## Posts Tagged ‘groups’

### On Tao’s talk and the 3-dimensional Hilbert-Smith conjecture

May 6, 2012

Last Wednesday Terry Tao briefly dropped by our little town and gave a colloquium. Surprisingly this is only the second time I hear him talking (the first one goes back to undergrad years in Toronto, he talked about arithmetic progressions of primes, unfortunately it came before I learned anything [such as those posts] about Szemeredi’s theorem). Thanks to the existence of blogs, feels like I knew him much better than that!

This time he talked about Hilbert’s 5th problem, Gromov’s polynomial growth theorem for discrete groups and their (Breuillard-Green-Tao) recently proved more general analogy of Gromov’s theorem for approximate groups. Since there’s no point for me to write 2nd-handed blog post while people can just read his own posts on this, I’ll just record a few points I personally found interesting (as a complete outsider) and moving on to state the more general Hilbert-Smith conjecture, very recently solved for 3-manifolds by John Pardon (who now graduated from Princeton and became a 1-st year grad student at Stanford, also appeared in this earlier post when he gave solution to Gromov’s knot distortion problem).

Warning: As many of you know I never take notes during talks, hence this is almost purely based on my vague recollection of a talk half a week ago, inaccuracy and mistakes are more than possible.

All topological groups in this post are locally compact.

Let’s get to math~ As we all know, a Lie group is a smooth manifold with a group structure where the multiplication and inversion are smooth self-diffeomorphisms. i.e. the object has:

1. a topological structure
2. a smooth structure
3. a group structure

It’s not too hard to observe that given a Lie group, if we ‘forget’ the smooth structure and just see it as a topological group which is a (topological) manifold, then we can uniquely re-construct the smooth structure from the group structure. From my understanding, this is mainly because given any element in the topological group we can find a unique homomorphism of the group $\mathbb{R}$ into the manifold, sending $0$ to identity and $1$ to the element. resulting a class of curved through the identity, a.k.a the tangent space. Since the smooth structure is determined by the tangent space of the identity, all we need to know is how to ‘multiply’ two such parametrized curves.

The way to do that is to ‘zig-zag’:

Pick a small $\varepsilon$, take the image of $\varepsilon$ under the two homomorphisms, alternatingly multiplying them to obtain a sequence of points in the topological group. As $\varepsilon \rightarrow 0$ the sequence becomes denser and converges to a curve.

The above shows that given a Lie group to start with, the smooth structure is uniquely determined by the topological group structure. Knowing this leads to the natural question:

Hilbert’s fifth problem: Is it true that any topological group which are (topological) manifolds admits a smooth structure compatible with group operations?

Side note: I had a little post-colloquium discussion with our fellow grad student Sam Lewallen, he asked:

Question: Is it possible for the same topological manifold to have two different Lie group structures where the induced smooth structures are different?

Note that neither the above nor Hilbert’s fifth problem shows such thing is impossible, since they both start with the phase ‘given a topological group’. My *guess* is this should be possible (so please let me know if you know the answer!) The first attempt might be trying to generate an exotic $\mathbb{R}^4$ from Lie group. Since the 3-dimensional Heisenberg group induces the standard (and unique) smooth structure on $\mathbb{R}^3$, I guess the 4-dimensional Heisenberg group won’t be exotic.

Anyways, so the Hilbert 5th problem was famously solved in the 50s by Montgomery-Zippin and Gleason, using set-theoretical methods (i.e. ultrafilters).

Gromov comes in later on and made the brilliant connection between (infinite) discrete groups and Lie groups. i.e. one see a discrete group as a metric space with word metric, ‘zoom out’ the space and produce a sequence of metric spaces, take the limit (Gromov-Hausdorff limit) and obtain a ‘continuous’ space. (which is ‘almost’ a Lie group in the sense that it’s an inverse limit of Lie groups.)

Hence he was able to adapt the machinery of Montgomery-Zippin to prove things about discrete groups:

Theorem: (Gromov) Any group with polynomial growth is virtually nilpotent.

The beauty of the theorem is (in my opinion) that we are given any discrete group, and all that’s known is how large the balls are (in fact, not even that, we know how large the large balls grow), yet the conclusion is all about the algebraic structure of the group. To learn more about Gromov’s work, see his paper. Although unrelated to the rest of this post, I shall also mention Bruce Kleiner’s paper where he proved Gromov’s theorem without using Hilbert’s 5th problem, instead he used space of harmonic maps on graphs.

Now we finally comes to a point of briefly mentioning the work of Tao et.al.! So they adopted Gromov’s methods of limiting and ‘ultra-filtering’ to apply to stuff that’s not even a whole discrete group: Since Gromov’s technique was to take the limit of a sequence of metric spaces which are zoomed out versions of balls in a group, but the Gromov-Hausdorff limit actually doesn’t care about the fact that those spaces are zoomed out from the same group, they may as well be just a family of subsets of groups with ‘bounded geometry’ of a certain kind.

Definition: An K-approximate group $S$ is a (finite) subset of a group $G$ where $S\cdot S = \{ s_1 s_2 \ | \ s_1, s_2 \in S \}$ can be covered by $K$ translates of $S$. i.e. there exists $p_1, \cdots, p_K \in G$ where $S \cdot S \subseteq \cup_{i=1}^k p_i \cdot S$.

We shall be particularly interested in sequence of larger and larger sets (in cardinality) that are K-approximate groups with fixed $K$.

Examples:
Intervals $[-N, N] \subseteq \mathbb{Z}$ are 2-approximate groups.

Balls of arbitrarily large radius in $\mathbb{Z}^n$ are $C \times 2^n$ approximate groups.

Balls of arbitrarily large radius in the 3-dimensional Heisenberg group are $C \times 2^4$ approximate groups. (For more about metric space properties of the Heisenberg group, see this post)

Just as in Gromov’s theorem, they started with any approximate group (a special case being sequence of balls in a group of polynomial growth), and concluded that they are in fact always essentially balls in Nilpotent groups. More precisely:

Theorem: (Breuillard-Green-Tao) Any K-approximate group $S$ in $G$ is covered by $C(K)$ many translates of subgroup $G_0 < G$ where $G_0$ has a finite (depending only on $K$) index nilpotent normal subgroup $N$.

With this theorem they were able to re-prove (see p71 of their paper) Cheeger-Colding’s result that

Theorem: Any closed $n$ dimensional manifold with diameter $1$ and Ricci curvature bounded below by a small negative number depending on $n$ must have virtually nilpotent fundamental group.

Where Gromov’s theorem yields the same conclusion only for non-negative Ricci curvature.

Random thoughts:

1. Can Kleiner’s property T and harmonic maps machinery also be used to prove things about approximate groups?

2. The covering definition as we gave above in fact does not require approximate group $S$ to be finite. Is there a Lie group version of the approximate groups? (i.e. we may take compact subsets of a Lie group where the self-product can be covered by $K$ many translates of the set.) I wonder what conclusions can we expect for a family of non-discrete approximate groups.

As promised, I shall say a few words about the Hilbert-Smith conjecture and drop a note on the recent proof of it’s 3-dimensional case by Pardon.

From the solution of Hilbert’s fifth problem we know that any topological group that is a n-manifold is automatically equipped with a smooth structure compatible with group operations. What if we don’t know it’s a manifold? Well, of course then they don’t have to be a Lie group, for example the p-adic integer group $\mathbb{Z}_p$ is homeomorphic to a Cantor set hence is not a Lie group. Hence it makes more sense to ask:

Hilbert-Smith conjecture: Any topological group acting faithfully on a connected n-manifold is a Lie group.

Recall an action is faithful if the homomorphism $\varphi: G \rightarrow homeo(M)$ is injective.

As mentioned in Tao’s post, in fact $\mathbb{Z}_p$ is the only possible bad case! i.e. it is sufficient to prove

Conjecture: $\mathbb{Z}_p$ cannot act faithfully on a finite dimensional connected manifold.

The exciting new result of Pardon is that by adapting 3-manifold techniques (finding incompressible surfaces and induce homomorphism to mapping class groups) he was able to show:

Theorem: (Pardon ’12) There is no faithful action of $\mathbb{Z}_p$ on any connected 3-manifolds.

And hence induce the Hilbert-Smith conjecture for dimension 3.

Discovering this result a few days ago has been quite exciting, I would hope to find time reading and blogging about that in more detail soon.

### Haken manifolds and virtual Haken conjecture

November 21, 2011

Hi people~ My weekends have been unfortunately filled up with grading undergrad assignments for the last couple of weeks >.< I'll try to catch up on blogging by finding some other time slot during the week.

As a grand-student of Thurston's I feel obligated to end my ignorance regarding Haken manifolds. I guess it's a good idea to start by writing my usual kids-friendly exposition here.

In the rest of the post, $M$ is a compact (so perhaps with boundary), orientable, irreducible (meaning each embedded 2-sphere bounds a ball) 3-manifold.

Definition: A properly embedded oriented surface $S \subseteq M$ is incompressible if $S$ is not the 2-sphere and any simple closed curve on $S$ which bounds an embedded disc in $M \backslash S$ also bounds one in $S$.

Figure 1

In other words, together with Dehn’s lemma this says the map $\varphi: \pi_1(S) \rightarrow \pi_1 (M)$ induced by the inclusion map is injective.

Note that the surface $S$ could have boundary, for example:

Figure 2

Definition: $M$ is Haken if it contains an incompressible surface.

Okay, at this point you should be asking, what’s good about Haken manifolds? The beauty about it is that, roughly speaking, once you find one incompressible surface in the manifold, you can just keep finding them until the manifold is completely chopped up into balls by incompressible surfaces.

Theorem: (Haken) Any Haken 3-manifold $M$ contains a hierarchy $S_0 \subseteq S_1 \subseteq \cdots \subseteq S_n$ where

1.$S_0$ is an incompressible surface in $M$
2.$S_i = S_{i-1} \cup S$ where $S$ is an incompressible surface for the closure of some connected component $K$ of \$latex $M \backslash S_{i-1}$
3.$M \backslash S_n$ is a union of 3-balls

Sketch of proof: This is much simpler than it might appear to be. The point is (at least in my opinion), except for trivial cases as long as a manifold has boundary it must be Haken.

Lemma: If $\partial M$ has a component that’s not $\mathbb{S}^2$ then $M$ is Haken.

The proof of the lemma is merely that any such $M$ will have infinite $H_2$ hence by the sphere theorem it will contain an embedded surface with non-trivial homology, if such surface is compressible then we just cut along the boundary of the compressing disc and glue two copies of it. This does not change the homology. Hence at the end we will arrive at a non-trivial incompressible surface.

Figure 3

Now back to proving of the theorem, so we start by setting $S_0$ to be an incompressible surface given by $M$ being Haken.

Now since $M$ is irreducible, we cut along $S_0$, i.e. take the closure of each component (may have either one or two components) of $M\backslash S_0$. Those will have a non-spherical boundary component, hence by lemma containing homologically non-trivial incompressible surface.

This process continuous as long as some pieces has non-spherical boundary components. But since $M$ is irreducible, any sphere bounds a 3-ball in $M$, hence all components with sphere boundary are 3-balls. (In particular, the case where a component have multiple sphere boundary components cannot occur since the first boundary component bounds a 3-ball hence it can’t have any non-trivial incompressible surfaces on both sides.)

Now the only remaining piece is to show that this process terminates. We apply a standard ‘normal surface argument’ for this. Essentially if we fix a triangulation of $M$,

A normal surface in $M$ is one that intersects each 3-simplex in a disjoint union of following two shapes:

Figure 4

There can’t be infinitely many non-parallel disjoint normal surfaces in $M$ (in fact there can be no more than 6 times the number of 3-simplexes since each complementry component need to contain at least one non-I-bundle part from one 3-simplex).

Figure 5

However, if the above process do not terminate, we would obtain a sequence of non-parallel non-spherical boundary components:

Figure 6

They represent different homology classes hence can be represented by disjoint normal which results in a contradiction.

In general, this gives a way to prove theorems about Haken manifolds by using inductionL i.e. one may hope to just show the property trivially holds for 3-balls and is invariant under gluing two pieced along an incompressible surface. Note that the gluing surface being incompressible is in fact quite strong hence making the induction step possible in many cases.

For example, by applying an incredible amount of brilliant techniques, Thurston was able to prove his revolutionary result:

Hyperbolization theorem for Haken manifolds: Any Haken manifold $M$ with tori boundary components that does not contain incompressible tori admits a complete hyperbolic structure of finite volume in its interior.

In other words, this is saying that given a Haken manifold, we cut along any incompressible tori, the resulting manifold with tori boundary must have a complete hyperbolic structure with cusps near each boundary component,

This is the best we could hope for since manifolds with incompressible tori would have their fundamental group split over $Z^2$ which of course imply they can’t be hyperbolic.

Now the more manifolds being Haken means the better this theorem is. Many evidences show that in fact a lot of manifolds are indeed Haken, in perticular we have:

Virtual Haken Conjecture: $M$ is finitely covered by a Haken manifold as long as $\pi_1(M)$ is infinite.

We can see that together with Thurston’s hyperbolization theorem, this would give full solution to the geometrization conjecture for general 3-manifolds.

However, although now Perelman has proved the geometrization conjecture, the virtual Haken conjecture remains open. But in light of Perelman’s result now we are able to try to ‘back-solve’ the puzzle and only prove the virtual Haken conjecture for hyperbolic manifolds.

(to be continued)

### Graph of groups in relation to 3-manifolds

October 24, 2011

(some images might appear soon)

Somehow I decided to wake up at 6:30 a.m. every Thursday to attend Bruce Kleiner‘s 9:30 course in NYU this semester. So far it’s been fun~

I learned about this thing called graph of groups. If you have been reading posts on this blog regarding any geometric group theory stuff (especially those posts related to Kleiner), then warning: this ‘graph’ has nothing to do with the Cayley graph. It’s not much about geometry but a rather ‘category-theoretical’ thing. Well, at this point you may think that you hate those algebra prople and is ready to leave…just don’t do that yet, because I hated them too, and now I finally got a tiny bit of understanding and appreciation on what those abstract non-sense was all about! :-P

So how do we connect cool 3-manifold stuff (incompressible surfaces, loops, embedded discs Heegaard splittings etc.) to groups?

Well, one handy thing is of course the Dehn’s lemma:

Theorem: For 3-manifold $M$ with boundary, if the inclusion map $i: \pi_1(\partial(M)) \rightarrow \pi_1(M)$ is not injective, then there exists a simple non-trivial loop in $\partial M$ bounding an embedded disc in $M$.

Note: Dehn’s theorem was proved by Papakyriakopoulos, I talked about it in this pervious post, although not exactly stated in this form, we can see that Dehn’s lemma follows easily from the loop theorem.

This means we can say things about the 3-manifold by only looking solely at maps between groups!

That’s cool, but sometimes we find groups and just one map between two groups are not enough, and that’s when graph of groups comes in:

Definition: A graph of groups is a graph with vertice set $V$, edge set $E$, to each vertex $v$ we associate a group $G_v$ and to each edge $e$ (say connecting $v_1, v_2$) we also associate a group $G_e$, together with a pair of injective homomorphisms $f_1: G_e \rightarrow G_{v_1}$, $f_2: G_e \rightarrow G_{v_2}$.

In our context, we should think of this as gluing together a bunch of spaces and take the fundamental group of those spaces, along with their pairwise intersections, as our vertice and edge groups. Just note that we need to have injections from the edge group to vertice groups. For simplicity one may first restrict oneself to the case where all edge groups are trivial (say spaces glued along contractible spaces).

There is something called the fundamental group of a graph of groups which is essentially the fundamental group of the resulting space after you glued spaces according to the given graph of groups. Note that the injection associated to edges takes into account how gluing of different pairs interact with each other (that is to say, for example, on a homotopy level it knows about triple intersections, etc.)

Let’s look at an application in this paper of Kleiner and Kapovich which I also talked about in an earlier post. Continue from that pervious post, now we know that

Theorem: Any hyperbolic group (plus obvious conditions, namely torsion free and does not split over a finite cyclic group) with 1-dimensional boundary has $\partial_\infty G$ homeomorphic to $\mathbb{S}^1$, the Sierpinski carpet or the Menger curve.

When $\partial_\infty G = \mathbb{S}^1$, my wonderful advisor Dave Gabai proved that $G$ would act discrete and cocompactly on $\mathbb{H}^2$ by isometries. (i.e. it’s almost the fundamental group of some hyperbolic surface except for possible finite order elements which make the action not properly discontinuous.)

Now the next step is of course figuring out when does groups act on $\mathbb{H}^3$, we have:

Cannon’s conjecture: If hyperbolic group $G$ has boundary $\mathbb{S}^2$, then $G$ acts discretely and cocompactly on $\mathbb{H}^3$ by isometries.

This conjecture was also mentioned another pervious post. Turns our we do not know much about groups with $\mathbb{S}^2$ boundary. However, using graph of groups, they were able to show:

Theorem: If Cannon’s conjecture is true, then those hyperbolic groups with Sierpinski carpet boundary are fundamental groups of hyperbolic 3-manifolds with totally geodesic boundary.

i.e. the idea is to ‘extend’ the group with Sierpinski carpet boundary to a group having sphere boundary. Of course as sets we can embed the carpet into a sphere and start to ‘reflect it along the boundary of the ‘holes’, continue the process and eventually the union of all copies of the carpets is the entire $\mathbb{S}^2$. The problem is how to ‘reflect’ a group?

First, since the boundary is homeomorphic to the carpet, there are countably many well-defined ‘boundary circles’, the group $G$ acts on the set of boundary circles. They showed this action has only finitely many different orbits. (those orbits of boundary circles will eventually correspond to those totally geodesic boundary components of our resulting 3-manifold). We pick one boundary circle from each orbit and denote their stabilizers $H_1, \cdots, H_k$ each $H_i < G$.

Define a graph of groups $\mathcal{G}$ with two vertices both labeled $G$, with $k$ edges, all going from one vertex to the other. Let the edge groups be $H_1, \cdots, H_k$.

Now we can start to 'unfold' the graph： Let $X_G$ be a 2-complex associated to a set of generators and relations for $G$ and $X_i$ be 2-complexes associated to $H_i$. The inclusion map induces cellular maps $h_i: X_i \rightarrow X_G$. Hence we have

$\displaystyle h: \sqcup_{i=1}^n X_i \rightarrow X_G$

Let $X$ be the mapping cylinder of $h$. i.e. $X$ has boundary components $\sqcup_{i=1}^n X_i$ and $X_G$.

Let $DX$ be the complex obtained by gluing together two copies of $X$ along $\sqcup_{i=1}^n X_i$, take it’s universal cover $\widetilde{DX}$. Now the fundemental group $\hat{G}$ of $DX$ is, in some sense, the group obtained by doubling $G$ along each $H_i$. i.e. $\hat{G}$ is the fundamental group of the graph graph of groups $\mathcal{G}$.

Now by studying the 1-skeleton of the complex $\widetilde{DX}$, one is able to conclude that $\hat{G}$ is Gromov hyperbolic with $\mathbb{S}^2$ boundary, as expected.

Hence from groups with Sierpinski carpet boundary we are able to produce groups with sphere boundary. Now if Cannon’s conjecture is true, $\hat{G}$ is fundamental group of some hyperbolic 3-manifold, together with Gabai’s result that $H_i$ are fundamental groups of hyperbolic surfaces, we would have that $G$ is the fundamental group of a hyperbolic 3-manifold with $n$ totally geodesic boundary components.

Well, since now we don’t have Cannon’s conjecture, there is still something we can conclude:

Definition: A n-dimensional Poincare duality group is a group which has group cohomology satisfying n-dimensional Poincare duality.

Those should be thought of as fundamental groups of manifolds in the level of homology. Well, I know nothing about group cohomologies, luckily we have:

Theorem: (Bestvina-Mess)

$\Gamma$ is a n-dimensional Poincare duality group iff it’s torsion free and $\partial \Gamma$ has integral Cech cohomology of $\mathbb{S}^{n-1}$

Great! In our case $\partial \hat{G}$ IS the sphere! So it’s a 3-dimensional Poincare duality group~ Now we have a splitting of $\hat{G}$ over a bunch of 2-dimensional Poincare duality groups (namely $H_i$) it follows that $(G; H_1, \cdots, H_n)$ is a Poincare duality pair.

It is not known whether all such pairs can be realized as fundamental groups of 3-manifolds with boundary. If so, then by Thurston’s geometrization we can also obtain what we derived assuming Cannon’s conjecture.

### A remark on a mini-course by Kleiner in Sullivan’s 70th birthday

June 7, 2011

I spent the last week on Long Island for Dennis Sullivan’s birthday conference. The conference is hosted in the brand new Simons center where great food is served everyday in the cafe (I think life-wise it’s a wonderful choice for doing a post-doc).

Anyways, aside from getting to know this super-cool person named Dennis, the talks there were interesting~ There are many things I found so exciting and can’t help to not say a few words about, however due to my laziness, I can only select one item to give a little stupid remark on:

So Bruce Kleiner gave a 3-lecture mini-course on boundaries of Gromov hyperbolic spaces (see this related post on a piece of his pervious work in the subject)

Cannon’s conjecture: Any Gromov hyperbolic group with $\partial_\infty G \approx \mathbb{S}^2$ acts discretely and cocompactly by isometries on $\mathbb{H}^3$.

As we all know, in the theory of Gromov hyperbolic spaces, we have the basic theorem that says if a groups acts on a space discretely and cocompactly by isometries, then the group (equipped with any word metric on its Cayley graph) is quasi-isometric to the space it acts on.

Since I borrowed professor Sullivan as an excuse for writing this post, let’s also state a partial converse of this theorem (which is more in the line of Cannon’s conjecture):

Theorem: (Sullivan, Gromov, Cannon-Swenson)
For $G$ finitely generated, if $G$ is quasi-isometric to $\mathbb{H}^n$ for some $n \geq 3$, then $G$ acts on $\mathbb{H}^n$ discretely cocompactly by isometries.

This essentially says that due to the strong symmetries and hyperbolicity of $\mathbb{H}^n$, in this case quasi-isometry is enough to guarantee an action. (Such thing is of course not true in general, for example any finite group is quasi-isometric to any compact metric space, there’s no way such action exists.) In some sense being quasi-isometric is a much stronger condition once the spaces has large growth at infinity.

In light of the above two theorems we know that Cannon’s conjecture is equivalent to saying that any hyperbolic group with boundary $\mathbb{S}^2$ is quasi-isometric to $\mathbb{H}^3$.

At first glance this seems striking since knowing only the topology of the boundary and the fact that it’s hyperbolic, we need to conclude what the whole group looks like geometrically. However, the pervious post on one dimensional boundaries perhaps gives us some hint on the boundary can’t be anything we want. In fact it’s rather rigid due to the large symmetries of our hyperbolic group structure.

Having Cannon’s conjecture as a Holy Grail, they developed tools that give raise to some very elegant and inspring proofs of the conjecture in various special cases. For example:

Definition: A metric space $M$, is said to be Alfors $\alpha$-regular where $\alpha$ is its Hausdorff dimension, if there exists constant $C$ s.t. for any ball $B(p, R)$ with $R \leq \mbox{Diam}(M)$, we have:

$C^{-1}R^\alpha \leq \mu(B(p,R)) \leq C R^\alpha$

This is saying it’s of Hausdorff dimension $\alpha$ in a very strong sense. (i.e. the Hausdorff $\alpha$ measure behaves exactly like the regular Eculidean measure everywhere and in all scales).

For two disjoint continua $C_1, C_2$ in $M$, let $\Gamma(C_1, C_2)$ denote the set of rectifiable curves connecting $C_1$ to $C_2$. For any density function $\rho: M \rightarrow \mathbb{R}^+$, we define the $\rho$-distance between $C_1, C_2$ to be $\displaystyle \mbox{dist}_\rho(C_1, C_2) = \inf_{\gamma \in \Gamma(C_1, C_2)} \int_\gamma \rho$.

Definition: The $\alpha$-modulus between $C_1, C_2$ is

$\mbox{Mod}_\alpha(C_1, C_2) = \inf \{ \int_M \rho^\alpha \ | \ \mbox{dist}_\rho(C_1, C_2) \geq 1 \}$,

OK…I know this is a lot of seemingly random definitions to digest, let’s pause a little bit: Given two continua in our favorite $\mathbb{R}^n$, new we are of course Hausdorff dimension $n$, what’s the $n$-modulus between them?

This is equivalent to asking for a density function for scaling the metric so that the total n-dimensional volume of $\mathbb{R}^n$ is as small as possible but yet the length of any curve connecting $C_1, \ C_2$ is larger than $1$.

So intuitively we want to put large density between the sets whenever they are close together. Since we are integrating the $n$-th power for volume (suppose $n>1$, since our set is path connected it’s dimension is at least 1), we would want the density as ‘spread out’ as possible while keeping the arc-length property. Hence one observation is this modulus depends on the pair of closest points and the diameter of the sets.

The relative distance between $C_1, C_2$ is $\displaystyle \Delta (C_1, C_2) = \frac{\inf \{ d(p_1, p_2) \ | \ p_1 \in C_1, \ p_2 \in C_2 \} }{ \min \{ \mbox{Diam}(C_1), \mbox{Diam}(C_2) \} }$

We say $M$ is $\alpha$-Loewner if the $\alpha$ modulus between any two continua is controlled above and below by their relative distance, i.e. there exists increasing functions $\phi, \psi: [0, \infty) \rightarrow [0, \infty)$ s.t. for all $C_1, C_2$,

$\phi(\Delta(C_1, C_2)) \leq \mbox{Mod}_\alpha(C_1, C_2) \leq \psi(\Delta(C_1, C_2))$

Those spaces are, in some sense, regular with respect to it’s metric and measure.

Theorem: If $\partial_\infty G$ is Alfors 2-regular and 2-Loewner, homeomorphic to $\mathbb{S}^2$, then $G$ acts discrete cocompactly on $\mathbb{H}^3$ by isometries.

Most of the material appeared in the talk can be found in their paper.

There are many other talks I found very interesting, especially that of Kenneth Bromberg, Mario Bonk and Peter Jones. Unfortunately I had to miss Curt McMullen, Yair Minski and Shishikura…