomorphic function on this disc. (The sum is over positive multiples of $m$, of course.)
When $|z| < 1/|alpha_j|$ for
all $j$, the series can be computed to be
$$
sum_{j=1}^{d} frac{alpha_j^mz^m}{1-alpha_j^mz^m} =
sum_{j=1}^{d}frac{1}{1-alpha_j^mz^m} - d,
$$
so the rational function $sum_{j=1}^{d} 1/(1-alpha_j^mz^m)$ is holomorphic
on the disc $|z| < 1/B$. Therefore the poles of this rational function must have absolute value $geq 1/B$. Each $1/alpha_j$ is a pole, so $|alpha_j| leq B$ for all $j$. QED

Theorem.
The following are equivalent:

i$)$ $L(z)$ satisfies the Riemann hypothesis ($|alpha_j| = sqrt{q}$ for all $j$),

ii$)$ $N_n = O(q^{n/2})$ as $n rightarrow infty$,

iii$)$ for some $m geq 1$, $N_n = O(q^{n/2})$ as $n rightarrow
infty$ through the multiples of $m$.

Proof.
Easily i implies ii and ii implies iii.
Assuming iii, we get $|alpha_j| leq sqrt{q}$ for all $j$ by
Lemma 2, and this inequality over all $j$ is equivalent to i
by Lemma 1. QED

Brandon asked, after Rebecca's answer, if the inequality implies the Weil conjectures (for curves) and Dave also referred in his answer to the Weil conjectures following from the inequality. In this context at least, you should not say "Weil conjectures" when you mean "Riemann hypothesis" since we used the functional equation in the argument and that is itself part of the Weil conjectures. The inequality does not imply the Weil conjectures, but only the Riemann hypothesis (after the functional equation is established).

That the inequality is logically equivalent to RH, and not just a consequence of it, has some mathematical interest since this is one of the routes to a proof of the Weil conjectures for curves.

P.S. Brandon, if you have other questions about the Weil conjectures for curves, ask your thesis advisor if you could look at his senior thesis. You'll find the above arguments in there, along with applications to coding theory. :)

at.algebraic topology - Can one calculate the (co)homology of the loopspace of a Lie group from its Lie algebra?

Yes.

At least, rationally.

The result that you want starts on p68 of "Loop Groups" by Pressley and Segal. There, they prove surjectivity of H^*(L𝔤;ℝ) → H^*(LG;ℝ). The basic idea of the argument is as follows: for reasonably simple reasons, the cohomology of LG is easily obtainable from that of G. This yields specific formulae for generators of the de Rham cohomology of LG. Although these forms are not themselves left invariant, they are cohomologous to rational multiples of left invariant forms, and thus come from the cohomology of the Lie algebra, L𝔤. The proof of surjectivity is thus not hard.

The proof that it is an isomorphism is a little trickier and they defer that to section 14.6 (p299). That this is quite some way through the book is a good indication of how much you need to absorb to understand it.

Amazingly, parts of Loop Groups (including pages 68, 69, and 299, but not 300) are available on Google books. It says that it is a "Limited preview" but whether or not that is just limited in pages or limited in time, I do not know. However, Loop Groups is a great book for anyone interested in Lie Groups and infinite dimensional "stuff".

(Incidentally, this is all for G simply connected.)

lie algebras - 3/4-Lie superalgebras: how much of a theory can one develop?

Let $mathfrak{s} = mathfrak{s}_0 oplus mathfrak{s}_1$ be a real Lie superalgebra. (The ground field does not matter much, but at least one formula will not work as written if the characteristic is 2 or 3.) Recall that this means that there is a bilinear 2-graded bracket $[-,-]$ with three components

(a) $mathfrak{s}_0 times mathfrak{s}_0 to mathfrak{s}_0$ (skewsymmetric)

(b) $mathfrak{s}_0 times mathfrak{s}_1 to mathfrak{s}_1$

(c) $mathfrak{s}_1 times mathfrak{s}_1 to mathfrak{s}_0$ (symmetric)

satisfying the Jacobi identity, which splits into 4 components, which can be paraphrased as

(1) $mathfrak{s}_0$ is a Lie algebra under (a)

(2) $mathfrak{s}_1$ is an $mathfrak{s}_0$-module under (b)

(3) the map in (c) is $mathfrak{s}_0$-equivariant

(4) $[[x,x],x] = 0$ for all $x in mathfrak{s}_1$

The fact that the first three components can be written using words, whereas the fourth is easiest via a formula, suggests that they should perhaps be treated differently.

Indeed, over time I have come across many examples of superalgebras where the first three components of the Jacobi identity are satisfied but not the fourth. I'd like to call them 3/4-Lie superalgebras. I would like to know how far can this notion be pushed and in particular how much of the theory of Lie superalgebras still works in the 3/4 case.

To motivate this seemingly random question, let me end by pointing out one generic example where they arise. There are others, but they are lengthier to describe.

Let $mathfrak{g}$ be a metric Lie algebra; that is, a Lie algebra with an ad-invariant inner product $(-,-)$ and let $V$ be a symplectic $mathfrak{g}$-module; that is, one possessing a $mathfrak{g}$-invariant symplectic form $langle-,-rangle$. Now let $mathfrak{s} = mathfrak{g} oplus V$. Then maps (a) and (b) are obvious: given by the Lie bracket on $mathfrak{g}$ and the action of $mathfrak{g}$ on $V$, respectively. Map (c) is the transpose of map (b) using the inner products of both $mathfrak{g}$ and $V$; in other words, if $x,y in V$ then $[x,y] in mathfrak{g}$ is defined by
$$([x,y],a) = langle acdot x,yrangle$$
for all $a in mathfrak{g}$.

Then it is easy to see that $[x,y] = [y,x]$ and that (1)-(3) are satisfied, whereas in general (4) is not satisfied and instead defines a subclass of symplectic $mathfrak{g}$-modules.

linear algebra - Operation of GL_n(Z/bZ)

Any transformation
$$(v_1,ldots,v_n)mapsto (v_1,ldots,v_{j-1},v_j+av_k,v_{j+1},ldots,v_n)$$
for $jne k$ is achievable by means of some such matrix. It suffices to reduce
an admissible vector to $(1,0,ldots,0)$ by means of a sequence of such reductions.
I would do it in three stages

1. Make $v_n$ into a unit in $mathbb{Z}/bmathbb{Z}$;




2. Make $v_1=1$;




3. Make all $v_j=0$ for $j>1$.

Importance of Log Convexity of the Gamma Function

First, let me mention that log convexity of a function is implied by an analytic property, which appears to be more natural than log convexity itself. Namely, if $mu$ is a Borel measure on $[0,infty)$ such that the $r$th moment
$$f(r)=int_{0}^{infty}z^r dmu(z)$$
is finite for all $r$ in the interval $Isubset mathbb R$, then $log f$ is convex on $I$.

Log convexity can be effectively used in derivation of various inequalities involving the gamma function (particularly, two-sided estimates of products of gamma functions). It is linked with the notion of Schur convexity which is itself used in many applications.

An appetizer. Let $m=max x_i$, $s=sum x_i$, $x_i > 0$, $i = 1,dots,n$, then
$$[Gamma(s/n)]^nleqprodlimits_{1}^{n}Gamma (x_i)leq left[Gammaleft(frac{s-m}{n-1}right)right]^{n-1}Gamma(m).qquadqquadqquad (1)$$

(1) is trivial, of course, when all $x_i$ and $s/n$ are integers, but in general the bounds do not hold without assuming log convexity.

Edit added: a sketch of the proof. Let $f$ be a continuous positive function defined on an interval $Isubset mathbb R$. One may show that the function $phi(x)=prodlimits_{i=1}^{n}f(x_i)$, $xin I^n$ is Schur-convex on $I^n$ if and only if $log f$ is convex on $I$. Thus the function
$$phi(x)=prodlimits_{i=1}^n Gamma(x_i),quad x_i>0,qquad quadqquadqquadqquadqquadqquadquad (2)$$
is Schur-convex on $I^n=(0,infty)^n$. Since $x_ile m$, $i=1,dots,n$, and $sum x_i=s$, it is easy to check that
$$x prec left(frac{s-m}{n-1},dots,frac{s-m}{n-1},mright).$$
The latter majorization and the fact that $phi(x)$ defined by (2) is Schur-convex imply the upper bound (1). The lower bound follows from the standard majorization $xsucc (s/n,dots,s/n)$.

---

Have a look at the recent short article by Marshall and Olkin concerning this and related inequalities for the gamma function.

gt.geometric topology - Minimal-length embeddings of braids into R^3 with fixed endpoints

UPDATE.

I revisited the question and realized that verifying the local CAT(0) property is not that easy. When I wrote the original answer, I was under impression that removing any collection of codimension 2 subspaces (more precisely, their tubular neighborhoods) from $mathbb R^n$ leaves one with a locally CAT(0) space. This is true in $mathbb R^3$ but there are counterexamples in $mathbb R^4$. This particular subset might satisfy the condition but this does not seem to follow from any "generic" argument.

Further, there are some discouraging examples. First, approximating by arcs by "ropes" with fixed-size square sections does not actually work: in this case there are non-unique minimal configurations. So one really needs to deal with zero-width ones. Second, if you consider 3 arcs and allow two of them intersect while deforming the braid (but still disallow intersections with the 3rd one), then again, you can have two distinct minimal configurations in the same equivalence class.

Below is the original answer (and I suggest that it is unaccepted).

---

I can prove uniqueness of a local minimum for another length-like functional
(similar but not equal to the sum of lengths). I believe that it should work the same way
for the sum of lengths, but unfortunately the underlying geometric theory does not
seem to exist (yet?).

First let me reformulate the problem. Let $X$ denote the set of possible horisontal cross-sections of braids.
This is the set of $n$-tuples of distinct points in $mathbb R^2$. Geometrically this is $(mathbb R^2)^n$
minus a collection of codimension 2 subspaces corresponding to positions where some two points coincide.

Actually I prefer another formalization: the arcs forming the braid are ropes of nonzero width
and with square cross-sections.
More precisely, the horizontal section of every rope is an $varepsilontimesvarepsilon$ square (with sides parallel
to the coordinate axes), and these sections should not overlap. So $X$ is $mathbb R^{2n}$ minus a union of polyhedral
heighborhoods of codimension 2 subspaces. This formalization makes the local structure simpler,
and the original one is the limit as $varepsilonto 0$.

An embedded braid is a path $f:[0,1]to X$. We want to minimize the functional
$$
L = int_0^1 left(sqrt{(df_1/dt)^2+1}+dots+sqrt{(df_n/dt)^2+1}right) dt,
$$
over all paths between two given positions $a,bin X$, in a given connected component
of the space of such paths. Here $f_i=f_i(t)$ are 2-dimensional coordinates.
Another functional,
$$
L' = int_0^1 sqrt{(df_1/dt)^2+dots+(df_n/dt)^2+1} dt,
$$
is easier to deal with, because this is the Euclidean length of the corresponding
path $(f(t),t)$ in $Xtimesmathbb Rsubset mathbb R^{2n+1}$. Let us work with $L'$.

The connected component of the set of paths is the homotopy class. Fixing the homotopy class of a path is the same as
fixing endpoints of its lift to the universal cover of the space.
(In the case of zero-width ropes, you first take the universal cover and then the completion in order to add "boundary cases".)
And local length minimizers are called geodesics. So we want to show that, in the universal cover of $Xtimesmathbb R$,
every pair of points is connected by a unique geodesic.

Our space $X$ (and hence $Xtimesmathbb R$) is a locally CAT(0) space.
This can be shown using standard curvature tests for polyhedral spaces.
A globalization theorem says that a complete, simply connected, locally CAT(0) space is globally CAT(0),
in other words, it is a Hadamard space. So the universal cover is a Hadamard space.
Hadamard spaces feature uniqueness of geodesics, hence the result.

Now what about the original functional $L$? It is also a length functional, but for a non-Euclidean norm on $mathbb R^{2n+1}$.
So, to carry over the proof, one needs to develop an analog of CAT(0) spaces modelled after Finsler spaces rather than Riemannian. I think it should be possible - after all, all this CAT(0) business is about convexity of distance, and this convexity is there in all normed vector spaces.
But I have not heard of such generalizations (maybe this is just my ignorance).

ds.dynamical systems - When is an Anosov diffeomorphism mixing?

I don't have a proper answer to your main question beyond pointing to the list of equivalent properties in Pesin's book (your first reference), which you've obviously seen already. However, I'll point out that in Ruelle's paper (your third reference), the first main theorem (on page 3), which contains a statement on exponential decay of correlations, is proved under the hypothesis that the unstable manifolds are dense in the attractor. As in Pesin's book, that will imply mixing (on the attractor), so there's no mystery in the coexistence of the result stated by Ruelle and the open problem stated by Pesin.

As far as I know the best that you can say in general is that Anosov diffeos are mixing on the non-wandering set, or on the closure of an unstable manifold. So an equivalent question is, "Under what circumstances is every point non-wandering for an Anosov diffeo?" (Or, "Under what circumstances are unstable manifolds dense?")

computability theory - {transcendental numbers} {computable transcendental numbers}

Note: Answer is pending update per attached comments.

The difference, stated informally, is that that the non-computable transcendentals in their k-base digit representation are entirely random and non compressible. A computable transcendental, such as e, can be represented by a finite algorithmic description, such as a series expansion, which is a form of compression. For the non-computable numbers no such shorter representation exist. Their shortest computational description is their own infinite digit sequence.
You can read more about computational complexity here:
http://en.wikipedia.org/wiki/Kolmogorov_complexity.

There is a wealth of similar numbers to the Ω class of numbers. In general it is "easy" to come up with new definitions for such numbers. These all belong to the countably infinite set of non-computable definable numbers.

To make matters worse, what is left are the non-definable (and therefore also non-computable) numbers. They are the numbers that cannot be described in any way what-so-ever, other than by just iterating through their infinite non-compressible digit sequence. The set of all non-definable numbers is uncountable.

cv.complex variables - Adeles of Holomorphic Functions

In number theory, an adele is a restricted product of elements of the completion at each prime. For function fields, we take (a kind of) product of the completion at each point, and at non-singular points, this completion is a ring of formal power series. The adeles are intuitively defined so that we have an injective homomorphism from the ring of functions on a given open set (Zariski open for a number field) to the adeles for that open set.

My question is, what if we replace formal power series by convergent Laurent series (i.e. in the complex topology)? This is what I might call a holomorphic adele. There is a natural injective map from the ring of meromorphic functions on an open set to the ring of adeles on that set. We can also put a topology on these "holomorphic adeles" similar to how it's normally done. Is this construction ever considered? Might it be useful? Is there a connection between the adelic topology and the complex topology?

ag.algebraic geometry - The Jouanolou trick

In Une suite exacte de Mayer-Vietoris en K-théorie algébrique (1972) Jouanolou proves that for any quasi-projective variety $X$ there is an affine variety $Y$ which maps surjectively to $X$ with fibers being affine spaces. This was used e.g. by D. Arapura to (re)prove that the Leray spectral sequence of any morphism of quasi-projective varieties is equipped from the second term on with a natural mixed Hodge structure.

Here is a proof when $X$ is $mathbf{P}^n$ over a field $k$: take $Y$ to be the affine variety formed by all $n+1 times n+1$ matrices which are idempotent and have rank 1. This is indeed affine since it is given by the equations $A^2=A$, the characteristic polynomial of $A$ is $x^n(x-1)$. Moreover, $Y$ is mapped to $mathbf{P}^n(k)$ by taking a matrix to its image. The preimage of a point of $mathbf{P}^n(k)$ is "the set of all hyperplanes not containing a given line", which is isomorphic to an affine space.

The general (quasi-projective) case follows easily from the above. However, it is not clear how to generalize Jouanolou's trick for arbitrary varieties. Nor is it clear (to me) that this is impossible.

1. Is there an analogue of the Jouanolou lemma for arbitrary (not necessarily quasi-projective) varieties (i.e. reduced separated schemes of finite type over say an algebraically closed field)?




2. (weaker version of 1 over complex numbers) Is there, given a complex algebraic variety $X$, an affine variety $Y$ that maps surjectively to $X$ and such that all fibers are contractible in the complex topology? A negative answer would be especially interesting.




3. (the following question is a bit vague, but if it has a reasonable answer, then it would probably imply a positive answer to 2.) Is there a quasi-projective analog of the topological join of two projective spaces? I.e., if $P_1$ and $P_2$ are two complex projective spaces, is there a quasi-projective variety $X$ which "contains the disjoint union of $P_1$ and $P_2$ and is formed by all affine lines joining a point in $P_1$ with a point in $P_2$"?

Edit 1: in 1. and 2. the varieties are required to be connected (meaning that the set of closed points is connected in the Zariski topology; in 2 one could use the complex topology instead).

Edit 2: as Vanya Cheltsov explained to me, the answer to question 3 is most likely no.

ca.analysis and odes - Do upper-semicontinuous polyhedral multifunctions have Lipschitz continuous selections?

We are interested in the following question (definitions and references are given below):

Main Question: Given an upper-semicontinuous polyhedral multifunction $F:R^n rightarrow R^m$, is there always a Lipschitz continuous function $g:R^n rightarrow R^m$ such that $g(x) in F(x)$ for all $x in R^n$ ?

In general, upper semicontinuity is probably not the right property to guarantee continuous selections (see [HP-1], page 89). On the other hand, polyhedrality of the multifunction may help.

Related Question: Under what mild conditions is a positive answer to the above question guaranteed?

Motivation: Solutions of finite-dimensional variational inequalities (VI) or complementarity problems (CP) (see [FP-1]) typically have upper semicontinuous solution maps. Furthermore, if the functions and sets defining the VI or CP have affine or linear (or polyhedral) structure, their solution maps are polyhedral multifunctions. The main question posed above is then a natural question to ask in the study of differential inclusions $dot{x} in G(x)$ that have these solution maps appearing in the definition of $G(x)$.

Definitions:

1. A multifunction is simply a set-valued map.


2. A multifunction $F:R^n rightarrow R^m$ is said to be upper semicontinuous at a point $bar{x}$ if for every open set $mathcal{V}$ containing $F(bar{x})$, there exists an open neighbourhood $mathcal{U}$ of $bar{x}$ such that, for each $x in mathcal{U}$, $mathcal{V}$ contains $F(x)$.


3. A multifunction is said to be polyhedral if its graph is a polyhedral subset of $R^{n + m}$.

Referenecs:

1. [FP-1] F. Facchinei and J-S Pang, Finite-Dimensional Variational Inequalities and Complementarity Problems (vol. 1), pp. 138-139.


2. [HP-1] S. Hu and N.S. Papageorgiou, Handbook of Multivalued Analysis (vol. 1), p. 36 and p. 89.


3. [M-1] Ernest Michael, Continuous Selections I, The Annals of Mathematics, Vol. 63, (1956), pp. 361-382.


4. [M-2] Ernest Michael, Continuous Selections II, The Annals of Mathematics, Vol. 64, (1956), pp. 562-580.


5. [M-2] Ernest Michael, Continuous Selections III, The Annals of Mathematics, Vol. 65, (1957), pp. 375-390.

rt.representation theory - Is there an analogue of the Hive model for Littlewood-Richardson coefficients of types $B$, $C$ and $D$?

The recent work of Goncharov-Shen gives a good generalization of the hive model for any reductive group $ G$. They show that $n$-fold tensor product multiplicities for $ G $ are counted by positive tropical integral points of the space $ G^vee setminus ( G^vee / N)^n $. When $ G = GL_m$ and $n = 3$, this gives the Hive model. When $ G = GL_m $ and $ n = 4 $, this gives the octahedron recurrence.

Unfortunately, outside of type $ A$, it is hard to give a simple description of their set of positive tropical points. To do so requires some choices, and once one makes these choices, you end up with one of the Berenstein-Zelevinsky models which Allen mentioned.

Which computer algebra system should I be using to solve large systems of sparse linear equations over a number field?

This is related to Noah's recent question about solving quadratics in a number field, but about an even earlier and easier step.

Suppose I have a huge system of linear equations, say ~10^6 equations in ~10^4 variables, and I have some external knowledge that suggests there's a small solution space, ~100 dimensional. Moreover, the equations are sparse; in fact, the way I produce the equations gives me an upper bound on the number of variables appearing in each equation, ~10. (These numbers all come form the latest instance of our problem, but we expect to want to try even bigger things later.) Finally, all the coefficients are in some number field.

Which computer algebra system should I be using to solve such a system? Everyone knows their favourite CAS, but it's often hard to get useful comparisons. One significant difficulty here is that even writing down all the equations occupies a big fraction of a typical computer's available RAM.

I'll admit that so far I've only tried Mathematica; it's great for most of our purposes, but I'm well aware of its shortcomings, hence this question. A previous slightly smaller instance of our problem was within Mathematica's range, but now I'm having trouble.

(For background, this problem is simply finding the "low weight spaces" in a graph planar algebra. See for example Emily Peter's thesis for an explanation, or our follow-up paper, with Noah Snyder and Stephen Bigelow.)

inequalities - Another plausible inequality.

As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, I stand corrected and claim the inequality is valid at least for the case
$$
a_1b_1+xa_2b_2ge0
qquad(*)
$$
(and this seems to be a necessary condition as well).

Let me do some standard things. First let
$$
a_1=frac{1-u^2}{1+u^2}, quad
a_2=frac{2u}{1+u^2}, quad
b_1=frac{1-v^2}{1+v^2}, quad
b_2=frac{2v}{1+v^2}
$$
where $uvge0$.
Substitution reduces the inequality to the following one:
$$
((1-u^2)(1-v^2)+4xuv)^2
lebiggl(frac{(uv+1)^2-x(u-v)^2}{(uv+1)^2+x(u-v)^2}biggr)^2
((1-u^2)^2+4xu^2)((1-v^2)^2+4xv^2).
qquad{(1)}
$$
Now introduce the notation
$$
A=(1-u^2)(1-v^2)+4xuv, quad
B=(uv+1)^2, quad C=x(u-v)^2
$$
and note that $A,B,C$ are nonnegative; the inequality $Age0$ is equivalent to the above condition $(*)$. In addition,
$$
Ale B-C
qquad{(2)}
$$
because
$$
B-C-A=(1-x)(u+v)^2ge0.
$$
In the new notation the inequality (1) can be written more compact:
$$
A^2(B+C)^2le(B-C)^2(A^2+4BC)
$$
which after straightforward reduction becomes
$$
A^2le(B-C)^2,
$$
while the latter follows from (2).

sheaf theory - Sheaves as full reflective subcategories

To add to what Charles wrote, another reference is Mac Lane and Moerdijk's Sheaves in Geometry and Logic. They prove something a bit more general, involving Lawvere-Tierney topologies on a topos. For the purposes of understanding what I'm about to write, it's not necessary to know what a Lawvere-Tierney topology is.

Mac Lane and Moerdijk's book contains the following two results:

1. Let $mathcal{E}$ be a topos. Then the subtoposes of $mathcal{E}$ (i.e. the reflective full subcategories with left exact reflectors) correspond canonically to the Lawvere-Tierney topologies on $mathcal{E}$.




2. Let $mathbf{C}$ be a small category. Then the Lawvere-Tierney topologies on $mathbf{Set}^{mathbf{C}^{mathrm{op}}}$ correspond canonically to the Grothendieck topologies on $mathbf{C}$.

Result 1 is almost part of Corollary VII.4.7. The "almost" is because they don't go the whole way in proving the one-to-one correspondence, but I guess it's not too hard to finish it off. (Edit: it also appears as Theorem A.4.4.8 of Johnstone's Sketches of an Elephant, where Lawvere-Tierney topologies are called local operators.) Result 2 is Theorem V.4.1.

I agree with the point of view that Charles advocates. When I started learning topos theory I got bogged down in detailed stuff about Grothendieck topologies, and it all seemed pretty technical and unappealing. It wasn't until years later that I learned the wonderful fact that Charles mentions: an elementary topos is Grothendieck iff it's a subtopos of some presheaf topos. I wish someone had told me that in the first place!

career - Where are mathematics jobs advertised if not on mathjobs (e.g. in Europe and elsewhere)?

My impression is that in the US, there is a canonical place for finding math jobs, namely mathjobs.org. For those of us who live and apply for jobs elsewhere, life is more complicated, and searching for advertised academic mathematics jobs for example in Europe can be a real hassle, with loads of different sites, different systems, and some jobs apparently advertised only on the web page of the hiring institution, or one some obscure mailing list.

So, where are academic math jobs advertised when they for some reason are not or cannot be on mathjobs.org? Of course I know of a few such places, but I am sure there must be many more.

All answers welcome, this would help me and probably many others.

fourier analysis - estimates of exponential polynomials

Let $ p(t) = Sigma_{k=1}^n c_k e^{i lambda_k t}$ be an exponential polynomial.

In the paper "Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type" http://www.math.msu.edu/~fedja/Published/paper.ps Nazarov proves an estimate on the maximum value attained by the polynomial $p$ in an interval $I$, in terms of the maximum of $p$ in a subset $E subset I$.

To be precise he obtains the following estimate:-

$$ sup_{t in I} |p(t)| leq ( frac{A mu(I)}{mu(E)} )^{n-1} sup_{tin E} |p(t)|.$$

At one point he mentions that the result holds true for more general functions of the type $$p(t) = Sigma_{k=1}^n q_k(t) e^{i lambda_k t}$$ where $q_k(t)$ are algebraic polynomials of degree d_k$; by an "obvious" approximation argument.

It is not clear to me, what exactly is the argument he is suggesting ?

One of the method he uses to obtain an estimate as mentioned above is by using Turan's Lemma. Although in that case one gets exponent $2 n^2$ instead of $n-1$.

$underline {Turan's Lemma}$ Let $ z_1,dots,z_n$ be complex numbers, $|z_j|geq 1, j=1,dots,n.$ Let $ b_1,dots, b_n in mathbb C $ and $$S_j:= Sigma_{k=1}^n b_k z_k^j$$
Then $$|S_0| leq {frac{4 e (m+n-1)}{n}}^{n-1} max_{j=m+1}^{m+n} |S_j|.$$
As a simple consequence of this result when the value of an exponential polynomial (with constant coefficient) is known for $n$ consecutive term of an arithmetic progression, then one can get an estimate of the value of the polynomial along that arithmetic progression. i.e.,

Let $p(t)=Sigma_{k=1}^n c_k e^{i lambda_k t}$ and assume that the value of the polynomial $p(t)$ is known for $t_j=t_0+j delta$ for $ j= m+1,...,m+n$. Then substitute $b_k=c_k e^{i lambda_j t_0}$ and $z_k= e^{i lambda_k delta}$ and apply Turan's lemma.

The result now follows by Lebesgue's density theorem and some averaging argument.

We might want to get a similar result like Turan's Lemma for the more general type of exponential polynomial $p(t) = Sigma_{k=1}^n q_k(t) e^{i lambda_k t}$ where $q_k(t)$ are algebraic polynomials of degree d_k$.

But I doubt this is what he is suggesting here, as later in order to get the sharper result (the one with exponent $n-1$) he uses some weak type estimates it seems. (I have not read this part of the proof yet).

So, what exactly is the obvious approximation argument he is trying to suggest here?

Minimum enclosing rectangle of a convex polygon proof

Suppose you've got your polygon P, embedded in the Euclidean plane with some standard coordinate system. Then there's some rectangle R with horizontal and vertical sides which encloses P and is as small as possible among all such rectangles (just take horizontal lines through the topmost and bottommost points, and vertical lines through the rightmost and leftmost points). Translating the coordinate system doesn't change the rectangle or its area, but rotating it through an angle θ will; so the rectangle R_θ and its area A_θ are really functions of θ (and (π/2)-periodic functions at that). As such, A definitely has a minimum at some angle θ_min.

The meat of the theorem is that, for any angle φ where R_φ does not meet the conclusion, the function A does not even have a local minimum. To see this, note that R_φ intersects P at a finite set of points S — usually |S| = 4, but 3 or 2 are also possible. Now consider the rectangle R'_θ which is the smallest rectangle with sides parallel to the θ-axes and which contains S, and let A'_θ be its area. Note two things:

• For θ near φ, R'_θ = R_θ, and hence A'_θ = A_θ.


• Near φ, the function A' is concave downward. Why? Divide the rectangle R'_θ along the lines between the points of S. We get a convex polygon determined by S (and hence constant), plus |S|-many right triangles. Each triangle has a hypotenuse which stays constant as θ changes, and an interior angle which changes as θ/2, hence an area which is concave downward.

Under our assumption, φ can't possibly be a local minimum of A, let alone a global one; so wherever the global minimum is, it must have a side coincident with one of the sides of P.

enumerative combinatorics - Non-isomorphic graphs of given order.

See:

Combinatorial algorithms: an update,
Herbert S. Wilf, Albert Nijenhuis
SIAM, 1989. Chapter 8: Generating Random Graphs.

The chapter gives an algorithm for producing an undirected graph uniformly over all graphs of size $n$. It is based on Polya counting. Computing the enumerating polynomial depends on some group theory that is time consuming (I don't know the complexity class, but I'll just conjecture it is most likely exponential space on $n$). But it is a guarantee of uniform distribution. Unfortunately I don't know of a way (I haven't heard of a way) to derandomize this to create an unranking algorithm (to give a mapping from the naturals to the set of unlabeled graphs).

The algorithm presented in your link (by de Wet) is cute (I mean that in the sense that it is cleverly simple, does not lie, but doesn't really give the meat of it, what it means to have a list of non-isomorphic graphs). The graphs created there have a very particular structure (two paths with an arbitrary subset of edges between the paths, plus some small widgets on one end of each path to break symmetry. All subsets is a good trick but having two paths is pretty uncommon and $sqrt{T_n}over T_n$ goes to 0 as $n$ grows.

As to practicality, in addition to the suggestions of nauty and Sage, there's also Mathematica (commercial) which has a list (that you can manipulate) of graphs up to size 11.

Is a polynomial group law on $mathbb{R}^n$ automatically nilpotent?

I was told that a polynomial group law on (all of) $mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group.

Is it true? Where can I find a proof?

A counterexample for open subsets of $mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law.

subfactors - Short Introduction to Planar Algebras

Vaughan Jones' paper "The planar algebra of a bipartite graph" is short and example-focused. It's got the "right" definition (phrased with operads) and the mechanics of the bipartite graph planar algebra. While the bipartite graph planar algebra isn't a subfactor planar algebra, it's pretty darn close.

A subfactor planar algebra, by the way, is one which meets restrictions on the sizes of the spaces involved and has an inner product. Unsurprisingly, they give rise to subfactors (through machinery of Popa).

Is every finite-dimensional Lie algebra the Lie algebra of an algebraic group?

Sorry to tune in so late to this conversation, but I think it's worth pointing out some of the paper trail on the question (supplementing BCnrd's comment). This goes back to Chevalley's initial work in the 1950s, especially in volume II (in French) of his projected
six volume work Theorie des groupes de Lie. The first volume (in English) was published by Princeton Press, then II and III followed but no more; his 1956-58 Paris seminar changed the whole approach to linear algebraic groups and largely ignored the Lie algebras. In Section 14 of II, working over an arbitrary field of characteristic 0, Chevalley asks which Lie subalgebras $mathfrak{g} subset mathfrak{gl}(V)$ (with $dim V < infty$) can be Lie algebras of closed subgroups of the general linear group. He worked out a number of nice features of the unique smallest algebraic subalgebra containing $mathfrak{g}$: it has the same derived algebra as $mathfrak{g}$, for instance. In fact, the derived algebra of any $mathfrak{g}$ is algebraic.

Some of these ideas were written down by Borel (Section 7) and by me (Chap. V)
in our Springer graduate texts, working over an algebraically closed field of characteristic 0 (my treatment came from the earlier Bass/Borel notes). These sources include further references to papers by Hochschild and others along with the more scheme-theoretic treatment in the 1970 book by Demazure and Gabriel: II, Section 6, no. 2. They assume $k$ is a field and $mathfrak{G}$ is a " $k$-groupe localement algebrique" with an appropriate Lie algebra attached, then study possible algebraic subalgebras.

In prime characteristic the notion of algebraic Lie algebra becomes much more problematic. See Seligman's 1965 book Modular Lie Algebras, VI.2, for some
discussion. For example, the Lie algebra
$mathfrak{sl}(p,k)$
is usually simple modulo its one-dimensional center, but the quotient algebra can't be algebraic since then it would be the Lie algebra of a known simple algebraic group. Even more extreme are the extra simple Lie algebras of "Cartan type"
(and others for small primes), for which there are no corresponding groups. `

gt.geometric topology - Which manifolds admit a diffeomorphism of order $n$?

The Nielsen Realisation Problem asks when a (finite) subgroup of the mapping class group (the group of isotopy classes of diffeomorphisms) of a surface can be realised as a group of diffeomorphisms. Kerckhoff proved in the 80s that every finite subgroup of the mapping class group can be realised. (For infinite subgroups, there are various known obstructions, such as the Miller-Morita-Mumford characteristic classes.) Thus, Kerckhoff's theorem implies that a surface admits a diffeomorphism of order n if its mapping class group has an element of order n. Conversely, one can show that any diffeomorphism of a surface must have infinite order if it is isotopic to the identity, every diffeomorphism of order n gives an order n mapping class group element.

If you have a diffeomorphism of finite order on a surface then you can find a complex structure (or a Riemannian metric or a symplectic structure or a conformal structure) for which the diffeomorphism is an automorphism/isometry. This is accomplished by choosing an arbitrary metric and then averaging over all translates of it by powers of the diffeomorphism.

So the point of all this is that in dimension 2 finding an order n diffeomorphism of a genus g surface is the same as finding a complex curve with an order n isometry, or equivalently, a Z/n orbifold point in the moduli space. As Sam and Robin alluded to, there is a bound on the order of n relative to g. Hurwitz's theorem states that the order of the automorphism group of a genus g curve is less than or equal to 84(g−1). There are various other theorems that tell you about what sorts of finite subgroups you can find in mapping class groups.

In higher dimensions, it's harder to give a useful answer. If your manifold has a circle action then you are done because $S^1$ contains Z/n for any n. But there are plenty of manifolds around which do not admit circle actions, such as K3 surfaces. The $widehat{A}$-genus is an obstruction to admitting a circle action. Some nice things are known about the finite groups of automorphisms of K3 surfaces. In fact, I think they are pretty much completely classified into a finite list.

In general, by averaging over translates of a metric, you can still assume that a given finite order diffeomorphism acts by isometries for some metric. Generally, the isometry group of a compact Riemannian manifold will be a finite dimensional compact Lie group (I think this is a theorem of Yau).

gn.general topology - Smooth classifying spaces?

The answer to this does depend highly on the category in which you are prepared to work. If by "smooth structure" you mean "when is BG a finite dimensional manifold" then the answer is, as Andy says, "not many".

However if you are prepared to admit that there are more things that deserve the name "smooth" than just finite dimensional manifolds, then the answer ranges from "a few" to somewhere near "all".

To illustrate this with examples, the classifying space of ℤ is, of course, S¹ whilst the classifying space of ℤ/2 is ℝℙ^∞. Both are manifolds, but only the first is finite dimensional.

Here are some more details for the "somewhere near all". Take any topological model for BG. Then consider all continuous maps ℝ → BG. These correspond to G-bundles over ℝ. Amongst those will be certain bundles which deserve the name "smooth" bundles. By taking the corresponding curves, one determines a family of curves ℝ → BG which should be called "smooth". Using this one can define a Frölicher space structure on BG. (It is possible that you will get more smooth bundles than you bargained for this way. If that's a problem, you could work in the category of diffeological spaces but then you'd need to use all the ℝⁿs).

In the middle, one can consider infinite dimensional manifolds. Then as your group is discrete it would be enough to ensure that you have a properly discontinuous action on an infinite sphere (there's a question somewhere around here about that being contractible). Some would say that your sphere "ought" to be the sphere in some Hilbert space. Failing that, if you have a faithful action on a Hilbert space (or more generally Banach space) with one or two topological conditions then you can quotient the general linear group by your group. Indeed, if your group is discrete then take the obvious action on ℓ²(G) (square summable sequences indexed by G).

A good example, but which is about as far from your situation for discrete groups as possible, is that of diffeomorphisms on a manifold. The classifying space of this group is the space of embeddings of that manifold in some suitable infinite dimensional space.

For more on the categories behind all this, see the nlab entries starting with generalised smooth spaces and the references therein. Also, anything by Kriegl, Michor, or Frolicher in the literature is worth a look.

at.algebraic topology - Proving homotopy invariance of cellular homology by constructing a chain homotopy

I'm trying to follow an argument in Lück's "Algebraische Topologie: Homologie und Mannigfaltigkeiten" (to which there apparently doesn't exist an english translation). The aim is to check homotopy invariance of cellular homology by constructing a chain homotopy.

Let me sketch the argument. Let $hcolon (X,A)times [0,1]to (Y,B)$ be a cellular homotopy from $f_0$ to $f_1$. Using the CW-structure on $[0,1]$ with the two 0-cells {0}, {1} and one 1-cell, we identify
$$
C_n((X,A)times [0,1])=C_n(X,A)oplus C_n(X,A)oplus C_{n-1}(X,A).
$$
Then $C_n(h)$ is of the form $C_n(f_0)oplus C_n(f_1)oplus u_{n-1}$, where $u_{n-1}$ is some map $C_{n-1}(X,A)to C_n(Y,B)$.

Now we would like to compute the $n$th differential of $C_*((X,A)times [0,1])$ under the above identification. Since it is a map from $C_n(X,A)oplus C_n(X,A)oplus C_{n-1}(X,A)$ to $C_{n-1}(X,A)oplus C_{n-1}(X,A)oplus C_{n-2}(X,A)$, we can denote it by a 3x3-matrix.
My computation yielded
$$
begin{pmatrix}
c_n & 0 & 0 newline 0 & c_n & 0 newline -id & id & c_{n-1}
end{pmatrix},
$$
where $c_n$ is the $n$th differential of $C_*(X,A)$.
In the book, however, I find
$$
begin{pmatrix}
c_n & 0 & (-1)^n newline 0 & c_n & (-1)^{n-1} newline -id & id & c_{n-1}
end{pmatrix}.
$$

Question: What is meant by $(-1)^ncolon C_{n}(X,A)to C_{n-2}(X,A)$ and how can I
understand that the second matrix above is the correct expression?

---

Edit: Apparently, the correct form of the the matrix representing the $n$th differential of $C_*((X,A)times [0,1])$ is
$$
begin{pmatrix}
c_n & 0 & 0 newline 0 & c_n & 0 newline (-1)^{n+1}cdot id & (-1)^ncdot id & c_{n-1}
end{pmatrix}
$$
or
$$
begin{pmatrix}
c_n & 0 & 0 newline 0 & c_n & 0 newline (-1)^{n}cdot id & (-1)^{n+1}cdot id & c_{n-1}
end{pmatrix},
$$
depending on the orientation of the 1-cell in $[0,1]$.

cv.complex variables - Multiplicity of zero (higher dimensional analog)

Consider a sistem of n holomorphic equations with n unknowns in a neighborhood of zero. Suppose that a solution in a neighborhood of 0 is a k-dimensional manifold.
I want to associate to it some number n_k, its multiplicity, so that solution of any perturbation of the system has not more than n_k k-dimensional strata. Does such multiplicity exist and what is the right definition for it?

nt.number theory - modular eigenforms with integral coefficients [Maeda's Conjecture]

This is a (far too long) comment on Buzzard's comment about Hida's remark.

I think I can guess what Hida was saying. He was probably talking about non-vanishing of L-functions of Hecke eigenforms of level one and weight $k equiv 0$ (mod 4). This is a long-standing (folklore? ) conjecture in its own right, well-known among analytic number-theorists.

Here is how such a thing can be proven using Maeda's conjecture. There is a result of Shimura that says that the Galois group acts nicely on the central values (in fact any critical value) L-function of eigenforms. In particular, if one of them is zero then all the Galois twists are also zero and hence their sum is also zero. Now, even though it may be difficult to show that an L-function doesn't vanish at the centre, it is often easy to show that the sum of the central values of L-functions in a family is non-zero (see, for example, the work of Rohrlich and Rodriguez-Villegas on non-vanishing of L-functions of Hecke characters).

In the case in question, Maeda's conjecture will imply that if one central L-value is zero then the sum of all the central L-values over the whole basis must be zero and I think a contradiction will ensue after one uses the approximate functional equation to write the central value in terms of the Fourier coefficients and then using the Petersson formula ( I need to check this up).

Note 1: There is an article by Conrey and Farmer titled "Hecke operators and nonvanishing of L-functions" (Ahlgreen et al. (eds.), Topics in Number Theory, 1999) where they prove the above mentioned result along a different line.

Note 2: I think the following is easier. One can think of $frightarrow L(f,k/2)$ as a linear functional on the space of cusp forms $S_k(Gamma(1))$ and indeed it is possible to explicitly write a function $G$ such that

$L(f,k/2)=langle f,G rangle$

for all Hecke eigenform $f$ in $S_k(Gamma(1))$. Now Maeda + Shimura's result will imply that $G$ is orthogonal to the whole space and therefore zero. So it is just a matter of checking that $G$ is not identically zero, which shouldn't be too hard.

mg.metric geometry - Smallest dilation of a quadrilateral?

What is the smallest dilation of a quadrilateral in $mathbb{R}^d$?
This may be an open problem, which I know is verboten on MO.
So my question is: Is this indeed open?

It will take me some time to explain the terms.
The notion of dilation derives from Gromov, as far as I know
(He defines a version in
Metric Structures for Riemannian and Non-Riemannian Spaces,
p.11,
although he called it distortion).
I came upon it myself via $t$-spanners.

The version in which I am interested is this.
Let $P$ be a polygon (its boundary, not its interior), and $x,y$ two points on
$P$. You can think of $P$ in $mathbb{R}^2$, but also $mathbb{R}^3$ and $mathbb{R}^d$
for $d>3$ are interesting.
Define $delta(x,y)$ as the maximum (supremum) of $d_P(x,y) / | x y |$,
where $d_P(x,y)$ is the distance between $x$ and $y$ following
$P$ (the shortest path staying on the closed path that consitutes $P$),
and $|xy|$ is the Euclidean distance in $mathbb{R}^d$.
Thus $delta(x,y)$ measures how much $P$ dilates w.r.t. Euclidean distance.
I am interested in the minimum value $delta(P)$
of $delta(x,y)$ over all
$x,y in P$, for all $n$-gons $P$, for fixed $n$.

---

Example 1.
If $P$ is a unit square, then $delta(x,y)$
for $x,y$ opposite corners is $sqrt{2}$, but
$delta(P)=2$ because with $x,y$ midpoints of
opposite sides, $delta(x,y)= 2/1$.

Example 2.
If $P$ is an equilateral triangle, $delta(P)=2$, as shown in the figure.
In fact, the dilation of any triangle is $ge 2$ [Lemma 7 in the 2nd paper below].

Example 3.
It is known the the dilation of any closed curve $C$
satisfies $delta(C) ge pi/2$, with equality achieved
only by the circle. [Corollary 23 in the first paper below.] This is (apparently) due to Gromov.

---

So I finally come to my question. By reading these two
papers,
"Geometric Dilation of Closed Planar Curves: New Lower Bounds,"
and
"On Geometric Dilation and Halving Chords,"
it appears to me that the minimum dilation of a quadrilateral
in $mathbb{R}^2$ (and $mathbb{R}^d$) is not known.
I had heard this was the case three years ago in a seminar
in Brussels, but (a) I didn't quite believe it,
(b) it was hearsay, and (c) it is now out of date.
I am trying to clarify with the authors of these papers, but in parallel
I would appreciate any information on the status of
this question.
The latter paper cited above proves a lower bound of $4 tan(pi/8) approx 1.66$
(if I have interpreted it correctly).

Finally, if indeed open, this seems a potential
PolyMath undergrad project, as well as fun for anyone else!

Addendum. I don't want to close-out this question, but I have heard from one of the authors of
the above cited papers, and indeed it appears that the dilation of a planar quadrilateral is unknown.
So I have tentatively tagged this as an open-problem, and I will update if new information surfaces. Thanks for everyone's interest and input!