nt.number theory - Is there a canonical notion of "mod-l automorphic representation"?

This is in my mind a central open problem.

Here is an explicit example which I believe is still wide open. Serre's conjecture (the Khare-Wintenberger theorem) says that if I have a continuous odd irreducible 2-dimensional mod p representation of the absolute Galois group of the rationals then it should come from a mod p modular form. We have a perfectly good definition of mod p modular forms (sections of the usual line bundles on mod p modular curves).

But what is the "even" analogue of Serre's conjecture?

One problem is that in characteristic zero (i.e., for 2-dimensional continuous complex representations, which must have finite image) these guys are expected to come from algebraic Maass forms, which have, as far as anyone knows, no algebraic definition: they are genuinely analytic objects and it is a complete miracle that their Hecke eigenvalues are algebraic. So there's a big open problem: give an algebraic definition of an algebraic Maass form which doesn't use analytic functions on the upper half plane which are eigenvectors for the Laplacian with eigenvalue 1/4. I have no idea how to do this and I don't think that anyone else has either. The problem is that the forms are not holomorphic (so you can't use GAGA and alg geom) and they're not cohomological (so you can't use group cohomology either), and those are in some sense the only tricks we have (other than Langlands transfer, but I don't know any functorial transfer of these guys which (a) loses no information and (b) gives rise to a form which is cohomological or holomorphic). It might be an interesting PhD problem to check this out in fact. Blasius and Ramakrishnan once tried to go up to an im quad field and then to Sp_4 over Q but the resulting form isn't holomorphic. One might define an algebraic automorphic rep to be "accessible" if it's cohomological or perhaps limit of discrete series (perhaps these are the ones for which there's a chance of proving they're arithmetic), and then try and find examples of algebraic auto reps which should never transfer to an "accessible" rep on any other group.

But even after that, there's another problem, which is that as far as I know one does not expect a continuous even irreducible mod p representation of this Galois group to lift to a de Rham representation in characteristic zero (and indeed perhaps Frank Calegari might be able to give explicit counterexamples to this, after his recent observation that oddness sometimes follows from other assumptions). So even if one could give an algebraic definition of a Maass form, one wouldn't have enough Maass forms---they would all (at least conjecturally) give rise to Galois representations with images all of whose Jordan-Hoelder factors are cyclic or A_5 (by the classification of finite subgroups of PGL_2(C)). So there's another non-trivial sticking point.

In summary---nice question but I don't think that mathematics has a good answer yet (unless you're willing to just cheat and say that an automorphic representation "is" a Galois rep with some properties).

co.combinatorics - Average distance between numbers of the form $2^{a}3^{b}$

The point of this answer is simply to repeat Scott's answer in a more visible place. If he wants to post it himself, I'll delete my post:

Let $a_i$ be the sequence of these integers, sorted into order. Let $a_r leq N < a_{r+1}$. We want to estimate
$$frac{1}{r-1} sum (a_{i+1} - a_i) = (a_r-a_1)/(r-1).$$

As Scott explains,
$$r=(1/2) cdot (log N/log 2 + O(1)) cdot (log N/log 3 + O(1)) = (log N)^2/(2 log 2 log 3) + O(log N) .$$

Also, there is clearly a power of $2$ between $N$ and $N/2$, so $a_r sim N$.

So the average distance is $sim N/(log N)^2$. If you work harder, you can probably tighten up the bounds to show that it is $N (2 log 2 log 3)/(log N)^2 (1+O(1/log N))$

dg.differential geometry - What are CR manifolds like?

CR submanifolds of a complex manifold are defined as submanifolds M⊂X such that TM∩iTM⊂TX has constant rank (i is the imaginary unit). Note that the condition is automatically verified if M has codimension one; for higher codimension this is not true.

An abstract CR manifold is a real manifold M, with a distinguished subbundle HM⊂TM, corresponding to TM∩iTM, endowed with a linear endomorphism J with J²=-Id. The structure is furthermore required to satisfy a so called integrability condition:
For all sections X,Y of HM:

• [X,JY]+[JX,Y] is a section of HM




• ([X,Y]-[JX,JY]) + J([X,JY]+[JX,Y]) = 0

Not every abstract CR manifold can be realized as a CR submanifold.

ct.category theory - What are κappa-categories?

The intuitive explanation is that $kappa$-categories are to first-order functions what cartesian closed categories are to higher-order functions.

This all started with Lambek's work on polynomial categories; the best reference for that is
J. Lambek. Functional completeness of cartesian categories. Annals of Mathematical Logic, 6:251–292, 1973. That paper introduced the choice of the letter $kappa$. A polynomial category is what you get when you take a category with a terminal object, pick an object $X$, and then freely adjoin a new morphism $f:1to X$ and close under composition. This is very much like the ring of polynomials $R[X]$ over a ring $R$ arrived at by adjoining an indeterminate element and closing under the ring operations. Lambek shows that this property can be stated in universal terms -- as the unique category admitting a particular kind of functor from the original category. This is formally more satisfying but not the best route for beginners.

Later, Hasegawa developed this idea much further in M. Hasegawa. Decomposing typed lambda calculus into a couple of categorical programming languages. Lecture Notes in Computer Science, 953, 1995. He showed that just as the $lambda$-calculus can be used as a "syntax" for specifying morphisms in a cartesian closed category, so too can the $kappa$-calculus -- roughly the $lambda$-calculus without first-class functions -- be used as a "syntax" for specifying morphisms. I recommend studying the figure immediately after the first paragraph of section 3 in his paper (very carefully). It conveys both the essence of these categories and their relevance to the study of programming languages.

To wrap up, cartesian closed categories have been an immensely useful tool in understanding programming languages with first-class functions. Unfortunately they can only be used to study languages with the property that for every pair of types $B$ and $C$ there is also a type of functions from $BRightarrow C$ and the ways of getting a $BRightarrow C$ from an $A$ are in one-to-one-correspondence with the ways of getting a $C$ from a (cartesian) pair of an $A$ and a $B$. Lambek-Hasegawa $kappa$-categories are an elegant way to extend these techniques to languages in which this assumption does not hold.

Lastly, as a postscript, both Lambek and Hasegawa assume that the underlying monoidal structure of their categories is Cartesian. Some of the most fascinating results arise when you repeat their constructions in categories which are merely binoidal or premonoidal -- you'd be surprised how few modifications are required.

nt.number theory - Uniformly computable classes of graphs

This question sounds more like a research project than a definite problem. Part of the reason for me to say this is because this question is only interesting if you impose some (ill-defined) qualifications on what an interesting answer can be; and even then, it is not clear how one can answer the question without having the insights into the structure of the integers which you are hoping to find. Having said that, I'll hazard some elementary observations.

Throughout the following, I will refer to the integer which corresponds to a vertex in a graph as its index.

The problem of Gödel numbering

Your question is still about computable predicates. As long as you do so, without imposing further restrictions, there will still be too much room open for Gödel numbering to give solutions to sub-problems.

The predicate V₀ that I described before, while 'elaborate' (in that it would not be so simple in closed form), relied quite heavily on using prime factorization as a means of describing compound data structures --- specifically, in order to construct the indices of vertices to represent a sort of label for the vertex, its adjacency relations to other vertices, and even complete information about the entire graph.

In order to obtain "interesting" representations of graphs just by families of indices and relations on them, we obviously want to reduce reliance on Gödel numbering. We can do this by disallowing Gödel numbering in the indices,* i.e. in the first argument n to the predicate V and the first two arguments to the predicate E. But even if we do this, a very modest amount of Gödel numbering in the third argument μ to E gives away the whole game, as I show below.

^{* Of course, this pre-supposes that determining what constitutes Gödel numbering is a computable problem; I would argue that it isn't even well-defined. We can concievably describe a restriction to how much one can exploit the prime factorization of indices, but this would not address information being carried by the integer in e.g. representation of the index in base 1000, or any arbitrarily esoteric-but-computable representation of the index. Furthermore, even while prohibiting Gödel numbering, we still presumably want/need the index to bear some information; just not potentially arbitrarily complex information. But let's ignore that technical problem and suppose it can be done for the sake of argument.}

A "significantly less complex" universal set of predicates

It suffices to do the following:

• Set $nu$ to be a multiple of the first n primes, for a graph on n vertices.


• Set $V(n,nu) equiv big( nu in n mathbb N ;;&;; n ~text{prime}big)$.


• Set $mu = 2^{e_1} 3^{e_2} cdots p_m^{e_m}$, for a graph on m edges, where each integer e_j is a product of two primes which are the indices of adjacent vertices.


• Set $E(n,m,mu) equiv (exists p~text{prime}):big(mu in p^{nm}mathbb N ;;&;; mu notin p^{nm + 1} mathbb Nbig)$.

The sole explicit instance of "Gödel numbering" in this scheme is in the construction of μ. However, it suffices to admit very simple choices of V and E to represent an arbitrary graph. The number μ isn't even terribly esoteric: the only retriction on μ is that the exponents of its prime factorization are always square-free, and have exactly two prime factors when they are non-zero. This construction generalizes to hypergraphs and graphs with loops trivially.

The key to this construction, of course, is that the prime numbers serve as "easily distinguishable" indices for the vertices, to the extent that any number which is a multiple of two or more primes may be easily interpreted as a set; and then we store multiple sets by storing them in the exponents of a prime factorization of some integer.

I may have fallen afoul of some restriction you have in mind; for instance, it is likely that the set of μ of the form above (either in the explicit construction or the generalization for prime power indices) have small "measure", for many reasonable definitions of measures. From you examples e.g. of cycle graphs, however, I assume the fact that not all μ fall under this construction is not a problem.

Approaches to trees which avoid Gödel numbering

'Universal' predicates which rely on Gödel numbering are obviously boring, in that they indicate only what is possible. After one understands Gödel numbering as a means of representing data structures, they are not in themselves interesting. The challenge is then to see what one can do without using Gödel numbering (which I take in a practical sense to refer to using integers only to denote sequences or collections of other integers).

This is tricky for a graph class such as "arbitrary trees" (or even "arbitrary binary trees"), because there is little structure to deal with; it seems to me that natural representations of them will amount to something like lattices of subsets with added restrictions.

The trick is that for each index, you would like to identify a unique vertex (depending possibly on μ) which can be its "parent" (in the picture of rooted trees); or more generally, however one defines the set of allowed indices, at most one element of the set can be its parent, with exactly one vertex failing to have a parent.

This is where your question becomes ill-defined. Your motivation seems to be to somehow plumb the structure of the integers using graphs; but to do this, we must somehow already have a structure to hand to represent trees. This is not a research problem, so much as it is a very open-ended research programme.

Two potential approaches --- simple ones, and so likely not to be realizable --- occur to me.

1. Once more, we can consider prime factorizations. To avoid the temptation of Gödel numbering, we should not be too picky with the indices; the index-set should admit integers with non-trivial common divisors. A natural choice of edge-predicate is then
   $$ E(n,m,mu) equiv big(exists A,Bbig):Big({A,B} = {n,m} ;;&;; A/B ~text{is the largest prime factor of $A$}Big) ,$$
   that is, where transition from each vertex v to its parent corresponds to division by the largest prime factor of the index of v. A natural choice of predicate V is then to choose indices which are factors of some particular integer.

   
   It is not clear to me how general a class of trees we can get from this; we can easily obtain paths and stars, and some variety of other trees. It is also not clear to me how to get a reasonably varied class of binary trees.




2. We can consider ways of partitioning integers into a sum of smaller integers. Because addition in itself has so "little" structure, any structure would have to come from the allowed set of summands, which has to be given by the integer μ; this comes dangerously close to Gödel numbering. A simple solution is to use this integer to parameterize a family of summands, e.g. powers of μ. This too can lead to Gödel numbering of a different sort; a possible approach is to use ternary expansions of integers to represent the location of a node in a tree by left-right branching from the root, with log₃(n) representing the distance from the root, and the j^th digit representing whether one takes the left or right branch at the j^th level to reach the vertex in question. So there is a real question of what there is that one can do, under the constraint of doing something interesting.

   
   A compromise is to succumb to prime factorization, but to more or less ignore the exponents to avoid the temptation to do Gödel numbering. So, the possible transitions may be taken to be the (maximal) prime-power factors of μ. This still uses μ to denote a set of integers, but not a set with arbitrary elements. To identify a parent of an index, one may take the number which may be reached by subtracting the largest allowable transition.
   One may easily obtain paths (e.g. restrict to odd indices and take μ = 2) and arbitrary star graphs (e.g. restrict indices to 1 and p+1 for p prime, and take μ to be a suitable square-free integer). It's not obvious how to get arbitrary trees.

   
   One can think of generalizing this to allow transitions which are e.g. arbitrary prime-power factors, not just maximal prime-power factors. This can be used to get a wide variety of trees. For instance, we may consider restricting the indices to integers n which are a sum of at most k powers of two, and let the parent of each index be the one which is obtained by subtracting the largest power 2^r < n. This is pretty simple; but also dangerously close to the ternary scheme described above. It is not clear exactly how to delineate the boundary between 'interesting' and 'using Gödel numbering'. (Then again, perhaps this last decomposition isn't interesting anyway --- or perhaps it is merely the concept of 'the structure of an arbitrary tree' which is not especially interesting.)

In closing

Unless your objective is actually to embark on an open-ended research project, you might want to restrict your question a bit more. In particular, you should give some thought to imposing conditions which would prevent "trivial" solutions, especially using numerical tricks to embed the structures you are interested in into the indices and/or parameters.

Unless you somehow refine your question, it is just extremely open-ended. Not to say that it isn't a nice sort of project.

gr.group theory - Finite index normal subgroups of a free group.

Let $G= (mathbb{Z} bigoplus mathbb{Z}) star (mathbb{Z} bigoplus mathbb{Z})$, where $star$ denotes the free product, let F be the commutator subgroup of G, it is free by a theorem of Kurosh. Find a proper normal subgroup of F (other than the trivial one) such that it is of infinite index.

nt.number theory - Writing down minimal Weierstrass equations

Let $E$ be an elliptic curve over $mathbb Q_p$. It is possible that $E$ has bad reduction but then when you see $E$ as a curve over a finite extension $K$ of $mathbb Q_p$, it obtains good reduction. Let $v$ be the valuation defined on $K$ and $R$ its valuation ring. I was interested in checking $E$ has good reduction over $K$ by hand, using the Weierstrass equation. What that amounts to then is writing down the Weierstrass equation $y^2+a_1xy + a_3y = x^3 + a_2x^2+a_4x + a_6$ with the $a_i in R$ and considering changes of coordinates $x=u^2x' + r$ and $y=u^3y' + u^2sx' + t$ for $u,r,s,t in R$ in hopes of finding an equation with $v(Delta')$ minimized, subject to each $a_i'$ being in $R$. There are certain congruence conditions that guarantee minimality of the new equation, e.g. $v(Delta') < 12$, which only depend on the choice of $u$. However, guaranteeing the new equation has coefficients in $R$ requires solving other congruence relations depending on $r,s$ and $t$, e.g. you need $v(a_1+2s)geq v(u)$ (because $a_1' = u^{-1}(a_1+2s)$). The few times I have done this by hand, I have just had to look at the equations and make some choices until something worked out.

My question is whether or not there exists a general method for obtaining a good change of coordinates $u,r,s,t$ and if not, then how do people go about writing down minimal Weierstrass models. I can't imagine there should be general methods for solving the system of non-linear congruences (higher powers of $u,r,s$ and $t$ appear in the other congruences) in the ring $R$, but if there is then I would also be interested in understanding that as well.

nt.number theory - Class number measuring the failure of unique factorization

Theorem (Carlitz, 1960): The ring of integers $mathbb{Z}_F$ of an algebraic number field $F$ has class number at most $2$ iff for all nonzero nonunits $x in mathbb{Z}_F$, any two factorizations of $x$ into irreducibles have the same number of factors.

A proof of this (and a 1990 generalization of Valenza) can be found in $S 22.3$ of my commutative algebra notes.

This paper has spawned a lot of research by ring theorists on half-factorial domains: these are rings in which every nonzero nonunit factors into irreducibles and such that the number of irreducible factors is independent of the factorization.

To be honest though, I think there are plenty of number theorists who think of the class number as measuring the failure of unique factorization who don't know Carlitz's theorem (or who know it but are not thinking of it when they make that kind of statement).

Here is another try [edit: this is essentially the same as Olivier's response, but said differently; I think it is worthwhile to have both]: when trying to solve certain Diophantine problems (over the integers), one often gets nice results if the class number of a certain number field is prime to a certain quantity. The most famous example of this is Fermat's Last Theorem, which is easy to prove for an odd prime $p$ for which the class number of $mathbb{Q}(zeta_p)$ is prime to $p$: a so-called "regular" prime.

For an application to Mordell equations $y^2 + k = x^3$, see

http://math.uga.edu/~pete/4400MordellEquation.pdf

Especially see Section 4, where the class of rings "of class number prime to 3" is defined axiomatically and applied to the Mordell equation. (N.B.: These notes are written for an advanced undergraduate / first year grad student audience.)

The Mordell equation is probably a better example than the Fermat equation because:

(i) the argument in the "regular" case is more elementary than FLT in the regular case (the latter is too involved to be done in a first course), and

(ii) when the "regularity" hypothesis is dropped, it is not just harder to prove that there are no nontrivial solutions, it is actually very often false!

riemannian geometry - Prescribing Gaussian curvature

I'm not sure that this will help, but let me suggest thinking about the following: You are looking for a metric of the form $g = e^{2u(r)}(dr^2 + r^2 dtheta^2)$ where $u(r)$ is to be chosen so that the curvature of $g$ is a certain function $K(r)$ and so that $u$ tends to zero as $rtoinfty$. Now, I wouldn't have called this problem "specifying the curvature profile" just because $r$ won't represent the $g$-distance from the origin when you are done. Instead, the $g$-distance $s= h(r)$ from the origin will be given by solving $ds = e^{u(r)} dr$ with $s(0)=0$, and I would have called $Kbigl(h^{-1}(s)bigr)$ the 'curvature profile'.

Are you sure that you wouldn't have rather had the metric in the form $g = ds^2 + f(s)^2 dtheta^2$ where $f(0)=0$ and $f'(0)=1$ and then choose $f$ so that it satisfies the equation
$$
f''(s) + K(s) f(s) = 0
$$
where $K$ is your given function?

If this is really your problem (and I'm not saying it has to be, but...), then you can, indeed, solve for $f$ explicitly, in a sense, but its definition will be piecewise, of course. You'll have $f(s) = sin s$ for $0le s le 1$, but on the intervals $1le sle 3$ and $3le sle 4$, $f$ will be given in terms of translated Airy functions (different ones on the different intervals), and then, for $sge 4$, you'll have $f$ be a linear expression in $s$. Of course, determining the constants at the breakpoints so that $f$ is $C^2$ there is probably not going to be doable in any fully explicit fashion.

derived category - determinant of a perfect complex

As I understand the construction of the determinant of a perfect complex, this definition is quite straightforward, following from the fact that in a short exact sequence, say
$$ 0rightarrow Srightarrow Erightarrow Qrightarrow 0$$
defining the determinant of the sequence to be the alternating tensor is the canonical way to make it isomorphic to $mathbb{1}$.

Also, I think good references to this may be the original paper by Knudsen-Mumford, a book by Kato, and also a paper by Kings which are listed below:

Finn Faye Knudsen and David Mumford, The projectivity of the moduli space of stable curves. I. Preliminaries on ''det'' and ''Div'' (pdf). The part about determinants appears in Chapter I, but note that there is a typo defining the determinant, namely in the map of the transposition of tensor product, there should be $alphacdotbeta$ instead the sum of these two as a power of $-1$;

Guido Kings, An introduction to the equivariant Tamagawa number conjecture: the relation to the Birch-Swinnerton-Dyer conjecture (pdf)

There is a part about determinants in lecture 1 section 5, where there are not a lot of details but it provides a good view towards the construction of determinant.

Kazuya Kato, Lectures on the approach to Iwasawa theory for Hasse-Weil L-functions via $B_{dR}$, part I Springer LNM 1553 pp 50-163 (doi:10.1007/BFb0084729), which mentions determinant in 2.1.

higher category theory - Models for, and motivation for, (oo,n)-categories for general n

For me a "model of (∞,n)-categories" is something (e.g., a model category) from which one can extract "the" (∞,1)-category of (∞,n)-categories. One could make this more precise by choosing a preferred definition of (∞,n)-categories and asking for things equivalent to it. Of course it's currently less clear than for, say, models of spaces or spectra that there really is a unique "correct" (∞,1)-category which really is equivalent to everything we hope it's equivalent to. And indeed already for n = 1 there are related but distinct useful notions of category as I write here which manifest themselves in homotopy theory as Segal spaces and complete Segal spaces. But I think most everyone expects that there's a natural notion of (∞,n)-category up to n-categorical equivalence which is the right analogue of n-category (whatever that means).

The easiest examples are of course in the case n = 2, where we have (∞,2)-categorical analogues of the usual examples of 2-categories, for instance the (∞,2)-category of A_∞ ring spectra and bimodules, or the (∞,2)-category of (∞,1)-categories, or presentable (∞,1)-categories, or stable (∞,1)-categories, ... For instance if you wanted to understand the relationship between (∞,1)-categories and their stabilizations—(∞,1)-categories form an (∞,2)-category, and stable ones form some kind of subcategory, and you might ask whether there is something like an adjoint to the inclusion—there isn't quite, but maybe there's an adjunction if we view these (∞,2)-categories as objects of some other (∞,3)-category. So (∞,n)-categories do arise naturally in the study of (∞,k)-categories for k < n. These examples are in a sense "algebraic" objects, as opposed to bordism categories, which turn out to be algebraic too, in a sense, but a priori are given by geometric constructions.

As for models for (∞,n)-categories: Besides the iterated complete Segal space model we have Charles Rezk's Θ_n-spaces, and I think simplicial strict n-categories are also supposed to give the right notion. There's the "complicial sets" model, which seems to me to be more conjectural. I would also like to hear about results about equivalence of these models—as far as I know none have been written down yet, except in the case n = 2.

fa.functional analysis - rules for operator commutativity?

One obvious but important observation is that, for operators on a $n$-dimensional vector space over a field, if $1 < n < infty$, we have $AB neq BA$ generically. In other words, consider the commutativity locus $mathcal{C}_n$ of all pairs of $n times n$ matrices $A,B$ such that $AB = BA$ as a subset of $mathbb{A}^{n^2}$. This is clearly a Zariski closed set -- i.e., defined by the vanshing of polynomial equations. It is also proper: take e.g.
$A = left[ begin{array}{cc} 1 & 1 \ 0 & 1 end{array} right] oplus 0_{n-2}$ and $B =
left[ begin{array}{cc} 0 & 1 \ 0 & 1 end{array} right] oplus 0_{n-2}$. Since $mathbb{A}^{n^2}$ is an irreducible variety, $mathcal{C}_N$ therefore has dimension less than $N^2$. This implies that over a field like $mathbb{R}$ or $mathbb{C}$ where such things make sense, $mathcal{C}_N$ has measure zero, thus giving a precise meaning to the idea that two matrices, taken at random, will not commute.

One could ask for more information about the subvariety $mathcal{C}_N$: what is its dimension? is it irreducible? and so forth. (Surely someone here knows the answers.)

I would guess it is also true that for a Banach space $E$ (over any locally compact, nondiscrete field $k$, say) of dimension $> 1$, the locus $mathcal{C}_E$ of all commuting pairs of bounded linear operators is meager (in the sense of Baire category) in the space $B(E,E) times B(E,E)$ of all pairs of bounded linear operators on $E$.

Kevin Buzzard has enunciated a principle that without further constraints, the optimal answer to a question "What is a necessary and sufficient condition for $X$ to hold?" is simply "X". This seems quite applicable here: I don't think you'll find a necessary and sufficient condition for two linear operators to commute which is nearly as simple and transparent as the beautiful identity $AB = BA$.

Still, you could ask for useful sufficient conditions. Diagonalizable operators with the same eigenspaces, as mentioned by Jonas Meyer above, is one. Another is that if $A$ and $B$ are both polynomials in the same operator $C$: this shows up for instance in the Jordan decomposition.

oa.operator algebras - Relative Bicommutant

Let $Asubseteq B(H)$ be a subset, and let $text{alg}(A)$ be the algebra generated by $A$. Then it's easy to see that $$A' = text{alg}(A)'.$$ A similarly easy check shows that if $A$ and $B$ are subsets, then $$A' cap B' = (Acup B)' = text{alg}(Acup B)'.$$

So, for your question, pick some normal representation $Msubseteq B(H)$ (so that $M''=M$), and let $Asubseteq M$ be a subset. Set $$X={xin M:ax=xa (ain A)} = A'cap M = A'cap M'' = (Acup M')',$$ so your relative commmutatant is [{min M:xm=mx (xin X) } = X' cap M = X' cap M'' = (Xcup M')'.] So, yes, this is a von Neumann algebra.

In fact, as $Asubseteq M$, clearly $M'subseteq A'$ and so $A''cap M = (A'cup M')' = A''$. So as $Xsubseteq A'$, thus $A''subseteq X'cap M$.

Edit: There is probably an easier example than this... But, let $M=VN(mathbb F_2)$, say with canonical generators $a$ and $b$. Let $A$ be a star-algebra generated by $b$ and $a^{-1}ba$: so $A$ is just linear combinations of $b^n$ and $a^{-1}b^na$ for $ninmathbb Z$. A bit of combinatorics shows that $X=mathbb C1$ and so the the relative bicommutant is all of $M$. However, $A$ is not ultraweakly dense in $M$, because we cannot approximate the generator $a$.

Conclude: So, if I haven't messed up, this shows that the relative bicommutant is always a W*-subalgebra of $M$, but that it might be larger than the ultraweak closure of $A$ in $M$.

ag.algebraic geometry - Is the tangent space functor from PD formal groups to Lie algebras an equivalence?

The previous version of this question was rather badly broken, and I hope this version makes some sense.

There have been a few questions on MathOverflow about how much representation-theoretic information is lost when passing from a Lie group to its Lie algebra, e.g., away from the semisimple case, Lie algebras have many more representations. In the algebraic setting, there is an intermediate construction between an algebraic group and its Lie algebra, given by the formal group. One completes the algebraic group along the identity to get a formal scheme equipped with a group law, and one can pass from there to the tangent space to get the Lie algebra. In characteristic zero, the tangent space functor is an equivalence of categories from formal groups to Lie algebras, but in positive characteristic, formal groups form an honest intermediate category since the tangent space can lose a lot of information. For example, there is only one isomorphism class of one-dimensional Lie algebra, but one-dimensional formal groups have a rich arithmetic theory, with a moduli space stratified by positive integer heights. The completions at the identity of the additive group and the multiplicative group have very distinct formal group structures, and one way to explain the lack of isomorphism is by the presence of denominators in the usual logarithm and exponential power series.

It seems to me that in positive characteristic, there could be an intermediate construction between formal groups and Lie algebras, given by passing to PD rings and replacing the coordinate ring of the formal group with the divided power envelope of the identity section. If I'm not mistaken, this construction yields a group object in PD formal schemes.

Here is a bit of explanation for the uninitiated (see Berthelot-Ogus for more): PD rings are triples $(A,I,gamma)$, where $A$ is a commutative ring, $I$ is an ideal, and $gamma = { gamma_n: I to A }_{n geq 0}$ is a system of divided power operations. I think they arose when Grothendieck tried to get De Rham cohomology to give the expected answers for proper varieties in characteristic $p$, since the naïve definition tended to yield infinite dimensional spaces. There is a forgetful functor $(A,I,gamma) mapsto (A,I)$ from PD rings to ring-ideal pairs, and it has a left adjoint, called the divided power envelope. In characteristic zero, $gamma$ is canonically given as $gamma_n(x) = x^n/n!$, so both functors are equivalences in that case. The notion of PD ring can be sheafified and localizations have canonical PD structures, so one has notions of PD scheme and PD formal scheme.

Question: Do PD formal groups contain any more information than the underlying Lie algebra?

I have a suspicion that the answer is "no" and the answer to the title question is "yes". Vague word-association suggests that the divided power structure is exactly what one needs to get a formal logarithm, but maybe there is a more fundamental obstruction.

I was originally motivated by the question of how Gelfand-Kazhdan formal geometry would differ in charateristic $p$ if I switched between ordinary and PD structures (cf. David Jordan's question). Unfortunately, I was laboring under some misconceptions about formal completions, and I'm still a bit confused about the precise structure of the automorphism group of the completion (PD or ordinary) of a smooth variety at a point in characteristic $p$.

definitions - What is a proper stack?

I have seen the use of the word "proper Deligne-Mumford stack". Now, it is clear to me what it means for a morphism f of stacks to be proper: as usual it should be representable, and every morphism between schemes obtained from f by base change should be proper.

Now, first I guess that "proper" here actually means "complete". A scheme over a field is complete when the structural morphism to a point is proper. But it does not make sense for a stack to ask that the morphism to the point is proper. Indeed it would be in particular representable, and since a point is a scheme this would imply that the stack itself is a scheme.

Another possibility is that the sentence means "a stack with a proper atlas", so that one cannot speak of proper stacks, but only of proper Deligne-Mumford stacks.

So I am asking here what the standard terminology is.

co.combinatorics - Two [n] to [n] function families

$bf Note.$ This question had a bounty, so at the end I accepted the best (and only) answer but in fact it is still open. It is (hopefully) equivalent to this question, if you have any ideas, please post them there.

$bf Question.$
Fix n. We are interested in the biggest t for which there exist two families of functions, $P_i,Q_i$, of size t from [n] to [n] such that for any $i,j$ whenever we consider the infinite sequence $P_i(Q_j(P_i(Q_jldots P_i(3))ldots)$ (where the number of iterations tends to infinity), it contains no 2's and infinitely many 1's if $i=j$ and it contains no 1's and infinitely many 2's if $ine j$.

$bf A lower bound.$
I know a construction that shows that $tge 2^{frac n2-O(1)}.$ For every subset $S$ of [n] that contains exactly one of $2k$ and $2k+1$ for $2le kle frac n2-2$ we construct a pair of functions, $P_S,Q_S$ as follows. For any number m denote by $m^+$ the smallest element of $S$ that is bigger than m or if all elements of $S$ are at most m then define it to be 1.
$P_S(1)=1, P_S(2)=2$ and for bigger $m$'s $P_S(m)=m^+$, while $Q_S(1)=1, Q_S(2)=2$ and for bigger $m$'s $Q_S(m)=m$ if $min S$ and $Q_S(m)=2$ if $mnotin S$. This way we go through all the elements of S and end in 1 if the functions have the same index, but we are pushed to 2 if they differ.

$bf Upper bound.$
It is of course true that $tle n^n$. So can you do better than $2^n$?

ag.algebraic geometry - Any relationship of frobenius homomorphism and frobenius category?

I did not understand number theory or characteristic p-algebraic geometry at all. I just know a little about frobenius homomorphism between two schemes. On the other hand, when I learned something on triangulated category. I found there was also a definition of "frobenius morphism" the definition is as follows:

There are two categories C and D. f_:D--->C, f^ :C--->D is left adjoint to f_, we call f_ is a Frobenious morphism if there exists an auto-equivalence G of C such that composition f^* G is right adjoint to f_*.

First question is:is there any relationship between this two frobenius morphism?

Second question is:does frodenius category play roles in algebraic geometry?

All the comments related to this are welcomed.

nt.number theory - What's the Hilbert class field of an elliptic curve?

My question points in a direction similar to Qiaochu's, but it's not the same (or so I think). Let me provide you with a little bit of background first.

Let E be an elliptic curve defined over some number field K. The Tate-Shafarevich group of E/K consists of certain curves of genus 1, isomorphic to E over some extension, with points everywhere locally. In the simplest case of an element of order 2, such a curve has the form C: y²=f₄(x) for some quartic polynomial f₄(x) in K[X]; here, C does not have a K-rational point, but has points in every completion of K.

If we look at E in some extension L/K, this curve C still has points everywhere locally, but if it has a global point (in L) we say that the corresponding element in the Tate-Shafarevich group of E/K capitulates. Heegner's Lemma says that elements of order 2 cannot capitulate in extensions of odd degree, which is the analogue of the similarly trivial observation that ideals generating a class of order 2 cannot capitulate (become principal) in an extension of odd degree.

I gave a few talks on the capitulation of Tate-Shafarevich groups more than 10 years ago. A little later the topic became almost fashionable under the name of "visualizing" elements of Sha. I discussed the following question with Farshid Hajir back then, but eventually nothing came out of it. Here it is. For capitulation of ideal classes, there is a "canonical" extension in which this happens: the Hilbert class field. So my question is:

• may we still dream about the existence of a curve with all the right properties, or are there reasons why such a thing should not exist?

We also know that capitulation is not the correct notion for defining the Hilbert class field, which is the largest unramified abelian extension of a number field. These notions do not seem to make any sense for elliptic curves, but we can characterize the Hilbert class field also in the following way: among all finite extensions L/K for which the norm of the class group of L down to K is trivial, the Hilbert class field is the smallest.

Taking the norm of Sha of an elliptic curve defined over L down to K does make sense (just add the equivalence classes of the conjugate homogeneous spaces using the Baer-sum construction or in the appropriate cohomology group). So here's my second question:

• Has this "norm map" been studied in the literature?

(I know that the norm map from E(L) to E(K) was investigated a lot, in particular in connection with Heegner points).

Let me add that I do not assume that such a "Hilbert class curve" can be found among the elliptic curves defined over some extension field; if there is a suitable object, it might be the Jacobian of a curve of higher genus or an abelian variety coming from I don't know where.

nt.number theory - Why the search for ever larger primes?

I used to do very large computations of pi, and even wrote several programs to do it. I see this as analogous to computing (a) the primality of large numbers, and (b) finding very large primes. I computed pi because, well, it was fun! There was no practical reason to do so. Sure, I could do some statistical tests on the digits or search for my phone number and date of birth in the string of digits, but otherwise it was just an endeavor in pushing my hardware to its "computational limits" and combining what I know about programming and mathematics to make some integrated product (no pun intended).

So, while I can't speak for the prime number crunchers, I believe it is just done "just because". It also gives an avenue to try new and more efficient numerical algorithms for multiplication and related operations, since the programs that compute large primes use these algorithms extensively.

Edit: In response to the part about there being an infinite number of primes, while there certainly are an infinite number, generating the *N*th prime isn't a trivial task (while, for example, computing pi to a million decimal places is a trivial task). There are "prime number formulas" that can give one the *N*th prime directly and deterministically, but these algorithms/formulas are incredibly inefficient, usually derived from Wilson's theorem. Mill's theorem provides another prime number formula, but requires the value of a certain constant called "Mill's constant". But to find that constant, you must find primes beforehand (this doesn't make Mill's theorem irrelevant, just not useful for the computation of primes).

ag.algebraic geometry - Is there triangulated category version of Barr-Beck's theorem?

There isn't a descent theory for derived categories per se - one can't glue objects in the derived category of a cover together to define an object in the base. (Trying to apply the usual Barr-Beck to the underlying plain category doesn't help.)

But I think the right answer to your question is to use an enriched version of triangulated categories (differential graded or $A_infty$ or stable $infty$-categories), for which there is a beautiful Barr-Beck and descent theory, due to Jacob Lurie. (This is discussed at length in the n-lab I believe, and came up recently on the n-category cafe (where I wrote basically the same comment here..)
This is proved in DAG II: Noncommutative algebra. In the comonadic form it goes like this. Given an adjunction between $infty$-categories (let's call the functors pullback and pushforward, to mimic descent), if we have

1. pullback is conservative (it respects isomorphisms), and


2. pullback respects certain limits (namely totalizations of cosimplicial objects,
   which are split after pullback)

then the $infty$-category downstairs is equivalent to comodules over the comonad
(pullback of pushforward). (There's an opposite monadic form as well)
This can be verified in the usual settings where you expect descent to hold.
In other words if you think of derived categories as being refined to $infty$-categories (which have the derived category as their homotopy category), then everything you might want to hold does.

So while derived categories don't form a sheaf (stack), their refinements do:
you can recover a complex (up to quasiisomorphism) from a collection of complexes on a cover, identification on overlaps, coherences on double overlaps, coherences of coherences on triple overlaps etc.
More formally: define a sheaf as a presheaf $F$ which has the property that
for an open cover $Uto X$, defining a Cech simplicial object $U_bullet={Utimes_X Utimes_X Ucdotstimes_X U}$, then $F(X)$ is the totalization of the cosimplicial object $F(U_bullet)$. Then enhanced derived categories form sheaves (in appropriate topologies) as you would expect. This is of course essential to having a good theory of noncommutative algebraic geometry!

examples - Applications of the Chinese remainder theorem

The Chinese Remainder Theorem gives a way to compute matrix exponentials.

Indeed, let $A$ be a complex square matrix, put $B:=mathbb C[A]$. This is a Banach algebra, and also a $mathbb C[X]$-algebra ($X$ being an indeterminate). Let $S$ be the set of eigenvalues of $A$, $$mu=prod_{sin S} (X-s)^{m(s)}$$ the minimal polynomial of $A$, and identify $B$ to $mathbb C[X]/(mu)$.

The Chinese Remainder Theorem says that the canonical $mathbb C[X]$-algebra morphism $$Phi:Bto C:=prod_{sin S} mathbb C[X]/(X-s)^{m(s)}$$ is bijective.

Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $Phi$.

Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $Phi^{-1}(e_s)$. This element will be of the form $$f=frac{mu}{(X-s)^{m(s)}} g mbox{ mod }mu$$ with $f,ginmathbb C[X]$, the only requirement being $$gequivfrac{(X-s)^{m(s)}}{mu}mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

group cohomology - Coboundary Representations for Trivial Cup Products

Suppose $G$ is a pro-$p$-group, $p$ odd, and $mathbb{F}_p$ is given the trivial $G$-action. By skew-symmetry of the cup-product in degree 1, given $chiin H^1(G,mathbb{F}_p)$, we have $chicupchi=0in H^2(G,mathbb{F}_p)$. In fact, in this case, it's even possible to explicitly write $chicupchi$ as a coboundary -- $chicupchi=dleft(binom{chi}{2}right)$, the coboundary of "$chi$ choose 2".

In any case, my question is whether or not there anyone has seen any other tricks of this sort, i.e., for the explicit realization of a trivial cup product as a coboundary. In my specific case, I know a particular cup product is zero since I can force it, via the $G$-equivariance of the cup-product, to land in a known-to-be-trivial eigenspace of $H^2$. I was hoping there was some "eigenspace-averaging" trick similar to the construction of orthogonal idempotents to get my hands on an explicit pre-image, but really, I'd just like to be aware of any tricks for doing this.

ac.commutative algebra - Homological dimension of a graded ring which is like polynomial ring

I do not know the answer to your first question. As for the next two the answer
is positive; one need only slightly modify standard proofs for the usual
polynomial ring:

If $R$ is a graded ring then we define the twisted polynomial ring $R[x]_t$ as
the graded ring generated by $R$ and $x$ with $deg x=1$ and the relations
$rx=(-1)^{|r|}xr$ where $r$ is homogeneous of degree $|r|$. Your ring is then
$k[x_1]_t[x_2]_tldots$ (I assume you meant to have a field as base ring) and in
general an element of $R[x]_t$ can be uniquely written in the form
$sum_nr_nx^n$. We can now prove that $R[x]_t$ is (left) Noetherian if $R$
is. Let $Isubseteq R[x]_t$ be an ideal and let $I'subseteq R$ be the ideal of
top terms of $I$. Picking generators for $I'$ and representing polynomials for
them gives a finite number of element of $I$ such that any element of $I$ can be
reduced to a polynomial of fixed degree.

As for finite global dimension by general results it suffices to show that $k$
has a finite resolution. Again we can use induction and assume that $k$ has a
finite resolution as $R$-module and then it suffices to show that $R$ has a
finite resolution as $R[x]_t$-module. This is done by considering $R[x]_te to
R[x]_t$ with $e$ mapped to $x$. Carrying the induction through gives an explicit
resolution which is a sign-twisted version of the Koszul resolution: In
homological degree $k$ it has an $R[x]_t$-basis $e _{i_1}e_{i_2}cdots e_{i_k}$ where $1leq
i_1 < i_2 < cdots < i_kleq n$ with $d(e_{i_1}e_{i_2}cdots e_{i_k})=sum_r
x_{i_r}e_{i_1}cdots widehat{e_{i_r}}cdots e_{i_k}$ (Look Ma no signs!).

pr.probability - book for probability

Rick Durrett's book "Probability: Theory and Examples" is a very readable introduction to measure-theoretic probability, and has plenty of examples and exercises. This is the second text that I learned probability theory out of, and I thought it was quite good (I used Breiman first, and didn't enjoy it very much). As a bonus, there is a free .pdf version of the 4th edition (which will be published in a few months) available on his website for the time being.

A recent text covering similar material (which I admit I haven't read that fully) which looked good on a quick reading was "Probability Theory: A Comprehensive Course" by Klenke. It has a very nice selection of modern topics.

homological algebra - Convergence of spectral sequences of cohomological type

Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the homological situation.

An exact couple consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that

$$
begin{array}{rcl}A&xrightarrow{i}&Anewline {scriptsize k}nwarrow&&swarrow{scriptsize j}newline&E&end{array}
$$

is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=jcirc k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $ain mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees

$$
begin{array}{rcl}A^{(r)}&xrightarrow{(1,-1)}&A^{(r)}newline {scriptsize (-1,0)}nwarrow&&swarrow{scriptsize (-(a-1+r),(a-1+r))}newline&E^{(r)}&end{array}
$$

induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}circ k^{(r)}$. Here ${scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.

Here is the canonical example. Let $0=C_{-1}subseteq C_0subseteqldots C_psubseteqldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them:
$$
begin{array}{lcccccr}
to H_{p+q}(C_p,C_{p-1})&xrightarrow{k}&H_{p+q-1}(C_{p-1})&xrightarrow{i}&H_{p+q-1}(C_p)&xrightarrow{j}&H_{p+q-1}(C_p,C_{p-1})tonewline
to H_{p+q}(C_{p-1},C_{p-2})&xrightarrow{k}&H_{p+q-1}(C_{p-2})&xrightarrow{i}&H_{p+q-1}(C_{p-1})&xrightarrow{j}&H_{p+q-1}(C_{p-1},C_{p-2})tonewline
to H_{p+q}(C_{p-2},C_{p-3})&xrightarrow{k}&H_{p+q-1}(C_{p-3})&xrightarrow{i}&H_{p+q-1}(C_{p-2})&xrightarrow{j}&H_{p+q-1}(C_{p-2},C_{p-3})to
end{array}
$$

There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.

Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.

The spectral sequence $E^1(C_bullet)$ of homological type
converges to $H_{p+q}(C)$ if it is bounded (=only
finitely many non-zero entries on
every fixed diagonal $p+q$). One has
$$ E^infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})) .$$
($i$ denotes the image in the colimit $H_{p+q}(C)$.)

Proof. Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$
begin{array}{c}
to E^{r}_{p+r,q-r+1}xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}xrightarrow{i^{r}}A^{r}_{p+r,q-r}newline
xrightarrow{j^{r}}E^{r}_{p,q}xrightarrow{k^{r}}newline
A^{r}_{p-1,q}xrightarrow{i^{r}}A^{r}_{p,q-1}xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}to.
end{array}
$$

The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n<0$, the last three terms are $0$. Exactness implies the result.

Now the cohomological situation.

Let $0=C_{-1}subseteq C_0subseteqldots C_psubseteqldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(hom(C,mathbb{Z}))$, the homology of the dualized complex as one does in topology.

Again we have many long exact sequences:
$$
begin{array}{lcccccr}
to H^{p+q}(C_p,C_{p-1})&xrightarrow{k}&H^{p+q}(C_p)&xrightarrow{i}&H^{p+q}(C_{p-1})&xrightarrow{j}&H^{p+q+1}(C_p,C_{p-1})tonewline
to H^{p+q}(C_{p+1},C_p)&xrightarrow{k}&H^{p+q}(C_{p+1})&xrightarrow{i}&H^{p+q}(C_{p})&xrightarrow{j}&H^{p+q+1}(C_{p+1},C_{p})tonewline
to H^{p+q}(C_{p+2},C_{p+1})&xrightarrow{k}&H^{p+q}(C_{p+2})&xrightarrow{i}&H^{p+q}(C_{p+1})&xrightarrow{j}&H^{p+q+1}(C_{p+2},C_{p+1})to
end{array}
$$

One can again follow $d^1:E_{p,q}^1to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.

An exact couple with bidegrees

$$
begin{array}{rcl}A_{(r)}&xrightarrow{(-1,1)}&A_{(r)}newline {scriptsize (0,0)}nwarrow&&swarrow{scriptsize (a+r,-(a-1+r))}newline&E_{(r)}&end{array}
$$

induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}circ k_{(r)}$. Here ${scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.

Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:

The spectral sequence $E_1(C_bullet)$ of cohomological type
converges to $H^{p+q}(C)$ if it is bounded. One has
$$ E_infty^{p,q}=ker(H^{p+q}(C)to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)to H^{p+q}(C_{p})) .$$

Proof? Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$
begin{array}{c}
to E_{r}^{p-r,q+r-1}xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}newline
xrightarrow{j_{r}}E_{r}^{p,q}xrightarrow{k_{r}}newline
A_{r}^{p,q}xrightarrow{i_{r}}A_{r}^{p-1,q+1}xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}to.
end{array}
$$

The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n<0$, the second and the third term are $0$. Exactness implies that
$$
E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))to i(H^{p+q}(C_{p+r-1}))).
$$
I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:

How does Lemma 1.2. of Hatcher's text works in the cohomological case?

ag.algebraic geometry - The importance of EGA and SGA for "students of today"

At the time I write this, there are a number of wise words already written here, so I'll add just incremental thoughts. (Much of what I might say was already said by Matt Emerton.)

EGA and SGA are dangerous, because they are so powerful, and thus so tempting. It is easy to be mesmerized.

Their writing is roughly synonymous with the founding of modern algebraic geometry as a field. But I think their presence has driven people away from algebraic geometry, because they give a misleading impression of the flavor of the subject. (And even writing "flavor" in the singular is silly: there are now many vastly different cuisines.) But this is less true than a generation ago.

Here are some well-known negative consequences. Relatively few successful practicing researchers hold these views, but I have heard them expressed more than once by younger people.

• There is a common feeling that there is an overwhelming amount one has to know just to understand the literature. (To be clear: one has to know a lot, and there is even something that can be reasonably called a "canon" that will apply to many people. But I think the fundamentals of the field are more broad and shallow than narrow and deep. It is also true that much of the literature is not written in a reader-friendly way.)




• Those who can quote chapter and verse of EGA are the best suited to doing algebraic geometry. (To be clear: some of the best can do this. But this isn't a cause of them being able to do the kind of work they do; it is an effect.)




• If you can’t use “non-Noetherian rings” in a sentence, you can’t do algebraic geometry.
  (This is just as silly as saying that if you don’t know the analytic proofs underlying classical Hodge theory, then you can’t call yourself an algebraic geometer. It depends on what you are working on.)

Your goal is to (eventually) prove theorems. You want to get to “the front” as quickly as possible. You want to be able to do exercises, then answer questions, then ask questions, then do something new. You may think that you need to know everything in order to move forward, but this is not true. Learn what you need, do some reading for fun, and do no more.

Don't forget: EGA and SGA were written at the dawn of a new age.
The rules were being written, and these were never intended to be final drafts; witness the constant revisions to EGA, as the authors keep going back to improve what came before. These ideas have been digested ever since. Just because it is in EGA or SGA doens’t mean it is important. Just because it is in EGA and SGA and not elsewhere doesn’t mean it is not important. How will you know the difference?

So when should you read EGA or SGA?

• A very small minority can and should and will read them as students. But you have to be thinking about certain kind of problems, and your mind must work in a certain way. Most people who read EGA and SGA as students are not in this group.




• Some will read them later as they need facts, and will realize how beautiful they are.




• Some will read them later for "pleasure", like reading the classics.

In summary, you should read EGA and SGA only when you need to, where "need" can have many different meanings.

In fairness, I should say how much of EGA and SGA I've read:

A small part of EGA I've read in detail. I had a great time in a "seminar of pain" with a number of other people who were also already reasonably happy with Hartshorne (and more). Reading the first two books of EGA (with some guidance from Brian Conrad on what to skip) was quite an experience --- I had assumed that it would be like Hartshorne, only more so, with huge heavy machinery constantly being dropped on my head. Instead, each statement was small and trivial, yet they inexorably added up to something incredibly powerful. Grothendieck's metaphor of opening a walnut by soaking it in water is remarkably apt.

But other than that, I've read sections here and there. I'm very very happy with what I've read (I agree with Jonathan on this), and I'm also happy knowing I can read more when I need to, without feeling any need to read any more right now (I have better ways to spend my time). When I need to know where something is, I just ask someone. And as for my students: I'd say a third of my students have a good facility with EGA and possibly parts of SGA, and the rest wouldn't have looked at them; it depends on what they think about.

universal algebra - Defining 'free monoid' without Nat?

As I pointed out in the comments, the theory of free monoids is somewhat ill defined. It is still unclear what your logic is and what you really want, but you have two basic options which were proposed by Pete Clark and sigfpe. Here are a few additional remarks that may help you sort things out.

Universal Property à la Pete Clark. The free monoid functor $F:Setto Mon$ is left-adjoint to the forgetful functor $G:Mon to Set$. This adjunction completely determines the isomorphism classes of free monoids. Moreover, many of the essential properties of free monoids follow directly from properties of sets and adjoint functors.

Alternately, the bijection between $Set(A,GB)$ and $Mon(FA,B)$ can be described as Pete Clark suggested, without any reference to adjoint functors. This latter option preferable if your logic can handle the categories $Set$ and $Mon$, but not functors between them.

The obvious advantage of this approach is that this is the true freeness property of free monoids (the fact that word monoids are free is nothing but an interesting accident from this point of view). Other than the fact that $F1$ is a natural number object, there is no mention of natural numbers here. However, I always thought that there was a lot of unnecessary tedium to show that free monoids can be viewed as word monoids in this context.

Induction Axiom à la sigfpe. The idea here is to attempt to capture freeness via induction. Here is a variant that fixes the minor problem I pointed out in a comment to sigfpe's answer -- free monoids are structures $(F,A,e,{cdot})$ where $(F,e,{cdot})$ is a monoid and $A subseteq F$ (understood as a unary predicate) is a distinguished set of generators. The axioms to describe this setup are the usual monoid axioms together with the obvious modifications of sigfpe's two axioms

• $A(a) land A(a') land xcdot a = x' cdot a' to x = x' land a = a'$,


• $forall P,(P(e) land forall x,a,(P(x) land A(a) to P(xcdot a)) to forall x,P(x))$,

where $P$ varies over all unary predicates $P subseteq F$. In first-order logic, quantification over such $P$ is not possible and the induction axiom must be replaced by the corresponding induction scheme where $P$ varies over all formulas of the language instead. (In higher-order logics without predicate types, you can usually replace $P(x)$ by $f(x) = g(x)$ for appropriate $f$ and $g$.)

In the standard semantics for second-order logic, these axioms completely describe free monoids. However you must be careful since not all formulations of higher-logic are equal. Much like first-order logic, some of these formulations will implicitly allow for models of the above theory which are not free in the categorical sense. In truth, this is unlikely to be a problem for you but I am compelled to warn you of this possibility.

arithmetic geometry - current status of crystalline cohomology?

This is a "big-picture" question, but allow me to illustrate some recent progress by taking a small example close to my heart.

Let us adjoin to the field $mathbb{Q}_p$ a primitive $l$-th root of $1$, where $p$ and $l$ are primes, to get the extension $K|mathbb{Q}_p$. We notice that this extension is unramified if $lneq p$ but ramified if $l=p$. When we adjoin all the $l$-power roots of $1$, we get the $l$-adic cyclotomic character $chi_l:operatorname{Gal}(bar{mathbb{Q}}_p|mathbb{Q}_p)tomathbb{Q}_l^times$ which is unramified if $lneq p$ but ramified if $l=p$. But we cannot just say that $chi_p$ is ramified and be done with it. We have to somehow express the fact that $chi_p$ is a natural and a "nice" character, not an arbitrary character $operatorname{Gal}(bar{mathbb{Q}}_p|mathbb{Q}_p)tomathbb{Q}_p^times$, of which there are very many because the topologies on the groups $operatorname{Gal}(bar{mathbb{Q}}_p|mathbb{Q}_p)$, $mathbb{Q}_p^times$ are somehow "compatible".

The fact that $chi_p$ is a "nice" character is expressed by saying that it is crystalline. In general, we can talk of crystalline representions of $operatorname{Gal}(bar{mathbb{Q}}_p|mathbb{Q}_p)$ on finite-dimensional spaces over $mathbb{Q}_p$; the actual definition is in terms of a certain ring $mathbf{B}_{text{cris}}$, constructed by Fontaine, which can be understood in terms of crystalline cohomology.

My illustrative example is about the $l$-adic criterion for an abelian variety $A$ over $mathbb{Q}_p$ to have good reduction. For $lneq p$, this can be found in a paper by Serre and Tate in the Annals, and it is called the Néron-Ogg-Shafarevich criterion. It says that $A$ has good reduction if and only if the representation of $operatorname{Gal}(bar{mathbb{Q}}_p|mathbb{Q}_p)$ on the $l$-adic Tate module $V_l(A)$ is unramified.

What happens when $l=p$ ? It is too much to expect that $V_p(A)$ be an unramified representation when $A$ has good reduction; we have seen that even $chi_p$ is not unramified. What Fontaine proved is that the $p$-adic representation $V_p(A)$ is crystalline (if $A$ has good reduction). To complete the analogy with the case $lneq p$, Coleman and Iovita proved in a paper in Duke that, conversely, if the representation $V_p(A)$ is crystalline, then the abelian variety $A$ has good reduction.

I hope you find this enticing.

fa.functional analysis - prime ideals in C([0,1])

It is clear that each maximal ideal in ring of continuous functions over $[0,1]subset mathbb R$ corresponds to a point and vice-versa.

So, for each ideal $I$ define $Z(I) =$ {$xin [0,1]$ | $f(x)=0, forall f in I$}. But map $Ito Z(I)$ from ideals to closed sets isn't injection! (Consider ideal $J(x_0)=${$f$|$f(x)=0, forall xin$ some closed interval which contains $x_0$})

How can we describe ideals in C([0,1])? Is it true that prime ideal is maximal for this ring?

ca.analysis and odes - Are there Generalisations of a Limit (for Just-divergent Sequences)?

There are certain sequences such as

0, 1, 0, 1, 0, 1, 0, 1, ...

that do not converge, but that may be assigned a generalised limit. Such a sequence is said to diverge, although in this case a phrase such as has an orbit might be preferable.

One way to generalise a limit is by considering the sequence of accumulated means: given a sequence

a₁, a₂, a₃, a₄, ...

the accumulated mean sequence would be

a₁, (a₁+a₂)/2, (a₁+a₂+a₃)/3, (a₁+a₂+a₃+a₄)/4, ...

If this sequence has a limit, then the original sequence may be said to have that value as its generalised limit. In this way, the example sequence above has the generalised limit of 1/2; this seems natural as the sequence oscillates around this 'mean' value.

Is there a name for this kind of generalised limit? Are there other ways to define such a thing. Do you know of any good on-line references for this?

Thanks.

at.algebraic topology - Rational homotopy theory of a punctured manifold

This is an old question but I hope the following is still of interest.

If $M^n$ is a closed simply connected manifold then the inclusion $M^nbackslash { pt}hookrightarrow M^n$ is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that $M^nbackslash { pt}$ has zero homology in degrees above $n-2$. It's a general fact that given a minimal model up to dimension $k$ of a space whose cohomology vanishes in degrees above $k$ the rest of the minimal model is determined uniquely (and constructively) from the model up to degree $k$. This provides an easy recipe for computing the minimal model of
$M^nbackslash { pt}$ which can be more explicitly described as follows.

If $(Lambda V, d)$ is a minimal model of $M^n$ then consider the following dga $(A,d)=(Lambda VoplusLambda langle zrangle/(z^2), d)$ with $deg z=n-1, Vcdot z=0$ and $dz=[M]$ - the fundamental class of $M$. This is a model (non-minimal and even a non-free one!) of $M^nbackslash { pt}$. In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.

Here are a couple of examples.

Let $M^4=mathbb {CP}^2$. Its minimal model is $(Lambda langle x,yrangle,d)$ with deg x=2, deg y=5, dx=0, dy=x^3. Up to degree $n-2=2$ this is simply given by $Lambdalangle xrangle$ with dx=0. Next we need another generator to make $H^4=0$ (which is currently generated by $[M]=x^2$) so we add $z$ of deg $3$ such that $dz=x^2$. Now the model $(Lambda langle x,zrangle,d)$ with deg x=2, deg z=3, dx=0, dy=x^2 already has $H^i=0$ for $ige 4$ so we don't need to add anything else. The resulting model is easily recognized as the model of $mathbb S^2$ which is of course not surprising since $mathbb{CP}^2backslash{pt}$ is a Hopf disk bundle over $mathbb{CP}^1$.

A more interesting example: Let $M=mathbb S^3timesmathbb S^5$. Its minimal model is generated by $x,y$ with $deg x=3, deg y=5$ and $dx=0,dy=0$. Applying our recipe the model of $mathbb S^3timesmathbb S^5backslash { pt}$ will be the same through dimension 6. Next,
we need to kill off cohomology in degree 8 which is currently generated by $[M]=xy$. So we need another generator $z$ of degree 7 with $dz=xy$. However, adding such generator introduces more cohomology in degrees 10 and 12 generated by $xz$ and $yz$. So we need two more generators $a$ and $b$ with $da=xz, db=yz$. However, adding those introduces yet more cohomology and we need to keep adding more generators. This will continue forever because $mathbb S^3timesmathbb S^5backslash { pt}$ is rationally hyperbolic.

Lastly, let me mention that operations such as cell attachments (or in this case cell deletions) are usually easier handled by Quillen Lie algebra models which are better suited to work with cofibrations (while Sullivan models are better suited for fibrations). In this particular case it's especially easy. If $(mathbb L_V,d)$ is a minimal Quillen Lie model of $M^n$ then the model of $M^nbackslash { pt}$ is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.

complex geometry - Divisors, extensions of functions

To add to the excellent answers of jvp and David Speyer above:

Actually Hartog's theorem does even better: Given a domain $Usubsetmathbb{C}^{n{geq{2}}}$ and a compact $Ksubset{U}$ such that $Usetminus{K}$ is just connected, any holomorphic function on $Usetminus K$ extends holomorphically to $U$. Note that we don't need the codimension condition. This is actually a simple consequence of Cauchy's formula in several variables.

For algebraic geometry (resp. complex geometry) and dimension $geq{2}$: The analog of Hartog's theorem is that any regular (resp. holomorphic) function on the complement of an algebraic (resp. analytic) subset of codimension atleast $2$ in a normal algebraic(resp. analytic) variety, extends to the whole algebraic (resp. analytic) variety.

I would also like to point out the important difference between the affine (resp. Stein) case and the projective case as perhaps alluded to by jvp above. If $X$ is a projective algebraic variety over $mathbb{C}$ and you find a holomorphic function $f$ on the complement of a divisor $D$ then you will not be able to extend the function to all of $X$ however hard you try(!) since there are no global non-constant functions on projective varieties (similar statement for compact analytic vars). So to add to jvp's answer (which might seem like hair-splitting but is important IMHO): His second paragraph must refer to a non-projective neighborhood of the contractible curve that he discusses, otherwise the argument is still true but only vacuously! This is because there is no non-constant function to be found on a projective surface on the complement of a contractible curve and hence there is nothing to extend!

Moreover, it will not be possible in general to extend holomorphic functions from the complement of divisors whose examples can be easily constructed. So there is no hope in this direction without any local boundedness hypothesis.

ag.algebraic geometry - Is there any value in studying divisors with coefficients in a ring R?

As a rule, the various groups and quotients of the divisor group on a variety have coefficients in $mathbb{Z}$. That is, you take $mathbb{Z}$-linear combinations of Weil divisors or Cartier divisors, and then to construct other groups you take quotients.

However, in some cases, people tensor with $mathbb{Q}$ and $mathbb{R}$. So my question is:

Are these the only rings that people use as coefficients for divisors on a variety?

My vague intuition is that it probably is, because $mathbb{Z}$ is initial in commutative rings with identity, $mathbb{Q}$ is a field of characteristic zero, so we can use it to kill torsion, and $mathbb{R}$ is complete, so we can guarantee that there is an $mathbb{R}$-divisor, plus with orbifolds, rational coefficients seem to show up naturally. But is this it? More generally, what about for cycles and cocycles? There's an analogy with cohomology and the Chow ring, and we do sometimes take cohomology with coefficients either in an arbitrary ring or in some other rings (finite fields, for instance, when studying things like nonorientable manifolds), which is why I started wondering about this.

ag.algebraic geometry - What is the recent development of D-module and representation theory of Kac-Moody algebra?

I am not an expert in this but I would of course expect something like ind-scheme approach to be natural. Gerd Faltings used I think ind-schemes to treat Sugawara construction, algebraic loop groups and Verlinde's conjecture in

Gerd Faltings, Algebraic loop groups and moduli spaces of bundles.
J. Eur. Math. Soc. (JEMS) 5 (2003), no. 1, 41--68.

You might also like to work with versions of Kac-Moody GROUPS in analytic approaches.

You could also consult comprehensive and not that old Kumar's book (Kac-Moody groups, their flag varieties and representation theory, Birkhauser) which is written in geometric language.

As far as Frenkel is concerned, not only his work with Feigin but even more I think his paper with Gaitsgory must be relevant (see arxiv:0712.0788).

Semi-infinite cohomologies are important but still misterious thing. Some related homological algebra has been recently studied by Positelskii in great generality. Another important thing is relation between the geometry of representations of quantum groups at root of unity and of affine Lie algebras, like in the book of Varchenko and many papers later.

Edit: Frenkel himself I think does not claim (I talked to him at the time) to have intuitive explanation why only derived equivalence. But you should not expect for more: by the correspondence with quantum groups the situation should be like in affine case where one has problems with non-closedness of diagonal in noncommutative geometry what has repercussions on the theory of D-modules. How this reflects in the case of relevant ind-schemes for affine side I do not know but somehow it does.