Fandom

Scratchpad

090910qa

215,853pages on
this wiki
Add New Page
Discuss this page0 Share

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.

2009 September 10th - questions/answers

Navigation: 20 Questions home | view only questions | next week


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 Greg Kuperberg on the SBS Blog

The answer is no, not even when you look at the facets of a convex simple 4-polytope with small-diameter facets. A convenient example is the Cartesian product of two convex polygons.

To achieve small-diameter facets, it is convenient to dualize to simplicial polytopes. Then the above example is the join of two polygons. The point then is that whatever examples you find, you stick two together at a facet. You stick any number together in a tree pattern, and the result can be made convex. Some of the vertex types stay the same, while others get more complicated.

Maybe there is a way to replace “smaller” by “more primitive” in the convex case. Certainly it cannot mean fewer than the number of sides of a truncated octahedron.

SECOND ANSWER by David Speyer on the SBS Blog

David Speyer - September 10, 2009 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.

(Greg Kuperberg: It’s more of an example than a counterexample, because dodecahedra have fewer facets (12) than truncated octahedra (14).)

(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.)

FIRST ANSWER by Matthew Emerton on the SBS Blog.

If the abelian category is small enough, e.g. all the objects are finite length, then you can embed it into its own Ind category, and the latter has enough injectives.

e.g.: torsion Z-modules are the inductive limits of finite abelian groups, and in fact the category of torsion Z-modules is equivalent to the Ind category of finite abelian groups. (And to get injective resolutions of finite abelian groups, you only need torsion Z-modules.)

quasi-e.g.: on a reasonable scheme (maybe Noetherian is enough) any quasicoherent sheaf is an inductive limit of coherent sheaves. On the other hand, coherent sheaves are not finite length, so this doesn’t quite fit into the above framework. Nevertheless, this example illustrates a general principal, of which the precise statement above is an illustration, which is that to get enough injectives, you have to allow certain inductive limits to exist.

4 Yael

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

FIRST ANSWER by Greg Kuperberg on the SBS Blog

The only way that I can think of that a description of cosets of G/H might be “nice” is if H is a complemented subgroup. Which in the normal case is the same as saying that G is a semidirect product.

In some important cases, Inn(G) is complemented by Out(G) in Aut(G). For instance, if G is compact and semisimple, then you can let Out(G) be the group of Dynkin diagram automorphisms.

If G = A_n so that Aut(G) = S_n with n ≥ 6, then Inn(G) is complemented, but in more than one way. There is more than one conjugacy class of odd involutions.

In some other cases, I don’t think that Inn(G) is complemented. (E.g. A6, thanks to David Speyer.) For instance, suppose that G is the group generated by translations in \R^n and by scalar multiplication. Then the automorphism group is GL(n,\R) semidirect \R^n, while Out(G) is PGL(n,\R). I think that Aut(G)/Inn(G) doesn’t split.

SECOND ANSWER by Alan

A G-bitorsor is a space on which G acts simply and transitively from the left and right, with the actions commuting. There is a natural 1-1 correspondence between isomorphism classes (in an obvious sense) of G-bitorsors and elements of Out(G). The group product of two bitorsors is (represented by) the cartesian product (of representative bitorsors) divided by the "inner diagonal" action of G. The unit is (represented by) G itself.

The paragraph above is also a concrete description of the automorphism group of what is sometimes called "the stack BG".

5 Mike

X a banach space, f:X\to R convex. If X is infinite dimensional, what extra conditions guarantee that f is continuous?

FIRST ANSWER by Greg Kuperberg on the SBS Blog.

From a paper: “A convex functional on a convex domain of a topological vector space is continuous if it is bounded above in an open subset, and then it becomes locally uniformly continuous”. The author cites Bourbaki. I’m not sure if that totally answers the question or not.

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

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.

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_p. 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 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.

FIRST ANSWER by Greg Kuperberg on the SBS Blog.

X = an open ball and Y = a closed ball.

If you want X and Y to be compact (and metrizable, say), then I can’t think of any such X and Y that are manifolds. Part of the question is to what extent is a manifold with boundary determined by its interior. Many compact manifolds are.

SECOND ANSWER by Andrew Critch

X=(0,1) and Y=[0,1]. (A specific case of Greg's answer; easy to visualize.)

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?

FIRST ANSWER by David Speyer on the SBS Blog.

A finite set P has prime order if and only if it can be equipped with the structures (0,1,+,*,\leq) such that:

  • (0,1,+,*) is a field
  • \leq is a total order, and
  • if x\neq -1 then x \leq x+1.

Also on Fandom

Random wikia