co.combinatorics - Applications of infinite Ramsey's Theorem (on N)?

Beyond the infinite Ramsey's theorem on N, there is, of course, a kind of super-infinite extension of it to the concept of Ramsey cardinals, one of many large cardinal concepts.

Most of the large cardinal concepts, including Ramsey cardinals, generalize various mathematical properties of the countably infinite cardinal ω to uncountable cardinals. For example, an uncountable cardinal κ is a Ramsey cardinal if every coloring of finite subsets of kappa into 2 colors (or indeed, less than κ many colors) admits a homogeneous set of size κ. Such cardinals are necessarily inaccessible, Mahlo, and much more. The somewhat weaker property, that every coloring of pairs (or for any fixed finite size) from κ to 2 colors has a homogeneous set, is equivalent to κ being weakly compact, a provably weaker notion, since every Ramsey cardinal is a limit of weakly compact cardinals. Similarly, the concept of measurable cardinals generalize the existence of ultrafilters on ω, for an uncountable cardinal κ is said to be a measurable cardinal if there is a nonprincipal κ-complete ultrafilter on κ.

Ramsey cardinals figure in many arguments in set theory. For example, if there is a Ramsey cardinal, then V is not L, and Ramsey cardinals are regarded as a natural large cardinal notion just exceeding the V=L boundary. Another prominent result is the fact that every measurable cardinal is Ramsey (which is not obvious from first notions). Further, if there is a Ramsey cardinal, then 0^# exists. Indeed, this latter argument proceeds as a pure Ramsey style argument, using a coloring. Namely, if κ is Ramsey, then we may color every finite increasing sequence of ordinals with the type that they realize in L. By the Ramsey property, there must be a set of size κ, all of whose increasing finite subsequences realize the same type. That is, there is a large class of order indiscernibles for L. By results of Silver, this is equivalent to the assertion that 0^# exists.

The fact that Ramsey cardinals are strictly stronger than weakly compact cardinals suggests to my mind that there is something fundamentally more powerful about finding homogeneous sets for colorings of all finite subsets than just for pairs or for subsets of some fixed size. This difference is not revealed at ω, for which both are true by the infinite Ramsey theorem. But perhaps it suggests that we will get more power from Ramsey by using the more powerful colorings, since this is provably the case for higher cardinals.

Another point investigated by set theorists is that finding homogeneous sets in the case of infinite exponents---that is, coloring infinite subsets---is known to be inconsistent with the axiom of choice. However, in models of set theory where the Axiom of Choice fails, these infinitary Ramsey cardinals are fruitfully investigated. For example, under the Axiom of Determinacy, there are a great number of cardinals realizing an infinite exponent paritition relation.

ag.algebraic geometry - Cohomology of rigid-analytic spaces

Here's a first pass at your question; hopefully it will suggest something more definitive.

Let's imagine we were in the simplest case, where $X$ is a disk, with its smooth model
being the formal affine line over $R$, and that $Z$ was the sub-disk of elements of
absolute value less than or equal the absolute value of the uniformizer. Then we can find a semistable model in which $Z$ is one of the covering opens, by blowing up the formal affine line at the origin.

So in this test case, the answer seems to be yes .

Now in general, I think that Raynaud (and/or his collaborators or those who followed in his
tradition) will say that the open immersion $Z rightarrow X$ extends to an open immersion
of formal models. So we can blow up the smooth model of $X$ and the smooth model of $Z$
so that the latter sits inside the former. What I'm not very certain about is how much
you can control the nature of these blow-ups. (Presumably not at all in general, but you're starting in a fairly nice situation.)

Have you tried asking Brian Conrad yet?

Does a finite dimensional algebra having a Cartan matrix with determinant 1 imply finite global dimension (possibly with more hypotheses)?

Let $A$ be your favorite finite dimensional algebra, and $P_i$ be a sets of representatives for the indecomposible projectives (or PIMs, if you like). Then we have the Cartan matrix $C$ of the algebra, whose entries are $dim Hom(P_i, P_j)$. You can think of this as the matrix of the Euler form on the Grothendieck group $K^0(A-pmod)$ of projective $A$-modules.

Now, if the algebra $A$ has finite global dimension, then we can define the classes of the simple heads of these $L_i$ in $K^0(A-pmod)$ as integer linear combinations of the $P_i$'s, and $[P_i]$ and $[L_i]$ are dual bases in the Euler form. That is, the matrix $C$ is integer valued and has integer-valued inverse, i.e. it has determinant 1.

To what degree is the converse of this true? Is there a weaker hypothesis than finite global dimension itself such that $det(C)=1$ and that hypothesis will imply finite global dimension?

The application I have in mind for this is a little more complex. I'd like to consider a graded version of this question. So, let $A$ be a graded algebra such that each degree is finite dimensional (and let say the appearing gradings are bounded below). The the graded version of $C$ is well-defined in $mathbb{Z}((q))$, and similarly, if every simple has a resolution by projectives where only finitely many projectives generated in a given degree appear, this implies that this matrix has an inverse in $mathbb{Z}((q))$, that is determinant with leading coefficient 1.

The same question as above: can I use a hypothesis like the graded Cartan matrix having determinant with integral leading coefficient to conclude the existence of such a projective resolution?

ag.algebraic geometry - Matrix factorizations and physics

Indeed matrix factorizations come up in string theory. I don't know if there are good survey articles on this stuff, but here is what I can say about it. There might be an outline in the big Mirror Symmetry book by Hori-Katz-Klemm-etc., but I am not sure.

When we are considering the B-model of a manifold, for example a compact Calabi-Yau, the D-branes (boundary states of open strings) are given by coherent sheaves on the manifold (or to be more precise, objects of the derived category of coherent sheaves). Matrix factorizations come up in a different situation, namely, they are the D-branes in the B-model of a Landau-Ginzburg model. Mathematically, a Landau-Ginzburg model is just a manifold (or variety) $X$, typically non-compact, plus the data of a holomorphic function $W: X to mathbb{C}$ called the superpotential. In this general situation, a matrix factorization is defined to be a pair of coherent sheaves $P_0, P_1$ with maps $d : P_0 to P_1$, $d : P_1 to P_0$ such that $d^2 = W$. I guess you could call this a "twisted (or maybe it's 'curved'? I forget the terminology) 2-periodic complex of coherent sheaves". When $X = text{Spec}R$ is affine, and when the coherent sheaves are free $R$-modules, this is the same as the definition that you gave.

The relationship between matrix factorization categories and derived categories of coherent sheaves was worked out by Orlov: http://arxiv.org/abs/math/0503630 http://arxiv.org/abs/math/0503632 http://arxiv.org/abs/math/0302304

I believe that the suggestion to look at matrix factorizations was first proposed by Kontsevich. I think the first paper that explained Kontsevich's proposal was this paper by Kapustin-Li: http://arXiv.org/abs/hep-th/0210296v2

There are some interesting recent papers regarding the relationship between the open-string B-model of a Landau Ginzburg model (which is, again, mathematically given by the matrix factorizations category) and the closed-string B-model, which I haven't described, but an important ingredient is the Hochschild (co)homology of the matrix factorizations category. Take a look at Katzarkov-Kontsevich-Pantev http://arxiv.org/abs/0806.0107 section 3.2. There is a paper of Tobias Dyckerhoff http://arxiv.org/abs/0904.4713 and a paper of Ed Segal http://arxiv.org/abs/0904.1339 which work out in particular the Hochschild (co)homology of some matrix factorization categories. The answer is it's the Jacobian ring of the superpotential. This is the correct answer in terms of physics: the Jacobian ring is the closed state space of the theory.

Katzarkov-Kontsevich-Pantev also has some interesting stuff about viewing matrix factorization categories as "non-commutative spaces" or "non-commutative schemes".

Edit 1: I forgot to mention: Kontsevich's original homological mirror symmetry conjecture stated that the Fukaya category of a Calabi-Yau is equivalent to the derived category of coherent sheaves of the mirror Calabi-Yau. Homological mirror symmetry has since been generalized to non-Calabi-Yaus. The rough expectation is that given any compact symplectic manifold, there is a mirror Landau-Ginzburg model such that, among other things, the Fukaya category of the symplectic manifold should be equivalent to the matrix factorizations category of the Landau-Ginzburg model. For example, if your symplectic manifold is $mathbb{CP}^n$, the mirror Landau-Ginzburg model is given by the function $x_1+cdots+x_n + frac{1}{x_1cdots x_n}$ on $(mathbb{C}^ast)^n$. This is sometimes referred to as the Hori-Vafa mirror http://arxiv.org/abs/hep-th/0002222

I think that various experts probably know how to prove this form of homological mirror symmetry, at least when the symplectic manifold is, for example, a toric manifold or toric Fano manifold, but it seems that very little of this has been published. There may be some hints in this direction in Fukaya-Ohta-Oh-Ono http://arxiv.org/abs/0802.1703 http://arxiv.org/abs/0810.5654, but I'm not sure. There is an exposition of the case of $mathbb{CP}^1$ in this paper of Matthew Ballard http://arxiv.org/abs/0801.2014 -- this case is already non-trivial and very interesting, and the answer is very nice: the categories in this case are equivalent to the derived category of modules over a Clifford algebra. I quite like Ballard's paper; you might be interested in taking a look at it anyway.

Edit 2: Seidel also has a proof of this form of homological mirror symmetry for the case of the genus two curve. Here is the paper http://arxiv.org/abs/0812.1171 and here is a video http://www.maths.ed.ac.uk/~aar/atiyah80.htm of a talk he gave on this stuff at the Atiyah 80 conference.

co.combinatorics - Is the graph of a thresholded correlation matrix chordal?

It was described in this previous question how to obtain a correlation matrix whose entries come from the scalar product of certain vectors $u_1, u_2, dots,u_n$. If we let the vectors be $$u_i=(1, cos(frac{2pi i}{n}), sin(frac{2pi i}{n}),0,dots, 0)$$ we can set a high enough threshold so that the corresponding graph is a cycle of length $n$ and thus not chordal.

terminology - Translation of "le nilradicalisé de g"

I apologize for asking something that might well be found in a mathematical dictionary, but the similarity of the French word to an English one is frustrating my attempts to Google the answer (and the library is shut at time of typing). I suspect the answer should be obvious to those who, unlike me, know some basic Lie group/Lie algebra terminology.

Some context: I am reading an old paper of Dixmier from 1969, which has the following construction/definition. Let $mathfrak g$ be a Lie algebra (characteristic zero, finite-dimensional), let $mathfrak n$ be its largest nilpotent ideal -- the nilradical -- and put ${mathfrak h}=[{mathfrak g},{mathfrak g}]+{mathfrak n}$. Dixmier calls ${mathfrak h}$ "le nilradicalisé de ${mathfrak g}$".

Literal translation would surely be "the nilradicalised", but that sounds more like a mopey university indie band than a mathematical object. So what is the usual name for this object in English?

books - Famous exercise from Lang's Algebra

I got to this list since I was trying to remember whether the quote from Lang was true or not. Indeed, my copy, first edition I guess, does have that quote on p. 105. I don't know exactly how "positive"(ly) the term "abstract nonsense" is meant in that quote, though. I would imagine that Lang would have been annoyed had someone suggested they do that with any of his books.

Re: Spanier. Yes, the book is terse, and hard to read. I sincerely doubt that the intent was primarily to make a teaching text for beginning grad students; this was, and is, the standard reference book for basic algebraic topology. What makes it hard to read is the fact that every statement is made is as great a generality as he could. This is an advantage for a reference text. I sincerely doubt, however, that there was anything in that book that Spanier could not prove completely.

That being said, I certainly struggled through a course using it. I chose to take algebraic topology from someone else, to avoid taking it from Spanier himself, since I thought he would teach it as generally as was in the book. It turns out, though, that he did back down from that level of abstraction when teaching the course, and perhaps I missed something not taking the course from him.

st.statistics - Why isn't Likelihood a Probability Density Function?

A few questions were asked, so a few answers will be given. (main point: likelihood is not necessarily a product density, though this is the common interpretation.)

Frequently, the likelihood is the product of densities over some provided set of examples. The examples are drawn i.i.d., and therefore this product density is the density for the corresponding product measure over the product space. What I'm saying is that yes, from this perspective, you have constructed a product density.

Since you are dealing with densities, not probabilities, values are not constrained to [0,1], and your density can easily be greater than one. In fact, if you are dealing with
dirac measure (which puts all mass on one point on the real line), you essentially have "infinite" density. I put that in quotes since this is not a continuous probability measure, ie it does not have a density wrt to Lebesgue measure, let alone one with infinite mass on a point. (A quick fact check: the corresponding integral wrt lebesgue measure would have value zero since it is off zero only on a set of lebesgue measure zero, which means it is not a probability distribution; but it was, which contradicts this being its density.) perhaps a more apt example: any (continuous) distribution on [0,0.5] will have to have density greater than one on a set of nonzero lebesgue measure. (you can try
to construct a sequence of these which convergence to something which violates what i said, but that will be the density of something which is not continuous!)

things can get a little confusing because you can write discrete probability distributions
as densities wrt a measure putting 1 on each point in the support set of the probability (ie it is counting measure wrt that set). NOTE that this is a density wrt a measure which is NOT a probability measure. But anyway, the density values at each point are exactly the probability values. This allows an interchanging probability masses and densities, which can be confusing.

I'll close with some further reading. A good book on machine learning is "A probabilistic Theory of Pattern Recognition" by Devroye, Gyorfi, Lugosi. Chapter 15 is on maximum likelihood and you'll notice they do NOT define likelihood as being a product probability or density, but rather as a product of functions. This is because they are careful to encompass the differing interpretations; rather, they ignore the interpretations there and work out the math.

nt.number theory - Primes of the form a^2+1

The fact that the Riemann zeta function $zeta(s)$ and its brethren have a pole at $s=1$ is responsible for the infinitude of large classes of primes (all primes, primes in arithmetic progression; primes represented by a quadratic form). We cannot hope proving the infinitude of primes $p = a^2+1$ in this way because the series $sum 1/p$, summed over these primes, converges. This implies that the corresponding Euler product
$$ zeta_G(s)= prod_{p = a^2+1} frac1{1 - p^{-s}} $$
converges for $s = 1$. But if we could show that $zeta_G(s)$ has a pole at, say,
$s = frac12$, then the desired result would follow. Now I know that there are heuristics on the number of primes of the form $p = a^2+1$ below $x$ (by Hardy and Littlewood?)

Can these heuristics be explained by hypothetical properties of $zeta_G(s)$ (or a related Dirichlet series), or can the domain of convergence of $zeta_G(s)$ be derived from such asymptotics?

BTW, here's a little known conjecture by Goldbach on these primes: let $A$ be the set of all numbers $a$ for which $a^2+1$ is prime ($A = ${1, 2, 4, 6, 10, $ldots$}). Then every
$a in A$ ($a > 1$) can be written in the form $a = b+c$ for $b, c in A$. I haven't seen this discussed anywhere.

ag.algebraic geometry - When do equivariant sheaves on a formal neighborhood extend?

Suppose that $X$ is a variety (in char 0) with an action of an affine algebraic group $G$. Let $Y subset X$ be a subvariety fixed by $G$--the action map agrees with projection upon restriction to $Y$. Let $widehat{Y}$ be the formal completion of $X$ along $Y$. Furthermore let $widehat{G}$ be the the completion of $G$ at the identity, viewed as a formal group. There is a restriction functor $j^*$ from the $Qcoh^G(X)$, the category of $G$-equivariant quasicoherent sheaves on $X$, to $Qcoh^{widehat{G}}(widehat{Y})$, the category of $widehat{G}$-equivariant quasicoherent sheaves on $widehat{Y}$.

1) Is this situation considered in the literature? Where?
2) What tools are available to control this functor? How might one describe the essential image?

Although curious about this general package, I specifically care about the case $G =mathbb{G}_m$.

soft question - Are there elementary-school curricula that capture the joy of mathematics?

As far as a full curriculum goes, I don't believe there is one that does exactly what you want. Books (in the United States, at least) divide into two camps:

"Constructivist" (e.g. Everyday Math, Connected Math)

"Traditional" (e.g. Saxon, Singapore)

Now, any search you make that even has a whiff of these terms will summon up loud and angry missives (try this article from the New York Times for an idea).

Constructivist curriculum is an attempt to catch the "joy of mathematics" approach to learning; for example rather than a worksheet with addition problems there might be a question about all the different possible sets of numbers that add up to 20.

The downside (as pointed out by the article above) is that (especially when taught by teachers who aren't themselves strong in mathematics) it can lead to basic skills being missed.

This is a problem Lockheart's Lament acknowledges. He seems to think students won't miss anything important. This can be true if the person steering the education is a mathematician, but with a non-specialist (i.e. most elementary school educators and homeschoolers) things can go horribly wrong.

Now, it's possible to balance to pull off a fantastic curriculum, but the ones I know about (say, at the Russian School of Mathematics in Boston) are, as self-described by the teachers, not following a curriculum at all. That's great if the teachers are experts, but put homeschoolers in a quandry.

I think the world is still waiting for an inquiry-type elementary curriculum that can be followed by non-experts and doesn't shortchange basic skills. So for now I'd suggest:

a.) Pick a traditional curriculum (Singapore is fine, although do shop around).

b.) Supplement. This very question is filling with lots of suggestions.

rt.representation theory - missing Lie algebra action on vectors in p-adic case?

If one is doing smooth representations of the $p$-adic group $G$ on complex vector spaces,
then the action is locally constant (small open subgroups act trivially on vectors),
and so any interpretation of what it means to differentiate the action will give you
the trivial action. (Note also the the Lie algebra has $p$-adic coefficients, while
the representation $V$ has complex coefficients, so there is not much scope for a non-trivial action!).

On the other hand, if you mean not-necessarily-smooth actions of $G$ on a $p$-adic vector space, as one has in the theory developed by Schneider and Teitelbaum that is related to the $p$-adic Langlands program, then one can and does differentiate the action (provided it
is locally analytic), and one does obtain interesting Lie algebra actions.

harmonic analysis - The dual group of $mathbb Q$

In fact, uniform convergence on compact subsets of $mathbb{Q}subsetmathbb{R}$ induces the usual topology on its group of (continuous) characters $mathbb{R}simeq{tmapstoexp(ixt)}_{xinmathbb{R}}$.

Namely, consider $K={0}cup{1/n,ngeq 1}$. For $xinmathbb{R}$, the corresponding character is uniformly $epsilon$-close on $K$ to the trivial character iff $$|exp(ix/n)−1|<epsilon;;;; (*)$$ for all integers $ngeq1$. Then for $epsilon<1/sqrt{2}$, $x$ must be small : $|x|<2epsilon/pi$. Indeed, consider $kinmathbb{Z}$ such that $|x−kpi|leqpi/2$ , and take $n=|k|$; if $kneq 0$ we reach a contradiction in $(*)$. Hence $k=0$, and the claim follows easily.

This implies that uniform convergence on compact subsets of $mathbb{Q}$ (in fact the one compact subset $K$) induces the usual topology on $mathbb{R}simeqmathrm{Hom}(mathbb{Q},S^1)$.

riemannian geometry - Surjectivity of the normal exponential map

Given an isometric (in the Riemannian way) immersion $f:Nrightarrow M$ between complete, smooth riemannian manifolds, are there conditions on $M$, $N$, $f$, such that the normal exponential map $mathrm{exp}^{nu}:nu(N)rightarrow M$ is surjective?

I'm interested in the case of $f$ being not closed. An example of non surjectivity is given by $f:mathbb{R}rightarrowmathbb{R}^2$, where f is the logarithmic spiral. In this case, the normal exponential map misses the origin.

gn.general topology - Space whose product with paracompact space is paracompact

Is there a nice characterization of topological spaces with the property that the product with any paracompact space is paracompact?

All compact spaces have this property (this can be shown from the tube lemma). But somebody once gave me an example (that I cannot locate) of a non-compact space with the property. I didn't check the example carefully, so I cannot vouch for its accuracy.

If a characterization is too hard, an example of a non-compact space would also be great.

[NOTE: I don't assume Hausdorffness in my definitions of compact and paracompact, though it would be nice if the example were a Hausdorff space.]

ADDED LATER: I forgot to mention this, but a product of paracompact spaces need not be paracompact. The standard example is the Sorgenfrey line (the real line with the lower limit topology), which is paracompact, whose product with itself, the Sorgenfrey plane, is not paracompact.

dg.differential geometry - When are fiber bundles reversible?

My question, in its most general form is this:

Given a fiber bundle $Frightarrow Erightarrow B$, when is there a fiber bundle $Brightarrow Erightarrow F$?

Here, F,E, and B can lie in whichever category you wish, but I'm mostly interested in the case where all 3 are smooth closed manifolds.

Now, I realize that the initial answer is "unless E is a product, essentially never", so here is a more focused question (with background).

I've been studying a certain class of free actions of the 3-torus $T^3$ on $S^3times S^3times S^3 = (S^3)^3$. For each of these actions, by quotienting out by various subtori, I can show that the orbit space $E=(S^3)^3/T^3$ simultaneously fits into 2 fiber bundles:

$$S^2rightarrow E rightarrow S^2times S^2$$ and $$S^2times S^2rightarrow Erightarrow S^2$$ where the structure group for both bundles is $S^1$.

(In fact, the class of actions also gives rise to examples where either $S^2times S^2$ can independently be replaced with $mathbb{C}P^2sharp -mathbb{C}P^2$, the unique nontrivial $S^2$ bundle over $S^2$.)

By computing characteristic classes for (the tangent bundle to) E, I know that for an infinite sublcass of the actions I'm looking at, E is not homotopy equivalent to $S^2times S^2times S^2$, and each of the E are pairwise nondiffeomorphic.

I suspect the reason I could find so many E which fit into "reversible" fiber bundles is strongly related with the fact that the fiber and base are so closely related.

And so, I ask

For fixed manifold M, what is the relationship between bundles $Xrightarrow Erightarrow M$ and $Mrightarrow E'rightarrow X$ where $X$ is some $M$ bundle over $M$?

And just in case there is no general relationship,

Is there a reason I should have expected there to be a relationship in my examples, even though in general there isn't?

na.numerical analysis - Eigenvalues of A+B where A is symmetric positive definite and B is diagonal

The ambiguity in your question is the word 'rapidly'.

If you want to have an information on the eigenvalues of $A+B$, without any extra information besides those given in the question, then this is the problem raised By H. Weyl in 1912. The answer was conjectured in 1962 by A. Horn, and this conjecture was proved by A. Knutson and T. Tao in 1999. It is one of the works for which Tao received a Fields medal. So, the answer is that the spectrum may be any vector in a polytope in ${mathbb R}^n$ whose definition is given recursively in terms of the size $n$ of the matrix. A nice expository paper is R. Bhatia, Linear algebra to quantum cohomology: the story of A. Horn’s inequalities. Amer. Math. Monthly, 108 (2001), pp 289–318.

Historically, the interest in this question came from Quantum Mechanics.

Which Hadamard Products of Generating Functions Are Known?

The Hadamard product, Schur product, or entrywise product of two generating functions is computed as follows:

The Hadamard Product, H(x), given two generating functions f(x) and g(x) where

$$ f(x) = c_0 + c_1x + c_2x^2 + c_3x^3 + dots + c_nx^n + dots $$
$$ g(x) = d_0 + d_1x + d_2x^2 + d_3x^3 + dots + d_nx^n + dots $$

is defined as

$$ H(x) = c_0d_0 + c_1d_1x + c_2d_2x^2 + c_3d_3x^3 + dots + c_nd_nx^n + dots $$

Simply put, it is the result of multiplying individual coefficients of two generating functions.

Results detailing certain types of functions are known. I would like to compile a list of results that are known.

For example, if we have two generating functions of the form $(1-x)^a$ and $(1-x)^b$, we obtain:
${}_2F_1[-a,-b;1;x]$ where ${}_2F_1$ represents a hypergeometric function of Gauss, according to

"Singularity Analysis, Hadamard Products, and Tree Recurrences", by Jim Fill, Philippe Flajolet, and Nevin Kapur In Journal of Computational and Applied Mathematics, volume 174 (February 2005), pages 271--313 (around page 289)

Again, which Hadamard products involving generating functions are known or solved? Additionally, solutions to the equations would be greatly appreciated.

ag.algebraic geometry - the proof of "theorem of connectedness"

A typical context in which one has the condition $f_{*}mathcal O_X = mathcal O_Y$
in that in which $f$ is a birational morphism and $Y$ is normal. In this context, the proof
that the fibres are connected is due to Zariski (I believe that it's the original version
of his ``main theorem'') and certainly predates EGA methods. One can find the paper in his
collected works. (It's been a long time since I looked at it, but I would guess that his main
technical tool is valuation theory; I might check this when I get a chance.)

However, it is worth bearing in mind the evolution of Zariski's work on this kind of question:
his investiations of this sort of connectedness theorem culminated in his proof of his connectedness theorem, to the effect that a specialization of connected varieties is again
connected. His was the first purely algebraic proof of this result, I think. To give the proof, he invented his theory of formally holomorphic functions, which I believe was regarded
at the time as being the most difficult part of the algebraic theory of algebraic geometry
developed by Weil, Chevalley, and Zariski. This theory served as one of the inspirations for the theory of formal schemes, and is the precursor to Grothendieck's theorem on formal functions (the proof via formal schemes mentioned in the question).

If so great a geometer as Zariski was led to introduce these kinds of formal methods to study connectedness problems, it is probably reasonable to regard them as somewhat intrinsic to the problem.

Question about arxiv

Hello,
I have the following question about arxiv.org: is it possible to cross list a mathematical
paper (already posted) to one of the physics archives (hep-th for example).
It definitely used to be possible, but the current system doesn't seem to allow that (at least
not in the obvious way).

ac.commutative algebra - Lower bounds on the degrees of representatives of $u^n$ as $n to infty$

The following is inspired by/based on Felipe Voloch's excellent partial answer. It gives an affirmative answer, under a slightly stronger hypothesis of normality than that given in the question.

Note that the homomorphism $phi colon k[x_1, dotsc, x_n] to A$ that I assume given in the question is equivalent to giving a closed immersion $X = mathrm{Spec} A to mathbb{A}^n$. I am going to assume, not only that $A$ is integral and normal, but that the closure $overline{X}$ of $X$ in $mathbb{P}^n$ is normal. Although this does not quite answer the question I was asking, Donu Arapura's answer here shows that if we are given the freedom to choose $phi$, we can ensure that this condition is met. On the other hand, the proof does not require that $u$ be a unit, only that it be nonconstant.

Let $u in A$ be nonconstant. Then $u$ is a rational function on $overline{X}$. Moreover, any poles of $u$ must lie in $Y := overline{X} smallsetminus X$. Let $b(u)$ denote the order of the greatest-order pole of $u$ on $overline{X}$. If $u$ had no poles, then since $overline{X}$ is normal, $u$ could be extended to a regular function on $overline{X}$. Since $u$ is nonconstant, this is impossible, so $b(u) geq 1$.

Give $mathbb{P}^n$ homogeneous coordinates $T_0, dotsc, T_n$, where our embedding $mathbb{A}^n hookrightarrow mathbb{P}^n$ is given as $D_+(T_0)$. Then $T_0$ represents a global section of $mathcal{O}(1)$ on $overline{X}$. The set-theoretic union of the zeros of $T_0$ is $Y$. Let $c$ be the order of the highest-order zero of $T_0$ on $overline{X}$; clearly, $c geq 1$.

Claim: $b(u) leq c cdot deg(u)$.

If $deg(u) = d$, then $u = T_0^{-d} u'$ for some global section $u'$ of $mathcal{O}(d)$. Since $u'$ has no poles, the claim follows immediately.

Thus, we have
$$ deg(u^n) geq frac{1}{c} b(u^n) = frac{n}{c} b(u) to infty$$
as $n to infty$.

If am, of course, quite interested to see if anyone can find a way around the normality hypothesis (or show that it is necessary).

co.combinatorics - Combinatorial proof that large-girth graphs are sparse?

Theorem. Fix $epsilon > 0$; for sufficiently large n, any graph with n vertices and $epsilon binom{n}{2}$ edges contains many (nondegenerate) cycles of length 4.

The proof is simple; put an indicator variable $delta_{x, y}$ for each pair of vertices corresponding to whether or not there is an edge there; then start with

$n^8 epsilon^4 = (sum delta_{x, y})^4$

and apply Cauchy-Schwarz twice; finally, note that there are $O(n^3)$ "degenerate 4-cycles".

A basic corollary of this is the following fact:

Corollary. Any graph with girth at least 5 and n vertices has $o(n^2)$ edges.

This seems like it should be possible to prove without resorting to "analytic" machinery like Cauchy-Schwarz; indeed, it seems like it should be weak enough to prove almost by arguing "locally." But none of the obvious lines of reasoning seem to provide a proof.

Is it possible to get a good bound on the density of large-girth graphs without using Cauchy-Schwarz or equivalent?

pr.probability - The shortest path in first passage percolation

UPDATE: Fixed typing error in 2nd paragraph (greater than -> less than or equal to)

UPDATE2: Fixed typing errors pointed out by Gil Kalai

UPDATE3: I put off the detailed version from my webpage

UPDATE4: The solution is wrong, as pointed out to me by Nathanaël Berestycki. It is of course not enough to consider only the path that goes directly from the origin to (n,0). I didn't read the problem properly. Sorry.

I don't know whether this problem is still open, but I think I have found an elementary proof for the original question. It is almost too simple to be true, but I don't see any mistake. Here's the sketch:

All numbers here are natural numbers between $0$ and $n$, and $n$ is sufficiently large. Fix a (large) $K$. Let $x_l$ be the smallest $x < n/3$, such that for all $1le j le K$, the length of the path $(x,0)rightarrow (x,j) rightarrow (x+K,j)$ is less than or equal to the length of the path $(x,0)rightarrow (x+K,0)$. The arrow indicates that we take the direct path. For definiteness, set $x_l = lfloor n/3 rfloor + 1$ if such a number does not exist, but note that it exists with probability going to one as $nrightarrow infty$. Note further that since we took the smallest $x$ with the above property, conditioned on $x_l$, the lengths of the edges to the right of the vertical line $x=x_l+K$ are still independent, of the same law as before, and independent of the configuration to the left of this vertical line.

Now define $x_r$ by mirroring the above definition at the line $x=n/2$ (the largest $x>2n/3$, such that ....)

Then, the paths $(x_l+K,j)rightarrow(x_r-K,j)$ are independent for $0le jle K$, hence, with probability going to $K/(K+1)$, there exists $1le j le K$, such that the path $(x_l+K,j)rightarrow(x_r-K,j)$ is shorter than $(x_l+K,0)rightarrow(x_r-K,0)$.

Combining the above observations, with probability $K/(K+1) + o(1)$, there exist $x_l < n/3$, $x_r>2n/3$ and $1le j le K$ such that the path $(0,0) rightarrow (x_l,0)rightarrow (x_l,j)rightarrow (x_r,j) rightarrow (x_r,0) rightarrow (n,0)$ is shorter than the path $(0,0) rightarrow (n,0)$. Letting first $nrightarrow infty$, then $Krightarrowinfty$ finishes the proof.

pr.probability - Number of uniform rvs needed to cross a threshold

This problem has delighted me since I first encountered it before college. I wrote up a generalization for a less mathematical audience in a poker forum.

One way to look at the original is that if f(x) = E(N(X)), then $f(x) = 1 + int_{x-1}^x f(t) dt$, satisfying the initial condition that $f(x) = 0$ on $(-1,0)$. $f$ is 1 more than the average of $f$ on the previous interval.

We can get rid of the constant by using $g(x) = f(x) - 2x$ which satisfies $g(x) = int_{x-1}^x g(t) dt$. $g$ is equal to its average on the previous interval. Now the initial condition becomes nontrivial: $g(x) = -2x$ on $(-1,0)$, and it is from this $-2x$ that the asymptotic value of $2/3$ comes.

$g$ is asymptotically constant, and we can guess from geometry and verify that the weighted average
$int_0^1 2t~g(x+t)dt$ is independent of $x$, hence equal to the asymptotic value of $g$. The value at $x=-1$ is
$int_0^1 4t(1-t)dt = frac23$, so that is the asymptotic value of $g$.

The same technique works for more general continuous distributions supported on $mathbb R^+$ than the uniform distribution, and the answer turns out to be remarkably simple. Let's find an analogous conserved quantity.

Let $alpha$ be the density function. Let the $n$th moment be $alpha_n$. We'll want $alpha_2$ to exist.

Let $f(x)$ be the average index of the first partial sum which is at least $x$ of IID random variables with density $alpha$. $f(x) = 1 + int_{-infty}^x f(t) alpha(x-t) dt$, and $f(x) = 0$ for $xlt0$. (We could let the lower limit be 0, too.)

Let $g(x) = f(x) - x/alpha_1$. Then $g(x) = int_{-infty}^xg(t)alpha(x-t)dt$, with $g(x) = -x/alpha_1$ on $mathbb R^-$.

We'll choose $h$ so that $H(x) = int_{-infty}^x g(t) h(x-t) dt$ is constant.

$0 = H'(x) = g(x)h(0) + int_{infty}^x g(t) h'(x-t)dt.$

This integral looks like the integral equation for $g$ if we choose $h'(x) = calpha(x)$. $c=-1$ satisfies the equation. So, if $h(x) = int_x^infty alpha(t)dt$ then $H(x)$ is constant.

Let the asymptotic value of $g$ be $v$. Then the value of $H(x)$ is both $H(infty) = v~alpha_1$ and

$H(0) = int_{-infty}^0 g(t) h(0-t)dt$

$H(0) = 1/alpha_1 int_0^infty y~h(y) dy$

$H(0) = 1/(2alpha_1) int_0^infty y^2 ~alpha(y)dy~~$ (by parts)

$H(0) = alpha_2 / (2 alpha_1)$

$v~ alpha_1 = alpha_2 / (2 alpha_1)$

$v = alpha_2 / (2 alpha_1^2)$.

So, $f(x)$ is asymptotic to $x/alpha_1 + alpha_2/(2 alpha_1^2)$.

For the original uniform distribution on [0,1], $alpha_1 = frac12$ and $alpha_2 = frac13$, so $f(x) = 2x + frac23 + o(1)$.

As a check, an exponential distribution with mean 1 has second moment 2, and we get that $f(x)$ is asymptotic to $x+1$. In fact, in that case, $f(x) = x+1$ on $mathbb R^+$. If you have memoryless light bulbs with average life 1, then at time $x$, an average of $x$ bulbs have burned out, and you are on the $x+1$st bulb.

at.algebraic topology - Lifting a homeomorphism, always possible?

No. Take any homeomorphism that doesn't preserve the subgroup of $pi_1$ that lift to closed paths in the covering. For example, take the 2:1 covering $S^1to S^1$ take the product with the identity map on $S^1$. Let $h$ be the homeomorphism switching the factors.

In general, I believe a homeomorphism will lift if and only if the associated automorphism of $pi_1$ send the subgroup of the covering to a conjugate.

Another way of saying this is that the category of coverings is equivalent to the category of $pi_1$-sets, and a homeomorphism will lift if the corresponding twist of the $pi_1$-set preserves its isomorphism class.

nt.number theory - Prime divisors of numbers 2^n + 3

(Edited as the comments below suggest)

The ABC conjecture seemed to me like it would play a roll, however it comes up a little short:

"Are there infinitely many primes $p$ so that for each $p$ there is some integer $n$ with $p^2|2^n + 3?"$

If the ABC conjecture is true, then this answer to this question is almost "no", but still there is a problem at the end of the argument.

The ABC conjecture states that for any $epsilon > 0$ there is a constant $K_epsilon$ so that for any co-prime triple $A < B < C$ with $A+B = C$ then
$$C le K_epsilonprod_{p|ABC}p^{1 + epsilon}.$$

So, if there is such an infinite collection of primes, then for the corresponding infinite $n$ where this is true then $2^n + 3 = p^2C$ then
$$p^2C le K_epsilon(6Cp)^{1+epsilon}.$$

(Edited: The following sentence is incorrect "But this will clearly run into problems for sufficiently large $p.$" But I wanted to leave it so Kevin's comment makes sense.)

Note that as $C = C(p)$ is a function of $p$ then the $C^epsilon$ (when $C$ is square-free, or nearly square-free) term may still allow this inequality to work.

ag.algebraic geometry - algorithm for calculating the Chow groups of a variety over a finite field

I am not an expert, but let me point out that computing $CH^0(X)$ (which is freely generated by the irreducible components) is already quite hard. Algorithms do exist in this case, see page 206 of "Ideals, varieties and algorithms" by Cox, Little, O'Shea for references. I know of no way to compute the class groups (which can be identified with $CH^1(X)$ for smooth $X$) in general, but I will be very interested in what other people have to say about this.

Of course, in special situations, more is known. For example, the total Chow group of quadric hypersurfaces (at least up to tensoring with $mathbb Q$).

ag.algebraic geometry - Construction of the petit Zariski topos out of the gros topos of a scheme

I will deal with étale toposes because they behave much better in every possible way. They are also easier to define, althought they require substantially more commutative algebra to work with in practice:

The gros étale topos for $S$ is just $((Shv(Aff_{acute Et})downarrow S).$ We can construct from it the petit étale topos by considering $Shv(acute Et downarrow S)$, where $(acute Et downarrow S)$ is the subcategory of the gros étale topos consisting of étale morphisms $Ato S$ where $A$ is affine. This site is equipped with the induced topology.

Now for the ring object. For the petit topos, we let $mathcal{O}_S$ be defined simply the sheaf sending any affine scheme to its corresponding ring (exercise: Show that this is a sheaf). This defines a ring object in the category of sheaves on the small site (exercise: Prove this. (Hint: Think of the definition of a group object and recall that the Yoneda embedding is full.)). For the large topos, we just let it be the base change of the affine line. It's not hard to show that they agree on étale morphisms $Ato S$ for A affine.

It turns out that the gros and petit toposes have a geometric morphism induced by the inclusion of the small site into the large site. I don't know if there is a specific universal property, per se, but it turns out that they are "homotopy equivalent" in a suitable sense.

For an explanation of the homotopy condition, see

Mac Lane and Moerdijk - Sheaves in Geometry and Logic Chapter 7.

Edit: If I remember correctly, the statement about "homotopy equivalence" does not work in the fppf or fpqc topologies. The small flat sites are too small, in some sense.

graph theory - Conjugate vertices and distinguishing properties

Motivation (added)

A finite $n$-set is uniquely described (up to isomorphism) by a single population number $n$.

A finite $n$-set with $k$ predicates is uniquely described (up to isomorphism) by $2^k$ population numbers $n_i$, corresponding to the $2^k$ combinations of predicates, $sum_{i=1}^{2^k} n_i = n$

A finite $n$-set with 1 binary relation (a graph $G$) can be uniquely described (up to isomorphism) by $k$ population numbers $n_i$, corresponding to its $k$ $Aut(G)$-orbits, provided these are appropriately described, $sum_{i=1}^{k} n_i = n$.

I want to clarify what an appropriate description of the orbits might be and whether there is something like a canonical description of the orbits.

Definition 1

Let $v$, $w$ be vertices of a finite graph $G$. $v$ and $w$ are conjugate ($v sim w$) iff there is a $g in Aut(G)$ with $g(v) = w$.

Question 1 (postponed)

Is there an official and more common name for this equivalence? (Comment: I formerly called it "equivalent" but already changed this to "conjugate", thanks to Pete's hint.)

Definition 2 (third revision)

Let $G$ be a graph, $v in V(G)$ and $phi_v(x)$ be a first order vertex property (formulated in the first order language of graphs) such that $(forall w in V(G) )phi_v(w) Leftrightarrow v sim w$.

Let $lbrace phi_v(x)rbrace_{[v] in G/_sim}$ be a family of such properties.

If all graphs $H$ for which there is a bijection $f:V(G)rightarrow V(H)$ with

$phi_v(x) Leftrightarrow phi_v(f(x))$ for all $[v] in G/_sim$

are isomorphic to $G$, we call the family of properties $lbrace phi_v(x)rbrace_{[v] in G/_sim}$ distinguishing with respect to $G$.

(Comment: I added w.r.t $G$ to make things clearer and to be more precise.)

Claim

Given a graph $G$ and a distinguishing family of properties w.r.t. $G$ then the population function $f: lbrace phi_v(x)rbrace_{[v] in G/_sim} rightarrow mathbb{N}$ with $f(phi_v(x)) = |[v]|$ determines the graph up to isomorphism.

Definition 2a (added)

A distinguishing family of properties is minimal if its overall number of bound variables is minimal.

(Comment: Minimal distinguishing properties might serve as a canonical form of a graph.)

Question 2

Is there an official and more common name for distinguishing properties?

Question 3 (second revision)

Can a distinguishing family of properties be computed from the adjacency matrix in linear time? Or is this problem provably as hard as graph isomorphism?

Addendum: As pointed out in the second answer (Mariano's) it is straight forward to begin with a complete description of the graph (one existential formula, stating that there are exactly $n$ different vertices and their relations) and make successively each single variable free. In the resulting $|G|$ formulas one then has to find the orbits (= equivalent formulas) which is probably as hard as graph isomorphism.

Question 3a (added)

Can a minimal distinguishing family of properties be computed from the adjacency matrix in linear time? Or is this problem provably as hard as graph isomorphism?

Question 4 (postponed)

Which language $L$ is appropriate? Can it always be the language of first order logic with a graph specific signature?

Addendum: Example

The family of properties with the single member $phi(x)$ = "x has exactly one neighbour" together with the number of vertices that share this property - 2 - fix $K_2$

Addendum: More examples

Definitions:

1. Let a d-neighbour of x be a vertex d edges away from x.


2. Let $phi_d^n$ stand for x has exactly n d-neighbours.


3. Let $C_l^n$ be the graph consisting of n cycles of lenght $l, l geq 2$, $C_2^1 = K_2$.


4. Let $P_l$ be the path graph of length $l$.

(1)

Consider the vertex transitive graphs $C_2^n$.

For each $C$ of them the one-element (vertex transitivity!) family of properties $lbrace phi_1^1 rbrace$ is distinguishing w.r.t. $C$.

(2)

Consider the vertex transitive graphs $C_l^1, l > 2$, $l$ prime or $l = 4$.

For each $C$ of them the one-element family of properties $lbrace phi_1^2 rbrace$ is distinguishing w.r.t. $C$.

For $C = C_3^2$ (vertex transitive, too) $lbrace phi_1^2 wedge phi_2^0 rbrace$ is distinguishing w.r.t. $C$.

For $C = C_6^1$ (vertex transitive, too) $lbrace phi_1^2 wedge phi_2^2 rbrace$ is distinguishing w.r.t. $C$.

These examples seem to be easily expanded by combinatorical means.

(3)

Consider the path graphs $P_l$ with $lceil frac{l}{2} rceil$ conjugacy classes of vertices. Let $psi_d$ stand for x has a d-neighbour with degree 1. Then

$lbrace phi_1^1 rbrace cup lbrace phi_1^2 wedge psi_k rbrace_{k = 1,..,lceil frac{l}{2} rceil - 1}$

is a distinguishing family of properties (not necessarily minimal).

definitions - When is a classification problem "wild"?

Although your specific request for terminology in the case of Lie algebras has now been answered, there is a very interesting broader question underlying your inquiry. Namely, how can we understand in a precise general way the idea that a given classification problem is complicated? How are we to compare the relative difficulty of two classification problems?

These questions form the central motivation for the emerging subject known as Borel equivalence relation theory (see Greg Hjorth's survey article). The main idea is that many of the most natural equivalence relations arising in many parts of mathematics turn out to be Borel relations on a standard Borel space. To give one example, the isomorphism problem on finitely generated groups, but of course, there are hundreds of other examples. A classification problem for an equivalence relation E is really the problem of finding a way to describe the E-equivalence classes, of finding an E-invariant function that distinguishes the classes.

Harvey Friedman defined that one equivalence relation E is Borel-reducible to another relation F if there is a Borel function f such that x E y if and only if f(x) F f(y). That is, the function f maps E classes to F classes in such a way that different E classes get mapped to different F classes. This provides a classification of the E classes by using the F classes. The concept of reducibility provides a precise, robust way to say that one relation F is at least as complex as another E. Two relations are Borel equivalent if they reduce to each other, and we are led to the hierarchy of equivalence relations under Borel reducibility. By placing an equivalence relation into this hierarchy, we come to understand how complex it is in comparision with other equivalence relations. In particular, we say that one equivalence relation E is strictly simpler than F, if E reduces to F but not conversely.

It sometimes happens that one has a classification problem E and is able to provide a classification by assigning to each structure a countable list of data, such that two structures are equivalent iff they have the same data. This amounts to a reduction of E to the equality relation =, for two structures are E equivalent iff their data is equal. Such relations that reduce to equality are called smooth, and lay near the bottom of the hierarchy of Borel equivalence relations. These are the simplest equivalence relations. Thus, one way of showing that a relation is comparatively simple, is to show that it is smooth, and to show it is comparatively hard, show that it is not smooth.

The subject of Borel equivalence relation theory, as now developed by A. Kechris, G. Hjorth, S. Thomas and many others, is focused on placing many of the natural classification problems of mathematics into this hierarchy. Some of the main early results are the following interesting dichotomies:

Theorem.(Silver dichotomy) Every Borel equivalence relation E either has only countably many equivalence classes or = reduces to E.

The relation E₀ says that two binary sequences are equivalent iff they agree from some point onward. It is easy to see that = reduces to E₀, and an elementary argument shows that E₀ does not reduce to =. Thus, E₀ is strictly harder than equality. Moreover, it is a kind of next-step up in the hiearchy, in light of the following.

Theorem.(Glimm-Effros dichotomy) Every Borel equivalence relation E either reduces to = or E₀ reduces to E.

The subject continues with many interesting results that gradually illuminate more and more of the hierarchy of Borel equivalence relations. For example, the Feldman-Moore theorem shows that every Borel equivalence relation E having every equivalence class countable is the orbit equivalence of a countable group of Borel bijections of the space. The relation E_oo is the orbit equivalence of the left-translation action of the free group F₂ on its power set. This relation is complete for the countable Borel equivalence relations, in the sense that every countable Borel equivalence relation reduces to it. It's great stuff!

soft question - Famous mathematical quotes

Like many people, I am fascinated by the quote from Weyl (already listed
here), that

In these days the angel of topology and the devil of abstract algebra
fight for the soul of each individual mathematical domain.

But I can see why people are puzzled by the quote, so I'd like to add some
more information (too much to put in a comment) as another answer.

First, what is the context? The quote occurs in Weyl's paper Invariants
in Duke Math. J. 5 (1939), pp. 489--502, the first page of which can be seen
here. This page includes most of what Weyl has to say on algebra v.
geometry, though the quote itself does not occur until p.500. Then on p.501
Weyl explains his discomfort with algebra as follows

In my youth I was almost exclusively active in the field of analysis;
the differential equations and expansions of mathematical physics were
the mathematical things with which I was on the most intimate footing.
I have never succeeded in completely assimilating the abstract
algebraic way of reasoning, and constantly feel the necessity of translating
each step into a more concrete analytic form.

Second, why the image of angel and devil? According to V.I Arnold,
writing here, Weyl had a particular image in mind, namely, the
Uccello painting "Miracle of the Profaned Host, Episode 6", which can be
viewed here.

Arnold describes this painting as "representing an event that happened in
Paris in 1290." "Legend" is probably a better word than "event," but in
any case it is a very strange origin for a famous mathematical quote.

pr.probability - When does a probability measure take all values in the unit interval?

This is a property of $mu$, not that of $mathcal A$, and it is called being atomless. It is equivalent to not having sets $A in mathcal A$ of positive measure such that for all $B in mathcal A$, $B subseteq A$ the measure $mu(B)$ is either 0 or $mu(A)$.

edit: Wikipedia article, complete with the proof of the property you describe from atomlessness.

edit: yup, the comments are right and I'm wrong. The precise condition for finite measures composed entirely of atoms to have full range is $a_n leq sum_{j>n} a_j$ - it is clearly necessary as $a_n-varepsilon$ has to be produced somehow, and the greedy algorithm shows sufficiency.

ag.algebraic geometry - Puiseux series for roots of polynomials with smooth coefficients

If

$$p(x,y) = x^N + a_{N-1}(y)x^{N-1} + ldots + a_0(y), quad x,y in mathbf{C}$$

is a monic polynomial in $x$, and the coefficients $a_j$ are analytic functions of $y$, then the roots of $p$ have expansions in Puiseux series (in powers of $y^{1/m}$ for some $m$) which are convergent for $y$ sufficiently close to 0.

Is this true in an asymptotic sense when the $a_j$ are only assumed to be smooth? i.e. Do the roots have asymptotic expansions which are formal Puiseux series (not necessarily convergent)?

For $A$ to have an asymptotic expansion $A sim B_1 + B_2 + ldots$ with respect to some grading $mathcal{O}(n)$ means that $B_n in mathcal{O}(n)$, and for each $N$, $A - sum_{n=1}^N B_n in O(N+1)$. Here $mathcal{O}(n)$ means $mathcal{O}(|y|^{n/m})$ in the usual big-O notation, as $yto 0$.

ac.commutative algebra - Elementary / Interesting proofs of the Nullstellensatz

I have been thinking about the question "What is the best -- i.e., some combination of shortest, most natural, easiest -- proof of the Nullstellensatz?" recently on the eve of a commutative algebra course.

In my notes up to this point I had been following Kaplansky's treatment of Goldman domains and Hilbert-Jacobson rings. This places the problem in a more general context and allows for an attractively thorough analysis. At the end one comes out with the following results:

(1) The polynomial ring $k[t_1,ldots,t_n]$ is a Jacobson ring -- i.e., every radical ideal is the intersection of the maximal ideals containing it.

(2) (Zariski's Lemma): If $mathfrak{m}$ is a maximal ideal of $k[t_1,ldots,t_n]$, then $k[t_1,ldots,t_n]/mathfrak{m}$ is a finite degree field extension of $k$.

(That Hilbert's Nullstellensatz follows from (1) and (2) is an easy, standard argument that I won't discuss here.)

But it is well-known that to prove the Nullstellensatz one needs only (2), because then (1) follows by a short and easy argument that everyone seems to like: Rabinowitsch's Trick. So perhaps this is a sign that developing the theory of (Hilbert-)Jacobson rings to prove the Nullstellensatz is overkill.

So the question seems to be: what is the best proof of Zariski's Lemma?

After looking around at various proofs, here is what I think the answer is now: it is an easy consequence of the following result.

Theorem (Artin-Tate Lemma): Let $R subset T subset S$ be a tower of rings such that
(i) $R$ is Noetherian,
(ii) $S$ is finitely generated as an $R$-algebra, and
(iii) $S$ is finitely generated as a $T$-module.
Then $T$ is finitely generated as an $R$-algebra.

Proof: Let $x_1,ldots,x_n$ be a set of generators for $S$
as an $R$-algebra, and let $omega_1,ldots,omega_m$ be a set of
generators for $S$ as a $T$-module. For all $1 leq i leq n$, we may
write
begin{equation}
label{ARTINTATEEQ1}
x_i = sum_j a_{ij} omega_j, a_{ij} in T.
end{equation}
Similarly, for all $1 leq i,j leq m$, we may write
begin{equation}
label{ARTINTATEEQ2}
omega_i omega_j = sum_{k} b_{ijk} omega_k, b_{ijk} in T.
end{equation}
Let $T_0$ be the $R$-subalgebra of $T$ generated by the $a_{ij}$ and $b_{ijk}$. Since $T_0$ is a finitely generated algebra over the Noetherian ring $R$, it
is itself a Noetherian ring by the Hilbert Basis Theorem. indent
Now each element of $S$ may be expressed as a polynomial in the $x_i$'s
with $R$-coefficients. Making substitutions using the two equations above shows that $S$ is a finitely generated $T_0$-module. Since
$T_0$ is Noetherian, the submodule $T$ is also finitely generated as a $T_0$-module. This immediately implies that $T$ is finitely
generated as a $T_0$-algebra and then in turn that $T$ is finitely generated
as an $R$-algebra, qed!

Proof that Artin-Tate implies Zariski's Lemma:

It suffices to prove the following: let $K/k$ be a field extension which is finitely generated as a $k$-algebra. Then $K/k$ is algebraic. Indeed, suppose otherwise: let $x_1,ldots,x_n$ be a transcendence basis for
$K/k$ (where $n geq 1$ since $K/k$ is transcendental), put $k(x) = k(x_1,ldots,x_n)$ and consider the tower of rings

$k subset k(x) subset K$.

To be sure, we recall the definition of a transcendence basis: the elements $x_i$ are algebraically independent over $k$ and $K/k(x)$ is algebraic. But since $K$ is a finitely generated $k$-algebra, it is certainly a finitely generated $k(x)$-algebra and thus $K/k(x)$ is a finite degree field extension. Thus the Artin-Tate Lemma applies to our tower: we conclude that $k(x)/k$ is a finitely generated $k$-algebra. But this is absurd. It implies the much weaker statement that $k(x) = k(x_1,ldots,x_{n-1})(x_n)$ is finitely
generated as a $k(x_1,ldots,x_{n-1})[x_n]$-algebra, or weaker yet, that
there exists some field $F$ such that $F(t)$ is finitely generated as
an $F[t]$-algebra: i.e., there exist finitely many rational functions
${r_i(t) = frac{p_i(t)}{q_i(t)} }_{i=1}^N$ such that every rational function
is a polynomial in the $r_i$'s with $k$-coefficients. But $F[t]$ is a PID with infinitely many nonassociate nonzero prime elements (e.g. adapt Euclid's argument of the infinitude of the primes), so we may choose a nonzero prime element $q$ which does not divide $q_i(t)$ for any $i$. It is then clear that $frac{1}{q}$ cannot be a polynomial in the $r_i(t)$'s: for instance, evaluation at a root of $q$ in $overline{F}$ leads to a contradiction.

Note that this is almost all exactly as in Artin-Tate's paper, except for the endgame above, which has been made a little more explicit and simplified: their conclusion seems to depend upon unique factorization in $k[t_1,ldots,t_n]$, which does not come until later on in my notes.

Further comments:

(i) The proof is essentially a reduction to Noether's normalization in the case of field extensions, which becomes the familiar result about existence of transcendence bases. Thus it is not so far away from the most traditional proof of the Nullstellensatz. But I think the Artin-Tate Lemma is easier than Noether Normalization.

(ii) Speaking of Noether: the proof of the Artin-Tate Lemma is embedded in the standard textbook proof that if $R$ is a finitely generated $k$-algebra and $G$ is a finite group acting on $R$ by
ring automorphisms, then $R^G$ is a finitely generated $k$-algebra. In fact I had already typed this proof up elsewhere in my notes. Realizing that the Artin-Tate Lemma is something I was implicitly proving in the course of another result anyway was part of what convinced me that this was an efficient route to the Nullstellensatz. Note that the paper of Artin and Tate doesn't make any connection with Noether's theorem and conversely the textbooks on invariant theory that prove Noether's theorem don't seem to mention Artin-Tate. (However, googling -- Artin-Tate Lemma, Noether -- finds several research papers which allude to the connection in a way which suggests it is common knowledge among the cognoscenti.)

Added: It turns out this is the proof of Zariski's Lemma given in Chapter 7 of Atiyah-Macdonald. I had missed this because (i) they give another (nice) proof using valuation rings in Chapter 5 and (ii) they do not attribute the Artin-Tate Lemma to Artin and Tate, although their treatment of it is even closer to the Artin-Tate paper than mine is above. (In the introduction of their book, they state cheerfully that they have not attributed any results. I think this is a drawback of their otherwise excellent text.)

gr.group theory - Torsion-free and torsionless abelian groups

This question is motivated by my most spectacular answer on MO (:

Let $A$ be a module over $mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $nin mathbb Z, ain A$. $A$ is torsionless if the map $phi: A to A^{**}$ is injective (here ${}^*$ means $text{Hom}_{mathbb Z}(A,mathbb Z)$).

If $A$ is finite, then torsion-free and torsionless are equivalent. In general, it is not hard to see that being torsionless implies torsion-free. On the other hand, $A=mathbb Q$ is torsion-free but not torsionless since $A^*=0$. But the question and answers quoted above (which shows that for $A=mathbb Z[x]$, $phi$ is an isomorphism) raised the following:

Question: If $A$ is a finite $mathbb Z[x_1,...,x_d]$-module, are being torsion-free and torsionless equivalent?

ct.category theory - What is known about higher-categorical reconstruction theorems? (reference request)

The answer to my question is almost certainly "not much" — at least, I've asked a few people, and that was their answer. But I'd like to refine this answer, and MathOverflow seems like the best place.

I learned from David Ben-Zvi in an answer to this question the following theorem:

Let $mathcal C$ be a 1-category, and consider the category $operatorname{Rep}(mathcal C)$ of 1-functors $mathcal C to operatorname{1Vect}$. It is a (symmetric) monoidal category by "pointwise tensor product", i.e. pulling back along the diagonal map $mathcal C to mathcal C^{times 2}$. Conversely, we can consider some sort of "spec" of $operatorname{Rep}(mathcal C)$, namely the category of monoidal functors (and monoidal natural transformations) $operatorname{Rep}(mathcal C) to operatorname{1Vect}$. In fact, this "spec" is equivalent as a category to $mathcal C$.

Given this, it is natural to ask the following three questions (or combinations thereof):

• Recognition: which monoidal categories are of the form $operatorname{Rep}(mathcal C)$ for some $mathcal C$?


• Bump up $n$: modulo definitions, it is clear what the statement is with "$1$" replaced by "$n$". For example, the "$0$" version of the above says that a set is recoverable up to isomorphism from its algebra of all functions (the 0-category $operatorname{0Vect}$ is precisely the ground field).


• Internalize: is there a similar statement for "topological categories" and "continuous functors", for example? A version of in algebrogeometric land is in these questions (see also the answer here).

I'm not asking for definite answers to any of these directions, because I expect that telling the complete story is hard. But I am hoping for references to the existing literature. Hence: "What's already known (in the literature) about higher-categorical reconstruction theorems?"

modular forms - Why are functional equations important?

The simplest reason functional equations have importance, for someone learning this stuff for the first time (not knowing about modular forms, automorphic representations, etc.) is that they can be used to verify the Riemann hypothesis numerically up to some height!

To explain this, let's start off with a limitation of methods of complex analysis in detecting zeros of functions. There are theorems in complex analysis which tell you how to count zeros of an analytic function $f(s)$ inside a region by integrating $f'(s)/f(s)$ around the boundary (this is the argument principle). So we could integrate around the boundary of a box surrounding the critical strip up to some height and see there are, say, 10 zeros of the Riemann zeta-function up to that height (yeah, there's a pole on the boundary at $s = 1$ which messes up the argument principle integration, but don't worry about that right now). How can we prove the 10 zeros in the critical strip up to that height are on the critical line? Complex analysis provides no theorems that assure you an analytic function has zeros on a line!

The functional equation comes to the rescue here. I'll illustrate for the Riemann zeta-function $zeta(s)$. Its functional equation is most cleanly expressed in terms of

$$Z(s) = pi^{-s/2}Gamma(s/2)zeta(s)$$

and is the following:

$$Z(1-s) = Z(s).$$

We also need another "symmetry": $Z(s^*)^* = Z(s)$, where ${}^*$ means complex conjugation.
Where does this come from? For an entire function $f(s)$, the function $f(s^*)^*$ is also entire: in fact its local power series expansion at any point a is the one whose coefficients are complex conjugate to the coefficients of $f(s)$ at $a^*$. Or you could directly prove $f(s^*)^*$ is complex-differentiable when $f(s)$ is. The significance of this is that if $f(s)$ is real-valued for some interval of real numbers then $f(s^*)^* = f(s)$ on that interval, so by the rigidity of analytic functions we must have $f(s^*)^* = f(s)$ everywhere when it is true on a real interval (not one point intervals, obviously). Lesson: an entire function $f(s)$ that is real-valued on some interval of the real line satisfies the formula $f(s^*)^* = f(s)$ for all complex numbers $s$. By the way, this also applies to meromorphic functions on $mathbf C$ too (rigidity of meromorphic functions).

Let's now return to the zeta-function. Because $zeta(s), pi^{-s/2}$, and $Gamma(s/2)$ are real-valued for real $s > 1$, their product $Z(s)$ is real for $s > 1$, so $Z(s^*)^* = Z(s)$ for all complex $s$. In particular, for a number $s = frac12 + it$ on the critical line (here $t$ is real),
we have the key calculation

$$Z(s)^* = Z(1/2 + it)^* = Z(1 - (1/2 + it))^* = Z(1/2 - it)^* = Z((1/2 + it)^* )^* = Z(1/2 + it) = Z(s),$$

where we used $Z(s) = Z(1-s)$ in the second equation and $Z(s^*)^* = Z(s)$ in the second to last equation.

This tells us the function $Z(s)$ is real-valued on the critical line. The Riemann zeta-function is not real-valued on the critical line, but this modified (completed) zeta-function $Z(s)$ is. Moreover, because $Z(s)$ differs from $zeta(s)$ by factors that are finite and nonzero inside the critical strip ($pi^{-s/2}$ is a nowhere-vanishing entire function and $Gamma(s/2)$ is meromorphic with no zeros and only has poles at $s = 0, -2, -4, dots$), the zeros of $zeta(s)$ and $Z(s)$ inside the critical strip are the same thing. (In fact, nontrivial zeros of $zeta(s)$ are exactly the same thing as all zeros of $Z(s)$, which is one reason $Z(s)$ is a nicer object that $zeta(s)$: the Riemann hypothesis is a statement about all zeros of $Z(s)$!) So what? Well, we just showed in the key calculation above that the function $Z(1/2 + it)$ is real when $t$ is real, so by computing we can provably detect zeros of $Z(s)$ on the critical line Re($s$) = $frac12$ by looking for sign changes of $Z(1/2 + it)$ as t runs through the real numbers.

So here is a two-step procedure for proving the RH numerically up to height $ T$ (i.e., in the box in the critical strip from the real axis up to height $T$):

1. Use complex analysis (the argument principle) to count how many zeros $Z(s)$ has in the critical strip up to height $T$ by integrating $Z'(s)/Z(s)$ around a box surrounding that region. (If the poles of $Z(s)$ at $s = 0$ and $s = 1 $ bother you, recall the argument principle can account for poles or you might prefer to use $s(1-s)Z(s)$ in place of $Z(s)$ to be working with an entire function which satisfies the same functional equation as $Z(s)$ and is also real-valued on the critical line.)




2. Count sign changes for $Z(1/2 + it)$ when $0 le t le T$. There is a zero between any two sign changes, so $Z(s)$ has at least as many zeros on the critical line as the number of sign changes that were found. (Finding a sign change is a computable thing: if a function value at a point is approximately positive or negative then it is provably so by checking the error in your computation well enough, whereas proving a function value at a point is exactly zero with a computer is basically impossible.)

If the counts in steps 1 and 2 match, then voila: all zeros of $Z(s)$ up to height $T$ in the critical strip are on the critical line, which confirms the Riemann hypothesis up to height $T$.

This method will not work if there are any multiple zeros on the critical line: the argument principle counts each zero with its multiplicity, so if for instance there is a double zero then the argument principle may tell us $Z(s)$ has 10 zeros (with multiplicity!) up to some height while we find only 9 sign changes because one zero is a double zero so it doesn't give us a sign change. (Or if there were a triple zero we get 10 zeros with multiplicity from the argument principle but we find only 8 sign changes.) A graph may suggest that the mismatch in the numbers in the two steps is coming from a multiple zero, but it doesn't rigorously prove anything. Fortunately, this has never happened in practice with the Riemann zeta-function: the two counts always match. In fact the conjecture is that all nontrivial zeros of $zeta(s)$ are simple zeros.

What about more general L-functions $L(s)$? By multiplying $L(s)$ by suitable exponential and Gamma functions, you get a function $Lambda(s)$ whose functional equation is

$$Lambda(1-s^*)^* = wLambda(s),$$

where $w$ is a constant with absolute value 1. (For the Riemann zeta-function, $Lambda(s) = Z(s)$ and $w = 1$. For Dirichlet L-functions, $w$ is usually not equal to 1.) Let $u$ be one of the square roots of $w$, so $w = u^2$. Then using the above functional equation, the function

$$frac1u Lambda(s)$$

is real-valued on the critical line ($s = 1/2 + it$ for real $t$), so we can detect its zeros there by looking for sign changes. The same method described above for detecting zeros of $zeta(s)$ in the critical strip by using $Z(s)$ and its functional equation can be applied also to $L(s)$. This is basically the way all variants on the Riemann hypothesis are checked numerically (modulo important details of practical calculation that I don't get into), and the functional equation is an essential ingredient in justifying the method.

What is crucial here is not just the idea of sign changes, but also the expectation that the zeros are all simple (so we can find all the zeros by sign changes and the argument principle). As with the Riemann zeta-function, it is expected that the nontrivial zeros of Dirichlet $L$-functions are all simple. But there are examples of $L$-functions with a multiple zero on its critical line, thanks to ideas from the Birch and Swinnerton-Dyer conjecture. This does not wreck this approach to verifying the Riemann hypothesis for such L-functions, because such multiple zeros are supposed to happen only at one (known) point which we think we understand.

ag.algebraic geometry - Why does the algebraic condition of flatness on the structure sheaves give a good definition of family?

There is also the following (probably unhistorical) point of view (it is a version of Hailong Dao's answer). Namely, you don't have to work with flat families at all, so if you want, you can just declare all morphisms to be families'. The problem with this approach is that
this is a family ofderived' objects. Here's an example:

Let $S$ be a scheme, and let $F$ be a coherent sheaf on S. When is it a `family' of its fibers? If it is flat, it definitely deserves to be called a family of vector spaces (a vector bundle). But even if it is not flat, you can still view it as a family, but the family of what? The (derived) fibers of $F$ are no longer vector spaces, they are complexes of vector spaces (precisely because $F$ fails to be flat), so we can view $F$ as a nice family of complexes of vector spaces, even though $F$ itself is a sheaf, not a complex.

To summarize: by all means, let's forget about flatness and declare any morphism to be a family... of some kind of derived objects. If we now want members of the family to be actual objects (schemes, vector spaces, sheaves, or whatever it is we are trying to include in a family), flatness is forced on you more or less by definition.

real analysis - Rolle's theorem in n dimensions

EDIT: The following solution is incomplete. We need to make sure that if $F^{prime}left(tright)$, $F^{primeprime}left(tright)$, ..., $F^{left(n-1right)}left(tright)$ are linearly dependent vectors for every $t$, then the coordinate functions of $F^{prime}$ are linearly dependent on a sufficiently small interval. This follows from Wronskian considerations if the coordinate functions of $F^{prime}$ are sufficiently nice (i. e., locally real-analytic), so this solves the problem for this nice class of functions, but I can't use this ansatz further.

"SOLUTION".

IMPORTANT: I consider $F$ to be a map from $S^1$ to $mathbb R^n$, because a map from an interval with equal values at the ends is the same as a map from the circle. I will assume continuity of $F^{prime}$ (yes, this includes the two endpoints of the interval which I have glued together). So I don't claim I have 100% solved the original problem.

I will say that an $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ is in counterclockwise position if there is an orientation-preserving map $Phi:S^1to left[0,1right]$, continuous except at one point, such that $Phileft(t_1right)<Phileft(t_2right)<...<Phileft(t_nright)$. I need the following intuitively obvious fact:

(1) If an $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ is in counterclockwise position, and an $n$-tuple of distinct points $left(s_1,s_2,...,s_nright)in left(S^1right)^n$ is in counterclockwise position as well, then there exists a smooth way to move the points $t_1$, $t_2$, ..., $t_n$ along $S^1$ such that they occupy the places of $s_1$, $s_2$, ..., $s_n$ at the end, and such that they stay distinct at any time during the process.

I could formalize this if anyone asks me to.

As a consequence of (1) and the intermediate value theorem (the usual one, for functions $mathbb Rto mathbb R$), we have:

(2) If an $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ is in counterclockwise position, and an $n$-tuple of distinct points $left(s_1,s_2,...,s_nright)in left(S^1right)^n$ is in counterclockwise position as well, and $R:left(S^1right)^nto mathbb R$ is a continuous map that never takes the value $0$ on $n$-tuples of distinct points, then the reals $Rleft(t_1,t_2,...,t_nright)$ and $Rleft(s_1,s_2,...,s_nright)$ have the same sign.

In other words,

(3) If $R:left(S^1right)^nto mathbb R$ is a continuous map that never takes the value $0$ on $n$-tuples of distinct points, then for any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ in counterclockwise position, the real $Rleft(t_1,t_2,...,t_nright)$ has the same sign.

Now, let's solve the problem: Assume that the assertion is wrong, and thus $F^{prime}left(t_1right)$, $F^{prime}left(t_2right)$, ..., $F^{prime}left(t_nright)$ are linearly independent for any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$. Then, applying (3) to the continuous map

$R:left(S^1right)^nto mathbb R,$
$left(t_1,t_2,...,t_nright)mapsto detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$,

we obtain that:

(4) For any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ has the same sign.

Let's WLOG say that it is positive all the time (if its negative, just rewrite the proof with negative instead of positive...), i. e. we have:

(5) For any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ is positive.

Now, move $t_2$, $t_3$, ..., $t_{n-1}$ (not $t_n$) closer and closer to $t_1$ (while keeping the counterclockwise position, of course), while keeping $t_1$ and $t_n$ fixed. Then, $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ tends to $detleft(F^{prime}left(t_1right),F^{primeprime}left(t_1right),...,F^{left(n-1right)}left(t_1right),F^{prime}left(t_nright)right)$. So, we get:

(5) For any pair of distinct points $left(t_1,t_nright)in left(S^1right)^2$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{primeprime}left(t_1right),...,F^{left(n-1right)}left(t_1right),F^{prime}left(t_nright)right)$ is nonnegative.

Notice how "positive" became "nonnegative" due to the limiting process (the limit of positive reals needs not be positive, but is always nonnegative).

Now, any pair of distinct points on $S^1$ is in counterclockwise position (and in clockwise position, too), so (5) can be simply rewritten as follows:

(6) For any pair of distinct points $left(t,sright)in left(S^1right)^2$, the real $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ is nonnegative.

We can drop the "distinct" in (6), as well, because for $s=t$, the real is simply $0$. So we have:

(7) For any pair of points $left(t,sright)in left(S^1right)^2$, the real $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ is nonnegative.

But if we fix $t$ and integrate $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ over $sin S^1$, we must get zero (because $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ is linear in $F^{prime}left(sright)$, and the integral of $F^{prime}left(sright)$ over $S^1$ is zero). The integral of a continuous nonnegative function is zero only if the function itself is identically zero. Thus, $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ for all $sin S^1$. This means that $F^{prime}left(sright)$ lies in a fixed hyperplane for all $sin S^1$. Now, taking ANY $n$ points $t_1$, $t_2$, ..., $t_n$ on $S^1$, we get linearly dependent vectors $F^{prime}left(t_1right)$, $F^{prime}left(t_2right)$, ..., $F^{prime}left(t_nright)$, and this is of course a contradiction!

Or do we?

alexandrov geometry - Are isometries the only geodesic preserving maps in a CAT(0)-space?

Given any CAT(0) space $X$, we can define a map $s:Xtimes Xtimes [0;1]rightarrow X$, such that $s(x,y,-)$ is the constant speed geodesic from $x$ to $y$ . Any isometry $f$ of $X$ is compatible with that map in the sense, that $s(f(x),f(y),t)=f(s(x,y,t))$. Then one can ask, whether any self-homeomorphism of $X$, which is compatible with $s$ in the upper sense is already a isometry.

This is clearly wrong for $X=mathbb{R}^n$, as all affine maps are compatible with $s$. So the question is, whether these are the only examples.

For example I think I can show, that the $n$-dimensional hyperbolic space ($nge 2$) is rigid in that sense.

EDIT: Due to the big amout of counterexamples one could better ask the following question:

Are the spaces $mathbb{R}^n$ the only spaces, which have self homeomorphisms compatible with $s$ (in the upper sense), that are not self-similarities ?

gr.group theory - Is a normal subgroup of a finitely presented group finitely generated or normal finitely genrated?

If $G$ is the free group on two generators, then $N$ the commutator subgroup is not finitely generated.
If $H$ is any finitely generated, but not finitely presented group, then $H$ is the quotient of a finitely generated free group $G$, with kernel $N$ which is not normally finitely generated.

Steve

Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing?

The following question came up in the course on Quantum Groups here at UC Berkeley. (If you care, I have been TeXing uneditted lecture notes.)

Let $X,Y$ be (infinite-dimensional) Hopf algebras over a ground field $mathbb F$. A linear map $langle,rangle : Xotimes Y to mathbb F$ is a bialgebra pairing if $langle x,y_1y_2 rangle = langle Delta x,y_1otimes y_2rangle$ and $langle x_1x_2,yrangle = langle x_1otimes x_2,Delta yrangle$ for all $x,x_1,x_2 in X$ and $y,y_1,y_2 in Y$. (You must pick a convention of how to define the pairing $langle,rangle : X^{otimes 2} otimes Y^{otimes 2} to mathbb F$.) And we also demand that $langle 1,- rangle = epsilon_Y$ and $langle -,1rangle = epsilon_X$, but this might follow from the previous conditions. (See edit.)

A bialgebra pairing is Hopf if it also respects the antipode: $langle S(x),y rangle = langle x,S(y)rangle$. A pairing $langle,rangle : Xotimes Y to mathbb F$ is nondegenerate if each of the the induced maps $X to Y^*$ and $Y to X^*$ has trivial kernel.

Question: Is a (nondegenerate) bialgebra pairing of Hopf algebras necessarily Hopf? (Does it depend on whether the pairing is nondegenerate?)

My intuition is that regardless of the nondegeneracy, the answer is "Yes": my motivation is that a bialgebra homomorphism between Hopf algebras automatically respects the antipode. But we were unable to make this into a proof in the infinite-dimensional case.

Edit: If $langle,rangle: Xotimes Y to mathbb F$ is nondegenerate, then it is true that as soon as it satisfies $langle x,y_1y_2 rangle = langle Delta x,y_1otimes y_2rangle$ and $langle x_1x_2,yrangle = langle x_1otimes x_2,Delta yrangle$, so that the induced maps $X to Y^*$ and $Y to X^*$ are (possibly non-unital) algebra homomorphisms, then it also satisfies $langle 1,- rangle = epsilon_Y$ and $langle -,1rangle = epsilon_X$, so that the algebra homomorphism are actually unital. But I think that this does require that the pairing be nondegenerate. At least, I don't see how to prove it without the nondegeneracy assumption. So probably the nondegeneracy is required for the statement about antipodes as well.

ds.dynamical systems - Switching function for Bang-Bang nagivation

I'm attempting to develop an equation to determine the "switching time" for a control system. I've managed to work out a specific solution for when starting and ending velocities are are the same, yet am unable to adapt this for determining the switching time given an arbitrary starting velocity, and an arbitrary ending velocity.

More formally, given the acceleration of an object:

$a(t) = left\{
begin{array}{lr}
MaxAccel & : 0 lt t le switchTime\
-MaxAccel & : switchTime lt t lt stopTime\
0 & : otherwise
end{array}
right.$

determine $switchTime$, such that the object starting at $velocityInitial$, will travel $distance$, and arrive there travelling at $velocityFinal$. The calculation of $stopTime$ is trivial.

My current specific solution for starting and ending at the same velocity is:

$-velocityInitial + sqrt{velocityInitial^2+MaxAccel*distance} over a$

Sorry for the lack of a more formalized mathematical definition, I'm doing my best. =)

For those curious, yes this is related to my other navigation question. I'm trying to now integrate proportional navigation for small facing adjustments, and a limited bang-bang algorithm for increasing velocity towards the object. I'll post the whole algorithm once I get it working.

(Side note: This is the first time I'm actually using LaTeX for something "real" instead of fiddling around. Awesome!)

lo.logic - Properties of collections (functions) that make them proper classes (uncomputable)

There are collections too big to be a set, e.g. the collection of all sets (in ZFC), and there are collections that cannot be sets for "pure" logical reasons, e.g. the collection of sets that do not contain themselves.

There are functions growing too fast to be computable, e.g. the busy beaver function, and there are functions that cannot be computed for "pure" logical reasons, e.g. the halting function.

In the final end it is of course shown by logical means that being too big (growing too fast) prohibit a collection to be a set (a function to be computable), but those properties are not "purely" logical (they are about sizes and growth rates), opposed to the "pure" logical reasons mentioned above.

Is there a simple lesson to be learned
from these observations? Are there
other not "purely" logical properties
of collections (functions) that
prohibit them to be sets (computable)?

Edit: It's from the very definition of a collection -- $lbrace x | x = x rbrace$ or $lbrace x | x notin x rbrace$ -- that it is shown by logical means, that it cannot be a set. And not, firstly, from a hard to define meta-property of "defining too big a collection" or "being intrinsically inconsistent".

Then it's at least somehow astonishing, that some of those definitions cohere with the meta-property of "defining too big a collection".

Cannot - after all - the meta-property of "defining too
big a collection" be rigorously
defined such that it can be shown that no definition
of a collection with this
meta-property defines a set?

(The same should go - mutatis mutandis - for functions and computability.)