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.

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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.