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1 Scott
In R^2, you can tile the plane with hexagons. However any closed trivalent graph has a face that’s smaller than a hexagon. You can tile R^3 with vertex-truncated octahedrons. Say we have a “generic” closed finite cell-complex (every edge has 3 incident faces, every vertex has 4 incident edges). Is there something “smaller” than a vertex-truncated octahedron (or the other polytopes that give generic tilings)?

FIRST ANSWER by David Speyer on the SBS Blog

David Speyer - September 10, 2009 1 Since this isn’t a precise question, I don’t have a precise answer. But why should I let that stop me?

Two thoughts: (a) Would you consider the 120-cell a counter-example to what you are asking for? All of its faces are dodecahedra.

(b) I expect you want to ask your cell complex to be a 3-sphere, just as you wanted your graph in the two dimensional example to be embedded in a 2-sphere. (It is possible to tile a genus 3 curve with heptagons.) But there is an important difference between the 2-sphere and the 3-sphere: the 2-sphere has Euler characteristic 2, greater than the 2-torus, while the 3-sphere and 3-torus have the same Euler characteristic, namely 0. Since the bit with the hexagons can be thought of as a discrete analogue of the Gauss-Bonnet theorem, I would expect it to work differently in the 3 dimensional case.

2 Critch
Is there a space with trivial homology, non trivial homotopy? (Anton: isn’t there a result that say that first nontrivial homology and homotopy agree?)

FIRST ANSWER by Matthew Kahle on the SBS Blog.

I also wondered this as a graduate student and eventually found out that the answer is, “Yes, certainly.” They are called acyclic spaces. Probably the easiest example is a punctured Poincare sphere. The first nontrivial homology group (with integer coefficients) is the abelianization of the fundamental group. So an acyclic space has a perfect fundamental group. As is frequently the case in topology, everything comes down to the fundamental group. It is not too hard to show, using the Hurewicz theorem, that a simply connected acyclic CW-complex is contractible.

3 Yuhao
Let A be an abelian category, that might not have enough injectives? Can you embed into another abelian category with enough injectives? Is there a universal way? e.g. finite abelian groups embeds into Z-modules e.g. coherent sheaves embeds into quasi-coherent sheaves (Anton: the Freyd embedding theorem says every abelian category embeds in Z-mod. But this doesn’t help universality.)

4 Yael
Out(G) = Aut(G)/Inn(G). Is there a nice description of cosets, beyond that they’re cosets?

5 Mike
X a banach space, f:X->R convex. If X is infinite dimenionsal, what extra conditions guarantee that f is continuous?

6 Darsh
Take a triangle in R^2 with coordinates at rational points. Can we find the smallest denominator point in the interior? (Take the lcm of the denominators of the coordinates.) (Hint: you can do the 1-d version using continued fractions.)

7 Jakob
Take a “sparse” (every vertex has reasonably small degree) graph. Consider a maximal independent set for the graph (a maximal set of disconnected vertices). Can we make a new graph, with vertices the set, and whatever edges we like, that is ``as topologically similar to the original graph as possible''? (What does this mean?)

8 Andrew D
What’s the deal with algebraic geometry? Just kidding! Consider the sequence x0=0, x1=1, x_{n+2} = a x_{n+1} + b x_n, generalizing the Fibonacci sequence. Fix p a prime. If k is minimal so p|x_k and p|x_l implies k|l, then v_p(x_nk) = v_p(x_k) + v_p(n). (Here v_p(z) is the power of p dividing z.) Is there some framework that makes this sort of result obvious? Andrew only knows strange proofs.

FIRST ANSWER by David Speyer on the SBS Blog:

Let r and s be the roots of r^2=ar+b. Then x_n is (r^n-s^n)/r-s.

For the purposes of this question, think of r and s as living in the algebraic closure of Q. The integer k is minimal such that r^k=s^k mod p. Expanding (r^{kn} - s^{kn})/(r-s) as a p-adically convergent power series in n should imply the result in question, and many similar ones.

9 Anton
Take I=[0,1), the half open interval. Do there exist topological spaces X and Y, with X and Y not homeomorphic, but XxI and YxI are homeomorphic? E.g., if instead I=[0,1], the closed interval, you can take X=mickey mouse=disc with two discs removed, Y=cross-eyed frog=disk with two linked bands glued on the boundary.

10 Pablo
x^x^x^x … converges if x \in [e^{-1}, e^{1/e}], e.g. with x=\sqrt{2}, this converges to 2. Given a sequence (a_i), when does the “power tower sequence” converge?

11 Andrew D again
Can you define the set of all primes with a first-order theory?

Given a first-order theory, let S(T)={|M| | M is a finite model of T}. You can get all prime powers with the field axioms. Is there some T so S(T)={primes}.