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# 2009 September 10th - questions

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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)?

#### 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?)

#### 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 $\mathbb{Z}$-modules e.g. coherent sheaves embeds into quasi-coherent sheaves (Anton: the Freyd embedding theorem says every abelian category embeds in $\mathbb{Z}$-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, $f:X\to R$ convex. If $X$ is infinite dimensional, 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.) (You can do the 1D 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

Consider the sequence $x_0=0, x_1=1, x_{n+2} = a x_{n+1} + b x_n$, generalizing the Fibonacci sequence. Fix $p$ a prime. If $k$ is minimal such that $p|x_k$ 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.

#### 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 $X\times I$ and $Y \times I$ 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 finitely-axiomatized first-order theory?

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