qa.quantum algebra - Basis of quantum SU(n)

[edit: Following John's helpful comments below, I made this answer much more complete.]

Yes, this is the statement that $O_q(G)$ is a flat deformation of $O(G)$ for any semi-simple group G. See the book by Klimyk and Schmuedgen, "Quantum Groups and Their Representations" for a proof of this: on page 311 they state the relevant theorem for $Mat_q(n)$ (although the proof is just a reference to the original source). In the following section, they prove that det_q is central, which allows us to identify $O_q(SL_N)$ with $Mat_q(n)/(det_q-1)$. The OP asked about $SU(N)$, but in the context of algebraic groups one studies SL_N, which has a compact real form $SU(N)$, and morally the same representation theory.

In general we have to be careful when either inverting or specializing to a scalar any element in a noncommutative algebra, because this can in general drastically change the size of the algebra relative to what you'd expect from the commutative situation (it is bigger in the former case and smaller in the latter than expected). For inverting, you need the element to lie in a "denominator set", which assures that you don't have to add too many more things to invert it (imagine inverting $y$ in the free algebra $k((x,y))$ on two generators x and y: it would be a lot bigger than the vector space $k((x,y))[1/y]$). [edit: I can't get carot's or braces to work, hence the awkward symbol for free algebra; I hope it's clear.] For specializing, your element should honestly lie in the center of the noncommutative algebra, since it's image in the quotient will be a scalar (thus central). For instance, if you take $A^2_{q}=k((x,y))/(yx=qxy)$, this has the same basis as $A^2=k[x,y]$. However, quotienting A_q by y-1 forces x=0, which doesn't happen in A.

So far as I remember, the standard proof of the PBW theorem in this example (and many examples) relies on a technical lemma called the diamond lemma, Lemma 4.8 from KS, which gives an ordering on the monomials of O_q(G) compatible with the defining relations, allowing one to prove the existence of PBW basis.

lo.logic - term equality in algebraic theories

For algebraic theories how relevant is the underlying logic? Is it possible that two terms $s$ and $t$ can be shown to be equal with respect to one set of logical axioms but not necessarily so with another set of logical axioms? From my limited knowledge it seems that the logical axioms shouldn't matter much since playing with terms only requires substitution and reduction using the axioms of the theory.

My question is motivated by the syntactic category construction. I'm reading some notes on categorical logic and the syntactic category has a notion of morphism equality that depends on the theory but no mention is made of the underlying logic and I'm wondering why.

For example: Take the empty theory. The only terms are just variables so the objects of the syntactic category will be contexts and the morphisms will be tuples of variables, e.g. $x_1:[x_1,x_2]to[x_1], x_2:[x_1,x_2]to[x_1,x_2], (x_1,x_2):[x_1,x_2]to [x_1,x_2]$, etc. Now what can I claim about the arrows $x_1:[x_1,x_2]to[x_1]$ and $x_2:[x_1,x_2]to[x_1]$? Is it true that $x_1 = x_2$? If I assume the law of excluded middle then I should be able to claim something about $x_1 = x_2$ but if I don't assume the law of the excluded middle then it seems that I can't make a positive or negative claim about the status of $x_1 = x_2$ since the theory doesn't imply $x_1 = x_2$ so can I infer from this that $x_1 neq x_2$? I'm probably over-thinking it.

ag.algebraic geometry - Does smooth target space and smooth fibers imply smooth total space?

I apologize for answering such an old question, but it seems fundamental. A classical counterexample occurs for the abel map of a Prym variety with exceptional singularities on the theta divisor. The point is that the fibers of the abel prym map X-->Y of the double cover C'-->C are included among those for the abel map of C', hence are all smooth. (A map obtained by restricting another map over a subvariety of the target has the same fibers.)

Nonetheless X is singular at any exceptional divisor. (see lemma 2.13 of http://www.math.uga.edu/%7Eroy/sv2rst.pdf).

The point of the previous paper was that generalizing the Riemann - Kempf singularity theorem to prym varieties is easy when X is smooth. But when X is singular it is considerably harder:

http://www.math.uga.edu/%7Eroy/sv5rst2.pdf

http://annals.math.princeton.edu/2009/170-1/p05

For a detailed discussion of the case of the abel prym map for a prym variety isomorphic to the intermediate jacobian of the cubic threefold, see:

http://www.math.uga.edu/%7Eroy/onparam.pdf

The answer is yes however if the target Y is a smooth curve, since X is smooth at any point lying on a smooth cartier divisor, (compare Mumford, chap.7, Prop. 2, redbook.)

p-adic L-functions

The following is more a long comment than an answer per se.

One thing to keep in mind when discussing $p$-adic $L$-functions is that to a given algebraic automorphic representation $pi$ or Galois representation $rho$ is potentially attached several objects which could reasonably called the $p$-adic $L$-function of $rho/pi$. Largely for historical reasons, when one speaks of the $p$-adic $L$-function of $rho$ without further comment, one generally speaks of the $p$-adic $L$-function coming from the cyclotomic $mathbb Z_{p}$-extension, as I assume you do in your question. The most natural object from a strictly mathematical point of view seems to me to be the $p$-adic $L$-function attached to the universal deformation ring of $bar{rho}$ (at least when this universal deformation ring exists).

Even restricting yourself to the simplest case of the cyclotomic $p$-adic $L$-function, the case of $GL_{n}$ over $mathbb Q$ has not been done (that I know of) and I doubt (euphemism) that it will follow from the work of Eischen, Emerton, Harris, Li and Skinner (Emerton claims nothing of the sort). Unless I am very much mistaken, the cyclotomic case for $GL_{n}$ over $mathbb Q$ would be an extremely impressive progress. Somehow, the case of the anticyclotomic $mathbb Z_{p}$-extension of a CM field is sometimes easier because one can use the Rankin-Selberg method to prove that special values are algebraic and the Rankin-Selberg method is quite amenable to $p$-adic methods. I imagine that this is an ingredient in the work of EEHLS (but I know nothing about it, so please M.Emerton correct me if I'm wrong).

Leaving the real world for a second: conjecturally, cyclotomic $p$-adic $L$-functions are now constructed for any motive over $mathbb Q$ (though you will have a really hard time finding this in the literature, as one has to combine an impressive series of very involved papers). Of course, the conjectural construction would not tell you much in way of an actual construction (the conjectural construction gives you an element in some local cohomology group and you will have somehow to identify it as a global element), even though I admit I have been more than mildly impressed by an answer of Idoneal to a question here on MO about $p$-adic $L$-functions here which seems to indicate that analytic argument allows you to do just that in the case of modular forms.

Kevin Buzzard, sure the anticyclotomic $p$-adic $L$-function of an elliptic curve is part (technically, a specialization) of a two-variable $p$-adic $L$-function. In this setting and at least in the ordinary case, this has been known for more than 25 years (it was done in his thesis by S.Haran and later widely expanded by H.Hida in his Invent. Math. 79 paper). And further, this two-variable $p$-adic $L$-function is a specialization of a three-variable $p$-adic $L$-function taking into account variation of the weight in the Hida family passing through this elliptic curve. Even in the finite slope non-ordinary case, I think this three-variable $p$-adic $L$-function is known to exist by the work of A.Panciskin.

characteristic p - What is the correct formulation of the CDE triangle?

To revisit Bruce's earlier question, it might be useful to suggest a more sceptical alternative to Ben's answer and the related comments. I doubt that there will be a "correct" version of the CDE-triangle that has enough breadth to take in the variety of analogues that have emerged in representation theory, though it's obviously important to make the assumptions precise about underlying fields, local rings, etc. See also the conference paper: MR2184010 (2006g:17027) 17B67 (17B10)
Ekedahl, Torsten (S-STOC),
Kac-Moody algebras and the cde-triangle.
Noncommutative geometry and representation theory in mathematical physics, 49–58, Contemp.
Math., 391, Amer. Math. Soc., Providence, RI, 2005.

For me the essential ingredients to start with are suitable module categories (or more general analogues) in which three distinctive types of objects play a leading role: (1) simple modules, which are "small" and basic but often hard to get at directly; (2) indecomposable projectives (or injectives ...) which are also naturally present in suitable categories and usually have just finitely many composition factors, but are typically "large" and messy to study; (3) intermediate objects, easier to construct and often having known dimensions (if finite) or "formal" characters. The idea then is to express (say) a projective cover of a simple module formally, perhaps via a filtration, in terms of the intermediate objects. To make this interesting, when all composition factor multiplicities are finite or otherwise controllable, one wants a kind of "reciprocity" between multiplicities of simples in intermediate objects and multiplicities of the latter in the big modules. Here is some brief history followed by a sample of more recent instances. (It does get a bit long...)

(1) Elie Cartan actually studied what we now call finite dimensional associative algebras and emphasized the importance of knowing the composition factor multiplicities of indecomposable projectives (now dubbed Cartan invariants). For a finite group with $r$ classes of elements having orders not divisible by a given prime $p$, you get an $r times r$ matrix $C$ (over a large enough field).

(2) Work by Richard Brauer and his Toronto student Cecil Nesbitt after 1937 introduced in the finite group setting (for a prime $p$ dividing the group order and large enough fields) an $s times r$ decomposition matrix $D$ showing how to express the $s$ ordinary irreducible characters (= number of classes) as formal sums of $p$-modular irreducible characters (counting composition factor multiplicities in reduction mod $p$ for any suitable lattice in a module). Then $C = D^{t} D$ (so $C$ is symmetric), where the transpose of $D$ shows how to express a projective (lifted to characteristic 0) as a combination of ordinary characters.
These ideas were exposed by Curtis-Reiner (1962) in Section 83, etc. Brauer was studying $p$-blocks, which show up in the block decompositions of the matrices.

(3) Following Swan's formalism, Serre (1971) formulated the more abstract cde-triangle in his part III, using homomorphisms between various Grothendieck groups. This was further codified by Curtis-Reiner in their later 1981 book, Section 18, with a lot of attention to the rings and fields involved.

(4) Having studied the old CR book in a 1963-64 course at Yale taught by Jacobson (!), I later tried to adapt $C = D^{t} D$ to modular representations of Lie algebras of simple algebraic groups (working just in prime characteristic). The f.d. restricted enveloping algebra imitates a finite group algebra in some ways. Here the "intermediate" objects were f.d. analogues of Verma modules. At first I studied blocks and multiplicities but not filtrations. This rough version appeared in J. Algebra (1971) and inspired Verma's introduction of affine Weyl groups relative to $p$, as well as much more sophisticated work by Jantzen treating filtrations of projectives, plus action of a maximal torus in the group.

(5) Work by Bernstein-Gelfand-Gelfand in the early 1970s was partly inspired by Jantzen's work and by my 1971 paper, leading to their "BGG category" and "BGG reciprocity" in 1976. Then Kazhdan-Lusztig theory for finite and affine Weyl groups came into play, etc.

(6) Eventually some but not all of the ideas spread elsewhere in representation theory, including the work by Alvany Rocha and her thesis advisor Nolan Wallach
and much other work on Kac-Moody algebras (recently by Arakawa-Fiebig for the mysterious critical level in the affine case). Plus early work by Dan Nakano on other modular Lie algebra settings, recent work on rational Cherednik algebras, and so on.

(7) The most fruitful general formulation for some purposes was given in a 1988 Crelle paper by Cline-Parshall-Scott on highest weight categories and quasi-hereditary algebras. Other general settings were proposed by Ron Irving and by Apoorva Khare. It's hard though to find just one common framework.

ag.algebraic geometry - Is there a category-theoretic definition of the arithmetic Grothendieck group

Let $X$ be a regular scheme which is flat over $mathbf{Z}$. The arithmetic Grothendieck group $hat{K}(X)$ is defined to be the quotient of $hat{G}(X)$ by $hat{G}^prime(X)$. This is actually quite a length definition which I added below for the sake of completeness.

In the classical case, for $X$ any noetherian scheme, the Grothendieck group $K_0(X)$ is defined to be the Grothendieck group of the category of vector bundles on $X$. That is, one applies the notion of a Grothendieck group for an additive subcategory of an abelian category. (In our case the abelian category is the category of coherent sheaves on $X$.) This means just modding out by short exact sequences.

I would like to know if there is a categorical type of definition for this group. Thus, first one needs to decide what kind of categories we're talking about (objects are pairs in some sense) and then the notion of exact sequence should coincide in some sense with the below definition.

Probably there is no such thing. I just ask this question in order to understand the arithmetic Grothendieck group better.

Note. Let me sketch the definition of the arithmetic Grothendieck group as given in Faltings. In the above $hat{G}(X)$ is the direct sum of "the free abelian group generated by all vector bundles which have a hermitian metric on $X_{mathbf{C}}$ which is invariant under complex conjugation $F$" and the abelian group $widetilde{A}^ast(X)$. Here $widetilde{A}^ast(X)$ is generated by all $p$-forms $alpha^p$ such that $F^ast alpha^p = (-1)^p alpha^p$. Furthermore, $hat{G}^prime(X)$ is the subgroup generated by elements of the form $E_2 - E_1-E_3 - widetilde{ch}(E)$, where $E$ is the short exact sequence $$0rightarrow E_1 rightarrow E_2 rightarrow E_3 rightarrow 0$$ and $widetilde{ch}(E)$ is the secondary Chern form.

soft question - What are some examples of colorful language in serious mathematics papers?

The paper "Division by three" by Peter Doyle and John Conway has a wealth of colorful language including:

"If the arrows are good, straight, American arrows, it is very natural for each arrow to dream of marrying the arrow next door."

and

"Not that we believe there really are any such things as infinite sets, or that
the Zermelo-Fraenkel axioms for set theory are necessarily even consistent.
Indeed, we’re somewhat doubtful whether large natural numbers (like $80^{5000}$
, or even $2^{200}$) exist in any very real sense, and we’re secretly hoping that
Nelson will succeed in his program for proving that the usual axioms of
arithmetic—and hence also of set theory—are inconsistent. (See Nelson [6].)
All the more reason, then, for us to stick with methods which, because of their
concrete, combinatorial nature, are likely to survive the possible collapse of
set theory as we know it today."

ca.analysis and odes - What functions are not represented by their power series?

Functions are represented by their power series if and only if they are analytic, i.e. complex differentiable.

Power series are easier to understand as functions of complex variables. For example, there's no apparent reason why the power series of 1/(1 + x^2) centered at 0 should have radius of convergence 1. It's infinitely differentiable everywhere. But as a function of a complex variable, it has a singularity at x = i, and that's why the radius of convergence is 1.

at.algebraic topology - Twistings for other cohomology theories

Twisted forms exist for all multiplicative generalized cohomology theories. A nice paper which discusses a modern point of view for twists of homology, K-theory, and TMF is the following paper Twists of K-theory and TMF by Ando-Blumberg-Gepner.

If E is a generalized cohomology theory, represented by a spectrum also denoted E, then the E-cohomology of X coincides with the homotopy classes of maps

$$[ Sigma^{-i} X, E] $$

i.e. the "E-valued functions on X". Morally, if E is a ring spectrum then we can talk about E-module spectra, and about E-lines: those E-module spectra which are equivalent to E, but not necessarily canonically so. With a sufficiently robust theory, we should be able to talk about bundles of spectra over X, and in particular E-line bundles over X. Then the usual E-cohomology of X can be thought of as the sections of the trivial E-line bundle over X. An E-twist is a possibly non-trivial E-line bundle over X, and twisted E-cohomology consists of the sections of this line bundle.

The tricky part is making this philosophical picture into something mathematically precise. The above paper is one way to do this.

In general the E-lines are classified by the space $BGL_1(E)$, which is the classifying space of the $A_infty$-space $GL_1(E)$. This space is defined by the pull-back diagram

GL_1(E) --> &Omega&infinE
   |        |
   v        v
 &pi0(E)x --> &pi0(E)

From this you can read off the homotopy groups of $BGL_1(E)$ and you see that for $n geq 2$ they agree with those of E, but are shifted in degree.

More generally, when E is a commutative ring spectrum, one can study the larger class of "E-lines" which are invertible E-modules. This requires a robust theory of spectra where you have a good notion of smash product over E. This leads to a larger classifying space of E-lines whose zeroth homotopy group is the Picard group Pic(E). Even more generally, you could consider bundles of general E-modules (not necessarily invertible) to be twists. There are probably applications of this, but I don't recall any off-hand.

As far as geometric descriptions go, you might be asking for too much. Even for K-theory it is only the simplest kinds of twists corresponding to the bottom few homotopy groups of $BGL_1(K)$ which appear to have a clear geometric description (e.g. in terms of super gerbes and clifford algebras). The higher twists of K-theory are more subtle and it is not a priori clear that they have a purely geometric description.

ds.dynamical systems - Do there exist Markov partitions with (nearly) uniform SRB measures?

I wanted to elaborate on coudy's original answer, but I also think something is wrong with this argument. In the absence of any other answers I will make the bounty available for an explanation of what goes wrong here...

---

If $mathcal{P}$ is a partition of $M$, write $mathcal{P}_m^vee := bigvee_{j=0}^m T^j mathcal{P}$. (Because $T$ preserves the SRB measure $mu$ we don't have to consider, e.g., $j<0$ terms.)

A Markov partition $mathcal{R}$ is generating, so the supremum over partitions in the Kolmogorov-Sinai entropy is realized by it: the KS entropy is $h_mu^{KS}(T) = -lim_m m^{-1}sum_{R in mathcal{R}_m^vee} mu(R) log mu(R)$. Meanwhile the topological entropy is $h(T) = lim_m m^{-1} log # mathcal{R}_m^vee$.

So if $mu$ is also the measure of maximal entropy (as is the case, e.g. when $T$ is a hyperbolic toral automorphism or the Poincaré map for a geodesic flow on a surface of negative curvature), then $h_mu^{KS}(T) = h(T)$ and in the limit we have that $-sum_{R in mathcal{R}_m^vee} mu(R) log mu(R) sim log # mathcal{R}_m^vee$, so that $mu$ is asymptotically uniform on $mathcal{R}_m^vee$.

LaTeX based document editors

On linux, I usually use gummi. You type your code on the left and the document is compiled using pdflatex in real time and shown on the right. It's handy if you're not doing anything too long since you can see where you've gone wrong as soon as you type it.

For longer documents that might take a while to compile (ie longer than a second or two, since you will notice this in gummi), I'd use Kile. The only time I've ever really noticed this though is if I have a good few graphics written with xy-pic to compile in the document, but in that case, you can use OnlyOutlines to remove that delay while you're working on the text.

On that note though, if you're writing a large document, you can set up a bare-bones environment that will be used throughout your document, write each chapter individually and then just input{} them into a master document in order as each one is finished, meaning your compile for each section you're working with should be fast enough to use gummi.

It's still in early development, so it doesn't have any frills like project support, or any way of editing more than one document at once without running another instance of it, but I still love it.

ag.algebraic geometry - Which statements in section 5 of BBD will fail if we consider $mathbb{Q}_l$-adic sheaves there?

A stupid question: which statements in section 5 of BBD will fail if we replace $overline{mathbb{Q}_l}$-sheaves by just $mathbb{Q}_l$-ones? I am especially interested in Proposition 5.1.15.

BBD = Beilinson A., Bernstein J., Deligne P., Faisceaux pervers, Asterisque 100, 1982, 5-171.

gt.geometric topology - Properties of the n-dimensional Stereographic Projection

The circle and the north pole (or wherever the origin of the stereographic projection is) span a 3-dimensional subspace generically, such that the restriction to this subspace is the 2-dimensional stereographic projection. If the circle goes through the north pole, then it is actually sent to a line under stereographic projection, and this is in some sense a reduction to the 1-dimensional case.

Yana Mohanty has a nice proof that stereographic projection sends circles to circles.

A more sophisticated approach is to notice that stereographic projection is the restriction of inversion through a sphere orthogonal to $S^n$ in $R^{n+1}subset S^{n+1}$. Then one needs to see that inversions send circles to circles, or more generally that Mobius transformations of $S^n$ do. The group of Mobius transformations of $S^n$ is $PO(n+1,1)$ or $Isom(mathbb{H}^{n+1})$, the isometry group of hyperbolic $n+1$-space. This groups preserves the cone $x_0^2+x_1^2 +cdots - x_{n+1}^2=0$. The sphere at infinity (in the projectivization) of this cone is $S^n$, and the action is by the Mobius group. A circle is the intersection of the projective closure of a 3-dimensional subspace with the sphere at infinity. Since $PO(n+1,1)$ consists of linear transformations, it permutes 3-dimensional subspaces of $R^{n+1,1}$, and therefore sends circles to circles in the projectivization.

reference request - Galois groups via cohomology

I don't have a reference, but it does not seem too hard.

1. Assume $L/k$ normal, and take $sigma in G$, which can be extended to an element of $Gal(L/k)$. Then $sigma(sqrt{mu})/sqrt{mu} in L$, and if $gamma$ is the nontrivial element of $Gal(L/K)$, $sigma^{-1} gamma sigma$ is trivial on $K$, and being nontrivial on $L$ it has to be equal to $gamma$. So $gammaleft( sigma(sqrt{mu})/mu right) = sigma(gamma(sqrt{mu}))/gamma(sqrt{mu}) = sigma(sqrt{mu})/sqrt{mu}$, so we can take $alpha_{sigma} = sigma(sqrt{mu})/sqrt{mu}$, it is an element of $K$.
   Now for the other way, take $tilde{sigma} in Gal(bar{L}/k)$. Denote the restriction of $tilde{sigma}$ to $K$ by $sigma$. Then $sigma(sqrt{mu})/sqrt{mu}= pm alpha_{sigma}$ (this equality is in $bar{L}$), so $sigma(sqrt{mu}) = pm alpha_{sigma} sqrt{mu} in L$, so $L$ is normal.




2. Consider a set-theoretic section $sigma mapsto tilde{sigma}$ for the surjective morphism $Gal(L/k) rightarrow G$. Then the (up to a coboundary) 2-cocycle $beta_0$ associated to the group extension is given by the formula $tilde{sigma} tilde{tau} = beta_0(sigma,tau) widetilde{sigma tau}$. Evaluating at $sqrt{mu}$ gives the equality between $beta$ and $beta_0$, if for every $sigma in G$, $alpha_{sigma} = tilde{sigma}(sqrt{mu})/sqrt{mu}$. You can always choose your section so that it is the case (change of section = associating a sign to each $sigma in G$).

EDIT: There's a left/right action problem, because $beta_0(sigma,tau) = sigma(alpha_{tau}) alpha_{sigma} alpha_{sigma tau}^{-1}$ with what I wrote. I think it has to do with the fact that you use exponential notation, so somehow your action is on the right? Maybe you define $x^{sigma} = sigma^{-1}(x)$? Otherwise the definition of $beta$ doesn't make it a 2-cocycle, with the definitions I know. Could you clarify?

ds.dynamical systems - Is the composition of non-wandering maps still non-wandering?

I am wondering if there is an example $(f,X)$ and $nge2$ with $Omega(f)neqOmega(f^n)$.
It is interesting to know such examples.

There is an observation that for a homeo $f:Xto X$, if $xinomega(x,f)$, then $xinomega(x,f^n)$ for each $nge1$. The proof is:

Let $nge2$ be given. Note that $omega(x,f)=bigcup_{0le k< n}omega(f^kx,f^n)$. So there exists $k$ with $xinomega(f^kx,f^n)$.

If $k=0$ we are done.

Otherwise let $l=n-kin[1,n-1]$. Then $f^lxinomega(x,f^n)$. We show inductively $f^{jl}xinomega(x,f^n)$ for each $jge1$. Since $omega(x,f^n)$ is strictly $f^n$-invariant and $f^{nl}xinomega(x,f^n)$, we get $xinomega(x,f^n)$, too.

$f^{(j+1)l}x=f^l(f^{jl}x)in f^lomega(x,f^n)=omega(f^lx,f^n)subsetomega(x,f^n)$, where $in$ is from induction hypothesis and $subset$ is from the forward invariance of $omega$-sets.

riemannian geometry - Is the wedge product of two harmonic forms harmonic?

Generically, the wedge product of two harmonic forms will not be harmonic. It is harder to find examples than counter-examples. For example, on compact Lie groups with a bi-invariant metric or, more generally, on riemannian symmetric spaces, harmonic forms are invariant and invariance is preserved by the wedge product. In general, though, this is not the case.

According to Kotschick (see, e.g., this paper) manifolds admitting a metric with this property are called geometrically formal and their topology is strongly constrained. He has examples, already in dimension 4, of manifolds which are not geometrically formal.

st.statistics - Probability estimates for pairwise majority votes

This is related to the rank aggregation question I asked previously.

I have items $I_1, ldots, I_N$ and the observations of a number of pairwise trials which pit pairs $I_i$ and $I_j$ against eachother and select a "winner". Let $W_{ij}$ be the number of times i beats j.

Note that the number of trials between i and j is very much dependent on i and j: In some cases there may be none, in some there may be very many.

I am trying to estimate a matrix $P_{ij}$ corresponding to the probability that i beats j in a trial (I consider $P_{ii} = frac{1}{2}$ for convenience reasons). My current, somewhat unprincipled, approach is a bayesian average

[ P_{ij} = frac{ frac{1}{2} S + W_{ij} }{S + W_{ij} + W_{ji}} ]

where S is some smoothing constant (I currently have S = 5). This corresponds to a bayesian approach with a prior for $P_{ij}$ of $beta(frac{1}{2}S, frac{1}{2}S)$ and then taking the expected value of the posterior distribution.

My problem with this is the following:

This is effectively treating each pair i, j as independent, whereas in fact we "believe" that there is a consistency between them. In particular if i tends to beat j and j tends to beat k, this should count as evidence that i tends to beat k even in the absence of pairwise trials between i and k.

There may be circumstances where we have very many trials for i, j and j, k and conclude that both $P_{ij}$ and $P_{jk}$ are high, but we have very few trials for i, k and thus conclude that $P_{ik}$ is very close to $frac{1}{2}$ (possibly even concluding it's less than $frac{1}{2}$ if e.g. there was only one trial and it had a surprising result). This is non-optimal.

So I'd like some sort of reasonably principled way of introducing intermediate results as evidence that the majority prefers one to the other. There are various plausible sounding things I could try, but I'd like to do this "properly" if at all possible, and most of my ideas involve more hand waving than solid mathematics.

One example of something plausible but possibly nonsensical I'm considering trying is iterating an expand/collapse process of:

Expand: $P to P^2$

Collapse: $P to Q$, where $Q_{ij} = frac{P_{ij}}{P_{ij} + P_{ji}}$

The idea being that we inflate probabilities where there are a lot of large intermediate results and then collapse down to the symmetry condition that $P_{ij} + P_{ji} = 1$.

This seems to produce semi-tolerable results (I've not tested extensively yet), but it's not actually clear to me that this process converges or why it should work.

Suggestions?

Edit:

On having thought about this a little more carefully, I think the following may capture what I am trying to achieve:

I want to assume that there is some distribution on the permutations of 1..N, with a strong prior belief that this distribution is close to uniform, and that each pairwise trial consists of sampling from this distribution and comparing the positions of i and j.

pr.probability - statistical approach to multinomial distribution

The classical approach is to build a Neyman-Pearson style hypothesis test (warning: incredibly ugly mathematics, in desperate need of replacement, but ubiquitous).

Say you rolled your die $N$ times to produce $X$. Let the multinomial distribution have parameters $(p_1, p_2, ..., p_6)$, where $sum_i p_i = 1$. Then construct a one dimensional measure such as $Q = | X/N - p |$, using your favorite $p$-norm. Calculate the probability distribution of $Q$.

Your null hypothesis in this case is $p_i = frac{1}{6}$ for all $i$. For a test of level of significance $alpha$ (conventionally 0.05 or 0.01), there is a region $[a,b]$ such that $int_a^b p(Q = x) dx = 1 - alpha$. Actually, there are many such, and there are other criteria to choose among them. In your case, invariance might be a good one: you expect the whole problem to be symmetric if you let $Q$ go to $-Q$, in which case the interval should be symmetric about 0, i.e., $[-a,a]$.

For a given value of $Q$ from your data, you do the integral over $[-Q,Q]$ and get $1 - alpha$. That $alpha$ is the lowest level of significance at which the observed data will be significant.

As I said, classical hypothesis testing is a very ugly theory. There are other approaches, such as minimax tests which you can construct via Bayes priors, since the set of all Bayes priors contains but is usually not much larger than the set of all admissible statistical procedures.

gr.group theory - An explicit example of a finitely presented group containing a subgroup isomorphic to $(mathbb Q,+)$.

Every recursively presented (even infinitely generated) group can be effectively embedded into a finitely generated recursively presented group either by using HNN extensions (as in Higman-Neumann-Neumann original paper) or by using small cancelation quotients of the free group. Every finitely generated recursively presented group can be effectively embedded into a finitely presented group by the Higman embedding theorem. The finite presentation is explicitly constructed using any Turing machine recognizing the set of relations of the finitely generated group. This means that one can explicitly write down a finite presentation of a group containing $mathbb Q$. The number of generators and the number of relations will depend (linearly) on the number of commands in the Turing machine recognizing the defining relations of $mathbb Q$. The real question is to find a "natural" finitely presented group containing $mathbb Q$. That is not known so far. There are finitely presented groups containing close relatives of $mathbb Q$. The Baumslag-Solitar group $BS(1,d)$ contains the group of $d$-adic rationals. And the R.Thompson group $V$ contains the group ${mathbb Q}/{mathbb Z}$.

Update 1. The first step can be simplified for $mathbb Q$. For every $nge 2$ take the group $G_n=BS(1,n)=langle a_n,b_nmid b_n^{-1}a_nb_n=a_n^nrangle $. The direct product $Pi G_n$ contains $mathbb Q$. Add two generators $t,s$ to the presentation of the direct product and all relations $t^{-1}a_it=a_{i+1}$, $s^{-1}b_is=b_{i+1}$ for all $ige 1$. That is a finitely generated (by $a_1,b_1, t,s$) recursively presented group containing $mathbb Q$. The presentation of that group can be easily recognized by a Turing machine. Then the Higman construction gives a presentation of a finitely presented group containing $mathbb Q$. The presentation will contain something like a 100 generators and 100 relations (I did not compute exact numbers).

Update 2. In Valiev, M. K.
Universal group with twenty-one defining relations.
Discrete Math. 17 (1977), no. 2, 207–213, Valiev constructed an explicit presentation with 21 defining relations of a group containing all finitely presented groups, hence containing $mathbb Q$ (earlier a 26-relator example was constructed by Boone and Collins). The difference with the example in Update 1 is that it is hard to describe an embedding of $mathbb Q$ in that group. That embedding is defined by the Turing machine describing a presentation of $mathbb Q$.

ra.rings and algebras - Calculating norms over a finite field (orthogonal groups).

I'm trying to work through calculating the order of orthogonal groups in characteristic $neq 2$. However there is one proof by induction used that i can't quite follow. Could someone help me understand where the formula for $z_{m+1}$ comes from and how we know $U$ must contain $2q-1$ vectors with norm $0$ and $q-1$ vectors of every non-zero norm in the following extract:

Let $z_m$ denote the number of non-zero isotropic vectors in an orthogonal space with dimension $2m$ or $2m+1$. We claim that:

$z_m = q^{2m}-1$ for dimension $2m+1$

$z_m = (q^m-1)(q^{m-1}+1)$ for plus type with dimension $2m$

$z_m = (q^m-1)(q^{m-1}-1)$ for minus type with dimension $2m$

For our inductive step we look at a $n+2$ dimensional space $V$ to ensure all spaces remain of the same type. Split V into the direct sum of $U$ and $W$ where $U$ is a $2$-dimensional space of plus type and $W$ is an $n$-dimensional space with the same type as $V$. Any isotopic vector in $V$ can be written as $u+w$ for isotropic vectors $uin U$ and $win W$. Either $u$ and $w$ both have norm $0$ (with one being non-zero) or $u$ has norm $lambda neq 0$ and $w$ has norm $-lambda$. Since $U$ contains $2q-1$ vectors of norm $0$ (including the zero vector) and $q-1$ vectors of every non-zero norm we count:

$z_{m+1}=(2q-1)(1+z_m)+(q-1)(q^n-1-z_m)-1$

The other three cases are similar.

Thanks

nt.number theory - Separable and algebraic closures?

First of all, there is not the algebraic/separable closure. Choices have to be made. However, if an algebraic closure $k^{mathrm{alg}}$ of $k$ is fixed, inside it there is a unique separable closure $k^{mathrm{sep}}$ of $k$, namely the subfield consisting of the separable elements over $k$.

Ignoring the failure of uniqueness, you can regard $k^{mathrm{alg}}$ as the biggest algebraic extension of $k$, whereas $k^{mathrm{sep}}$ is the biggest galois extension of $k$. The latter is because $k^{mathrm{sep}}$ is easily seen to be normal. In particular, you can apply Galois theory and relate the group theory of the absolute Galois group $mathrm{Gal}(k^{mathrm{sep}}/k)$ with the field theory of Galois extensions of $k$. The algebraic closure is too big to make Galois theory work.

Obviously $k$ is perfect if and only if $k^{mathrm{alg}} = k^{mathrm{sep}}$. Finite fields and fields of characteristic $0$ (in particular number fields) are perfect. But what is the difference in the other cases? Let $p = mathrm{char}(k) > 0$. Then $k^{mathrm{alg}} / k^{mathrm{sep}}$ is purely inseparable, i.e. for every $a in k^{mathrm{alg}}$ there is some $n geq 1$ such that $a^{p^n} in k^{mathrm{sep}}$. In other words, this field extension is given by adjoining all $p^n$-th roots. A consequence of this is that the restriction map $mathrm{Aut}_k(overline{k}) to mathrm{Gal}(k^{mathrm{sep}}/k)$ is an isomorphism.

Actually one can show that the canonical map $k^{mathrm{sep}} otimes_k k^{mathrm{perf}} to k^{mathrm{alg}}$ is an isomorphism, where $k^{mathrm{perf}}=cup_{n geq 0} k^{1/p^n}$ is the perfect hull of $k$.

rt.representation theory - What is Extreme/Extremal vector according to some weights

I know this might be a very elementary question. But I could not find the original definition of Extreme(or Extremal)vectors of some weights $lambda$(fixed the $win W$,where $W$ is Weyl group)

I am looking for definition for Extreme vector for finite dimensional Lie algebra and Affine Lie algebra. I found a paper saying :"Extreme vector verifies Weyl Character formula" What does it mean?

I am looking for reference talking about this concept. Thanks in advance

EDIT: I guess it is just highest weight vector, but I am not sure

ap.analysis of pdes - Can the "physical argument" for the existence of a solution to Dirichlet's problem be made into an actual proof?

Well, I don't understand the electrostatics, but here is another physical heuristic:

Impose a temperature distribution at the exterior, and measure (after some time has passed) the temperature in the interior. This gives a harmonic function extending the exterior temperature. [What's the electrostatic analogue? Formerly I had written "charge density",
but now I am not sure if that's right.]

I think this strongly suggests a mathematically rigorous argument: We are naturally led to model the time-dependence of temperature in the interior.
This satisfies a diffusion (or heat) equation, but in words:

"After a time delta, the new temperature is obtained by averaging the old temperature along a circle of radius sqrt{delta}."

This process converges under reasonable conditions, as time goes to infinity, to the solution of the Dirichlet problem. Anyway, we are led to the Brownian-motion proof of the existence, which I personally find rather satisfying. Another personal comment: I think one should always take "physical heuristics" rather seriously.

[In response to Q.Y.'s comments below, which were responses to previous confused remarks that I made: neither the electric field nor the Columb potential is a multiple of the charge density on the boundary: the former is a vector, and in either
case imagine the charge on the boundary to be concentrated in a sub-region; neither the electric field nor the potential will be constant outside that sub-region.]

cv.complex variables - Elementary functions with zeros only at the positive integers

Does there exist a (meromorphic) elementary function $f(z)$ that is zero at all the positive integers $z = 1, 2, 3, ldots$ and only at those points?

Edit: an elementary function can be written as a finite composition of constants, rational functions, exponentials and logarithms.

Obviously a function with those zeros can be constructed using the gamma function or a Weierstrass product, but the question is whether there is an elementary function.

reference request - Approximation by analytic functions

Dear all.

Let
$$
f(x) = sum_{k in mathbb{Z}} hat{f}(k) exp(2pi mathrm{i} kx)
$$
be a function given by usual fourier series.

Since my original question hasn't got any answer yet, and I came across another related question, I am just adding it. Denote by $T_n$ the set of all trigonometric polynomials of degree $n$, that is $gin T_n$ if
$$
g(x) = sum_{k=-n}^{n} hat{g}(k) exp(2pi mathrm{i} kx).
$$
So now what is $min_{g in T_n} |f - g|_{infty}$ and what is the optimal $g$?

Since the Fourier series of a continuous function must not converge, I expect that the answer isn't $g(x) = sum_{k=-n}^{n} hat{f}(k) exp(2pi mathrm{i} kx)$ but something else. However, the other choice the Fejer kernel
$$
g(x) = sum_{k=-n}^{n} frac{n - |k|}{n} hat{f}(k) exp(2pi mathrm{i} kx)
$$
seems to give worse estimates on $min_{g in T_n} |f - g|_{infty}$ once $hat{f} in ell^2$.

Thanks,
Helge

Original question:

I am interested in the question of how well one can approximate $f$ by functions that are analytic in some strip. The naive approach yields for example that if one sets
$$
f_R(x) = sum_{|k|leq R} hat{f}(k) exp(2pi mathrm{i} kx)
$$
and assumes $f in C^{n+1}$ then $f_R(x)$ has an extension to a strip of width $frac{n log(k)}{2pi k}$ on which $f_R$ is bounded by $|hat{f}|_{ell^1}$.

This seems like a pretty natural question so I expect it to be well studied, but I don't know where... Does anybody has references? I am also interested in stronger regularity assumptions than $C^n$...

co.combinatorics - String of integers puzzle

I apologize for not have the math background to put this question in a more formal way.
I'm looking to create a string of 796 letters (or integers) with certain properties.

Basically, the string is a variation on a De Bruijn sequence B(12,4), except order and repetition within each n-length subsequence are disregarded.
i.e. ABBB BABA BBBA are each equivalent to {AB}.
In other words, the main property of the string involves looking at consecutive groups of 4 letters within the larger string
(i.e. the 1st through 4th letters, the 2nd through 5th letters, the 3rd through 6th letters, etc)
And then producing the set of letters that comprise each group (repetitions and order disregarded)

For example, in the string of 9 letters:

A B B A C E B C D

the first 4-letter groups is: ABBA, which is comprised of the set {AB}
the second group is: BBAC, which is comprised of the set {ABC}
the third group is: BACE, which is comprised of the set {ABCE}
etc.

The goal is for every combination of 1-4 letters from a set of N letters to be represented by the 1-4-letter resultant sets of the 4-element groups once and only once in the original string.

For example, if there is a set of 5 letters {A, B, C, D, E} being used
Then the possible 1-4 letter combinations are:
A, B, C, D, E,
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE,
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE,
ABCD, ABCE, ABDE, ACDE, BCDE

Here is a working example that uses a set of 5 letters {A, B, C, D, E}.

D D D D E C B B B B A E C C C C D A E E E E B D A A A A C B D D B

The 1st through 4th elements form the set: D
The 2nd through 5th elements form the set: DE
The 3rd through 6th elements form the set: CDE
The 4th through 7th elements form the set: BCDE
The 5th through 8th elements form the set: BCE
The 6th through 9th elements form the set: BC
The 7th through 10th elements form the set: B
etc.

* I am hoping to find a working example of a string that uses 12 different letters (a total of 793 4-letter groups within a 796-letter string) starting (and if possible ending) with 4 of the same letter. *

Here is a working solution for 7 letters:

AAAABCDBEAAACDECFAAADBFBACEAGAADEFBAGACDFBGCCCCDGEAFAGCBEEECGFFBFEGGGGFDEEEEFCBBBBGDCFFFFDAGBEGDDDDBE

reference request - What is the proper initiation to the theory of motives for a new student of algebraic geometry?

The word "motive" has a lot of different (although highly related) meanings. I suggest you go ahead to learn about "pure Chow motives" first, before looking at the more complicated theory of mixed motives.

For motivation, it is necessary to have seen at least one Weil-cohomology theory, so you might want to have a look at the Weil conjectures, too.

For technical stuff, you should know what an abelian category is (and then learn the rest along the way).

Related to the theory of motives are also: K-Theory, (stable) homotopy theory of schemes, intersection theory (Chow groups). If you have an interest in any of these topics, it might be good to look at a treatment that covers the relationsship between this and motives, to give a little bit more motivation.

Since there is no abelian category of mixed motives yet, but instead what "feels like" it's derived category, you might want to learn a little bit about derived categories and triangulated categories before walking to (Voevodsky's theory of) mixed motives.

Of course, there is also the AMS Notices article What is ... a motive? by Barry Mazur.

lo.logic - Is there a formula phi s.t. phi and not-phi have a stronger consistency?

Now that I know the answer, I've found my own simple proof. Probably it's interesting to someone else, so I post it:

I want to show that Con(Σ) is equivalent to ( Con(Σ + φ) OR Con(Σ + not φ) )

Proof:
Con(Σ + φ) OR Con(Σ + not φ) iff

( Σ doesn't prove [φ -> FALSE] ) OR
( Σ doesn't prove [not φ -> FALSE] ) iff

Σ doesn't prove [(not φ -> FALSE) AND (φ -> FALSE)] iff

Σ doesn't prove [FALSE], which is Con(Σ).

soft question - Good papers/books/essays about the thought process behind mathematical research

Not mentioned so far is Bill Thurston's On proof and progress in mathematics (1994). With more than three hundred citations, it surely qualifies as a classic ... it is a permanent left-column link on Terry Tao's weblog, for example.

Thurston's essay is unique, relative to other such essays, in that it describes (in Section 6, "Some Personal Experiences") not one path, but two distinct paths relating to thought processes in mathematical research:

• a solitary path associated to Thurston's early work on foliations


• a social path associated to Thurston's later work on the Geometrization Conjecture

Thurston's latter approach is the topic of much research today, under various rubrics that include "social media", "social networks", and "roadmapping".

The foresighted points -- by 17 years -- of Thurston's essay include:

• social elements of research can be consciously chosen by individuals


• fundamental mathematics can provide uniquely strong foundations for social enterprises


• healthy mathematical communities make faster progress, and also, a better environment for nurturing the next generation of young mathematicians.

A recent well-respected essay that amounts to a consensus abstraction of Thurston's ideas is the International Roadmap Committee (IRC) More-than-Moore White Paper. For modern-day systems engineers especially, it is very instructive to read-out the main themes of Thurston's 1994 essay from the IRC's 2010 white paper, and thus to appreciate that Thurston's ideas were far ahead of their time.

In particular, the IRC's five consensus preconditions for successful roadmapping are anticipated with near-perfection by Thurston's essay ... and this is why Thurston's essay no doubt will continue to gather new citations through decades to come.

proof theory - Independence of PA implies independence of PA union all true $Pi_1$ statements

The claim you have asked us to prove is not true. If PA is consistent, then by the Incompleteness Theorem there are $Pi_1$ statements that are independent of PA, such as Con(PA), which can be seen to be $Pi_1$ when expressed in the form "no number is the code of a proof of a contradiction in PA". Thus, if PA is consistent, then Con(PA) is a statement that is independent of PA but provable in $PA_1$, so it is a counterexample to your claim.

Perhaps a more striking counterexample would be $negtext{Con(PA)}$, which is independent of PA, but refutable in $text{PA}_1$. More generally, any statement having complexity $Sigma_1$ or $Pi_1$ that is independent of PA will be a counterexample to your claim, since such statements are settled by $text{PA}_1$.

Perhaps the folklore result you meant to ask about is the following?

Theorem. If a $Pi_1$ statement is independent of PA, then it is true.

Proof. If a $Pi_1$ statement $sigma$ is independent of PA, then it is true in some model $Mmodels PA$. The standard model $mathbb{N}$ is an initial segment of $M$, and since the statement $sigma$ is $Pi_1$, it has the form $sigma = forall n varphi(n)$, where $varphi$ has only bounded quantifiers. Since $sigma$ holds in $M$, it holds for all standard $n$ in $M$ and hence $sigma$ is true in the standard model. In other words, it is true. QED

Note that the proof that $sigma$ is true is not a proof in PA, but rather in a theory, such as ZFC, that is able to theorize about models of PA. So another way to view the theorem is as the claim that if ZFC can prove that a given $Pi_1$ statement is independent of PA, then ZFC can also prove that it is true.

co.combinatorics - A decision problem in graph coloring

Some thoughts.

I like the idea of inverting the graph. I think the problem can be converted, to coloring the (inverted) graph for which each color appears exactly twice, or just once but then only on selected vertices (a joker vertex).

If the original inverted graph contains a 3-clique, then one of the vertices of the clique can be selected and be allowed (but not necessarily) to be colored with a color not appearing anywhere else in the graph. You can repeat this step, until the graph does not contain any 3-clique anymore, that does not have a selected vertex. I think it is not difficult to prove that a coloring in the converted problem can be used to construct a coloring in the original problem and vice versa.

With the converted problem, you eliminate any vertex that has 1 or 2 edges. In case of 1 edge, you remove the vertex and its neighbor. In case of 2 edges, you contract. By contraction, you can create a new 3-clique. However, in the converted problem, you are not allowed to color that with one color (that is why the conversion is necessary, because it allows the contraction).

A cycle with an odd number of vertices, will end up in a single vertex, in which it becomes clear that a coloring is not possible. But not all impossible colorings will end up like that.

Finally, you can do a BFS. For the search-border, you have a set of possibilities. Each element of the set, specifies for every vertex on the border, whether it needs another vertex of the same color or not. You want the keep the search border small.

It might be NPC. For that, consider the vertices on the search-border as propositional variables and prove that any propositional expression can expressed as such graph (I don't know if that is possible).

Lucas

mathematics education - Depressed graduate student.

You have to ask yourself some basic questions.

1) Perhaps you lost your enthusiasm for a good reason. Maybe your initial enthusiasm was naive. Maybe you liked mathematics for unsustainable reasons? You're likely to go through far larger "down" cycles in the future if you stay in mathematics (we all do, it's a chronic problem in the field) so if you're going to stick with mathematics you have to find some kind of joy you can hold on to, through all kinds of messy situations.

2) Maybe you really do love mathematics but there's aggravating factors causing problems. Maybe you're not working on enough easy problems. Solving easy problems is fun, and if they're the right kind of problems you build up new skills. This is one of the reasons why I frequent this webpage.

I had a bunch of issues like (2) as a grad student. IMO I'm mostly better off for them. I'm talking about issues like solving a problem (or making progress on a problem) and finding out perhaps way too late that the problem had been solved by someone else. I found it pretty tricky to balance focus with awareness of what other people are doing. The math arXiv and MathSciNet are excellent resources that help with that.

Being in a very active place where you can talk to lots of people about various areas of mathematics helps. Being surrounded by enthusiastic people helps. Going to small conferences where you get to know people can help. Talking to people about what you're interested in helps. Barring external impetus, "computing the daylights out of things" is an excellent fall-back procedure. I know quite a few very successful mathematicians for which this is one of the main approaches to things. You start piling up enough computations on things that interest you and you notice patterns -- maybe not what you were looking for, but sometimes of interest to people for reasons you never expected. Sometimes publishable. :)

edit: After reading your recent edit I can say I saw some similar things as a grad student. Sometimes the most talented/bright/whatever grad students have a hard time completing a Ph.D. Some students have too high expectations of themselves. They give up because they realize they're not going to prove the Riemann hypothesis -- they want that great big creative insight. In that regard it's good to ensure such grad students are working on both big hard problems and medium-sized publishable work, so that they can complete a Ph.D even if they never prove the Riemann hypothesis or whatever. Basically, always make sure you have a managable goal in sight. If your goals are only huge enormous things, you're setting yourself up for a potentially horrible failure. On the other hand, some people want that kind of situation, and if they're conscious of it, IMO you might as well let them be. It's their life. If they prove a major theorem, we're all the better off for it. If they don't, well at least they tried.

ca.analysis and odes - Evaluation of the following Series

Hi there,

I was wondering if you guys could be able to find the sum of the following series:

$ S = 1/((1cdot2)^2) + 1/((3cdot4)^2) + 1/((5cdot6)^2) + ... + 1/(((2n-1)cdot2n)^2) $, in which ${ntoinfty}$ .

This question came to mind when I was looking at this (http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf) paper by Professor Anirban DasGupta. In the last section, a couple of specific examples of his 'unified' method to find the sums of infinite series is pressented. In equation (34), he states that the following series:

$ 1/(1cdot2) + 1/(3cdot4) + 1/(5cdot6) + ... 1/(2ncdot(2n-1)) = log(2) $ (Note that ${ntoinfty}$ again). I was wondering If it's possible to find the sum if the values of the denominators of the terms are squared.

Thanks in advance,

Max Muller

real analysis - Looking for an interesting problem/riddle involving triple integrals.

I do not know the exact location in his Collected Works but Dirichlet found the $n$-volume of

$$ x_1, x_2, ldots, x_n geq 0 $$
and
$$ x_1^{a_1} + x_2^{a_2} + ldots + x_n^{a_n} leq 1 .$$

For example with $n=3$ the volume is
$$ frac{ Gamma left( 1 + frac{1}{a_1} right) Gamma left( 1 + frac{1}{a_2} right) Gamma left( 1 + frac{1}{a_3} right) }{ Gamma left(1 + frac{1}{a_1} + frac{1}{a_2} + frac{1}{a_3} right) } $$

Note that this has some attractive features. The limit as $ a_n rightarrow infty $ is just the expression in dimension $n-1,$ exactly what we want. Also, we quickly get the volume of the positive "orthant" of the unit $n$-ball by setting all $a_j = 2,$ and this immediately gives the volume of the entire unit $n$-ball, abbreviated as
$$ frac{pi^{n/2}}{(n/2)!} $$

I think he also exactly evaluated the integral of any monomial
$$ x_1^{b_1} x_2^{b_2} cdots x_n^{b_n} $$ on the same set.

So the question would be: given, say, positive integers $a,b,c,$ find the volume of $x,y,z geq 0$ and $ x^a + y^b + z^c leq 1.$ If you like, fix the exponents, the triple $a=2, b=3, c=6$ comes up in a book by R.C.Vaughan called "The Hardy-Littlewood Method," page 146 in the second edition, where he assumes the reader knows this calculation.

This came back to mind because of a recent closed question on the area of $x^4 + y^4 leq 1.$

algorithms - The mathematics of Schellings segregation model

For those who don't know the model. You can read this pdf. I want to find what is the probability that 2 nodes are each others neighbors when the algorithm converges (i.e. when all nodes are happy).

Here's the model in a gist. You have a
grid (say 10x10). You have nodes of
two kind (red and green) 45 each. So
we have 10 empty spaces. We randomly
place the nodes on the grid. Now we
scan through this grid (Exact order
does not matter according to
Schelling). Each node wants a specific
percentage of people of same kind in
its Moore neighborhood (say b = 50%
for each red and green). We calculate
the happiness of each node (a = Number
of neighbors of same kind/Number of
neighbors of different kind). If a
node is unhappy (a < b) it moves to an
empty cell where it knows it will be
happy. This movement can change the
dynamics of old as well as new
neighborhood. Algorithm converges when
all nodes are happy.

PS - I am looking for links for any mathematical analysis of the Schelling's model. I had posted this question on stackoverflow and was advised to post it here.

soft question - Your favorite surprising connections in Mathematics

My favorite connection in mathematics (and an interesting application to physics) is a simple corollary from Hodge's decomposition theorem, which states:

On a (compact and smooth) riemannian manifold $M$ with its Hodge-deRham-Laplace operator $Delta,$ the space of $p$-forms $Omega^p$ can be written as the orthogonal sum (relative to the $L^2$ product) $$Omega^p = Delta Omega^p oplus cal H^p = d Omega^{p-1} oplus delta Omega^{p+1} oplus cal H^p,$$ where $cal H^p$ are the harmonic $p$-forms, and $delta$ is the adjoint of the exterior derivative $d$ (i.e. $delta = text{(some sign)} star dstar$ and $star$ is the Hodge star operator).
(The theorem follows from the fact, that $Delta$ is a self-adjoint, elliptic differential operator of second order, and so it is Fredholm with index $0$.)

From this it is now easy to proof, that every not trivial deRham cohomology class $[omega] in H^p$ has a unique harmonic representative $gamma in cal H^p$ with $[omega] = [gamma]$. Please note the equivalence $$Delta gamma = 0 Leftrightarrow d gamma = 0 wedge delta gamma = 0.$$

Besides that this statement implies easy proofs for Poincaré duality and what not, it motivates an interesting viewpoint on electro-dynamics:

Please be aware, that from now on we consider the Lorentzian manifold $M = mathbb{R}^4$ equipped with the Minkowski metric (so $M$ is neither compact nor riemannian!). We are going to interpret $mathbb{R}^4 = mathbb{R} times mathbb{R}^3$ as a foliation of spacelike slices and the first coordinate as a time function $t$. So every point $(t,p)$ is a position $p$ in space $mathbb{R}^3$ at the time $t in mathbb{R}$. Consider the lifeline $L simeq mathbb{R}$ of an electron in spacetime. Because the electron occupies a position which can't be occupied by anything else, we can remove $L$ from the spacetime $M$.

Though the theorem of Hodge does not hold for lorentzian manifolds in general, it holds for $M setminus L simeq mathbb{R}^4 setminus mathbb{R}$. The only non vanishing cohomology space is $H^2$ with dimension $1$ (this statement has nothing to do with the metric on this space, it's pure topology - we just cut out the lifeline of the electron!). And there is an harmonic generator $F in Omega^2$ of $H^2$, that solves $$Delta F = 0 Leftrightarrow dF = 0 wedge delta F = 0.$$ But we can write every $2$-form $F$ as a unique decomposition $$F = E + B wedge dt.$$ If we interpret $E$ as the classical electric field and $B$ as the magnetic field, than $d F = 0$ is equivalent to the first two Maxwell equations and $delta F = 0$ to the last two.

So cutting out the lifeline of an electron gives you automagically the electro-magnetic field of the electron as a generator of the non-vanishing cohomology class.

tqft - Can string topology be a open-closed TCFT with the full set of branes?

String topology studies the algebraic structure of the homology of the free loop space $LM = Map(S^1,M)$ of a oriented closed manifold. One aspect of this structure is that the pair $(H_ast(LM;mathbb{Q}),H_ast(M;mathbb{Q}))$ forms a open-closed HCFT with positive boundary (work of Godin). This means that there are operations coming from the homology of moduli space of Riemann surfaces with each connected component having a non-empty outgoing or free boundary. The conjecture (although Blumberg-Cohen-Teleman claim it is a theorem) is that in fact $H_ast(M;mathbb{Q})$ should be seen has $H_ast(P(M,M),mathbb{Q})$, where $P(M,M)$ is the space of paths starting and ending in $M$, and the HCFT-structure can be extended to include open-closed cobordisms which have open boundaries labelled with a closed oriented submanifold $N$ of $M$. This then gives operations not only for $H_ast(LM;mathbb{Q})$ and $H_ast(M;mathbb{Q})$, but also on $H_ast(P(N_1,N_2);mathbb{Q})$ for any two closed compact oriented $N_1$, $N_2$ in $M$. This is called the full set of branes. For $N_1 = N_2$ a single point and $M$ connected, $P(N_1,N_2) = Omega M$, the based loop space. In general $H_ast(P(N_1,N_2);mathbb{Q})$ will therefore not be finite-dimensional.

On the other hand, Costello has proven a classification theorem of open-closed TCFT. In this we don't work with homology, but chains in the moduli space of Riemann surfaces, and chain complexes instead of (graded) vector spaces. Costello has proven that a open-closed TCFT can be constructed from an open TCFT (cobordism without incoming or outgoing boundary components equal to the circle) and that an open TCFT is equivalent to a Calabi-Yau $A_infty$ category. One of the properties of a Calabi-Yau $A_infty$ category is that all hom-spaces are finite-dimensional, forced by a certain non-degenerate pairing.

One can construct a HCFT from a TCFT by applying homology everywhere. I think this HCFT will in fact be positive (or negative?) boundary, because the TCFT is defined from open-closed cobordisms where each connected component has at least one incoming boundary component. Is this correct?

Furthermore, Costello conjectures in his paper that string topology (with the full set of branes) can be constructed as a TCFT, and applying homology then reduces to the HCFT given by Godin. But I can think of two reasons which make this conjecture seems false:
1) The naive choice of $C_ast(P(N_1,N_2);mathbb{Q})$ as Calabi-Yau $A_infty$ category is impossible, because these spaces will certainly be infinite-dimensional.
2) But no choice will work, because the homology of a finite-dimensional cell complex is finite-dimensional and we know some branes must be assigned infinite-dimensional spaces in the HCFT structure.

So my question is: Is this reasoning enough to make my naive interpretation of Costello's conjecture false? If not, what is the mistake?

Trivial valuation

Maybe you meant for the extension $L/K$ to be algebraic, in which case it is true that any extension of the trivial valuation on $K$ to $L$ is trivial. This clearly reduces to
the case of a finite extension, and then -- since a trivially valued field is complete --
this follows from the uniqueness of the extended valuation in a finite extension of a complete field. Maybe you view this as part of the sledgehammer, but it's not really the heavy part: see e.g. p. 16 of

http://math.uga.edu/~pete/8410Chapter2.pdf

for the proof. (These notes then spend several more pages establishing the existence part of the result.)

Addendum: Conversely, if $L/K$ is transcendental, then there exists a nontrivial extension on $L$ which is trivial on $K$. Indeed, let $t$ be an element of $L$ which is
transcendental over $K$, and extend the trivial valuation to $K(t)$ by taking $v_{infty}(P/Q) = deg(Q) - deg(P)$. (The completion of $K$ with respect to $v_{infty}$ is the Laurent series field $K((t))$, so this is really the same construction as in Cam's answer.) Then I prove* in the same set of notes that any non-Archimedean valuation can be extended to an arbitrary field extension, so $v$ extends all the way to $L$ and is certainly nontrivial there, being already nontrivial on $K(t)$.

*: not for the first time, of course, though I had a hard time finding exactly this result in the texts I was using to prepare my course. (This does use the sledgehammer.)

nt.number theory - Why Is $e^{pisqrt{232}}$ an Almost Integer?

The standard reason why $e^{pisqrt{N}}$ is a near integer for some $N$ is that there is some modular function $f$ with $q$-expansion $q^{-1} + O(q)$, such that substituting $tau = frac{1 + isqrt{N}}{2}$ (or perhaps $frac{isqrt{N}}{2}$) and $q = e^{2 pi i tau}$ into the $q$-expansion of $f$ yields a rational integer. If $N$ is sufficiently large, positive powers of $q$ are very small, so the initial term $q^{-1} = e^{-pi i (1 + isqrt{N})} = -e^{pisqrt{N}}$ is large and very close to the rational integer. We usually see the phenomenon with $f$ as the $j$-function, but as Frictionless Jellyfish pointed out, there are other choices.

You might ask why a modular function would yield an integer when fed a quadratic imaginary input, and the answer seems to come from the theory of complex multiplication, i.e., elliptic curves whose endomorphism rings are strictly larger than the integers. I'll start by outlining the usual picture with the $j$ function, and then switch to moduli of symmetrized diagrams of curves.

Class number one

Given an elliptic curve $E$ over the complex numbers, we can choose a lattice $Lambda subset mathbf{C}$ such that $E cong mathbf{C}/Lambda$ as a complex manifold (and as an analytic group). $Lambda$ is uniquely determined by this property up to complex rescaling, also known as homothety. The endomorphism ring of $E$ is therefore isomorphic to the endomorphism ring of $Lambda$, which is a discrete subring of $mathbf{C}$ and is either $mathbf{Z}$ or the integers in a quadratic imaginary extension $K$ of $mathbf{Q}$. In the latter case, $E$ is said to have complex multiplication (or "$E$ is a CM curve"), and $Lambda$ is a complex multiple of a fractional ideal in $K$. Two fractional ideals in $K$ yield isomorphic curves if and only if they are related by a rescaling, i.e., by a principal ideal. This yields a bijection between isomorphism classes of elliptic curves with complex multiplication by the ring of integers in $K$, and elements of the ideal class group of $K$. These sets are finite.

For any complex elliptic curve $E$ and any ring-theoretic automorphism $sigma$ of $mathbf{C}$, we can define $E^sigma$ as the curve you get by applying $sigma$ to the coefficients of the Weierstrass equation defining $E$. Since the $j$-invariant is a rational function in the coefficients of the Weierstrass equation, $j(E^sigma) = j(E)^sigma$. Since $E^sigma$ and $E$ have isomorphic endomorphism rings, the conclusion of the above paragraph implies the automorphism group of $mathbf{C}$ acts on the set of CM curves, and their $j$-invariants, with finite orbits. In particular, $[mathbf{Q}(j(E)):mathbf{Q}]$ is bounded above by the class number of $K$ (and with more work, we find that we have equality). The fact that $j(E)$ is an algebraic integer can be proved in several different ways: see section II.6 of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves. My preferred method is showing that $j$ is the solution to lots of modular equations (which yield monic polynomials).

We are left with the problem of finding CM elliptic curves whose endomorphism rings have class number one, but this is equivalent to finding lattices $Lambda subset mathbf{C}$ that are the rings of integers of class number one imaginary quadratic fields. By work of Heegner, Stark, and Baker, we get the usual list: $N = 163, 67, 43, 19, dots$, and the first few terms yield near-integers.

Symmetrized diagrams

For any prime $p$, there is an affine curve $Y_0(p)$ roughly parametrizing triples $(E, E', phi)$, where $phi: E to E'$ is a degree $p$ isogeny of elliptic curves. Equivalently, points on this space correspond to pairs of (homothety classes of) lattices, such that one is an index $p$ sublattice of the other. I use the term "roughly" because the presence of extra automorphisms prevents the formation of a universal family over the parameter space, so the affine curve is only a coarse moduli space. The Fricke involution switches $E$ with $E'$, and sends $phi$ to its dual isogeny. The quotient is the curve $Y_0^+(p)$, roughly parametrizing unordered pairs of elliptic curves, with dual degree $p$ isogenies between them. It is possible to consider level structures with more complicated automorphisms, but I'll stick with primes for now.

Based on the class number one discussion above, we want a function $f$ that attaches to each such unordered pair a complex number such that:

1. $f$ has $q$-expansion $q^{-1} + O(q)$.


2. For any ring-theoretic automorphism $sigma$ of $mathbf{C}$, we have the compatibility: $$f({E,E'},{phi, bar{phi} })^sigma = f({E^sigma,{E'}^sigma },{phi^sigma, bar{phi}^sigma }).$$ This implies the value of $f$ is algebraic when $E$ and $E'$ are CM.


3. $f$ should satisfy enough modular equations (or some other condition that yields integrality).

By a theorem of Cummins and Gannon, these conditions taken together imply that $f$ is a normalized Hauptmodul for a genus zero curve. In particular, $p$ must be one of the fifteen supersingular primes: 2,3,5,7,11,13,17,19,23,29,31,41,47,59,71.

Now, suppose we have a class number two field, such as $mathbf{Q}(sqrt{-58})$. We want a supersingular prime $p$, and an unordered pair of fractional ideals in the field, one index $p$ in the other, that is stable (up to simultaneous homothety) under the action of the automorphism group of $mathbf{C}$ on the Weierstass coefficients of the quotient elliptic curves. Ideally, we would like there to be only one homothety class of unordered pairs, so $operatorname{Aut} mathbf{C}$ will automatically act trivially.

For the case at hand, the fractional ideal $(2,-isqrt{58})$ has index 2 in the ring of integers, its square is $(2)$, and it is the only index 2 fractional ideal. If we take $p=2$, we find that the unordered pair ${ 1, (2,-isqrt{58}) }$ represents the only homothety class of pairs of fractional ideals of index 2. This yields the result that Frictionless Jellyfish pointed out, that if $f_{2A} = q^{-1} + 4372q + 96256q^2 + dots$ is the normalized Hauptmodul of $X_0^+(2)$, then $f_{2A}(frac{1+isqrt{58}}{2}) in mathbf{Z}$. When $q=e^{-pi sqrt{58}}$, the terms with positive powers of $q$ in the expansion of $f_{2A}$ are small enough to make $q^{-1}$ close to an integer.

Powers

We still need to figure out why $e^{pisqrt{232}}$, the square of $e^{pi sqrt{58}}$, is also near an integer. The easy answer is: if we square $f_{2A}$, we get $q^{-2} + 8744 + O(q)$, which is an integer when $q=e^{-pi sqrt{58}}$. The $O(q)$ terms here are still small enough to make $q^{-2} = e^{pi sqrt{232}}$ very close to an integer. You get a similar phenomenon with 88 and 148.

If you want to ask about $e^{3pi sqrt{58}}$, which is an integer minus $1.5 times 10^{-4}$, a more sophisticated answer can be extracted from Alison Miller's answer to this question. The normalized Hauptmodul for $X_0^+(2)$ is a replicable function, meaning its coefficients satisfy a certain infinite collection of recurrences, introduced by Conway and Norton when studying monstrous moonshine. These recurrences are equivalent to the existence of certain modified Hecke operators $T_n$, such that $n cdot T_nf_{2A}$ is an integer-coefficient polynomial in $f_{2A}$ with $q$-expansion $q^{-n} + O(q)$. The coefficients in the $O(q)$ part get big as $n$ increases, so powers of $e^{pi sqrt{58}}$ eventually drift away from integers.

(N.B.: The 2A in $f_{2A}$ refers to a conjugacy class in the monster simple group. There is a distinguished graded representation of the monster for which the trace of identity is $j-744$ and the trace of a 2A element is $f_{2A}$.)

nt.number theory - integer solutions to quadratic forms

Here is the standard geometric argument: after extracting common factors, you are asking for rational points on the quadric $Qcolon x^2-w^2=z^2-y^2$ in $mathbb{P}^3$, which is isomorphic to $mathbb{P}^1timesmathbb{P}^1$ (as Matt Young's comment explains). Projection from a point on $Q$ gives a birational map $p:Qto mathbb{P}^2$ hence a rational parameterisation of $Q$. Explicitly, projecting from $P_0=(0:1:0:1)$ to the $(w=0)$-plane gives the parameterisation
$$
p : (a:b:c)inmathbb{P}^2 mapsto (2ab:c^2-a^2+b^2:2bc:c^2-a^2-b^2) in Q
$$
of all the points except for the two lines on $Q$ passing through $P_0$ - i.e. the intersection of $Q$ with the tangent plane at $P_0$, which are $z=w$, $x=pm y$. Notice also that the points ${b=0,anepm c}$ all go to $P_0$, for the same reason.

The advantage of this method is that there is nothing particularly special about the quadratic form $x^2+y^2-z^2-w^2$ here; the same argument parameterises the rational zeros of any nonsingular quadratic form in 3 or more variables, as soon as it has one non-trivial zero.

Solvable transitive groups of prime degree

This is exercise 7.2.12 of Robinson's Course in the Theory of Groups, page 195 in the first edition.

A transitive subgroup of prime degree is primitive, and primitive solvable groups have a regular normal subgroup that is complemented by a unique conjugacy class of maximal subgroups. In particular, the Sylow p-subgroup C of order p is that regular normal subgroup, and the complement (being a permutation group) acts faithfully on it. In other words, the centralizer of the subgroup C is C itself. Hence G/C is a subgroup of Aut(C), so cyclic of order dividing p-1.

Since G is solvable it has a Sylow p-complement M, and by Hall's 1928 theorem, the number of
such Sylow p-complements is a divisor of p. If it is 1, then M is normal, so M centralizes C, so M=1, and G=C is cyclic.

This is summarized by saying that G is a subgroup of AGL(1,p) containing the translation subgroup. More generally every primitive solvable group is a subgroup of AGL(n,p) where p^n is the degree of the permutation action (but AGL(n,p) is no longer solvable itself).

noncommutative algebra - Semisimple-ish rings!

Let S be the class of all rings R which have 1 and satisfy this condition:

for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.)

All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings.

My questions:

Will this condition by itself force an element of S to have any (known, interesting) structure?

A more important question:

What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!)

nt.number theory - Is there a high-concept explanation for why characteristic 2 is special?

I think 2 is not special, we just see the weirdness at 2 earlier than the weirdness at odd primes.

For example, consider Ext_E(x)(F_p , F_p) where E(x) denotes an exterior algebra over F_p. If p=2 this is a polynomial algebra on a class x₁ in degree 1 and if p is odd this is an exterior algebra on a class x₁ tensor a polynomial algebra on x₂. I say these are the same, generated by x₁ and x₂ in both cases and with a p-fold Massey product < x₁,...,x₁>=x₂. The only difference is that a 2-fold Massey product is simply a product.

In what sense are the p-adic integers Z_p the same? One way to say it is that if you study the algebraic K-theory of Z_p you find that the first torsion is in degree 2p-3. If p=2 this is degree 1, and K₁(A) measures the units of A (for a reasonable ring A). If p is odd it measures something something more complicated. Another way to say it is that Z_p is the first Morava stabilizer algebra and there is something special about the n'th Morava stabilizer algebra at p if p-1 divides n. If you study something like topological modular forms, this means the primes 2 and 3 are special.

The dual Steenrod algebra is generated by xi_i at p=2 and by xi_i and tau_i at odd primes. But really it is generated by tau_i with a p-fold Massey product < tau_i,...,tau_i>=xi_i+1 at all primes, after renaming the generators at p=2. (Again a 2-fold Massey product is just a product.)

I could go on, but maybe this is enough for now.

ho.history overview - What is the oldest open problem in mathematics?

Zeno's paradoxes are among the oldest puzzles at the intersection of mathematics, philosophy, and physics (in alphabetical order). The traditional resolution of Zeno's paradoxes of motion involves modeling them in terms of the real line and interpreting the iterated procedure as an infinite series.

As pointed out in one of the comments, Heisenberg's uncertainty principle provides another way of accounting for the puzzle, by arguing that it has no physical meaning.

H. Jerome Keisler in his article "The hyperreal line" (207–237) in the collection

Real numbers, generalizations of the reals, and theories of continua. Edited by Philip Ehrlich. Synthese Library, 242. Kluwer Academic Publishers Group, Dordrecht, 1994

provides a different mathematical resolution of the puzzle in terms of the hyperreal continuum.

More recently (2013), Terry Tao notes the mathematical significance of these paradoxes by noting that they "make the important point that real analysis cannot be reduced to a branch of discrete mathematics, but requires additional tools in order to deal with the continuum" (see http://www.ams.org/mathscinet-getitem?mr=3026767).

In a review of Graham Oppy's book, John H. Mason makes the following intriguing comment, indicative of the richness of the issues involved: Have you ever briefly called upon Zeno's paradoxes when introducing the notion of limit to students? For example, the fact that Achilles really does catch the tortoise is only because he crosses distances halving in length in intervals of time also halving in length; the arrow does actually get to its target, even though it has to surmount an infinite number of decreasingly small intervals. This book addresses these and many other paradoxes involving infinitely large and infinitely small quantities with philosophical precision and reasoning. It reveals that there are much larger issues at stake than are perhaps commonly recognised, and certainly than are `dismissed' with the Cauchy-Weierstrass formalism of limits. See http://www.ams.org/mathscinet-getitem?mr=2238333

calculus of variations - Maximizing the volume in a family of subsets of a cube

Starting from a question in probability, one is eventually lead to the following optimization problem.

Let $I:=[0,\, 1],$ and let $A$ be a Lebesgue measurable subset of the $n$-dimensional cube, $Asubset I^n.$ Consider, correspondingly, the set
$$hat A:= {xin I^{n+1}:\, (x_1,dots,x_n)in A,\, (x_2,dots,x_{n+1})notin A}=Atimes I\,cap\, I times A^c.$$

Problem. Maximize the $(n+1)$-dimensional Lebesgue measure
of $hat A$ over all measurable
$Asubset I^n$:
$$lambda_n:=sup_{Asubset I^n}verthat Avert.$$

If $n=1,$ we have $|hat A|=|A|(1-|A|),$ whence $lambda_1=1/4.$ For $n=2$ the maximizing set is the triangle below the diagonal, giving $lambda_2=1/3.$ The sequence $lambda_n$ is increasing, and converges to $1/2.$ If $n$ is even, one finds $$lambda_n=frac{1}{2}left(1-frac{1}{n+1}right).$$
(I will edit and provide the details of the computation at request). However, as a consequence of a computation by Trotter and Winkler (Ramsey theory and sequences of random variables, Probability, Combinatorics and Computing 7 (1998), 221-238), the formula can't hold true for all odd $n,$ for one has $lambda_5>frac{1}{2}left(1-frac{1}{6}right)=5/12.$

I would be very grateful for any suggestion or reference useful to shed light on the case of odd $n.$

mg.metric geometry - Set of vectors separated by at least a specified angle

Suppose theta and d are given.

How big can a set of d-dimensional vectors be such that no pair of them are at angle less than theta?

I particularly want an upper bound; that is, an n=n(theta,d) such that given n d-dimensional vectors, there must be at least 2 with angle less than theta between them.

Of course, the question can be rewritten in all sorts of ways, for example, coverings of the surface of the d-dimensional sphere by (d-1)-dimensional caps of given radius etc.

The bound doesn't need to be tight. Something out by a factor of (constant)^d might be fine
(although something more exact would be interesting too).

at.algebraic topology - Spectral sequence

When doing things related to free resolutions you may want the actual computation or you may want something highly structured. For the former you use the koszul resolution, it is nice and little and small. If you want a lot of structure you use the Bar resolution. This is sort of a philosophical things, so when you actually want to compute something you use the Koszul resolution since it is pretty small and you know that making this choice wont affect your answer.

soft question - How to know if somebody else is also working on your problem?

As others have indicated, the only 100% effective method of preventing getting "scooped" or finding out that your result already exists in the literature is that of complete abstinence: i.e., not trying to do any research at all.

Obviously this method is far too draconian for most of us on this site. I want to support statements of Gowers and Nielsen: finding out that what you have just proven has already been proven by someone else is far from the worst thing in the world. (Finding out that what you've proven, or published, is false, is much much worse, for instance.) On the contrary, for a mathematician who is making her own way and working on problems of interest to her, if you are doing any good work at all it is inevitable that you will duplicate some past research. This can be very encouraging: when I was younger, I often lacked confidence that some things which were of interest to me were of sufficient interest to anyone else (all I knew at that point was what people near to me were doing).

I remember in particular that as a first year graduate student, I discovered that every profinite group is, up to isomorphism of topological groups, the automorphism group of some Galois extension. This seemed neat but I thought, "Well, if anyone really cared, I would have heard about it before." Wrong -- this result has been published several times; off the top of my head by Leptin and by Waterhouse, but I know this list is not complete -- and in some texts (just not the ones I knew about at the time) it appears with due respect and appreciation. When I found out that someone had written and published a paper containing exactly the same mathematics that I had done, it was very encouraging.

nt.number theory - Is a "non-analytic" proof of Dirichlet's theorem on primes known or possible?

I was struggling to find an elementary proof of Dirichlet's theorem using another interesting technique.
I came finally to a proof from an entirely different direction but as i found out Erdos came first before many years.
The proof is not well known (I never understood why anyone mentions it!)and it uses Chebyshev type estimates. Here is the proof: http://kam.mff.cuni.cz/~klazar/ln_antcII.pdf
I hope this will be helpfull!

We are looking for prime numbers of the form $a+nm$.$(a,m)=1,n=1,2,...$

The general plan is:
$n!$ divides the product of $n$ consecutive terms of the arithmetic proggression $a+m$,$a+2m$,$...,$$a+nm$ with $(a,m)=1$ (if we disregard the factor of $n!$ which includes divisors of $m$)

For example, consider the proggression $1+3m$:

$4cdot7cdot10cdot13cdot16cdot19cdot22$ is divided by $7!/3^2$
(The factor that we "disregard" is $3^2$)

It is easy to prove in analog with Legendre's Lemma at binomial coefficients, that the highest power of a prime $p$ which divides $frac{(a+m)cdot(a+2m)...cdot(a+nm)}{n!}$ does not exceed $a+nm$.

The problem is that such a prime $p$ could not be of the form that is wanted.For example turning back to $frac{4cdot7cdot10cdot13cdot16cdot19cdot22}{7!/3^2}$ we find 11 as a divisor which is a prime of the form $2+3m$.

We wish to simplify the unwanted primes from the "big" fraction which Erdos calls $Pn(a,m)$.
In order to do this we divide $Pn(a,m)$ with a fraction of the same kind but from another proggression $a'+nm$ with $(a',m)=(a,m)=1$.

(The "other" proggression for $1+3m$ is $2+3m$ since only 1 and 2 are the only numbers coprime to 3 and less than 3).

But we find that in $Pn(a,m)$ every prime of the form $a'+km$ which is greater than $n$ exists exactly once, so (and here is the big idea) dividing $Pn(a,m)$ with $P(n/h)(a',m)$ ($h$ is the number with the property $a'h=a(modm)$ ) every prime of the form $a'+km$ which is greater than $n$ cancels.

Continuing like this you can cancel every unusefull prime that exceeds n and have only "small" unusefull primes whose product is significally smaller than $Pn(a,m)$.

With this,you prove that primes of the form $a+nm$ have a product which tends to infinity and so they are infinite.

I hope this is helpfull.

(note)We use $h$ because the smallest term of the proggression $a+nm$ that is divided by a prime $p$ of the form $a'+km$ is the term $a+km=hcdot p$ .

ac.commutative algebra - Factorial Rings and The Axiom of Choice

Lang uses Zorn's lemma only in the step that nonzero nonunits in a PID admit irreducible factorizations (not the uniqueness of irreducible factorizations, once we know such factorizations exist). The way he uses Zorn's lemma, I think, is excessive. What follows is how I work out the existence of irreducible factorizations when I teach the abstract algebra class.

Claim: In a PID which is not a field, any nonzero nonunit is a product of irreducibles.

We will use the following lemma (which is not Zorn's lemma).

Lemma: If $R$ is an integral domain and $a in R$ is
a nonzero nonunit which
does not admit a factorization into irreducibles then
there is a strict inclusion of principal ideals
$(a) subset (b)$ where $b$ is some other nonzero nonunit which
does not admit a factorization into irreducibles.

Proof of lemma: By hypothesis $a$ is not irreducible, so (since it is
neither 0 nor a unit either) there is some factorization
$a = bc$ where $b$ and $c$ are nonunits (and obviously are
not 0 either). If both $b$ and $c$ admitted
irreducible factorizations then so does $a$, so
at least one of $b$ or $c$ has no irreducible factorization.
Without loss of generality it is $b$ which has no
irreducible factorization. Since $c$ is not a unit, the inclusion
$(a) subset (b)$ is strict. QED lemma.

Now we can prove the claim.

Proof of claim: Suppose there is an element $a$ in the PID which
is not 0 or a unit and has no irreducible factorization.
Then by the lemma there is a strict inclusion
$$
(a) subset (a_1)
$$
where $a_1$ has no irreducible factorization.
Then using $a_1$ in the role of $a$ (and the lemma again)
there is a strict inclusion
$$
(a_1) subset (a_2)
$$
where $a_2$ has no irreducible factorization.
This argument (repeatedly applying the lemma to
the generator of the next larger principal ideal)
leads to an infinite increasing chain of principal ideals
$$
(a) subset (a_1) subset (a_2) subset (a_3) subset cdots
$$
where all inclusions are strict. (At this step I suppose you may say we need the Axiom of Choice to get an infinite ascending chain, but it's only countably many choices, so really not the full thrust of Zorn's lemma and in any case it feels like a less pedantic use of Zorn's lemma than the way Lang does this.) Such a chain of ideals is impossible in a PID.

Indeed, suppose a PID contains an infinite strictly increasing chain of
ideals:
$$
I_0 subset I_1 subset I_2 subset I_3 subset cdots
$$
and set
$$
I = bigcup_{n geq 0} I_n.
$$
This union $I$ is an ideal. The reason is that
the $I_n$'s are strictly increasing, so any finite set
of elements from $I$ lies in a common $I_n$.
Therefore $I$ is closed under addition and arbitrary multiplications from
the ring since each $I_n$ has these properties. Because we are in a PID, $I$ is principal:
$I = (r)$ for some $r$ in the ring. But because
$I$ is the union of the $I_n$'s, $r$ is in some $I_N$.
Then $(r) subset I_N$ since $I_N$ is an ideal, so
$$
I = (r) subset I_N subset I,
$$
which means
$$
I_N = I.
$$
But this is impossible because the inclusion $I_{N+1} subset I$
becomes $I_{N+1} subset I_N$ and we were assuming
$I_N$ was a proper subset of $I_{N+1}$.
Because of this contradiction, nonzero nonunits in a PID without
an irreducible factorization do not exist. QED

Lang's argument only uses the axiom of choice in a countable way, as above, but the way he pulls it in makes the application of Zorn's lemma feel a lot more fussy.

ct.category theory - Presheaves as limits of representable functors?

You mean "colimit of representable presheaves", not limit. Any limits that C has are preserved by the Yoneda embedding. So if C is, say, a complete poset like • → •, so that it is small and has all limits, you won't be able to produce any non-representable presheaves by taking limits of representable ones.

The way to write any presheaf as a colimit of representables is, like all things Yoneda-related, somewhat tautological, and should be worked out for oneself; but anyways it's explained to some extent at this nlab page. Rather than write out formulas, I usually think of the example of simplicial sets: every simplicial set X can be formed as a colimit of its simplices, i.e., a diagram of representables which is indexed on the "category of simplices of X", whose objects are pairs (n, x) where n is in the indexing category and x is an object of X_n. The same works in any presheaf category.

tag removed - is there a solution for equation $arcsin((1-x)^{1/2})=arccos(x^{1/2})$ in which $x$ is rational number

This may look elementary -- but it is most definitely not. This is because there are some nasty branch cuts involved, and making sure that the identity actually holds at all is not easy. The first thing to do is to look at what happens if one expands these functions into their simpler representation using logarithms:
$$ arcsin left( sqrt {1-x} right) =-iln left( sqrt {x}+isqrt {1-x} right) $$
and
$$ arccos left( sqrt {x} right) =1/2pi +iln left( sqrt {1-x}+isqrt {x} right) $$

For essentially all real $x$ outside $(0,1)$, these two quantities are complex. But, as it turns out, there are solutions. The simplest next step is to figure out 'where', and this is best done by splitting into cases. It can be written as
$$cases{i left( -ln left( sqrt {-x}+sqrt {1-x} right) -ln left( sqrt {1-x}-sqrt {-x} right) right) & xleq 0 cr
-1/2pi +arctan left( {frac {sqrt {1-x}}{sqrt {x}}} right) +arctan left( {frac {sqrt {x}}{sqrt {1-x}}} right) & xin(0,1) cr
i left( -ln left( sqrt {x}-sqrt {-1+x} right) -ln left( sqrt {-1+x}+sqrt {x} right) right) & x ge 1}
$$
(the point being that all arguments of the square roots are now positive).

With some extra work, it is then possible to in fact show that this is indeed $0$ everywhere (with a minor quibble about $x=0$ itself). From here the manipulations are indeed relatively straightforward.

ac.commutative algebra - Atiyah-MacDonald, exercise 2.11

I posted this question on a different site a couple of years ago. Eventually I found that a book of T.Y. Lam has a very nice treatment. Here is the writeup I posted on the other site:

---

After paging through several algebra books, I found that T.Y. Lam's GTM Lectures on
Rings and Modules has a beautiful treatment of this question.

The above property of a (possibly noncommutative) ring is called the "strong rank
condition." It is indeed stronger than the corresponding statement for
surjections ("the rank condition") which is stronger than the isomorphism version
"Invariant basis number property". However, in fact it is the case that all
commutative rings satisfy the strong rank condition. Lam gives two proofs
[pp. 12-16], and I will now sketch both of them.

First proof:

Step 1: The result holds for (left-) Noetherian rings. For this we establish:

Lemma: Let $M$ and $N$ be (left-) $A$-modules, with $N$ nonzero. If the direct sum
$M oplus N$ can be embedded in $M$, then $M$ is not a Noetherian $A$-module.

Proof: By hypothesis $M$ has a submodule
$M_1 oplus N_1$, with $M_1 cong M$ and $N_1 cong N$. But we can also
embed $M oplus N$ in $M_1$, meaning that $M_1$ contains a submodule $M_2 oplus N_2$ with $M_2 cong M$ and $N_2 cong N$. Continuing in this way we construct an ascending
chain of submodules $N_1$, $N_1 oplus N_2$,..., contradiction.

So if A is (left-) Noetherian, apply the Lemma with $M = A^n$ and $N = A^{m-n}$.
$M$ is a Noetherian $A$-module, and we conclude that $A^m$ cannot be embedded in $A^n$.

Step 2: We do the case of a commutative, but not necessarily Noetherian, ring.
First observe that, defining linear independent subsets in the usual way, the
strong rank condition precisely asserts that any set of more than $n$ elements in $A^n$
is linearly dependent. Thus a ring $A$ satisfies the strong rank condition iff: for
all $m > n$, any homogeneous linear system of $n$ linear equations and $m$ unknowns has a
nonzero solution in $A$.

So, let $MX = 0$ be any homogeneous linear system with coefficient matrix $M = (m_{ij}),
1 leq i leq n, 1 leq j leq m$. We want to show that it has a nonzero solution in $A$. But the subring $A' = mathbb{Z}[a_{ij}]$, being a quotient of a polynomial ring in finitely many variables over a Noetherian ring, is Noetherian (by the Hilbert basis theorem), so by Step 1 there is (even) a nonzero solution $(x_1,...,x_m) in (A')^m$.

This makes one wonder if it is necessary to consider the Noetherian case separately,
and it is not. Lam's second proof comes from Bourbaki's Algebra, Chapter III, §7.9, Prop. 12, page 519. [Thanks to Georges Elencwajg for tracking down the reference.] It uses the following elegant characterization of linear independence in free modules:

Theorem: A subset ${u_1,...,u_m}$ in $M = A^n$ is linearly independent iff: if $a in A$ is such that $a cdot (u_1 wedge ldots wedge u_m) = 0$, then $a = 0$.

Here $u_1wedge ldots wedge u_m$ is an element of the exterior power $Lambda^m(M)$.

(I will omit the proof here; the relevant passage is reproduced on Google books.)

This gives the result right away: if $m > n$, $Lambda^m(A^n) = 0$.

reference request - Removing intersections of curves in surfaces

Let $C_1, dots, C_n$ be a family of disjoint simple curves in a surface $Sigma$. If $C$ is any simple curve in $Sigma$, it turns out that we can map $C$ to a curve $C'$ (via a homeomorphism of $Sigma$) such that $C'$ only intersects each $C_i$ at most twice. I believe I have a proof of this result that works for all surfaces, but I'm pretty sure this is classical stuff.

Indeed, in John Stillwell's book Classical Topology and Combinatorial Group Theory he mentions that the above result was proven by Lickorish (1962) for orientable surfaces. In the case of orientable surfaces, the homeomorphism can in fact be achieved via Dehn twists and isotopies. Unfortunately, the Lickorish proof doesn't work for non-orientable surfaces.

Question: Can someone please provide a reference of the above result for non-orientable surfaces?