nt.number theory - What is the best known upper bound for the number of twin primes?

J Wu, Chen's double sieve, Goldbach's conjecture, and the twin prime problem, Acta Arith 114 (2004) 215-273, MR 2005e:11128, bounds the number of twin primes above by $2aCx/log^2x$, with $C=prod p(p-2)/(p-1)^2$, and $a=3.3996$; I don't know whether there have been any improvements.

ac.commutative algebra - Do n-th Witt polynomials generate {P | P' is divisible by n} ?

EDIT: Proved it on my own. It easily follows from the Witt integrality theorem. Sorry for posting.

Let $Pinmathbb{Z}left[Xiright]$ be a polynomial (where $Xi$ is a family of symbols that we use as indeterminates, for instance $Xi=left(X_1,X_2,X_3,...right)$). Let $ninmathbb{N}$.

Prove or disprove that $displaystylefrac{delta}{deltaxi}Pin nmathbb{Z}left[Xiright]$ for every $xiinXi$ if and only if there exist polynomials $P_dinmathbb{Z}left[Xiright]$ for all divisors $d$ of $n$ such that $displaystyle P=sum_{dmid n}dP_d^{n/d}$.

A few remarks on this: The $Longleftarrow$ direction is trivial. I can prove the $Longrightarrow$ if $n$ is a prime power.

PS. No, this does not help in proving the Witt integrality theorem, even if it is true.

noncommutative geometry - Gelfand duality in NCG

On surjectivity: No, not every representation comes from a state; only the cyclic ones. Every nondegenerate representation of a C*-algebra is a direct sum of cyclic representations (Zorn), and every cyclic representation comes from a GNS construction. But yes, every irreducible representation (which is also cyclic) comes from a GNS construction.

On injectivity: To even consider this question you probably want to identify representations if they are unitarily equivalent. But no, because for example two unit cyclic vectors in the Hilbert space of a representation will often yield different vector states, but these vector states will yield the same representation.

Which space? That would depend on your applications, and I don't have any absolute answers to that. Instead I'll tell you what comes to mind in the hope that it will help orient you. (Then I'll wait with you to be enlightened by another answer.)

The space of (unitary) equivalence classes of irreducible representations is often called the spectrum of a C*-algebra; a closely related space is the set of kernels of irreducible representations, called the primitive ideal space. The primitive ideal space is given the hull kernel topology and is a quotient of the spectrum. I recommend the book by Raeburn and Williams; see Appendix A to learn about these spaces, and see the rest of the book for how they're used.

As for the space of pure states, I'm more accustomed to the idea of studying the space of all states, in which the pure states are the extreme points. There is a nice characterization of the state spaces of C*-algebras in E. Alfsen, H. Hanche-Olsen and F.W. Shultz: State Spaces of C∗-Algebras, Acta Math. 144 (1980) 267–305, which came up at this other question. However, there is yet another option which I learned a little about from Pedersen's book, where the quasi-state space Q is used. A quasi-state is a positive functional of norm at most 1. (Other than 0, the extreme points of Q are also the pure states.) A C*-algebra can be studied as affine functions on Q; see section 3.10 of Pedersen for details.

at.algebraic topology - Cyclic spaces and S^1-equivariant homotopy theory

I don't know if this is exactly what you're looking for (and there's a good chance you already know what I'm going to write) but let me give it a try:

The realization functor of cyclic sets (not spaces!) to $S^1$-spaces can be made part of a Quillen equivalence for two of the three commonly desired model structures on $S^1$-spaces: The model structure that gives you "Spaces over $BS^1$" is given in a 1985 paper of Dwyer-Hopkins-Kan, while a model structure that gives you the equivalences that you want (i.e., checked on fixed sets for finite subgroups) is given in a 1995 paper "Strong homotopy theory of cyclic sets" by Jan Spalinksi.

(Irrelevant to your question, but along the same lines: A recent paper of Andrew Blumberg describes how one can throw in some extra--still combinatorial--data and obtain a combinatorial model of the third desirable model structure on $S^1$-spaces, namely where equivalences are those that induce equivalences on fixed sets for all closed subgroups.)

Spalinksi's model structure depends on the following construction of $|X_{.}|^{C_n}$ (as a space-over-$BS^1$) in terms of the subdivision construction: The simplicial set $(sd_r X)_n = X_{r(n+1)-1}$ has an action of $C_r$--since $C_r$ is a subgroup of the copy of $C_{r(n+1)}$ acting on $X_{r(n+1)-1}$; taking fixed points (in sSet) and then realizing gives $|X_{.}|^{C_n}$.

This suggests (though I haven't checked too carefully) that remembering each $X_n$ as a $C_{n+1}$-space (in the sense you suggest, with subgroups) is enough, as you expected.

---

Now begins the speculative (and probably wrong) part of this answer: I have nothing too certain to say about writing this as a functor category, but it doesn't seem too unreasonable (to me, right now, at least) based on the above simplicial subdivision construction that we might be able to construct a reasonable candidate: some sort of mix of the cyclic category and the orbit categories for the cyclic groups. Purely combinatorially, this seems to get tricky.

But, I think we can realize this geometrically: Let $(S^1)_r$ be the circle equipped with a $mathbb{Z}$ action given by the rotation by $2pi/r$. We could try to define $Hom'([m-1]_r, [m'-1]_{r'})$ along the lines of "(htpy classes of) degree $r'/r$, increasing $mathbb{Z}$-equivariant maps $S^1 to S^1$ sending the $mr$-torsion points to the $m' r'$-torsion points". This should correspond to taking all the $r$-cyclic categories and sticking them together, and in particular is bigger than what we want. But, the $mathbb{Z}$-action on the circles should induce one on the $Hom'$-sets and the composition should respect it. Taking the quotient, we seem to get something that looks like a reasonable candidate. For each fixed $r$, we should be getting a copy of the cyclic category. And, e.g. $Hom([m-1]_r, [mr-1]_1)$ should contain $Hom_{orbit}(Z/mr, Z/r)$. (Disclaimer: It's late and I haven't checked any of this too carefully!)

ct.category theory - In what category is the sum of real numbers a coproduct?

None (except trivially).

It's an elementary (though maybe not obvious) lemma that if $X$ and $Y$ are objects of a category and their coproduct $X + Y$ is initial, then $X$ and $Y$ are both initial.

Suppose there is some category whose objects are the real numbers, and such that finite coproducts of objects exist and are the same as finite sums of real numbers. In particular (taking the empty sum/coproduct), the real number $0$ is an initial object. Now for any real number $x$ we have $x + (-x) = 0$, so by the lemma, $x$ is initial. So every object is initial, so all objects of the category are uniquely isomorphic, so the category is equivalent to the terminal category 1.

If you just want non-negative real numbers then this argument doesn't work, and I don't immediately see an argument to take its place. But I don't think it's too likely that an interesting such category exists.

I wonder if it would be more fruitful to ask a slightly different question. Product and coproduct aren't the only interesting binary operations on a category. You can equip a category with binary operations (as in the concept of monoidal category). Sometimes this is a better thing to do.

For example, there is on the one hand the concept of distributive category, which is something like a rig (=semiring) in that it has finite products $times$ and finite coproducts $+$, with one distributing over the other. On the other hand, there is the concept of rig category, which is a category equipped with binary operations $otimes$ and $oplus$, with one distributing over the other. Distributive categories are examples of rig categories. Any rig, seen as a category with no morphisms other than identities, is a rig category. Any ordered rig can be regarded as a rig category (just as any poset can be regarded as a category): e.g. $[0, infty]$ is one, with its usual ordering, $otimes = times$, and $oplus = +$.

na.numerical analysis - Approximating a set with fixed number of elements

This is the $k$-center problem (or in your notation, the $n$-center problem). you're given a set $S$ of points, and you want to find a set $R$ of $n$ points such that the set of balls of radius $r$ around each point in $R$ cover all of $S$, and $r$ is minimized.

Your metric space is the line, so this problem is relatively easy to solve. Here's a two-step approach: First, "guess" the optimal solution r (ie. pick some value of r). Now go from left to right, assigning centers greedily, which is to say, as far away from the previously placed center as possible, while covering all points. If you use up $n$ points before covering all of $S$, your guess was wrong, and you need to restart with a larger value of r. Else, you're done.

Now of course $r$ is a real number, but there are only discretely many "guesses", since the optimal r must be such that there are two points at distance exactly $r$ from a center (otherwise r is not optimal). so the total set of choices of r is merely the set constructed from measuring the pairwise distances and halving them.

All of this assumes you're in algorithms-land, which means that you have reasonable ways of representing points and comparing them.

p.s this algorithm is well known (not original).

mg.metric geometry - Uniformly Sampling from Convex Polytopes

Rejection sampling definitely works if you are able to find a superset $Q$ of the polytope $P$ from which you can sample. If you sample a point from that superset, the probability that it gets accepted is equal to the ratio $frac{text{Vol}(P)}{text{Vol}(Q)}$, so $Q$ should be as small as possible. For instance, it is sample to sample from $Q$ if it is a box or a ball.

In the case where the polytope is specified as a list of inequalities, finding the smallest enclosing ball can be quite hard.

Contrary to what Simon Barthelmé mentions, Boyd and Vandenberghe do not deal with this problem. Actually they deal with the case where the vertices of the polytope are available. Going from the list of inequalities to the set of vertices is also hard (I am actually looking for a MATLAB implementation of that).

One possible approach is to find a small box enclosing the polytope. The box is defined by a set of coordinates $(b_i^{text{min}},b_i^{text{max}}), i=1dots n$, and each coordinate can be found by :
$$ b_i^{text{min}} = arg min_x x_i quad text{subject to } A x leq b $$
$$ b_i^{text{max}}= arg max_x x_i quad text{subject to } A x leq b $$
Those are linear programs for which you can use your favourite solver.

ag.algebraic geometry - theta divisor on a principally polarized abelian variety

This is true. For $A_{/mathbb{C}}$ an abelian variety, $L$ an ample line bundle on $A$, then any line bundle
$M in operatorname{Pic}^0(A)$ -- over $mathbb{C}$, this is equivalent to having first Chern class zero -- is of the form $T_x^{*} L otimes L^{-1}$ for some $x in A$. (e.g. Theorem 1 on p. 77 of Mumford's Abelian Varieties).

Applying this theorem with $L = L(D_2)$, $M = L(D_1) - L(D_2)$, we get that

$L_1 - L_2 = T_x^*(L_2) - L_2$, so

$L_1 = T_x^*(L_2)$.

So $D_1$ and $x+D_2$ (meaning translation of $D_2$ by $x$!) must be linearly equivalent, but by your assumption $h^0(L(D_1)) = h^0(L(D_2)) = 1$, they are each the unique effective
divisors in their linear equivalence classes, so we must have $D_1 = x + D_2$.

How exactly is Hochschild homology a monad homology?

This is partly in response to Reid, but also intended as general clarification.

As I understand it, Peter's original question was:

-- here is the Hochschild chain complex for an algebra $A$ and bimodule $M$, as defined in Hochschild's original papers;
-- it is the chain complex associated to a certain simplicial object as defined on the Wikipedia page;
-- one is told that this object comes from the bar construction (or standard resolution) associated to some monad;
-- where/what is the monad?

The last one seems to be Reid's underlying point/question. Tyler says you can get it, up to a dimension-shift, from the adjunction between k-modules and k-algebras (at least when $A=M$). My earlier recollection was that this more naturally leads to cyclic homology a.k.a. additive K-theory as defined by Feigin and Tsygan, but I have yet to check this against a copy of their paper. (The point is that in characteristic zero, the cyclic homology of a free tensor algebra on a given k-module, coincides with the cyclic homology of the ground field, so one can take free resolutions of a given $k$-algebra and then use spectral sequence arguments.) On reflecting a bit more, because the Hochschild homology of a free (=tensor) algebra is confined to degrees 0 and 1, perhaps one can also obtain $H_n(A,M)$ as Tyler suggests, by taking the free algebra resolution of A (in the category of k-algebras) and then hitting the resulting simplicial object with a suitable functor - but this seems trickier than in the commutative case (Andre-Quillen) and I can't get hold of a copy of Quillen's paper at the moment.

Alors. As I understand it, following Weibel's book (and the papers of Barr & Beck et al), the simplicial object (in the category of $k$-modules) that yields the Hochschild chain complex, arises by applying a certain Hom-functor (namely ${}_A{rm Hom}_A( cdot ,X)$ ) to another simplicial object, say $beta(A)$, in the category of $A$-bimodules.

Now $beta(A)$ is not contractible in the category of $A$-bimodules, in general, and doesn't come from a (co)monad on that category. However, $beta(A)$ can be identified with another simplicial object $F(A)$, which lives in the category of $A$-modules.

What is $F(A)$?

Well, take a step back and consider the adjunction between $k$-modules and $A$-modules (maybe you need $k$ to be a field at this point, maybe not). That gives rise to a bar construction in $A$-mod, namely for any given $M$ in $A$-mod one obtains a simplicial object $F(M)$ which is given in each degree by

$$ F_{-1}(M)=Mquad,quad F_n(M) = M otimes A^{otimes n+1} {rm for } n geq 0. $$

Note that this is contractible in $A$-mod by the general machinery of the bar resolution associated to a monad. There was nothing to stop us taking $M=A$, that's a perfectly good $A$-module; and on doing so, lo and behold, we get the same simplicial object $F(A)$.

Thus, Hochschild homology, regardless of the choice of coefficients, can be thought of as "coming from" a comonad - namely, that induced on $A$-mod by the forgetful functor from $A$-mod to $k$-mod. In my opinion, that is probably the (co)monad they are talking about.

It so happens that, since $F(A)$ is contractible in $A$-mod and hence a fotiori in $k$-mod, the "chain-complex-ification" of $beta(A)$ is, as a chain complex in $R$-bimod, a resolution of $R$ by $k$-relatively projective $R$-bimodules -- and hence applying ${}_R{rm Hom}_R( cdot ,X)$ to it and taking homology coincides with taking $k$-relative Tor of $R$ and $X$ as R-bimodules. Hence the point of view that Hochschild homology is a special case of relative Tor.

Finally, I actually agree with Reid that this is not the best example to motivate (co)monad (co)homology. Group cohomology with coefficients in the ground field; or indeed André-Quillen cohomology, which is given by a "free algebra" adjunction but only for commutative algebras, or sheaf cohomology, would be better. (No originality in my choices; I've cribbed them out of Weibel Section 8.6).

(Apologies for the length and the tediousness, by the way.)

career - When your paper makes a borderline case for a top journal

If you've written an excellent paper, that's still no guarantee that it'll get into the prestigious general journals (such as those listed in your question). For example, the editors or referees might declare your topic to not be of general interest. If this turns out to be the case, you should consider publishing in a prestigious specific journal (not necessarily a less prestigious journal).

The worst possibility is the long rejection -- that is, having a paper refereed for a year or so, only to be rejected. This seems to happen a lot with the general journals as the referees are trying to maintain a high standard (and the editors can't always tell if a paper is worthy or not). The refereeing process is confidential, so the only downside is time wasted, and you might get some excellent feedback too.

For job applications, it's much easier for a potential employee to gauge the merit of a general journal (LMS, AMS, etc.) then a specific journal. They can be 100% a result published in these general journals is decent, regardless of which field they are in. Whereas, it can be difficult to explain the importance of a specific journal.

I heard from one university that they had 600+ applications for one position. If the hiring committee looked at one application per minute, they would still take 10 hours. They won't have time to look at arXiv, chat to colleagues, etc. for the vast majority of applicants. Moreover, even if the paper is on the arXiv, unless the committee are experts in your field also, they won't know whether or not your paper is decent. You can explain the merits of your work in-person at a job interview.

Having a paper on the arXiv (vs. not having it on the arXiv) counts for nothing in job applications. Generally, you want not-yet-accepted papers listed as either "in preparation" or "submitted".

ag.algebraic geometry - How do you explicitly compute the p-torsion points on a general elliptic curve in Weierstrass form?

Here is an attempt to answer my own question, using the "division
polynomials" of Kevin's and Jared's answers. It is probably the maximally naive idea, and I do not claim it
works, though it's not clear to me that it can't.
I've community wiki-ed this answer, as it's probably junk anyway ...

Fix $A,Bin mathbb{Z}$, obtaining an elliptic curve $E$ over $S=Spec
mathbb{Z}[Delta^{-1}]$, and a prime $pgeq 5$ not dividing the
discriminant $Delta=Delta(A,B)$. The
division polynomial $psi_p(t)$ is supposed to be a polynomial of
degree $d=(p^2-1)/2$ over $mathbb{Z}$, whose roots are the values
$t(P)=x(P)/z(P)$ as $P$ ranges over the points of exact order $p$ in
$E$.

Turn $psi_p$ into a homogeneous polynomial $g$ of degree $d$ in
$mathbb{Z}[x,y,z]$, so that $g(x,y,1)=psi_p(x)$. The polynomial
deterimes a curve $C=(g)$ in $P^2/S$, and thus a closed subscheme
$D=Ecap C$ of $E$. Over $mathbb{Z}[Delta^{-1},p^{-1}]$, $D$ should
consist of
the points of exact order $p$ on $E$ (with multiplicity $1$), together
with the basepoint of $E$ with multiplicity $d$.

Claim. $D$ is an effective Cartier
divisor on $E/S$, of degree $3d$.

Proof. I don't know. I need to prove that $Dto
S$ is flat, the main concern being the behavior over
$mathbb{Z}_{(p)}$. I don't even know if this is really plausible in general.

Let's pretend we somehow know $D$ is an effective Cartier divisor on $E$
relative to the base $S$. There is another relative Cartier
divisor, namely
$$ D' = E[p] quad + quad (d-1)[0]$$
where $E[p]$ is the $p$-torsion subgroup scheme of $E$, and $[0]$ is the
degree one divisor of the basepoint of $E$.
It seems clear that away from characteristic $p$, the divisors $D$
and $D'$ are equal. Equality of effective divisors on a smooth curve
is a closed condition, so they should be equal over all of
$S$.

The divisor I want is thus $D''=D-d[0]=D'-d[0]$ (which is still
effective). Then it's really easy to find a homogeneous polynomial
$h$ of degree $2d/3$ which defines $D''$; because of the form of the
Weierstrass equation, you can produce it from $g$ by hand, and if my claim is true you can produce it globally, i.e., with coefficients in $mathbb{Z}[A,B]$.

I've carried out this out in the case $p=5$, and it appears to "work".
That is, I get an answer which appears sane for general $A$ and $B$, and
which for some explicit cases I've tried of $A,Bin mathbb{Z}$ appears to give me a flat $Dto S$.
For instance, if $E/mathbb{Z}[6^{-1}]$ is the curve with $(A,B)=(0,1)$ (which reduces to a supersingular curve at $p=5$), I find
$$h= 729z^{8}-1350x^4z^4+360x^6z^2+5x^8.$$

Measure between the counting measure and the Lebegue measure

There are subsets of the real line that has infinite counting measure, but Lebegue measure 0, so the Lebegue measure is used for measuring larger sets than the counting measure. My question is: Is there a translation invariant measure m such that for some sets with Lebegue measure 0 the m-measure is infinite and for some sets with infinite counting measure, the m-measure is 0?

I have found one example: m(A)=0 if A is countable, and m(A)=infinite otherwise. So I will require that the measure can take the value 1.

If such a measure exist, can we find a measure between this and the counting measure? and between this and the Lebegue measure? and so on.

ag.algebraic geometry - Points on algebraic stacks

I'm a bit confused concerning a definition in Laumon--Moret-Bailly. Perhaps someone could shed some light on the following.

It concerns the definition of (closed) point in Chapter 5. More precisely, in 5.5 they define generization and specialization of points. But what are they really saying there? I mean if both x and y are closed points, then how can one be the generization of the other? What am I missing here?

/Daniel

gr.group theory - "Remove a vertex" map for right-angled Artin groups

No. Let $Gamma$ be the graph with two vertices and no edges - the non-abelian free group of rank two - and let $g$ be the commutator of the two generators $s_1$ and $s_2$. Then $g$ is certainly non-trivial, but $g$ dies whenever you kill $s_1$ or $s_2$.

UPDATE:

For an example with a connected graph, let's take $Gamma$ to be the straight-line graph with four vertices $a,b,c,d$ (so $[a,b]=[b,c]=[c,d]=1$). Now consider $g=[[c,a],[b,d]]$. Clearly this dies when you kill any generator. On the other hand,

$g=cac^{-1}a^{-1}bdb^{-1}d^{-1}aca^{-1}c^{-1}dbd^{-1}b^{-1}$

and a well-known solution to the word problem in right-angled Artin groups tells you that $g$ is non-trivial.

dg.differential geometry - mean curvature and polar tangential angle

To elaborate a little on the other answers, here's one approach: If you know the tangential angle $psi$, then you know the angle $pi/2 - psi$ between the point and the outer unit normal, i.e. the Gauss map. Using $theta$, $u(theta)$, and $pi/2 - psi(theta)$, you can write down an explicit formula for the Gauss map. Differentiate it to get the second fundamental form and take its trace (ADDED: You'll need to compute the first fundamental form to do this).

But this appears to require derivatives of both $u$ and $psi$. It's possible that $u'$ disappears from the final formula, but I wouldn't know. (ADDED: I see from a comment that $u'$ is allowed in the formula. In that case, it all definitely works.)

ADDED: Forget what I wrote above. It's easier than that. As Anton points out, the principal curvature directions are the obvious ones. Therefore, it is easy to figure out the principal curvatures in terms of the generating curve (and the circle carved out by each point on the curve). The final answer depends on $psi'$, $u$, and $u'$.

I now see why Anton was so brief. You really should work it all out yourself. It's a straightforward exercise.

lo.logic - cardinal equivalence: for each boolean formula, |quantifications| = |assignments|.

I think Daniel might be asking about the following proposition:

Let $f:{0,1}^nto{0,1}$ be any function (i. e., an "n-ary boolean function"). The number of true formulas $$ Q_1 x_1 ldots Q_n x_n : f(x_1,ldots,x_n) = 1,$$ where each $Q_i$ is a quantifier $forall$ or $exists$, is equal to the number of $(x_1,ldots,x_n)$ for which $f(x_1,ldots,x_n) = 1$.

The proof is very easy (by induction on $n$). It's an amusing proposition, no doubt, but I don't know of any applications. It might make an interesting advanced exercise in a discrete mathematics course, though.

I've implicitly answered the question, but explicitly:

Does anyone know the theorem by any other name?

I'm not aware of such.

Have you or anyone you know ever heard of this equivalence?

I discovered it a few years ago, apparently about five years after you did. Nobody I tried to tell seemed interested by it, though.

Do you prefer any other name, for casting into stone? (imo This theorem belongs in at least one major book...)

I prefer no name, actually. I don't think it's important enough to have the status of "theorem" (which is why I've been calling it a "proposition"), but I'm willing to be convinced otherwise.

mg.metric geometry - Diameter of m-fold cover

Here's a proposed sketch of an approach. I hope it actually works... [EDIT: it doesn't, as it stands. I guess the main take-away from the rough outline below is that whatever the answer is for graphs should carry over to manifolds].

First, we can prove an appropriate analog in the category of graphs. Let $G$ be a base graph and $tilde{G}$ a connected $m$-cover of $G$ in the combinatorial sense (the mapping takes vertices to vertices and edges to edges, and preserves local neighborhoods). It's useful to visualize $tilde{G}$ this as a set of discrete fibers over the vertices of $G$, the vertices of which can be aribtrarily numbered ${1,ldots, m}$. Now the edge-fibers correspond to permutations in $S_m$. Also notice that we may relabel the vertex fibers in order to make certain edge fibers "flat", meaning the corresponding permutation is the identity. This can simultaneously be done for a set of edges of $G$ which contain no cycle, such as a path (or a tree).

Given two vertices $tilde{x}, tilde{y}$ in $tilde{G}$, there's a path $P$ of length at most $d$ between their projections $x,y$ in $G$. We may assume that the permutations over the edges in $P$ are trivial. A path from $tilde{x}$ to $tilde{y}$ can now be formed by navigating across the floors (at most $d$ steps in each trip [EDIT: could be worse, since as you move to a new floor you're not guaranteed to land on the path]) and among the floors (at most $m$ steps overall), yielding $md+m$ steps in total. Sorry this is so vague but it's really quite simple if you draw a picture.

Now $m(d+1)$ is a bit too large (we want $md$) but this can't be helped in the category of graphs: for example, the hexagon (diameter 3) is a 2-cover of the triangle (diameter 1). But this is just because the triangle misrepresents the true diameter of the underlying geometry, which is really $3/2$. To resolve this nuisance, apply the procedure above to a fine subdivision of $G$ (and $tilde{G}$), which make $d to infty$ and the ratio is brought back to the desired $m$.

Next, consider simplicial complexes of higher dimension. It seems to me that if $X$ is a sufficiently nice topological space triangluated by a simplicial complex $K$, then the diameter of $X$ can be well approximated by the diameter of the 1-skeleton of a sufficiently fine subdivision of $K$. Is this true? Given two points in $X$ and a long path between them, if the path is close to a PL one than this should be the case. I hope that if $X$ is not too pathological, its diameter is represented by a tame path.

Finally, I would hope that a general Riemannian manifold (or some other kind of space for which we need to prove this) can be effectively triangulated, although this extends beyond my off-the-top-of-my-head knowledge.

Can something like this work?

rt.representation theory - Explicit description of all morphisms between symmetric groups.

Bret Benesh and Ben Newton determined all pairs $(m,n)$ such that $S_m$ contains a maximal subgroup isomorphic to $S_n$. They are either $(n+1,n)$ with the obvious inclusion (or mapping $S_5$ into the image of a point stabilizer under the outer automorphism of $S_6$); $(binom{n}{k},n)$, coming from the action of $S_n$ on the subsets of $k$ elements of ${1,2,ldots,n}$; and $((kr)!/(r!)^k k!, kr)$ with $1lt k,r$, with $S_{kr}$ acting on the the right cosets of a maximal subgroups of the wreath product $S_kwr S_r$. This appears in A classification of certain maximal subgroups of symmetric groups, J. Algebra 304 (no. 2) pp. 1108-1113, MR2265507.

Bret later also determined all pairs $(m,n)$ such that $S_m$ has a maximal subgroup isomorphic to $A_n$; such that $A_m$ has a maximal subgroup isomorphic to $S_n$; and such that $A_m$ has a maximal subgroup isomorphic to $A_n$. This appears in the book Computational Group Theory and the Theory of Groups, Contemporary Mathematics 470 (L-C Kappe, R. F. Morse, and me as editors), AMS 2008; the paper is A classification of certain maximal subgroups of alternating groups, pp. 21-26, MR2478411.

As pointed out by Jack, this does exhaust all possible embeddings of $S_n$ into $S_k$ (presumably you are okay with the maps that are not embeddings...)

co.combinatorics - Counting colored rook configurations in the cube - when is it even?

This can be phrased as a problem concerning Latin squares. Eg. a "rook set" is equivalent to a Latin square. For example:

123       100 010 001
231  <->  001 100 010
312       010 001 100

A colouring of the Latin square is a partition of its entries (corresponding to a coloured rook set).

We can therefore readily construct colour profiles P with c=n² such that N(P)=1 (that is, by partitioning some Latin square into n² parts). But we can do much better...

A defining set is a partial Latin square with a unique completion. A critical set is a minimal defining set. Let scs(n) be the size of a smallest critical set for Latin squares of order n. From a Latin square L containing a critical set of size scs(n), we can choose a partition (i.e. a c-colouring) such that the entries in the critical set are in parts of size 1 and the remaining entries of L are in a single part. This also will give rise to a colour profile P in which N(P)=1. It has been shown that scs(n)≤n²/4 for all n. (see J. Cooper, D. Donovan and J. Seberry, Latin squares and critical sets of minimal size, Australasian J. Combin 4 (1991), 113–120.) Hence we can deduce that for some c<=n²/4+1 we have N(P)≡1 (mod 2). But with a more intelligent choice of the partition, we can do better...

In fact, we can use the constructions in Cooper et al. to construct a colour profile P with c=n for which N(P)=1. I will only be able to prove this by example here (but it should be clear it can be readily generalised): Partition the "back circulant" Latin square of order 6 as follows.

123456     100000   020000   003000   000000   000000   000456
234561     000000   200000   030000   000000   000000   004561
345612 <-> 000000 + 000000 + 300000 + 000000 + 000000 + 045612
456123     000000   000000   000000   000000   000000   456123
561234     000000   000000   000000   000004   000000   561230
612345     000000   000000   000000   000040   000005   612300

Now observe that the three colour profiles are:

100005  111003  111003
020004  011004  011004
003003  001005  001005
000204  000006  000006
000015  000105  000105
000006  000114  000114

Together these form P. Given P, we can observe that any Latin square of order 6 with colour profile P must contain the following partial Latin square:

123...
23....
3.....
......
.....4
....45

Which is a critical set -- and therefore admits a unique completion (that is, to the "back circulant" Latin square of order 6). Therefore, if c=n there exists a color profile P for which N(P)=1, not just N(P)≡1 (mod 2).

EDIT: I presented this problem to our research group at Monash and we improved the upper bound to c=1 (which was a bit surprising!). The idea came from the following critical set (call it C) of the back-circulant Latin square, which is related to the one above, but contains more entries than the one above (but this doesn't matter for this problem).

12345.
2345..
345...
45....
5.....
......

We observed that if every entry in the critical set above is assigned one colour, and the remaining entries another colour, then there is a unique Latin square with that colour profile. That is, we derive the colour profile from in the following way.

123456     12345.   .....6
234561     2345..   ....61
345612 <-> 345... + ...612
456123     45....   ..6123
561234     5.....   .61234
612345     ......   612345

which gives rise to the following colour profile P:

15   51   51
24   42   42
33   33   33
42   24   24
51   15   15
06   06   06

We can deduce that any Latin square with the colour profile P will, in fact, contain the critical set C. Hence, N(P)=1.

To see how we deduce the critical set from the colour profile, we note that for the first colour, it contains 5 entries in the first row and column, 4 entries in the second row and column, and so on (and the same for columns). This selection of cells can take only one shape -- that is, it is unique. Then placing the symbols 1..5 in it can only be achieved in one way (to preserve the Latin property).

Since this can be generalised for all n≥2, the answer to your question "what is the largest c such that for all colour profiles P, N(P)≡0 (mod 2)" is c=1.

EDIT 2: We also discussed at our research meeting the complexity side of the problem.

Instance: colour profile P

Question: is N(P)≥1?

In fact, there is an easy way to see that this problem is NP-complete, since (a) we can embed instances of the problem of partial Latin square completion in the above problem and (b) a Latin square together with its colouring can be used as a certificate.

The problem of partial Latin square completion was shown to be NP-complete in: C. J., Colbourn, The complexity of completing partial Latin squares, Discrete Appl. Math. 8 (1984), no. 1, 25--30.

math philosophy - Were Bourbaki committed to set-theoretical reductionism?

I'm really upset because I wrote up a whole answer to your question and lost it because my connection dropped and it never registered =(.

Okay, Bourbaki's structuralism is effectively using categories, but only restricting yourself to concrete categories. One must remember that at the beginning of the writing, category theory had not yet been discovered, and by the time the first two chapters had been published, the work of Grothendieck and Lawvere hadn't even begun to discover topos theory. In terms of formal mathematics, set theories were the only game in town for formal exposition (and are still very much the prevailing model). That is, without first constructing a theory of metamathematics (chapter 1 section 1), logic (chapter 1), a proof calculus (chapter 1), and set theory (chapter 2), one was unable to be completely formal.

Bourbaki's global choice operator $tau$ allows you to find a distinguished object satisfying a proposition unless no object satisfies in which case it returns any object (this is by axiom scheme S7 of Bourbaki also called the axiom scheme of epsilon extensionality by Hilbert and his school). This effectively lets us talk about objects that are identical in terms of some structure, without worrying about the underlying set.

As for Bourbaki's reductionism in the later version of the book (I've read the older version, in fact [this is the source of the english translation]), I can say, having read the older version of the book, that the newer definition of an ordered pair is much easier to use to define the first and second projections, which is an exercise in painful tautology in the first edition (I just found and read the section in a copy of the french second edition, and the discussion is easier to understand even though I don't speak French). However, even in that book, the kuratowski structure is used once and then thrown away, never to be seen again. I would say that the change between editions was merely to make the page easier to read. Here is the reason why: The axiom of the ordered pair was redundant, since the ordered pair provably exists. Perhaps one could have defined the ordered pair (x,y) to be any object satisfying the axiom of the ordered pair (axiom 3 in Bourbaki Theorie des Ensembles 1. ed.), but this is really an unimportant point, and if you've read the book before, no time is wasted on unimportant details.

My conclusion on their reductionism in this case is that it was for simplicity of exposition and parsimony, because, as I've said above, why would one take as an axiom what one can prove?

[I have edited the following paragraph to maintain a positive tone and make clear that certain pronouncements are opinions rather than facts. -- Pete L. Clark]

[I edited it a little more because I didn't like the style, but the paragraph below is my opinion -- Harry Gindi]

Also, I find Mac Lane's criticism is a little strong. Bourbaki is a standard reference on elementary abstract algebra and general topology (if one wants to find the most general version of a theorem known to date in one of those subjects, a good place to start is Algebre or Topologie General by Bourbaki). One of the best places to learn about uniform spaces (which have come up on MO a striking number of times in the past few months) is in Bourbaki. Bourbaki proofs are also incredibly clear and really wonderful to read (once you have the mathematical maturity to do so). Again, Bourbaki on Topological Vector Spaces is again a standard reference on topological vector spaces. Their book on integration theory may only include Radon measures, but their section on the Haar measure is a standard reference on the subject. Their commutative Algebra book is one of the most in-depth books on commutative algebra currently around (rivaled, I would say, only by Matsumura [not so old] and Zariski-Samuel [which is really ancient]), and don't forget about the masterpiece that is Lie Groups and Lie Algebras, which is the only Bourbaki book that I've seen assigned as a class text rather than a reference. Anyone who's read SGA will see that Bourbaki actually wrote a number of sections (who participated isn't exactly clear, but the citation is to Bourbaki). Mac Lane has made great contributions to the world of mathematics, but I respectfully disagree with his assessment of Bourbaki. Bourbaki was a landmark in the style of mathematical exposition, with its emphasis on formalism, rigour, and clarity, in a way, ignoring the words of Goedel, and taking Hilbert's program of formalism as far as it could go.

Avoiding Minkowski's theorem in algebraic number theory.

For a course on algebraic number theory, you certainly can prove the finiteness of the class group without Minkowski's theorem. For example, if you look in Ireland-Rosen's book you will find a proof there which they attribute to Hurwitz. It gives a worse constant (which depends on a choice of $mathbf Z$-basis for the ring of integers of the number field; changing the basis can shrink the constant, but it's still generally worse than Minkowski's) but it is computable and you can use it to show, say, that $mathbf Z[sqrt{-5}]$ has class number 2.

As for the history of the proof of the unit theorem, it was proved by Dirichlet using the pigeonhole principle. If you think about it, Minkowski's convex body theorem is a kind of pigeonhole principle (covering the convex body by translates of a fundamental domain for the lattice and look for an overlap). You can find a proof along these lines in Koch's book on algebraic number theory, published by the AMS. Incidentally, Dirichlet himself proved the unit theorem for rings of the form $mathbf Z[alpha]$; the unit theorem is true for orders as much as for the full ring of integers (think about Pell's equation $x^2 - dy^2 = 1$ and the ring $mathbf Z[sqrt{d}]$, which need not be the integers of $mathbf Q(sqrt{d})$), even though some books only focus on the case of a full ring of integers. Dirichlet didn't have the general conception of a full ring of integers.

One result which Minkowski was able to prove with his convex body theorem that had not previously been resolved by other techniques was Kronecker's conjecture (based on the analogy between number fields and Riemann surfaces, with $mathbf Q$ being like the projective line over $mathbf C$) that every number field other than $mathbf Q$ is ramified at some prime.

ag.algebraic geometry - A necessary and sufficient condition for a curve to have an $A_k$ singularity.

Although the above answer involving the local algebra and the Milnor number is correct, it is often very hard to apply in real situations. Especially if you have a general function with arbitrary coefficients. You can perform a rather messy iterative process to check for an $A_k.$ In general the condition is far too ugly to want to, or be able to, write down.

You have a curve in the plane given by $f(x,y) = 0.$ The Taylor series, with respect to $x$ and $y$ is what you're really interested in. Let's assume we are only interested in the origin. If the linear terms vanish then you know that you have a singular point (a critical point of $f$). In that case you consider the quadratic part. If the quadratic part is non-degenerate, i.e. not a perfect square, then you have Morse singularity. These are $mathscr{A}$-equivalent to $x^2 pm y^2,$ and give the so-called $A_1^{pm}$-singularity types.

(Notice that $mathscr{A}$-equivalence has no relevance to the $A$ in $A_k.$ $mathscr{A}$-equivalence is also called $mathscr{RL}$-equivalence. You allow diffeomorphic changes of coordinate in the source and target (right and left sides of the commutativity diagram.)

If $f$ has a zero linear part and a degenerate quadratic part, we complete the square on the quadratic part. Then take a change of coordinates that turns the quadratic part into $tilde{x}^2$. The condition for exactly an $A_2$ is that $tilde{x}$ does not divide the new, post-coordinate change, cubic term. If not then $f$ is $mathscr{A}$-equivalent to $tilde{x}^2 + tilde{y}^3.$ (There is no $pm$ because $(x,y) mapsto (x,-y)$ changes the sign of the cubic term.

If $tilde{x}$ does divide the new cubic term, then you can complete the square on the three jet, i.e. on the quadratic and cubic terms as a whole. You take a change of coordinates so that this completed square become, say $X^2$. The condition for an $A_3^{pm}$ is that $X$ does not divide the new, post-coordinate change, quadric terms. If not then $f$ is $mathscr{A}$-equivalent to $X^2 pm Y^4.$

In general you follow the same pattern. Complete the square, take a formal power series change of coordinates so that the perfect square becomes $x_{new}^2.$ Check if $x_{new}$ divides the next set of fixed order terms. If not then stop. If $x_{new}$ didn't divide the order $n$-terms then $f$ is $mathscr{A}$-equivalent to $x_{new}^2 pm y_{new}^n.$

You just repeat the pattern: Is the quadratic part degenerate? If so then change coordinates (by a formal power series of low, but sufficient order) so that the degenerate part becomes $x_{new}^2 + O(3).$ Check if $x_{new}$ divides the cubic terms. If not then you have $x^2 + y^3.$ If so then complete the square on the new 3-jet and change coordinates so that you have $x_{new}^2 + O(4)$. Does $x_{new}^2$ divide the quartic terms? If not then you have $x^2 pm y^4.$ If so then complete the square on the new 4-jet and change coordinates. Just keep completing the square, checking divisibility, changing coordinates.

The conditions on the coefficients soon spiral out of control. To check an $A_6$ you'll need a computer program. For a general polynomial it's impossible without a computer. (Except for very special cases!) I wrote a program in Maple to calculate the conditions up to $A_k$ once, but the output was so messy then I gave up. Having said that, for an explicit polynomial it's child's play.

at.algebraic topology - relationship between borromean rings and hanging-a-picture-from-three-nails puzzle?

I recently heard the following puzzle: There are three nails in the wall, and you want to hang a picture by wrapping a wire attached to the picture around the nails so that if any one nail is removed the picture still stays but if any two nails are removed then the picture falls down. An answer, schematically, is given by abca^-1b^-1c^-1 (i.e., wrap it clockwise around each of the three nails in some order, then counterclockwise around each of the three nails in the same order).

This reminded me very much of the Borromean rings, where three rings are linked but when any one ring is removed the other two become unlinked. So I was trying to figure out if there might be some way to transform one situation into the other. My first instinct was to put the rings in S³ and have one of them pass through ∞, but that isn't really right. What seems to be tripping me up is that with the picture there's an extra object (the wire) that doesn't show up with the Borromean rings, but I have a vague idea that perhaps we could change the former situation by saying that we make the loop in the complement of three unlinked rings, and then perhaps "pulling really hard on the wire" would somehow thread the rings together. Maybe my issue is that what's really going on with the picture is just that we're making a loop in the complement of three points in plane, and I'm confounding phenomena of different dimensions...

Does anyone have any ideas?

von neumann algebras - When does a conditional expectation preserve some trace?

In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo & Isola,1996], etc.) define the index of a conditional expectation of a von Neumann algebra M onto a vN-subalgebra N (here, a conditional expectation is a normal, faithful N-N bimodule map fixing the subalgebra pointwise). An inclusion is said to have finite index if there exists a conditional expectation that has finite index. However, in the case where M is finite we might be interested in restricting ourselves to the conditional expectations that preserve some trace on M.

This leads us to the question: For a given (normal, faithful, finite) trace on M, Umegaki gives us a unique trace preserving conditional expectation E:M->N. Are there any nice necessary and sufficient conditions for a conditional expectation to arise in this manner? What if we allow the trace to be semifinite?

Since subfactors give rise to more than one conditional expectation, it is certainly not the case that all conditional expectations come from traces. A necessary condition is that E(xy)=E(yx) whenever x or y is an element of the relative commutant $N^prime cap M$. This is not sufficient, however.

geometry - How to transform a plane into a sphere? [SOLVED]

Given a 2-dimensional array of MxN heights, how to transform it to a sphere? Every element of this array is just a 3D point (x,y,z) where z represents some height. One has to transform this array into a sphere, twisting it around the origin so, that only minimal distortions will happen.

Representing it by spherical coordinates is not very good, because of the severe distortions. It's probably better if there is no direct one-to-one mapping from 2D plane to a surface of 3D sphere - many plane's points will not be involved. But what is the best possible mapping and how to transform involved points (elements of array)?

This is for a 3D-planet terrain simulation. First, fractal landscape is produced, then, it is to be transformed to 3D sphere.

Thanks in advance!

SOLUTION: Map projection

polynomials - what was Hilbert's geometric construction in his 17th problem?

Hilbert's 17th problem asked if a nonnegative real polynomial is the sum of squares of rational functions. It was answered affirmative by Artin in around 1920. However, in his speech, he also asked if the rational functions could have coefficients over Q rather than over R. Here is the relavant part of his speech

"At the same time it is desirable, for certain questions as to the possibility of certain geometrical constructions, to know whether the coefficients of the forms to be used in the expression may always be taken from the realm of rationality given by the coefficients of the form represented."

Does anyone know what these "certain geometrical constructions" are?

It seems maybe to me that Hilbert was attempting to embed rational projective space into higher dimensional rational projective space via these polynomials. Briefly, given a nonnegative homogeneous function $f(x_0,ldots, x_n)$ with rational coefficients induces a metric on $QP^n$. Suppose Hilbert's dream holds that $f=p_0^2+cdots + p_N^2$ where $p_i$'s are polynomials with rational coefficients. Then the map $p: QP^nto QP^N$ where $p(x)=(p_0(x),ldots, p_N(x))$ is an isometric embedding (almost!) where the metric induced by $f$ is the pullback back of the Euclidean metric on $QP^N$.

The above is just my hazard. But I would be delighted if anyone is aware of what exactly Hilbert's intended "geometrical constructions" are.

langlands conjectures - examples of admissible representations of $GL_{n}(K)$ over p-adic field

I second L Spice's recommendation of the book by Bushnell and Henniart, called "The local Langlands conjectures for GL(2)."

After you master the principal series representations, it's not too hard to tinker with some supercuspidals. Easiest among these are the tamely ramified supercuspidals. To construct these, let's start with the unramified quadratic extension $L/K$, with corresponding residue fields $ell/k$. Choose a character $theta$ of $L^times$ which has these properties:

(a) The character $theta$ is trivial on $1+mathfrak{p}_L$, so that $thetavert_{mathcal{O}_L^times}$ factors through a character $chi$ of $ell^times$.
(b) $chi$ is distinct from its $k$-conjugate. (In other words, $chi$ does not factor through the norm map to $k^times$.)

It's a standard fact that there's a corresponding representation $tau_chi$ of $text{GL}_2(k)$, characterized by the identity $text{tr}tau_chi(g)=-(chi(alpha)+chi(beta))$ whenever $gintext{GL}_2(k)$ has eigenvalues $alpha,betainellbackslash k$. (This is somewhere in Fulton and Harris, for instance.)

Inflate $tau_chi$ to a representation of $text{GL}_2(mathcal{O}_K)$, and extend this to a representation $tau_theta$ of $K^timestext{GL}_2(mathcal{O}_K)$ which agrees with $theta$ on the center. Finally, let $pi_theta$ be the induced representation of $tau_theta$ up to $text{GL}_2(K)$; then $pi_theta$ is an irreducible supercuspidal representation.

By local class field theory, our original character $theta$ can be viewed as a character of the Weil group of $L$. In the local Langlands correspondence, $pi_theta$ lines up with the representation of the Weil group of $K$ induced from $theta$. All the supercuspidals of $text{GL}_2(K)$ arise by induction from an open compact-mod-center subgroup, but the precise construction of these is a little more subtle than the above example.

Applications of string topology structure

Hossein Abbaspour gave an interesting connection between 3-manifold topology and the string topology algebraic structure in arXiv:0310112. The map $M to LM$ given by sending a point $x$ to the constant loop at $x$ allows one to split

$mathbb{H}_*(LM)$ as $H_*(M) oplus A_M$.

He showed essentially that the restriction of
the string product to the $A_M$ summand is nontrivial if and only if $M$ is hyperbolic. There are some technical details in the statements in his paper, but it was written pre-Perelman and I believe the statements can be made a bit more elegant in light of the Geometrization Theorem.

Philosophically, Sullivan has said that he his goal in inventing string topology was to try to find new invariants of smooth structures on manifolds. His original idea was that if you have to use the smooth structure to smoothly put chains into transversal positions to intersect them then you might hope that the answer will depend on the smooth structure. Unfortunately, we now know that the string topology BV algebra depends only on the underlying homotopy type of the manifold (there are now quite a few different proofs of various parts of this statement).

The string topology BV algebra is only a piece of a potentially much richer algebraic structure. Roughly speaking, $mathbb{H}_*(LM)$ is a homological conformal field theory. This was believed to be true for quite some time but it took a while before it was finally produced by Veronique Godin arxiv:0711.4859. She constructed an action of the PROP made from the homology of moduli spaces of Riemann surfaces with boundary. Restricting this action to pairs of pants recovers the original Chas-Sullivan structure.

Unfortunately, for degree reasons, nearly all of the higher operations vanish. In particular, any operation given by a class in the Harer stable range of the homology of the moduli space must act by zero. Hirotaka Tamanoi has a paper that spells out the details, but it is nothing deep.

Furthermore, it seems that the higher operations are homotopy invariant as well. For instance Lurie gets this as a corollary of his work on the classification of topological field theories.

Last I heard, Sullivan, ever the optimist, believes that there is still hope for string topology to detect smooth structures. He says that one should be able to extend from the moduli spaces of Riemann surfaces to a certain piece of the boundary of the Deligne-Mumford compactification. I've heard that the partial compactification here is meant to be that one allows nodes to form, but only so long as the nodes collectively do not separate the incoming boundary components from the outgoing boundary. Sullivan now has some reasons to hope that operations coming from homology classes related to the boundary of these moduli spaces might see some information about the underlying smooth structure of the manifold.

Remembering arrows' directions in basic Category Theory

I like to think of such properties in the forms they take for sets. For instance, I remember being rather surprised that the fibered product was actually a rather familiar thing (ordered pairs that lie over the same point in the base) after I read the section on it in Hartshorne, which just talked about the universal property, Yoneda's lemma means that you can restrict yourself to the case of Sets in proving results about them (Grothendieck spends some time on this in EGA 0 and gives plenty of detail).

For co things, though, you have to hom out of them. A map out of a quotient A/B is the same thing as a map out of A that vanishes on B. So, more generally, maps out of co things satisfy nice properties.

ag.algebraic geometry - Can projective hypersurfaces contain linear spaces? How big?

I am in this, rather friendly, situation:

I have a complex projective space $mathbb{P}^n$, and there i have a (possibly non-smooth) hypersurface $S$ defined by one irreducible polynomial $P$ of degree $d$.

What i want is to get information about the existence or not of linear subvarieties of $S$, and their maximal dimension $m$.
I seem to remember the existence of some ways to get bounds on $m$, given $n$ and $d$, but i don't remember anymore and i don't know where to look..

at.algebraic topology - Is the universal covering of an open subset of $mathbb{R}^n$ diffeomorphic to an open subset of $mathbb{R}^n$ ?

I don't have an answer, only a heuristically inspired hunch. If we think of the figure eight, we can thicken it slightly to an open connected set in the plane. The universal cover is the universal TV antenna times an open interval. But this can be put into the plane by narrowing the branches of the thickened UTVA as one moves out from the center, and since one can do this arbitrarily fast, even tiny branches very far out can be prevented from colliding.

Now there is more room in higher dimensions, so the above kind of argument should actually be easier to carry through than in the plane. Perhaps if one excludes torsion in the fundamental group at least, the countability would be enough if the dimension is at least 3. Just visualize the countably many generating loops, and wiggle them very slightly (there is room enough) so they don't intersect. Then hopefully one can proceed as with the figure eight above. That there could be countably many branches at the forks does not seem to be an essential difficulty.

The above argument does not work generally in the plane, but for the plane the desired statement follows from (a special case of) the uniformization theorem of complex analysis:
Every simply connected open Riemann surface is conformally equivalent (and thus diffeomorphic) to the whole plane or the open upper half plane.

EDIT: I think it must be more complicated than this. Otherwise any open connected subset with torsionfree fundamental group, of a manifold of dimension at least three, would have a universal cover diffeomorphic to an open connected set in the same manifold. Surely this is wrong? (By the way, there are open connected sets in Euclidean space with torsion in their fundamental groups).

EDIT: I doubt there is room enough to make this work in dimension 3, maybe in dimension 4.

ag.algebraic geometry - Is an algebraic space group always a scheme?

Over a field, the reference is indeed "Algebraization of formal moduli, I".

Over a base more general than a field, the answer may be no. For example, Artin proved in "Algebraization of formal moduli, I" that under some natural conditions the Picard functor is representable by an algebraic space (it is a group algebraic space of course). There are several sufficient conditions for it to be a scheme, due to Artin and Mumford. You will find a discussion, if not the proof, of them in Bosch-Lutkebohmert-Raynaud's "Neron models". But in general case is not known, I think.

On the other hand, abelian algebraic spaces over a scheme S (i.e. smooth proper with irreducible geometric fibers) is an abelian scheme over S. See p.3 of Faltings-Chai's "Degenerations of abelian varieties", where this is attributed to Raynaud and Deligne.

ca.analysis and odes - Transforming a multivariable integral to make it separable

In the following I will omit requirements of smoothness, extent of domain, finiteness, etc, both to simplify the exposition and because I don't know exactly what the requirements are. Please imagine that such requirements are stated correctly.

Let's say that a function $F : mathbb{C}^ntomathbb{C}$ separates (where $mathbb{C}$ is the complex numbers) if we can factorize it like
$F(z_1,ldots,z_n)=F_1(z_1)timescdotstimes F_n(z_n)$. In this case we can factorize the integral of $F$ like $iiint F(z_1,ldots,z_n) ~dz_1cdots dz_n = int F_1(z_1)dz_1times cdots timesint F_n(z_n)dz_n$.

In case $F$ does not separate, we might be able to change variables to make it separate. If $g : mathbb{C}^ntomathbb{C}^n$ is nice enough and $Delta$ is its Jacobian determinant (or maybe its inverse depending on which way you like to define it), then
$iiint F(z_1,ldots,z_n) ~dz_1cdots dz_n = iiint G(w_1,ldots,w_n) ~dw_1cdots dw_n$, where $G(w_1,ldots,w_n)=F(g(w_1,ldots,w_n))Delta(w_1,ldots,w_n)^{-1}$.

My question is: for which $F$ can $g$ be chosen so that $G$ separates?

An example everyone knows is $F(z_1,ldots,z_n)=exp(Q(z_1,ldots,z_n))$, where $Q$ is a quadratic form. Then $g$ can be chosen to be a linear transformation that diagonalizes the quadratic form. In general, a non-linear transformation will be required.

dg algebras - Derived algebraic geometry via dg rings?

Dear Kevin,

This is more or less an amplification of Tyler's comment. You shouldn't take it too seriously, since I am certainly talking outside my area of expertise, but maybe it will be helpful.

My understanding is that homotopy theorists are extremely (perhaps primarily) interested in torsion phenomena. (After all,
homotopy groups are often non-trivial but finite.) TMF, for example, involves quite subtle torsion phenomena. Coupled with Tyler's remark that homotopy theorists have no fear of $E_{infty}$ rings, and so are (a) happy to identify them
with dg-algebras in char. zero, and (b) don't feel any psychological need to fall back on
the crutch of dg-algebras, this makes me suspect that your assumption (1) is likely to be wrong. (I share your motivation (2), but this is a psychological weakness of algebraists that
homotopy theorists seem to have overcome!)

In particular, one of Lurie's achievements is (I believe) constructing equivariant versions of TMF,
which (as I understand it) involves (among other things) studying deformations of $p$-divisible groups of derived elliptic curves. It seems hard to do this kind of thing
without having a theory that can cope with torsion phenomena.

Also, when Lurie thinks about elliptic cohomology, he surely includes under this umbrella TMF and its associated torsion phenomena. (So your (3) may not include all the aspects
of elliptic cohomology that Lurie's theory is aimed at encompassing.)

soft question - Most striking applications of category theory?

First, a comment on `studying category theory for its own sake': this slur was very much setting up a straw man. Those accessing the category theory discussion list will know that the discussion there ranges very widely, and actually discusses issues in mathematics, in contrast to other email discussion lists I access.

Second, I have found some elementary facts from category theory very useful; examples are `left adjoints preserve colmits, right adjoints preserve limits'. Many years ago, listening to Albrecht Dold on half exact functors made me realise how I could cut down considerably a proof from my thesis by using the basic idea of representable functor: this automatically led to the existence of a homotopy equivalence making a diagram commutative. Again, the theory of ends and coends does make life simpler in discussing geometric realisations.

Third, I have fairly recently realised that the general framework of fibred and cofibred categories is specially useful for discussing pullbacks and pushouts for certain hierarchical structures with which I have dealt. A basic example here is the bifibration (Groupoids) $to$ (Sets) given by the object functor.

I wish I had a good application in my work of some of the deeper theorems!