at.algebraic topology - understanding Steenrod squares

The Steenrod square is an example of a cohomology operation. Cohomology operations are natural transformations from the cohomology functor to itself. There are a few different types, but the most general is an unstable cohomology operation. This is simply a natural transformation from $E^k(-)$ to $E^l(-)$ for some fixed $k$ and $l$. Here, one regards the graded cohomology functors as a family of set-valued functors so the functions induced by these unstable operations do not necessarily respect any of the structure of $E^k(X)$.

Some do, however. In particular, there are additive cohomology operations. These are unstable operations which are homomorphisms of abelian groups.

In particular, for any multiplicative cohomology theory (in particular, ordinary cohomology or ordinary cohomology with $mathbb{Z}/2mathbb{Z}$ coefficients) there are the power operations: $x to x^k$. These are additive if the coefficient ring has the right characteristic. In particular, squaring is additive in $mathbb{Z}/2mathbb{Z}$ cohomology.

Given an unstable cohomology operation $r: E^k(-) to E^l(-)$ there is a way to manufacture a new operation $Omega r: tilde{E}^{k-1} to tilde{E}^{l-1}(-)$ using the suspension isomorphism (where the tilde denotes that these are reduced groups):

$E^{k-1}(X) cong E^k(Sigma X) to E^l(Sigma X) cong E^{l-1}(X)$

This is quite straightforward and is a cheap way of producing more operations. When applied to the power operations it produces almost nothing since the ring structure on the cohomology of a suspension is trivial: apart from the inclusion of the coefficient ring all products are zero.

What is an interesting question is whether or not this looping can be reversed. Namely, if $r$ is an unstable operation, when is there another operation $s$ such that $Omega s = r$? And how many such are there? Most interesting is the question of when there is an infinite chain of operations, $(r_k)$ such that $Omega r_k = r_{k-1}$. When this happens, we say that $r$ comes from a stable operation (there is a slight ambiguity here as to when the sequence $(r_k)$ is a stable operation or merely comes from a stable operation).

One necessary condition is that $r$ be additive. This is not, in general, sufficient. For example, the Adams operations in $K$-theory are additive but all but two are not stable.

However, for ordinary cohomology with coefficients in a field, additive is sufficient for an operation to come from a stable operation. Moreover, there is a unique sequence for each additive operation. This means that the squaring operation in $mathbb{Z}/2mathbb{Z}$ cohomology has a sequence of "higher" operations which loop down to squaring. These are the Steenrod squares.

The sequence stops with the actual squaring (rather, becomes zero after that point) because, as remarked above, the power operations loop to zero.

One important feature of these operations is that they give necessary conditions for a spectrum to be a suspension spectrum of a space. If a spectrum is such a suspension spectrum then it's $mathbb{Z}/2mathbb{Z}$-cohomology must be a ring. That's not enough, however, it must also have the property that, in the right dimensions, the Steenrod operations act by squaring. (Of course, this is necessary but not sufficient.)

pr.probability - diameter of a graph with random edge weights

The diameter is defined as $$d(G') = sup_{x,y in G} inf_{gamma} sum_{e in gamma} w_{e},$$ where the infimum is over all paths $gamma$ connecting $x$ to $y$, and the sum is over the edges $e$ which $gamma$ crosses. The graph structure is very crucial to this problem. If the graph is very connected (as in your example), then the diameter is essentially the maximum value of the variables ${w_i}$. A large fluctuation of a single $w_i$ will cause the diameter to be very large.

On the other hand, if you're dealing with something more like a lattice (e.g., a large, finite subset of $mathbb Z^2$), then the diameter is a combination of many independent random variables. Large fluctuations of any single variable will be muted and not affect the diameter much, and limit theorems will apply. A variant of Kingman's subadditive ergodic theorem will show that $$d(G') sim d(G).$$

Questions like this are very much in the realm of first-passage percolation. Check out this survey by Howard, as well as this one by Blair-Stahn on the arXiv.

at.algebraic topology - Group completion theorem

Well, if $pi_0=pi_0(M)$ is already a group, then $H_*(M)approx H_*(M)[pi_0^{-1}]$. So $M$ and $Omega B M$ have the same homology in this case. This isn't quite enough on its own, but if you can produce a map $Mto Omega BM$ which induces this homology isomorphism, then the result follows using the Hurewicz theorem.

What McDuff-Segal actually do is show that if $M$ is a topological monoid which acts on a space $X$, in such a way that every $min M$ induces a homology equivalence $xmapsto mxcolon Xto X$, then you can produce a "homology fibration" $f:X_Mto BM$ with fiber $X$. "Homology fibration" means that the fibers of $f$ are homology-equivalent to the homotopy fibers of $f$.

If $pi_0M$ is an abelian group, you can find an $X$ such that $X_M$ is contractible, and the fiber of $f:X_Mto BM$ is $X$. This gives the homology equivalence you want, since the homotopy fibers of $f$ look like $Omega BM$.

Take a look at McDuff and Segal's paper, it's nice. There is a also a treatment in terms of simplicial sets in Goerss-Jardine, *Simplicial Homotopy Theory".

Added: The functor $Mmapsto Omega BM$ is the "total derived functor of group completion". The only convincing explanation of why this is so (that I'm aware of) is in Dwyer-Kan, Simplicial Localizations of Categories, JPAA (17) 267-283. Though they work simplicially, and work more generally (with categories in place of monoids), they show that $M$ is a cofibrant simplicial monoid, then the simplicial monoid $M[M^{-1}]$ is weakly equivalent to the space $Omega |BM|$.

nt.number theory - Transcendence of $log 2$

I am not number theorist, forgive me if this is a stupid question.

Recently I was curious about the ideas behind the transcendence of $log 2$.

For the number $e$, It seems that the transcendence can be obtained by a argument of fast convergence of the Taylor expansion but the same ideas do not apply to $log 2$. Talking with number theorists I got an explanation based on the results of Baker. Unfortunately this person can not point me towards a survey or a paper discussing the ideas behind this transcendence so I come here to ask:

What are the ideas behind the transcendence of this number?

Any text (paper,book, blog) suitable for non number theorists is also welcome as an answer.

conferences - What to expect from attending an ICM?

Regarding 1, ICM's are a place to acquire new ideas. With rare exception, there will be no "Introduction to ..." type material, just "Motivation for..." and "Teaser to ..." and
"Of course, since we all know ..., the following will be obvious ...", which is true for me after a few pages and perhaps months of study. However, even the last type may contain an idea I can use, even if I don't know what "we all know".

Regarding 2, there are times, places, and opportunities to meet; see below.

I hope one day to compare the lists of attendees from all the ICMs since the one in Berkeley in 1986. I know I am in all of them, and I suspect that I share that distinction with at most 10 other people.

The one in 1986 convinced me to go to Berkeley for graduate school; the others have given me a quick exposure to various fields in mathematics and usually too much to think about. Your mileage may vary, but I found the following routine helpful to me.

1) Each night, plan your lecture tour for the following day. You will need to get a list of programme changes (usually needed only for Short Communications sessions), and select your A, B and C list choices for each hour. (Sometimes your A choice and B choice get rescheduled or canceled.) If you have friends attending and want to split up to cover more lectures, plan with them. At the first few conferences, I had mild sessions of agony because there were more lectures I wanted to attend than slots available. I now have restricted my interests in the afternoons to three or four sections (currently logic, algebra, computing, and combinatorics), to make my planning easier.

2) When you suffer from lecture burnout, try a tour or even just walk around in the part of the city near the conference hall. One of my treats from the 2006 ICM was looking at the inspired architecture in that part of Madrid. Or check out the array of posters.
(Ideally, ask permission before photographing a poster. In any case, don;'t use the
photo beyond personal use without arrangments with the author of the poster.)

3) If you are very social (I was not), figure out how to work that into your visit. Lunch breaks and afternoon breaks are the most opportune for this. I did manage to chat up a few attendees (including Tim Gowers and Greg Kuperberg) during moments in between lectures. There will be a few social occasions, and in most cases informal dress is acceptable (but check the invitations). During the
lecture sessions, there will be limited or no time for questions, so make your question count. Also, unless you find someone to walk to the same lecture with you, it is not likely that you will have much time to chat between lectures.

4) Find the information booth and help desk. If any problems arise, they are your first step towards resolution, even if it is visit related and not conference related. Also,
avoid the temptation to check your email during the conference. You can easily get sucked in to spending too much time on the computer. (You get to choose how much time to devote to computer versus sleep, however.)

Regarding your a and b choices, I would say both are true. In point of a, if I had made a persistent ass of myself, I might have gotten more than a few words with Tim Gowers, but as he had been awarded the Fields Medal recently, I chose to think his time with me would be very limited and in high demand. Thus I did not persist, and being in lecture-attending mode most of the time, I often have the perception that there is little to no time for socializing.

Towards point b., when I relax, I find I can talk to a number of people and create opportunities for socializing quite easily, if I put my focus on it. There will be some inspiring talks, and after hearing a few of these, you can use one or more points from them as icebreakers. I will thus say it depends on what kind of experience you will want to make of your ICM.

Gerhard "Collectors edition ICM T-shirts for sale" Paseman, 2010.04.08

nt.number theory - What would a "moral" proof of the Weil Conjectures require?

I am by no means the expert on algebraic geometry. But maybe I can say a little bit. Kontsevich seems ever wrote a book"Beyond Number" There is one paragraph:

“Very often a mathematician considers his colleague from a different domain with disdain -- what kind of a perverse joy can this guy find in his unmotivated and plainly boring subject? I have tried to learn the hidden beauty in various things, but still for many areas the source of interest is for me a complete mystery.

My theory is that too often people project their human weakness/properties onto their mathematical activity.

There are obvious examples on the surface: for instance, the idea of a classification of some objects is an incarnation of collector instincts, the search for maximal values is another form of greed, computability/decidability comes from the desire of a total control.

Fascination with iterations is similar to the hypnotism of rhythmic music. Of course, the classification of some kinds of objects could be very useful in the analysis of more complicated structures, or it could just be memorized in simple cases.

The knowledge of the exact maximum or an upper bound of some quantity depending on parameters gives an idea about the range of its possible values. A theoretical computability can be in fact practical for computer experiments. Still, for me the motivation is mostly the desire to understand the hidden machinery in a striking concrete example, around which one can build formalisms.

..... In a deep sense we are all geometers."

I think what Kontsevich mean is that not only the result should be correct, but also the method to get the result should be elegant and natural. Just as he mentioned, the most interesting things to him is the hidden machinery behind the striking examples. For example, say Atiyah-Singer index theorem. Rosenberg ever mentioned in the class, this theorem should have the machinery living in the abelian categories or even exact categories instead of triangulated categories(where Grothendieck Riemman Roch is now living). I guess what they are thinking about is that one should use some universal constructions ,universal theory"(in some sense). They always make emphasis on one sentence "Mathematics should be simple" which might means that the proof of some big theorem should be simple. That is to say, one does not pay much "brain thinking" because "brain of human are weak"

However, I agree with Emerton that this is very personal feelings

soft question - Famous mathematical quotes

Grothendieck comparing two approaches, with the metaphor of opening a nut: the hammer and chisel approach, striking repeatedly until the nut opens, or just letting the nut open naturally by immersing it in some soft liquid and let time pass:

"I can illustrate the second approach with the same image of a nut to be opened. The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!
A different image came to me a few weeks ago. The unknown thing to be known appeared to me as some stretch of earth or hard marl, resisting penetration... the sea advances insensibly in silence, nothing seems to happen, nothing moves, the water is so far off you hardly hear it... yet it finally surrounds the resistant substance."

Grothendieck, of course, always pioneered this approach, and considered for example that Jean-Pierre Serre was a master of the "hammer and chisel" approach, but always solving problems in a very elegant way.

fa.functional analysis - Radii and centers in Banach spaces

I believe that the property does not hold for all Banach spaces, but my counterexample is a little involved. If you've the patience then follow me through...

Let $V=bigoplus_{n=1}^infty ell^n_2$ where $ell^p_2$ is $mathbb{R}^2$ with
norm $lVertcdotrVert_p$ (Note: $n$ is taking the role of $p$). For $igeq1$
and $jin{0,1}$ we have $e_{i,j}$, the $j^{th}$ standard basis vector of
$ell^i_2$ in $V$.

Give $V$ the norm $lVert vrVert=sup_nlVert v_nrVert_n$.

Let $W={vin V:lVert v_nrVert_nto 0}$. I assert that $W$ is a Banach space.
Certainly every $e_{i,j}in W$.

Let $A={e_{k,0}+e_{k,1}, e_{k,0}-e_{k,1}:kgeq 1}$.

Fact: Let $r(A)$ be the infimum of radii of balls containing $A$. Then $r(A)leq1$

Proof:

Let $c_N=sum_{i=1}^n e_{i,0}$. We wish to compute the distance of each point
of $A$ from $c_N$.

For $kleq N$ we have
$lVert c_N-e_{k,0}-e_{k,1}rVert$
$=lVertsum_{i=1 (inot=k)}^Ne_{i,0}-e_{k,1}rVert$
$=sup{lVert e_{i,0}rVert_i:ileq N,inot=k}cup{lVert-e_{k,1}rVert_k}$
$=1$
and similarly for $lVert c_N-e_{k,0}+e_{k,1}rVert$.

For $k>N$ we have
$lVert c_N-e_{k,0}-e_{k,1}rVert$
$=max(lVert c_NrVert,lVert e_{k,0}+e_{k,1}rVert_k)$
$= max(1,(1+1)^frac{1}{k})$
$= 2^frac{1}{k}$
$leq 2^frac{1}{N}$
and similarly for $lVert c_N-e_{k,0}+e_{k,1}rVert$.

Thus $Asubseteq overline{B}(c_N,2^frac{1}{N})$ and so $r(A)leq2^frac{1}{N}$. Letting
$Ntoinfty$ we have $r(A)leq 1$.

QED

Fact: $A$ is not contained in a ball of radius $1$.

Proof:

Suppose $Asubseteq overline{B}(c,1)$. Then in particular for every $n$ we have
$lVert c-e_{n,0}-e_{n,1}rVertleq 1$ and thus $lVert c_n-e_{n,0}-e_{n,1}rVert_nleq 1$. Similarly $lVert c_n-e_{n,0}+e_{n,1}rVert_nleq 1$.

Simple consideration of $ell^n_2$ shows that this implies $c_n=e_{n,0}$.
Thus $lVert c_nrVert=1notto0$ and $cnotin W$, contradicting the assumption.

QED

ag.algebraic geometry - Approximation of stacks / algebraic spaces

The answer is no, even for the sheaf-case.

First of all, you would have to make some assumptions such as assume that X_λ→ Spec(B_λ) is locally of finite presentation. For simplicity also assume that the algebraic space X is of finite presentation over Spec(B). Then if U→X is an étale presentation, it descends to a morphism U_λ→X_λ. Now the main problem is that even if this morphism is étale after base change to X, we cannot guarantee that it eventually becomes étale. Indeed, this is related to the openness of versality in Artin's algebraization theorem.

For example, let B₀ be a ring and let B=colim B_λ be a direct limit of essentially étale algebras such that B is henselian.
If openness of versality holds for X_λ, then U_λ→X_λ is smooth in a neighborhood of the image of U→U_λ and (at least if B₀ is noetherian) it follows that U_λ→X_λ is smooth for sufficiently large λ.

But ... openness of versality does not hold for general X_λ. Artin has given a nice bunch of examples where everything but this holds (see "The implicit function theorem in algebraic geometry", S5). For example, let B₀=k[x] be the affine line and let B be the henselization (or localization) at the origin. Let X₀ be a "bad" sheaf, e.g., the following (Ex. 5.10):

Let X₀=colim_k X_0,k where X_0,i=Spec(k[x,y]/y(x-1)(x-2)...(x-k)) — the union of the x-axis and k vertical lines. This is an example of an ind-space, so by definition:

  X₀(T)=colim_k X_0,k(T)

for any scheme T. Then clearly, X=X₀×_B0 Spec(B) is isomorphic to Spec(B) but X_λ is not an algebraic space for any λ.

If you assume that openness of versality holds for X_λ and that the Spec(B)→Spec(B_λ) are smooth, then the answer to your question is likely "yes". However, openness of versality is the most subtle of Artin's criteria so it is probably difficult to assert.

ra.rings and algebras - Are there countable index subrings of the reals?

Simon Thomas's approach answers Question 2 too. The answer is that for any nonzero real number $x$ there exists such a subring (possibly without 1) not containing $x$.

Proof: Let $K$ be the Puiseux series field $overline{mathbf{Q}}((t^{mathbf{Q}}))$, let $A$ be its valuation ring, and let $mathfrak{m}$ be its maximal ideal.

If $x in mathbf{R}$ is not algebraic over $mathbf{Q}$, then choose an identification $mathbf{C} simeq K$ sending $x$ to the transcendental element $1/t$, and use the subring $mathbf{R} cap A$.

If $x in mathbf{R}^times$ is algebraic over $mathbf{Q}$, then choose an identification $mathbf{C} simeq K$ again, and use $mathbf{R} cap mathfrak{m}$ (a subring of $mathbf{R}$ without $1$). $square$

Remark: If one insists on using subrings with $1$, then the answer is that such a subring not containing $x$ exists if and only if $x notin mathbf{Z}$.

Proof: Repeat the argument above, but in the case where $x$ is algebraic (and outside $mathbf{Z}$), use $mathbf{R} cap (mathbf{Z} + mathfrak{m})$.

Kronecker product definition

Some authors (v.g. the creators of Matlab, Campbell, Lo, MacKinlay (1997) in The Econometrics of Financial Markets) define the Kronecker product of two vectors as one single column vector containing the crossproduct of each lement of the first vector with each element of the second vector. This is not the usual definition in Wikipedia nor Mathworld nor other software like Mathematica. What's about that definition?, is it correct? (it seems to work in computations in Campbell, Lo, MacKinlay (1997), why does it work?, is it useful? Maybe if someone knows enough about this, he can update the Wikipedia page, otherwise I will do it with the information I receive. (This question seems not to be related at first sight with "vectorization", since vectorization of the standard definition is not equal to the nonstandard definition apparently). Many thanks in advance.

rt.representation theory - Proof of Steinberg's tensor product theorem

The 1980 CPS paper is short but not easy to read without enough background.
They gave the first conceptual alternative to Steinberg's somewhat opaque
and computational proof of the tensor product theorem in 1963 (which built
on the 1950s work of Curtis on "restricted" Lie algebra representations
coming from the algebraic group plus the older work of Steinberg's teacher
Richard Brauer on rank 1). Steinberg relied quite a bit on working with
covering groups and projective representations.

CPS already realized the importance of
getting beyond the Lie algebra by using Frobenius kernels. The best modern
source is the large but well-organized 1987 book by J.C. Jantzen,
Representations of Algebraic Groups (expanded AMS edition in 2003). Here
the foundations are worked out thoroughly and the CPS proof is given an
efficient treatment in part II, 3.16-3.17. While CPS had in mind the
analogy with Clifford theory for finite groups, Jantzen gives a self-contained
treatment avoiding use of projective representations or Skolem-Noether.

Apart from sources, the essential goal is to single out the finitely many
"restricted" simple modules for the Lie algebra among the infinitely many
simple (rational) modules for the ambient algebraic group, then realize the
latter modules as twisted tensor products of the former. This requires a notion of
Frobenius morphism for each power of the prime, Frobenius kernels being
infinitesimal group schemes. The Lie algebra just plays the role of
first Frobenius kernel (a normal subgroup scheme), so the Clifford theory
analogue developed by Ballard and CPS makes sense here.

ac.commutative algebra - Is weak normality stable under completion?

Here is a partial solution: modulo a problem of constructing "sufficiently generic" elements in the maximal ideal of a reduced noetherian local ring of dimension > 1 (in a sense made precise at the end in terms of associated primes and vanishing loci, and which might require some care when the residue field is finite), I claim it suffices to handle passage to residually trivial local-'etale extensions (or what comes to the same thing, as we'll see below, henselization instead of completion). I'm not sure if this is any easier than completion, but it certainly can't be harder! I will use Artin-Popescu approximation (which wasn't known in 1980, the time of the Greco-Traverso paper). Let $R$ be a weakly normal reduced excellent local ring. Its henselization $R^{rm{h}}$ not only inherits reducedness, but also inherits excellence (EGA IV$_4$, 18.7.6).

Let's reformulate the definition of weak normality in more convenient terms: the condition is that any finite radiciel surjection $X rightarrow Y := {rm{Spec}}(R)$ that is an isomorphism over a dense open is an isomorphism when $X$ is reduced. Now consider a finite radiciel surjection $X_1 rightarrow Y_1 := {rm{Spec}}(R^{rm{h}})$ that is an isomorphism over a dense open, with $X_1$ reduced. Since $R^{rm{h}}$ is a direct limit of residually trivial local-'etale neighborhoods of $R$, by "spreading out" techniques from EGA IV$_3$, sections 8 & 9 there is a residually trivial local-'etale cover $Y' rightarrow Y$ and finite radiciel surjection $X' rightarrow Y'$ that is an isomorphism over a dense open and induces $X_1 rightarrow Y_1$ after base change, with $X'$ reduced. Thus, if all such $Y'$ were weakly normal then $X' rightarrow Y'$ would be an isomorphism and so $X_1 rightarrow Y_1$ would be too. In other words, if we knew weak normality were inherited by residually trivial local-'etale extensions then the henselization would be weakly normal. I will now assume we knew such inheritance does hold, so we can replace $R$ with its henselization.

Suppose the completion $widehat{R}$ (which we know is reduced, and of course excellent) were not weakly normal. Then there would be a finite radiciel surjection $T rightarrow widehat{Y} := {rm{Spec}}(widehat{R})$ that is an isomorphism over a dense open but not an isomorphism, with $T$ reduced. Thus, there is some infinitesimal fiber whose structure map to the base is not an isomorphism, say $n$th-order with $n ge 1$. By A-P approximation for the excellent henselian $Y$, there is a local-smooth cover $Y' rightarrow Y$, a $Y$-map $s:widehat{Y} rightarrow Y'$, and a finite radiciel surjection $T' rightarrow Y'$ which is an isomorphism over a dense open $U' subseteq Y'$ such that $T'$ is reduced and the pullback $T'_{widehat{Y}}$ coincides with $T$ on $n$th-order fibers. Thus, if we pull back $T'$ along a $Y$-point $s'$ of $Y'$ which approximates $s$ to $n$th order (as we may certainly do, from the local structure theorem for smooth morphisms) then we arrive at a finite radiciel surjection $T'' rightarrow Y$ which is an isomorphism over the $s'$-pullback of $U'$ and not an isomorphism on some infinitesimal special fiber. Since $Y$ is reduced and in fact weakly normal, to get a contradiction it suffices to arrange that ${s'}^{-1}(U')$ is not only dense in $Y$ (i.e., contains all generic points) but also that the corresponding open in $T''$ is schematically dense (so $T''$ inherits the reducedness from that open subscheme).

In other words, we are now faced with a problem having nothing to do with weak normality and everything to do with making a good choice for $s'$ to not lose contact with "dense open" downstairs and "reducedness" upstairs. Since $T' rightarrow Y'$ is a finite radiciel surjection onto a reduced local ring with $T'$ also reduced, we can find a global function $f$ on $Y'$ that is non-vanishing at all generic points of $Y'$ and $T'$, hence nowhere a zero-divisor in both places. The issue is to find the "slice" $s'$ of the local-smooth (residually trivial) $Y$-scheme $Y'$ so that $s'$ is specified to $n$th order (from the $widehat{Y}$-point) but $s'$-pullback preserve that property of $f$ being nowhere a zero-divisor both downstairs and upstairs.

If the base is a field there is nothing to do in the original problem, so we can assume the base $Y$ has positive dimension, so the local-smooth $Y'$ has dimension at least 2 (otherwise $Y' = Y$ and again there's nothing to do). In a reduced noetherian local ring with dimension > 1,
If we build up the slice $s'$ one step at a time (going through the "smooth coordinate" specializations into the base $Y$ approximating the given $widehat{Y}$-point one at a time), standard arguments with associated primes and symmetry of ${rm{Tor}}^1$ turn this into a task of prime avoidance: since we're allow to tweak things (for approx. purposes) beyond $n$th order in the closed point of the base, we are left having to solve the following general problem: if $A hookrightarrow B$ is a module-finite radiciel extension of reduced noetherian local rings of dimension at least 2 and with the same total rings of fractions, we need to find $a in A$ which is a non-zero-divisor in both $A$ and $B$ such that (i) its associated primes in both rings lie outside some specified nowhere-dense closed subset $Z$ of their "common" Spec and (ii) for any other given $a'$ in the maximal ideal of $A$ and chosen generating set ${a_1,dots,a_n}$ of an ideal for $Z$ a "generic" linear combination of the $a_i$ using a fixed set of representatives for the residue field can be added to $a'$ to get an element non-vanishing at the generic points and the associated points for $a$ viewed in each of the two rings. (So you see, the case of finite residue field needs more care; but when you unravel where this is coming from, you may find a better way to express it to avoid that problem.)

soft question - What are some famous rejections of correct mathematics?

The Mordell-Weil theorem, when submitted by Mordell to the London mathematical society's journal, was rejected.

This theorem was the start of the whole set of investigations on elliptic curves, and indeed on arithmetic geometry. Andre Weil in his Ph. D. thesis created the subject of arithmetic of algebraic varieties and Galois cohomology, to prove his strengthened version of this theorem and to understand Mordell's calculations. I also believe that for him the motivation to re-write the foundations of algebraic geometry was also motivated by the desire to give the Mordell-Weil theorem a cleaner form, thoughs the officially stated motivation is for putting his proof of Riemann hypothesis for function fields over finite fields on a firm ground. And, the subject grew, flowered, through greats like Grothendieck, and one must remark the work of Faltings on Mordell conjecture on the same direction proposed in the same paper, which could be proved only so many years later, after Weil failed in his Ph. D. time. Indeed, Fermat's last theorem proof also belongs to the same subject. Looking back, rejection of Mordell's groundbreaking paper is so unbelievable.

Excerpt from source:

Mordell submitted his subsequent work on indeterminate equations of the third and fourth degree when he became a candidate for a Fellowship at St John's College, but he was not successful. His paper on this topic was rejected for publication by the London Mathematical Society but accepted by the Quarterly Journal. Mordell was bitterly disappointed at the way his paper had been received. He wrote at the time on an offprint of the paper:-

This paper was originally sent for publication to the L.M.S. in 1913. It was rejected ... Indeterminate equations have never been very popular in England (except perhaps in the 17th and 18th centuries); though they have been the subject of many papers by most of the greatest mathematicians in the world: and hosts of lesser ones ...

Such results as [those in the paper] ... marks the greatest advance in the theory of indeterminate equations of the 3rd and 4th degrees since the time of Fermat; and it is all the more remarkable that it can be proved by quite elementary methods. ... We trust that the author may be pardoned for speaking thus of his results. But the history of this paper has shown him that in his estimation, it has not been properly appreciated by English mathematicians.

The details of Weil's work can be found in his autobiography, "Apprenticeship of a mathematician"..

nim - The game of "nimble" with no stacking

The game of Nimble is played as follows. You have a game board consisting of a line of squares labelled by the nonnegative integers. A finite number of coins are placed on the squares, with possibly more than one coin on a square. A move consists of picking up one of the coins and placing it on a square somewhere to the left of its original position. Players alternate moves. The game ends when no moves are possible, i. e. when all the coins are on the square labelled 0. The last player to move wins.

As described, this is just Nim in disguise, where a coin in position $n$ corresponds to a pile containing $n$ coins. The winning strategy is well-known. Define the nim-sum of some positive integers as the result when they're written in binary and addition is performed without carrying. Then the only winning positions are those where the nim-sum of all the pile sizes (or coin positions) are zero.

But what is the winning strategy if we don't allow coins to be stacked? This corresponds to Nim where we don't allow two piles of the same size, but this restriction feels a lot more natural in Nimble than in Nim.

There are two reasonable terminal positions. One is to allow stacking on the square marked zero only, and say the game terminates when all the coins are at zero; the other is to not allow stacking and say the game terminates when coins are on $0, 1, ldots, k-1$ if there are $k$ coins. Either one would be of interest.

If someone gives me an answer by Friday, I might talk about it in my class.

homotopy theory - geometry of null homotopies

Your question should follow from: given a homotopically trivial map $g:S^nto S^m$, is there a homotopy $h: D^{n+1} to S^m, h_{| partial D^{n+1}} = g$ such that $textrm{stretch}(h) leq textrm{stretch}(g)$? Or maybe there exists a reasonable function $L_n:[0,infty) to [0,infty)$ such that $textrm{stretch}(h) leq L_n(textrm{stretch}(g))$. I think you're claiming that there exists $L_n$ which is computable. It seems to me that since $textrm{stretch}(f^k)leq kcdottextrm{stretch}(f)$, the fact that $g=f^k$ doesn't really matter in the problem.

Interpreting the problem this way, one approach is to try to find for $zin S^m$ a factorization $g': S^n to T_z S^m cong mathbb{R}^m$, such that $g=exp_zcirc g'$ using the fact that $g$ is homotopically trivial. If one could control the diameter of the image of $g'$, then one ought to be able to bound the stretch of the trivial filling by coning to the origin in $T_z S^m$ (this amounts to filling in $g$ by coning by geodesics to a point). I don't know when such a lift is possible though, since obviously $exp_z: mathbb{R}^m to S^m$ is far from being a fibration at concentric spheres about the origin.

Here's a possible approach: Take a map $h: D^{n+1} to S^m, h_{| partial D^{n+1}} = g$ where $h(0)=p$. Think of this a map $H: S^{n} to [([0,1],0), (S^m,p)]$, that is a map from $S^{n}$ to the space of intervals at p. One may show that we may homotope $h$, keeping $h_{| S^{n}=partial D^{n+1}}$ fixed, to a map sending each interval $[0,1]x, xin S^{n}$ to a piecewise linear path. This is the process described in Theorem 17.1 of Milnor's book. The idea then would be to try to homotope all of the intervals to be piecewise geodesics of bounded length (above I suggested homotoping all of the intervals to geodesics, which is absurd), and then try to show that the resulting map has bounded stretch. Like the fundamental theorem of Morse theory, I would expect to be able to homotope things down onto $n+1$-cells which correspond to index $n+1$ geodesics, and therefore have bounded length, by something like cellular approximation. But I don't quite know how to complete the argument yet.

spherical geometry - Find the subset of a line on a sphere "far" from a set of points on the sphere.

I have some code where the "hot part" relies on an inefficient solution to this problem.

Problem: I have 3 inputs:
a. A collection of N points on the surface of a sphere.
b. A line segment on the sphere.
c. A distance X (distance can be on the surface or in 3D as it's trivial to map between them)

Output:
Find the subset of the line segment which is more than distance X from the collection of points.

(My problem actually involves a curve on the sphere, but I can reduce it to a line segment by chopping it up into smaller pieces.)

At present, I parametrize the line and created a distance function, subtract X then slam it into a method that finds the times when a function is positive. VERY slow.

Also, precomputations count in this algorithm. That is, the set does change over time, but it changes less frequently than I need this result for different line segments. Maybe 5 queries to every set change? That's worst case.

X is fixed over time.

gn.general topology - Different forms of compactness and their relation

Given a topological space X one can define several notion of compactness:

X is compact if every open cover has a finite subcover.

X is sequentially compact if every sequence has a convergent subsequence.

X is limit point compact (or Bolzano-Weierstrass) if every infinite set has an accumulation point.

X is countably compact if every countable open cover has a finite subcover.

X is σ-compact if it is the union of countably many compact subspaces.

X is pseudocompact if its if its image under any continuous function to $mathbb{R}$ is
bounded.

X is paracompact if every open cover admits an open locally finite refinement (i.e. every point of X has a neighborhood small enough to intersect only finitely many members of the cover).

X is metacompact if every open cover admits a point finite open refinement (i.e. if every point of X is in only finitely many members of the refinement).

X is orthocompact if every open cover has an interior preserving open refinement (i.e. given an open cover there is a open subcover such that at any point, the intersection of all open sets in the subcover containing that point is also open).

X is mesocompact if every open cover has a compact-finite open refinement (i.e. given any open cover, we can find an open refinement such that every compact set is contained in finitely many members of the refinement).

So, there are quite a few notions of compactness (there are surely more than those I quoted up here). The question is: where are these definitions systematically studied? What I'm interested in particular is knowing when does one imply the other, when does it not (examples), &c.

I can fully answer the question for the first three notions:

Compact and first-countable --> Sequentially compact.

Sequentially compact and second-countable --> Compact.

Sequentially compact --> Limit-point compact.

Limit point compact, first-countable and $T_1$ --> Sequentially compact.

but I'm absolutely ignorant about the other cases. Has this been systematically studied somewhere? If so, where?

co.combinatorics - Asymptotic growth of a certain integer sequence

Here is an easier way to see this result:

• If f(x)=$sum a_kx^k$ is a polynomial divisible by $(x-1)^{n+1}$ then $sum a_kg(k)=0$ for any polynomial of degree n (or less).

It is enough to verify this for a basis of the set of polynomials of these degrees. Since x=1 is a root of the jth derivative $f^{(j)}$ the claimed equality holds for $g(k)=k(k-1)(k-2)cdots(k-j+1)$.

So $(x-1)(x^2-1)(x^4-1)=x^7-x^6-x^5+x^4-x^3+x^2+x-1$ and hence $sum_{[0,3,5,6[}k^j=sum_{[1,2,4,7]}k^j$ for $j=0,1,2$. Then a(3)=7 because nothing smaller works for the restricted problem $sum_A k^j=sum_Bk^j$ just for $j=2$ but with $A,B$ a partition of the positive integers to some point. Note that the partition is according to the parity of the sum of the bits in the the binary form of m. The similar construction shows that the integers from 1 to b^{n+1}-1 can be partitioned into b sets of size b^{n) with the sets agreeing on all sums of jth powers j=0..n.

$(x-1)(x^2-1)(x^3-1)$ yields $sum_{[0,4,5]}k^j=sum_{[1,2,6]}k^j$ for $j=0,1,2$, but there is that gap at 3. The paper The Prouhet-Tarry-Escott problem revisited Enseign. Math. (2) 40 (1994) by Peter Borwein and Colin Ingalls gives a wealth of information about this amazing (other) problem of equal sums of powers jth powers j=0,1,2,...,n . It is usually available on the website http://www.cecm.sfu.ca/ but at the moment I could not get there.

The value a(3)=12 comes from the partition {1,2,4,8,9,12},{3,5,6,7,10,11} and there is the polynomial (shifting down) $$1+x-x^2+x^3-x^4-x^5-x^6+x^7+x^8-x^9-x^{10}+x^{11}=$$ $$(x^3-1)(x^8-x^7-x^6+2x^5-2x^3+x^2-x-1)$$

The value a(4)=16 comes from the partition {5, 6, 7, 13, 14, 15}, {1, 2, 3, 4, 8, 9, 10, 11, 12, 16} and there is the polynomial (shifting down) $$x^{15}-x^{14}-x^{13}-x^{12}+x^{11}+x^{10}+x^9+x^8+x^7-x^6-x^5-x^4+x^3+x^2+x+1=$$
$$(x^7-x^6-x^5-x^4+x^3+x^2+x+1)(x^8+1)$$ which is intriguing. However I did not find anything that factored for a(9) (where the sets are listed at the OEIS) and didn't do the search for the sets giving the other a(n).

It may be that you can find constructions giving solutions with nice patterns that are not minimal but do establish an upper bound. However I can't at the moment. I'd try polynomials either shifting k to $x^{k-1}$ or else putting a 1 for $x^0$ into one set or the other.

motivation - Why are spectral sequences so ubiquitous?

Qiaochu links to a really nice article by Timothy Chow that says a lot about the mechanics of how to go from a filtered complex to its spectral sequence. Two questions that remain are, (1) why do filtered complexes show up so much, and (2) is there anything that you could do with a filtered complex other than compute its spectral sequence?

VA gives a very general motivation for why filtered complexes show up so much. Probably it is just more general than what I was going to say. To discuss things more concretely, (I am told that VA's interesting answer is not a generalization of this paragraph.) For question (1), here are two general ways that filtered complexes show up: (a) you might get a filtered complex if you are interested in a chain complex that comes from a filtered object. For instance it could be a stratified topological space. A CW complex is stratified, but it has a special structure so that its chain complex doesn't need a spectral sequence. (It is an acyclic resolution as VA says.) If you have a stratification that you wish was CW but isn't, then you just have a general filtered complex. (b) There are some cases where just a single map or similar gives you a filtered complex. These may all be special cases of what Grothendieck described. For example, suppose that you are interested in the de Rham cohomology of a manifold. (This is then a certain resolution of the sheaf of smooth constant functions on the manifold.) Suppose further that the manifold is fibered over another manifold, or maybe is just foliated. Then differential forms are filtered by how parallel they are to the foliation.

Concerning question (2), there is an interesting result for filtered complexes over a field coming from representation theory. (I learned this some years ago in a discussion with Michael Khovanov.) The theorem in this case is that there is no more information in a filtered complex than in its spectral sequence. For simplicity, let's look at filtrations of length $n$. Then a filtered vector space is a representation of the $A_n$ quiver with all arrows in the same direction. (Each term of the filtration is assigned to a vertex of the quiver, and the arrows correspond to the inclusions.) The quiver also has other representations, but filtered vector spaces are the projective modules. A filtered complex is thus a projective complex over the $A_n$ quiver algebra. It is just like homological algebra over any other ring; you usually look at projective complexes. Since the $A_n$ quiver has projective dimension 1, it is easy to identify the indecomposable chain complexes of filtered vector spaces. There are two types, with two terms and with one terms:
$$0 to k^{(i)} to k^{(j)} to 0 qquadqquad 0 to k^{(i)} to 0.$$
In this notation, $k$ is the ground field and $k^{(i)}$ is $k$ in degree $i$. The first type of complex is valid if $i ge j$. If you compute the spectral sequence of an indecomposable complex, you will see that detects the first type of term at the $(i-j)$th page, and kills it on the next page. The second type of complex is of course the surviving homology. You can also go backwards and reconstruct the filtered complex from its spectral sequence.

Of course it is simplistic to only discuss filtered complexes and spectral sequences over a field. Nonetheless, roughly speaking spectral sequences are no more than a framework for analyzing filtered complexes.

---

VA asked for more details about the indecomposable modules, which is a fair request because I was quite cryptic about the relation. In particular, I use non-standard indexing in my own thinking about this. Unfortunately, I'm not sure that I can convert to standard indexing without making a mistake, so I won't convert. But I did change one thing above: I fixed the filtration degrees so that they are correct.

Suppose that $C = (C_n)$ is a complex of filtered vector spaces. Say that the filtration is increasing and indexed by $mathbb{Z}_ge 0$. The complex has two degrees: The chain degree $n$ and the filtration degree $k$. Suppose that $partial$ is a differential with chain degree $-1$ and filtration degree $0$. (This is where the numbering begins to be non-standard, although it makes sense from the point of view of quiver representations.) Then the theorem is that over a field, the two kinds of indecomposable complexes are those listed above, where the term $k^{(i)}$ has chain degree $n$, and the term $k^{(j)}$ (if present) has chain degree $n-1$. To be precise about what $k^{(i)}$ means, it is a filtration of the field $k$ in which the degree $j$ subspace is $0$ when $j < i$ and $k$ when $j ge i$.

The page $E^0$ is the associated graded complex. The page $E^r$ has a differential of degree $(-1,-r)$. When $r = i-j$, the differential $partial^r$ of the first type of indecomposable complex connects the "tip" of $k^{(i)}$ to the "tip" of $k^{(j)}$ and kills them both on the next page. The other kind of indecomposable complex has a vanishing differential, so the induced differential on every page also vanishes.

Can we realize Weyl group as a subgroup?

In general it is not possible to embed the Weyl group $W$ in the group $G$: already you can see this for $SL_2(mathbb C)$, where the Weyl group has order $2$: if the torus fixes the lines spanned by $e_1$ and $e_2$ respectively, you want to pick the linear map taking $e_1$ to $e_2$ and $e_2$ to $e_1$, but this has determinant $-1$. A lift of $W$ to $N(T)$ must be an element of order $4$ not $2$, say $e_1 mapsto -e_2$ and $e_2 mapsto e_1$.

In fact, Tits has shown that this is essentially the only obstruction: the Weyl group can always be lifted to a group $tilde{W}$ inside $G$ which is an extension of $W$ by an elementary abelian $2$-group of order $2^l$ where $l$ is the number of simple roots. If I recall correctly, this lift is then unique up to conjugation.

dg.differential geometry - A polynomial map from ℝ^n to ℝ^n mapping the positive orthant onto ℝ^n?

Question: Is there a polynomial map from ℝⁿ to ℝⁿ under which the image of the positive orthant (the set of points with all coordinates positive) is all of ℝⁿ?

Some observations:

My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.

Of course it is relatively easy to show that the answer is 'no' when n=1. (In fact it seems like a nice homework problem for some calculus students.) But I can't seem to get any traction for n>1.

This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of ℝⁿ!

ag.algebraic geometry - Corank 4 hypersurface singularities

I'm not sure that the definition at the beginning is correct. Let's assume that we have a holomorphic function germ $f : (mathbb{C}^n,0) to (mathbb{C},0)$. Let $mathscr{O}_n$ denote the local ring of holomorphic function germs $(mathbb{C}^n,0) to (mathbb{C},0).$ Furthermore, let $J_f$ denote the Jacobian ideal generated by the partial derivatives of $f$, i.e. $J_f := (partial f/partial z_1,ldots,partial f/partial z_n).$ The local algerba of $f$, denoted $A_f$, is given by

$A_f := mathscr{O}_n/J_f .$

The local algebra is a complex vector space. The dimension is finite if and only if $f$ has an isolated critical point at $0 in mathbb{C}^n.$ The dimension $dim_{mathbb{C}}left(A_fright)$ is called the Milnor Number of $f$ at $0 in mathbb{C}^n$ and is equal to the absolute value of the Poincaré-Hopf index of the gradient vector field $nabla f$ at $0 in mathbb{C}^n.$

The set of $z in mathbb{C}^n$ such that $f(z) = 0$ has a hypersurface singularity if the Milnor Number, $mu > 0.$ The corank one singularities are given by the $A_{mu}$ series where $mu ge 1$ ($mu = 1$ gives a Morse type singularity, i.e. a non-degenerate critical point). The corank two singularities are given by the $D_{mu}$ series where $mu ge 4.$ There are also three $E_{mu}$ singularity types where $mu = 6, 7, 8.$ The $A_{mu}$, $D_{mu}$, $E_6$, $E_7$ and $E_8$ are the only simple singularities.

That means that under the action of $mathscr{A}$-equivalence (diffeomorphic changes of coordinate in the source and the target), any sufficiently small neighbourhood of a function germ representative of a simple singularity intersects only a finite number of other orbits. It's related to the modality of Lie group actions.

The book only deals with corank $le 3$. Anything higher and the classification becomes too wild (and there aren't really any applications because these functions would be horribly non-generic). To get some idea, the set of function in $n$ variables with corank $k le n$ has codimension $frac{1}{2}k(k+1)$ in the space of function germs with critical points at $0.$ That means that the set of functions with corank four has codimension 10. You would need a generic 10-parameter family of functions to except a single member to have a corank four singularity.

There is no series of singualrities called O. That notation is used in adjacency diagrams to represent some variable series. After the simple singularities (the countable families) there a uni-modular, bi-modular, ..., even functionally modular families. A modular family means there are a continuum of non-equivalent members of the family. Take a look at pages 242 - 257. There's a whole zoo of singularity types, but no O's .

The bottom line is that, as far as I know, if you can classify the corank four singularities then you'd write a book in the process and you'd be very famous. (You'd have done what Arnol'd and his school couldn't or wouldn't do.) Even for small corank you have moduli. So you'd have families with uncountably many non-$mathscr{A}$-equivalent members, i.e. functions. The only hope would be to use topological equivalence in order to reduce the moduli, but then you lose the point.

rt.representation theory - classification of irreducible admissible representations of GL(n)

Let me focus on the case when $K$ is non-archimedean; the archimedean case is somewhat easier.

There is a coarse classification, valid for any reductive group, into supercuspidals,
and all the others --- the others are ones that can be parabolically induced from supercuspidals of proper Levi's, and so in principle are understood by induction,
while the supercuspidals are the basic building blocks. There is a subtley that
certain parablic inductions of irreducibles are not themselves irreducible, which leads
to so-called special representations (EDIT: and in the general, ie. non-$GL_n$ case,
so-called packets), but at least in the $GL_n$ case these are well-understood too.
(EDIT: In particular the packets are actually just singletons).

So everything comes down to the supercuspidals. (This is explained in the introduction
to Harris and Taylor's book, among many other places.) This coarse classification is also
compatible in a natural way with the local Langlands correspondence.

For $GL_n(K)$ the supercuspidals are completely classified. (This is the difference between
$GL_n$ and most other groups.) In fact there are two forms of the classification.

(1) Via the local Langlands correspondence (a theorem of Harris and Taylor).

(2) Via the theory of types (a theorem of Bushnell and Kutzko).

The first classification relates them to local $n$-dimensional Galois reps. The second
relates more directly to the internal group theoretic structure of the representations.

As far as I know, the two classifications are not reconciled in general (say for large $n$,
where large might be $n > 3,$ or something of that magnitude), and this is an ongoing topic of investigation by experts in the area. (Any updates/corrections to this statement would be welcome!)

The difference in the archimedean case is that there are no supercuspidals, so everything comes down to inducing characters of tori, and understanding the reducibility of these parabolic inductions.

soft question - How would You encourage graduate students to learn algebraic geometry and/or complex analysis?

Hello,

I am the 3rd year undegraduate student of mathematics.
After I obtain a bachelor degree I want to study maths at graduate level, especially algebraic geometry and complex analysis.
This fields of mathematics are well-represented at my univeristy, so at the first glance this plan looks fine.

Unfortunately almost all of my collegues are not interested in this fields (they think that AG and CA are too technically involved; most of them are interested mainly in functional analysis and topology).
It will have such unpleasant effect on my studies, that most probably the most (or all) of courses in AG and CA in the upcoming year won't start at my university.

So I'll end up learning alone, from books.
It's not that it's a big problem to learn from the book, it's not a problem at all. But I think such learning have no comparison with the regular course, where I could discuss problems and see another approches of other (more gifted) students.

To avoid such situation I may try to convince my collegues to study CA and AG, but I don't have many arguments since I'm still an ignorant in this fields.
And here arise my questions:

1. How would You encourage graduate students to learn algebraic geometry?


2. How would You encourage graduate students to learn complex analysis?

noncommutative geometry - Why is it called *spectral* triple?

Well, it uses the spectral properties of the Dirac operator $D$ in the spectral triple quite extensively.
Also, in the article where he (essentially) introduces the notion of spectral triples ( http://www.alainconnes.org/docs/reality.pdf ) Alain Connes writes about the naming:

"We shall need for that purpose to adapt the tools of the differential and integral calculus
to our new framework. This will be done by building a long dictionary which relates the usual
calculus (done with local differentiation of functions) with the new calculus which will be done
with operators in Hilbert space and spectral analysis, commutators.... The first two lines of the
dictionary give the usual interpretation of variable quantities in quantum mechanics as operators in
Hilbert space. For this reason and many others (which include integrality results) the new calculus
can be called the quantized calculus’ but the reader who has seen the word “quantized” overused
so many times may as well drop it and use “spectral calculus” instead."

universal algebra - Terminology: Name for a homomorphism from the free object?

Is there a standard name for taking a homomorphism from the free object over an algebraic structure? Roughly speaking, this should amount to evaluation of any element of the free object under the homomorphism.

For example, take the free commutative monoid over the set of natural numbers; elements of this monoid are finite multisets of natural numbers. By taking commutative monoid homomorphisms we can express many common operations, such as cardinality (map every generator to 1 and take sum as the monoid operation), sum (take every generator to itself and map the monoid operation to sum of natural numbers), product (take product of natural numbers rather than sum) etc.

Since these operations are all abelian monoid operations, we can expect a sensible result which will not depend on our multiset being non-empty or listing its elements in a specific order.

Does this idea generalize to other algebraic structures? Any known references?

nt.number theory - Interpreting Euler's Criterion for Idoneal Numbers

I think that we have had a previous thread related idoneal numbers here. The known idoneal numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 21, 22, 24, 25, 28, 30, 33, 37, 40, 42, 45, 48, 57, 58, 60, 70, 72, 78, 85, 88, 93, 102, 105, 112, 120, 130, 133, 165, 168, 177, 190, 210, 232, 240, 253, 273, 280, 312, 330, 345, 357, 385, 408, 462, 520, 760, 840, 1320, 1365, and 1848 there is possibly one more. If there is an additional idoneal number it must be greater than 8436. If the generalized Riemann hypothesis holds the above sequence is complete. I got this number from the wikipedia article there is also a online integer sequence for these numbers at A000926.

co.combinatorics - Derangements with repetition

Hi all,

given (a1,...,an) formed by distinct letters, it's a well known problem to count the number of permutations with no fixed element.

I've been trying to solve a generalization of this problem, when we allow repetition of the letters.

I was able only to partially solve the problem when we have only repetition of a single letter.

If we have n letters and only one of them is repeated p times, then the number O(n,p) of permutations with no fixed element is given by the following recursive relation:

$O(n,0)=O(n,1)=mbox{Derangement}(n)$

$O(p+1,p)=dots=O(2p-1,p)=0, O(2p,p)=p!$

$O(n,p)={n-pchoose p} p!sum_{k=0}^p{p choose k}O(n-p-k,p-k)$

Does anybody know where this problem have been studied before? Does anybody know a general solution for this problem?

Thanks in advance.

quantum groups - Establishing the Co-Quasi- Triangular Structure of FRT Algebras

Let me try to rephrase everything in more modern terms.

First of all $Rinmathrm{End}(Votimes V)$. Then let me denote by $U$ the $mathbb{C}langle u^i_j rangle$-valued matrix with entries being $u^i_j$'s ($Uin{rm End}(V)langle u^i_j rangle$).

Now the two-sided ideal $mathcal J(R)$ is generated by the entries of the matrix
$$
M(U):=U_1U_2R-RU_2U_1.
$$
This expression lies in ${rm End}(Votimes V)langle u^i_j ranglecong{rm End}(V)otimes{rm End}(V)otimes mathbb{C}langle u^i_j rangle$. Viewing elements of this space as tensors with $3$ components some people rewrite it as follows:
$$
U_{1,3}U_{2,3}R_{1,2}-R_{1,2}U_{2,3}U_{1,3}.
$$

REMARK: it seems that Klimyk and Schmudgen chose a similar but different $M(U)$, but it is not very important.

Now we want to define $r:mathbb{C}langle u^i_j rangleotimes mathbb{C}langle u^i_j rangleto mathbb{C}$. It is sufficient to define it on generators, and the best way to organize the corresponding coefficients is to give an expression for
$$
r(Uotimes U)in{rm End}(V)otimes{rm End}(V)cong{rm End}(Votimes V).
$$
We define naively $r(Uotimes U):=R$.

We then need to check that the elements $r(M(U)otimes U)$ and $r(Uotimes M(U))$, lying in ${rm End}(V^{otimes3})$, vanish. Let me try with the second one.

First part:
$$
r(Uotimes U_1U_2R)=r(Uotimes U_1U_2)R_{2,3}=r(Uotimes U_1)r(Uotimes U_2)R_{2,3}=R_{1,2}R_{1,3}R_{2,3}
$$

Second part:
$$
r(Uotimes RU_2U_1)=R_{2,3}r(Uotimes U_2U_1)=R_{2,3}r(Uotimes U_2)r(Uotimes U_1)=R_{2,3}R_{1,3}R_{1,2}
$$

So there is no problem here since we find Yang-Baxter.

Let me try now with the first one.

First part:
$$
r(U_1U_2Rotimes U)=r(U_1U_2otimes U)R_{1,2}=r(U_1otimes U)r(U_2otimes U)R_{1,2}=R_{1,3}R_{2,3}R_{1,2}
$$

Second part:
$$
r(RU_2U_1otimes U)=R_{1,2}r(U_2U_1otimes U)=R_{1,2}r(U_2otimes U)r(U_1otimes U)=R_{1,2}R_{2,3}R_{1,3}
$$

There seems to be a problem here since we find an expression which is not Yang-Baxter:
$$
R_{1,3}R_{2,3}R_{1,2}-R_{1,2}R_{2,3}R_{1,3}
$$
But there is no: applying the flip $tau_{1,2}$ we get
$$
R_{2,3}R_{1,3}R_{2,1}-R_{2,1}R_{1,3}R_{2,3}
$$
which is an avatar of Yang-Baxter (as far as $R^{op}=R^{-1}$).

ca.analysis and odes - How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem?

I'm studying the proof of the Riesz Representation Theorem as it appears in Ch. 6 of Royden's Real Analysis. When I looked on the web I noted there are a few different theorems that go by the name "Riesz Representation Theorem" so I'll state the one I'm looking at:

Let F be a bounded linear functional
on $L^p$, $1 leq p < infty$. Then
there is a function $g in L^q ni$,
$F(f) = int fg$.

The proof starts by showing that g exists for the characteristic functions $chi_s = chi_{[0,s]}$. Then we can write any step function as the sum of $chi_{s_{i}}$. So based on an earlier theorem, if f is a bounded measurable function in [0,1] we can write a sequence of step functions, $<psi_n>$ that converge to f almost everywhere. This is the sentence that confuses me:

"Since the sequence $<|f- psi_n|^p>$
is uniformly bounded and tends to zero
almost everywhere, the bounded
convergence theorem implies that
$||f-psi_n||_p rightarrow 0$."

But, when I look at the bounded convergence theorem, it would require $mathop{lim}limits_{n to infty} |f-psi_n|^p = 0$. Period. Not just almost everywhere, to get $mathop{lim}limits_{n to infty} int_{[0,1]}|f-psi_n|^p = int_{[0,1]} 0 = 0$.

So, that's where I'm stuck. I just don't see how the bounded convergence theorem can work here. (Side question: I also, don't feel I really know what Royden means by "uniformly bounded" is that just saying there is one bound that works for the whole set? How is that different from regular bounded?)

nt.number theory - Prime numbers that lead to relatively prime

The answer is no. As the Wikipedia article in my comment states, the counterexample $p = 17, q = 3313$ was found by Stephens in 1971, but the stronger question of whether one can ever divide the other is a famous open problem because its solution would greatly simplify a step in the proof of the Feit-Thompson theorem.

nt.number theory - Does 2^m = 3^n + r have finitely many solutions for every r?

Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves.

Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+kphi(M)} equiv 3^{n+kphi(M)} + r (mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.

T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der
Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad.
Oslo, 12 (1937), 1–16.

Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 equiv 3^2 + 11 mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)

co.combinatorics - Random Alternating Permutations

Below is Mathematica code based on Igor Pak's answer. To get a random downup permutation on $[n]$, start by choosing the first entry $p_1$ with the appropriate probability; then randomly choose an updown permutation of size $n-1$ from the updown permutations with first entry $< p_1$; then join them together (incrementing entries $ge p_1$ in the updown permutation).

To implement this method, we actually need code to generate a random downup permutation with first entry $ge$ a specified number $k$ and the code below does so. It uses the ComplementPermutation operation to interchange updown and downup permutations.

(* e[n,k] is the Entringer number *)

e[0,0] = 1;
e[n_,0]/;n>=1 := 0;
e[n_,k_]/;k>n || k<0 := 0
e[n_,k_] := e[n,k] = e[n,k-1] + e[n-1,n-k]

ComplementPermutation[perm_] := Module[{n=Length[perm]}, n+1-perm];

incrementSpecifiedAndUp[perm_,k_]:=perm/.{i_/;i>=k :> i+1};

partialSums[list_] := Drop[FoldList[Plus,0,list],1];

RandomUpDownPermFirstEntryAtMostk[n_,k_]/;k==n :=
RandomUpDownPermFirstEntryAtMostk[n,n-1];
RandomUpDownPermFirstEntryAtMostk[n_,k_]/;1<=k<n :=
ComplementPermutation[RandomDownUpPermFirstEntryAtLeastk[n,n+1-k]]

RandomDownUpPermFirstEntryAtLeastk[1,1]={1};
RandomDownUpPermFirstEntryAtLeastk[2,2]={2,1};

RandomDownUpPermFirstEntryAtLeastk[n_,k_]/; n>=3 && 2<=k<=n :=
Module[{keys,m,i,firstEntry,restOfPerm},

(* pick first entry using the Entringer distribution *)
keys=partialSums[Table[e[n-1,j],{j,k-1,n-1}]];
m=Random[Integer,{1,Last[keys]}];
i=1;
While[Not[ m<=keys[[i]] ],i=i+1];
firstEntry=k-1+i;

(* choose restOfPerm uniformly from updowns with their first entry < firstEntry *)
restOfPerm=RandomUpDownPermFirstEntryAtMostk[n-1,k-2+i];

(* amalgamate firstEntry and restOfPerm *)
Join[{firstEntry},incrementSpecifiedAndUp[restOfPerm,firstEntry]] ]

RandomDownUpPerm[1]={1};
RandomDownUpPerm[n_]/;n>=2 := RandomDownUpPermFirstEntryAtLeastk[n,2]

Sample output:
In[264]:=RandomDownUpPerm[15]
Out[264]=
{8, 2, 4, 1, 15, 6, 7, 3, 10, 9, 13, 11, 14, 5, 12}

mg.metric geometry - Problem equivalent to "largest square in a cube"

When studying this problem in the general case, i.e. $m$-dimensional cube inside $n$-dimensional cube for $m<n$, it might make sense to look at small examples to gain some intuition. From the small examples (largest square in (hyper)cubes) one could get the impression that the coordinates of the optimal solution are always rationals and hence the optimal edge length are always square root of rationals. In this post I argue that this is not the case in general.

Here is how your formulation as an optimization problem generalizes: Let $S$ be a set of half of the $2^m$ vertices of the m-cube of diagonal lenght 2 centered at the origin such that $S cup {-s : sin S}$ is the entire set of vertices. Now find $2^{m-1}$ unit-vectors in $mathbb{R}^n$ such that each pair of them has the same vector product as a corresponding pair in $S$, and such that the absolute value of their coordinates in minimized.

For example for m=3 one can take as $S$ every other vertex of the cube, such that they form a regular tetrahedron (see picture by Kepler), and then for each pair there vector product would have to be $-frac{1}{3}$

One potential problem with this formulation as optimization problem in practice is that the number of variables is $n2^{m-1}$ and there are $frac{2^{m-1}(2^{m-1}+1)}{2}$ quadratic equations.

Together with this optimization formulation and methods described in http://arxiv.org/abs/1407.0683 I calculate some cases $f(m,n)$ for small $m$ and $n$. There is some numerical optimization involved, so it is not a formal proof that the following result is optimal, but I obtain exact symbolic values for the coordinates and it is certainly a lower bound.

On mathworld the cases $f(1,n)$ and $f(2,n)$ are explained, so the smallest cases that seem to be unknown are $f(3,n)$. The cases $f(3,4)$ and $f(3,5) $ are also mentioned in Sloane's oeis: A243309, A243313 .

---

The 8 vertices of a largest 3-cube inside $[0,1]^4$ have the following coordinates:

$$(1, 0, 1-a, b),(1, 0, b, 1-a), (0,c,1-d, 0), (0,c, 0, 1-d),(1, 1-c, 1,d), (1, 1-c,d, 1), (0,1, 1-b, a),(0, 1, a,1-b)$$

where $a,b,c,d$ are algebraic numbers of degree $4$, with the these minimal polynomials and decimal approximations:

$$begin{align*}aquad&16x^4 + 8x^3 - 23x^2 + 14x - 2& 0.204901553506651293143\
bquad&16x^4 - 24x^3 + 25x^2 - 14x + 1& 0.082734498297453867827\
cquad&8x^4 - 32x^3 + 45x^2 - 30x + 1& 0.035139649420907685891\
dquad&4x^4 - 4x^3 - x^2 + 4x - 1& 0.287636051804105160970
end{align*}$$

This gives a edge length s, which hast the following minimal polynomial:

$$4x^8 - 28x^6 - 7x^4 + 16x^2 + 16$$

and $1.007434756884279376098253595231$ as decimal approximation.

---

The 8 vertices of a largest 3-cube inside $[0,1]^5$ have the following coordinates:
$$(1, e, 0, e, 0), (1, e, 0, 1, 1-e), (0, 0, e, 0, e), (0, 0, e, 1-e, 1), (1, 1, 1-e, e, 0), (1, 1, 1-e, 1, 1-e), (0, 1-e, 1, 0, e), (0, 1-e, 1, 1-e, 1)$$

where $e=sqrt{frac{3}{2}}-1approx 0.224744871391589049098$

This gives as edge length $sqrt{11-8sqrt{frac{3}{2}}}approx 1.09637631717731280407593110$.

---

The 8 vertices of a largest 3-cube inside $[0,1]^6$ all lie on the vertices of (a diagonal section of) the 6-cube:

$$(0, 1, 1, 0, 1, 0), (0, 0, 0, 1, 1, 1), (1, 0, 1, 0, 0, 1), (1, 1, 0,
1, 0, 0), (1, 0, 0, 1, 0, 1), (1, 1, 1, 0, 0, 0), (0, 1, 0, 1, 1, 0),
(0, 0, 1, 0, 1, 1)$$

and this gives an edge length of $sqrt{2}approx 1.414213562373095048$.

at.algebraic topology - De Rham cohomology and antiderivatives

This probably doesn't answer your question, but maybe it is close.

The (Extended) Poincare lemma says that if U is a contractible smooth $n$-manifold and
$omega$ is a closed differentiable form meaning $domega=0$ then there is a form $phi$
with $dphi=omega$. So, $phi$ is an antiderivative of $omega$.

More to the point, if $omega$ is a closed form on the smooth manifold $M$ then $[omega]in H^*(M)$, and $omega$ has an antiderivative if and only if $[omega]=0$.

More specifically, suppose that $M$ is an oriented smooth $n$-manifold so that integration
is defined, and $omega$ is a smooth $n$-form, then $omega$ has an antiderivative if and only if $int_Momega=0$.

If $M$ is in addition compact, and $omega$ is a closed $i$-form then $omega$ has an antiderivative if and only if for all closed $n-i$ forms $eta$, $int_M omegawedge eta=0$.

If $omega$ is not closed it can never have an antiderivative.

cv.complex variables - Bernstein inequality for multivariate polynomial

Tung, S. H. Bernstein's theorem for the polydisc. Proc. Amer. Math. Soc. 85 (1982), no. 1, 73--76. MR0647901 (83h:32017)

(from MR review): Let $P(z)$ be a polynomial of degree $N$ in $z=(z_1,cdots,z_m)$; suppose that $|P(z)|leq 1$ for $zin U^m$; then $|DP(z)|leq N$ for $zin U^m$ where $|DP(z)|^2=sum_{i=1}^m|partial P/partial z_i|^2$.

Here $U^m$ is the polydisc. Same author proved Bernstein-type inequality for the ball,
Tung, S. H. Extension of Bernšteĭn's theorem. Proc. Amer. Math. Soc. 83 (1981), no. 1, 103--106. MR0619992 (82k:32013)

linear algebra - Extremum under variations of a traceless matrix

Sorry for my precedent tentative, I was a little hasty:

Ok, I think I'd better put the original problem:

I have an action of three fields: $A$ which is the spin-connection, $B$ an skew-symmetric 2-form and $Phi$ which is traceless and skew-symmetric scalar field. These fields take their values on some algebra, index their components in this algebra by $i,j,k,... = 1,2,3$

I want to implement a certain condition on B by using equations of motion of $Phi$, the action is:

$S=int (B_i wedge F^i + Lambda B_i wedge B^i + Phi_{ij} B^i wedge B^j) $

Now for me equations of motions are simply:

$B^i wedge B^j=0$

perhaps with the condition that all diagonal elements are equal (as jc showed) but this is automatically satisfied for a skew-symmetric matrix (here $B^i wedge B^j$).

But in all papers I find:

$B^i wedge B^j - frac{1}{3}delta^{ij}B_kwedge B^k = 0$

So I see that they all took the traceless part of the matrix representing equations of motion, necessarily it has a relation with the traceless character of $Phi$ but I do not see which one.

In addition, this expression is not antisymmetric in $i,j$.

Would anyone have an idea?

ca.analysis and odes - Notation for eventually less than

I agree with Joel Hamkins's answer, but I don't entirely agree with his comment on that answer. I generally use asterisks to mean "with finitely many exceptions" or "modulo finite sets", so I'd use $fleq^*g$ and $Asubseteq^*B$ as Joel says. But when working modulo some ideal $I$ other than the ideal of finite sets, I'd ordinarily avoid asterisks and instead write $fleq_Ig$ and $Asubseteq_IB$.

I'd like to protest vigorously against the use of $ll$ in this situation. To me, $fll g$ means that $f$ is a lot smaller than $g$ (at least eventually), whereas here you might have $f(n)=g(n)-1$ for all $n$.

terminology - Is there a name for this property of a topology?

In spaces where singleton points are closed, your property is equivalent to saying that the space has no isolated points. Or in other words, that it is perfect.

Clearly, no space with an isolated point can have your property. Conversely, when singletons are closed, then you can subtract one point from any open set and thereby have a proper open subset. So if U has at least 2 points x,y, then U = U-{x} union U-{y}, giving an instance with I of size 2.

However, your property does not imply that points are closed, since the space on reals R, where open sets have the form (-infty, a), has your property, but points are not closed in this space.

big list - One-step problems in geometry

Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:

There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?

To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)

P.S. When I was asked the question, I failed miserably!

st.statistics - Uniqueness of the function used in the Khinchine criterion for characteristic function

Khinchine criterion for $varphi$ being a characteristic functions states that it is a ch.f. if and only if $varphi(0)=1$ and

$varphi(t) = int_{-infty}^infty g(t+theta) overline{g(theta)} dtheta$

Is function $g$ uniquely defined for a given $varphi$?

vector bundles - Uniqueness of Chern/Stiefel-Whitney Classes

This question is closely related to this previous question.

Chern and Stiefel-Whitney classes can be defined on bundles over arbitrary base spaces. (In Hatcher's Vector Bundles notes, he uses the Leray-Hirsch Theorem, which appears to require paracompactness of the base space. The construction in Milnor-Stasheff works in general, as does the argument given by Charles Resk in answer to the above question. A posteriori, this actually shows that Hatcher's construction works in general too, since he really just needs $w_1$ and $c_1$ to be defined everywhere.)

The proof of uniqueness (as discussed in Milnor and Stasheff, or in Hatcher's Vector Bundles notes, or in the answers to the above question) relies on the splitting principle, and hence (it seems to me) requires the existence of a metric on the bundle in question. More precisely, if we have two sequences of characteristic classes satisfying the axioms for, say, Chern classes, and we want to check that they agree agree on some bundle $Eto B$, the method is to pull back $E$ along some map $f: B'to B$ (with $f^*$ injective on cohomology) so that $f^*E$ splits as a sum of lines. Producing the splitting seems to require a metric on $E$ (or at least on $f^*E$).

If $B$ is not paracompact, bundles over $B$ may not admit a metric (and may admit a classifying map into the universal bundle over the Grassmannian), so my question is:

Are Chern and/or Stiefel-Whitney classes unique for arbitrary bundles? If not, do $w_1$ and $c_1$ at least determine the higher-dimensional classes?

mg.metric geometry - Hausdorff measure question

Say we have some compact metrisable topological space $X$ with a measure $mu$ defined on the Borel sets of $X$. Then is there some way to determine whether $mu$ is the Hausdorff measure associated to some metric $d$ compatible with the topology of $X$? And if so, is there some process to recover a metric from the measure? I'd imagine that there would have to be some conditions placed on the space $X$, eg. that it's connected, and it might even be necessary to assume that it's some nice space such as a manifold, with a "gauge" metric $d_{0}$ relative to whose Hausdorff measure $mu$ is absolutely continuous, but I'd like to ask the question in the greatest generality possible, in the hope that there is an answer out there.