User matthew stover - MathOverflow

Answer by Matthew Stover for Reference for the geometry of horospheres

2010-05-31T15:20:10Z

I think you can find this in Chapter II.8 of Bridson and Haefliger.

Answer by Matthew Stover for It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-04-30T02:38:49Z

Let $T$ be a triangle in $\mathbb{H}^2$. Its area is $\pi - \alpha - \beta - \gamma$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles. You can find how slim this triangle is by considering an inscribed circle in $T$. The radius of this triangle, thus $\delta$, are bounded above by the area, so to find the $\delta$ that works for all triangles, you take the limit and consider an ideal triangle $T_\infty$. You can explicitly compute that the inscribed circle minimizing distance between the sides has length $4 \log \phi$, where $\phi$ is the golden ratio. (See here and here.)

Answer by Matthew Stover for Presentation for an infinite index subgroup of the braid group

2010-04-29T23:05:42Z

Unless you have a specific type of subgroup, like one that acts cocompactly on something, you're in deep trouble. Braid groups are incoherent (see Artin groups, 3-manifolds and coherence by Cameron Gordon). That is, there are finitely generated subgroups that are not finitely presented.

Answer by Matthew Stover for Reference for Unitary Group attached to $E/k$

2010-04-04T23:12:57Z

Scharlau's book, Quadratic and hermitian forms, gives the complete classification in Chapter 10.

Answer by Matthew Stover for Fundamental group of a compact space form.

2010-03-25T20:19:23Z

No, they are always different. Mostow Rigidity tells you that complete, finite-volume hyperbolic manifolds are determined by their fundamental groups. These groups never contain a $\mathbb Z \oplus \mathbb Z$, and any group acting on $\mathbb E^2$ contains this as a subgroup of finite index. Groups acting on the three-sphere are finite. This generalizes to all $n$: Mostow Rigidity remains true (and these fundamental groups never contain $\mathbb Z^n$; they are Gromov hyperbolic groups), Bieberbach Theorems say that the groups acting on $\mathbb E^n$ are virtually abelian, and the groups acting on the $n$-sphere are finite.

EDIT: I forgot to say compact above. Peripheral subgroups of a nonuniform lattice acting on $\mathbb H^n$ are abelian of rank $n-1$, so they aren't Gromov hyperbolic. However, they contain nonabelian free subgroups, so they still don't act nicely on $\mathbb E^n$ or the sphere.

Answer by Matthew Stover for How far can the analogy between a Cayley graph and a symmetric space be pushed?

2010-03-25T13:22:51Z

The Cheeger constants for graphs and Riemannian locally symmetric spaces are closely related. Via inequalities of Buser and Cheeger, these are also related to eigenvalues of the laplacians for each. This analogy led to the first construction of expander graphs, by Margulis, via Property (T). More recently, this analogy has been exploited by several people, notably Marc Lackenby, to study finite-sheeted coverings using Cayley graphs of finite quotients as a finite simplicial approximation.

The point, roughly, is the following. Let $\Gamma$ be a group with generating set $S$, and suppose $\Gamma = \pi_1(M)$ for some Riemannian manifold $M$. Then any finite quotient $F$ under a homomorphism $\phi$ has a generating set $\phi(S)$, so we can form the corresponding Cayley graph $\mathcal{G}(F, \phi(S))$. Properties of $\mathcal{G}(F, \phi(S))$ like girth, spectrum, expansion constants, Cheeger constant, and so forth are closely related to the analogous concept for the finite-sheeted covering $M_\phi$ of $M$ corresponding to the subgroup $\mathrm{kernel}(\phi)$ of $\Gamma$. This analogy is most potent when you consider a family {$\mathcal{G}(F_j, \phi_j(S))$} of Cayley graphs corresponding to a family $F_j$ of finite quotients of $\Gamma$.

References for all these concepts are the books On Property ($\tau$) by Lubotzky and Zuk (unpublished, but on Lubotzky's website), Discrete Groups, Expanding Graphs and Invariant Measures by Lubotzky, Elementary Number Theory, Group Theory and Ramanujan graphs by Davidoff, Sarnak, and Valette, and Marc Lackenby's paper Expanders, ranks and graphs of groups, Israel J. Math. 146 (2005) 357-370.

Answer by Matthew Stover for What are your favorite instructional counterexamples?

2010-03-02T21:07:52Z

The Poincaré homology sphere, a spherical 3-manifold with fundamental group the binary isosahedral group, was Poincaré's counterexample to the original formulation (in terms of homology) of his conjecture. Due to its countless descriptions -- as a spherical 3-manifold, via Dehn surgery, as the configuration space of an isosahedron, etc -- it's still a motivational example in geometry and topology.

Answer by Matthew Stover for Experimental Mathematics

2010-03-01T01:43:48Z

A fake projective plane is a smooth compact complex surface with the same Betti numbers as $\mathbb{P}^2(\mathbb{C})$. Mumford (MR0527834) built the first one in the late 70s using $p$-adic uniformization. Klingler (MR1990668) and, independently, Yeung (MR2128300 with erratum MR2559112), showed that they necessarily corresponded to arithmetic lattices in $\mathrm{PU}(2,1)$. (That they live in $\mathrm{PU}(2,1)$ follows from Yau's Theorem MR0451180; they do the arithmeticity.) Prasad and Yeung (MR2289867) then gave the list of arithmetic constructions which could possible create such a beast.

Using a computer experiment, Cartwright and Steger gave presentations for all the fundamental groups, giving a complete classification. They have recent announcement in Comptes Rendus #348, 11-13. Their presentations also allow one to show that the congruence subgroup property doesn't hold for certain commensurability classes of lattices for which it was previously unknown. In fact, these lattices are a few of the `simplest' examples for which the congruence subgroup problem remained mysterious. (Serre conjectures in his paper on $\mathrm{SL}_2$, MR0272790, that it fails for all $\mathbb{R}$-rank one groups.)

Answer by Matthew Stover for Do decidable properties of finitely presented groups depend only on the profinitization?

2010-02-27T02:22:46Z

Is it decidable from a presentation whether or not a group is large, i.e. admits a homomorphism onto the nonabelian free group on two letters? This seems totally unlikely, and surely either Henry or Daniel would know, but I like the following theorem anyway, so I'll advertise. Lackenby showed (`Detecting large groups', GR/0702571) that largeness is a property of the profinite completion of discrete finitely presented groups.

Answer by Matthew Stover for The finite subgroups of SL(2,C)

2010-02-22T04:04:10Z

Here are a few references for arithmetic Kleinian groups. One good reference is Chapter 12 of The Arithmetic of Hyperbolic 3-Manifolds (GTM 219) by Maclachlan and Reid, which is partially based on Chinburg and Friedman, The finite subgroups of maximal arithmetic Kleinian groups, Ann. Inst. Fourier (Grenoble) 50 no. 6 (2000), 1765--1798. Also, there's Vignéras, Arithmétique des Algébres de Quaternions, Lecture Notes in Math. 800. According to the notes at the end of Ch. 12 of Maclachlan--Reid, there's also a paper by V. Schneider in Math. Z. from `77.

Answer by Matthew Stover for What are some results in mathematics that have snappy proofs using model theory?

2010-02-18T15:38:35Z

You can prove Gromov's Theorem on groups of polynomial growth. See ven den Dries and Wilkie, Gromov's theorem on groups of polynomial growth and elementary logic, J Algebra, 89 (1984), 391-396.

Comment by Matthew Stover

2010-05-31T17:54:57Z

I thought it followed from what they do there, possibly throwing in strict convexity of the distance function. Perhaps I'm mistaken.

Comment by Matthew Stover

2010-05-19T21:59:36Z

You can be more explicit if you pick something with one boundary component and $\mathbb Z$ homology not coming from the boundary. For example, take a closed hyperbolic 3-manifold with a totally geodesic separating surface that doesn't carry all the homology, and cut the closed manifold in half. (Okay, and make sure there's homology supported on one half.) Pull-back from $\mathbb Z / n \mathbb Z$ quotients from that homology, and you get your covers with $n$ boundary components.

Comment by Matthew Stover

2010-05-13T15:00:42Z

Now that I have the references more handy, I can be more specific. I would recommend Chapter 8 of Scharlau for the basic question of the field splitting the algebra. He works in full generality. The book of involutions doesn't deal with this directly. However, for the functorial properties of the unitary group, their section on group schemes and Hopf algebras is quite good.

Comment by Matthew Stover

2010-05-12T14:56:13Z

You can find this in either Scharlau's book `Quadratic and hermitian forms' or in `The book of involutions'.

Comment by Matthew Stover

2010-04-30T04:21:28Z

Well, that's what I get for not checking it...

Comment by Matthew Stover

2010-04-30T03:05:02Z

Ahh, I see the confusion. I said radius instead of hyperbolic length. Edit forthcoming.

Comment by Matthew Stover

2010-04-30T02:50:31Z

If you make the ideal triangle in the disk model, the $4 \log \phi$ comes from connecting the midpoints. It's the minimum distance between the sides.

Comment by Matthew Stover

2010-04-30T01:32:02Z

Do you know them as words in standard generators? Perhaps you could experiment a bit with Magma or Gap using a known presentation.

Comment by Matthew Stover

2010-04-28T15:15:30Z

Sorry, now I see. I didn't have the paper handy last night, but did know that 6.5 is the heart of the paper. You do get finite generation from their argument. All I could recall is that 6.5 shows there's an open set $O$ for which $G(\mathbb{R}) = O G(\mathbb{Z}}) K O$, but also $|O O^{-1} \cap g_1 G(\mathbb{Z}) g_2| < \infty$ for $g_j \in G(\mathbb{Q})$. The last condition gives it to you.

Comment by Matthew Stover

2010-04-28T06:10:21Z

Well, first of all, $G(\mathbb{Z} \otimes \mathbb{R})$ might not be reductive, as in the examples given in the question. Perhaps I'm missing something, but I don't see why the argument works even in the semisimple setting. It is a lattice, which is not accidentally analogous to Dirichlet's unit theorem, but I still don't see how you get to finite generation for a generic discrete subgroup of finite covolume in $G(\mathbb{R})$.

Comment by Matthew Stover

2010-04-28T05:31:33Z

I think it's a bit more subtle. For cocompact lattices, I think you still need to use Weil's paper, and, as I recall, one of the reasons Kazhdan defined Property (T) was to prove that nonuniform lattices were finitely generated (a group with Property (T) is compactly generated, so discrete implies finitely generated).

Comment by Matthew Stover

2010-04-19T18:21:29Z

Ahh, right. This was my vague recollection. Excellent, thanks.

Comment by Matthew Stover

2010-04-19T17:13:13Z

Is there an easy example that isn't?

Comment by Matthew Stover

2010-04-19T16:21:39Z

Doesn't it suffice to prove that Diff(M) is path connected? I think Smale gave a simple proof of something along these lines, but I could well be mixing things up.

Comment by Matthew Stover

2010-04-09T17:11:30Z

Of course, you're right; one can figure out the automorphisms without the classical picture. I was merely saying that they're always drawn as they are because that's the picture for which the automorphisms stare you in the face. I only meant to argue for aesthetic optimization, not deeper mathematical content.