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

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.

Side note: I learned about this through the very detailed and well-presented course by Dave Gabai. (I thought I must have blogged about this, turns out I haven’t…)

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.

# Ultrametrics and the nonlinear Dvoretzky problem

Hi guys~ The school year here at Princeton is finally (gradually) starting. So I’m back to this blog :-P

In this past week before anything has started, Assaf Naor came here and gave a couple rather interesting talks, which I decided to make a note about. For the complete content of the talk please refer to this paper by Naor and Mendel. As usual, I make no attempt on covering what was written on the paper nor what was presented in the talk, just some random pieces and bits which I found interesting.

A long time ago, Grothendieck conjectured what is now known as Dvoretzky’s theorem:

Theorem: For any $D>1$, for any $k \in \mathbb{N}$, there exists $N$ depending only on $k, D$ such that any normed vector space of dimension $N$ or larger has a k-dimensional linear subspace that embeds (via a linear transformation) with distorsion at most $D$ into the Hilbert space.

i.e. this says in the case where $D$ is close to $1$ (let’s write $D = 1+\varepsilon$) for any given k, *any* norm on a vector space of sufficiently high dimension will have a $k$ dimensional subspace that looks almost Eculidean (meaning unit ball is round up to a multiple $1+\varepsilon$).

Well, that’s pretty cool, but linear. As we all know, general metric spaces can be much worse than normed vector spaces. So here comes a nonlinear version of the above, posted by Terrence Tao:

Nonlinear Dvoretzky problem: Given $\kappa>0, D>1$, does there exist a $\alpha$ so that every metric space of Hausdorff dimension $\geq \alpha$ has a subset $S$ of Hausdorff dimension $\kappa$ that embeds into the Hilbert space with distorsion $D$?

Indeed, we should expect this to be a lot more general and much harder than the linear version, since we know literally nothing about the space except for having large Hausdorff dimension and as we know metric spaces can be pretty bizarre. That why we should find the this new result of theirs amazing:

Theorem 1 (Mendel-Naor): There is a constant $C$ such that for any $\lambda \in (0,1)$, every compact metric space with dimension $\alpha$ has a closed subset $S$ of dimension at least $(1-\lambda) \alpha$ that admits an embedding into Hilbert space with distorsion $C/ \lambda$.

Note that, in the original linear problem, $N = N(k, D)$ of course blows up to infinity whenever $k \rightarrow \infty$ or $D \rightarrow 1$. (whenever we need a huge dimensional space with fixed ‘flatness’ or a fixed dimension but ‘super-flat’ subspace). That is, when we are looking for subspaces inside a random space with fixed (large) dimension $N$, the larger dimension we need, the less flat (more distorsion) we can expect it to be. i.e.

$k \rightarrow N$ necessarily forces $D \rightarrow \infty$ and
$D \rightarrow 1$ forces $k \rightarrow 0$.

We can see that this nonlinear theorem is great when we need large dimensional subspaces (when $\lambda$ is close to $0$), but not so great when we want small distorsion (it does not apply when distorsion is smaller than $C$, and blows up when it gets close to $C$).

In the original statement of the problem this gives not only that a large $\alpha$ exists, but $\alpha(\kappa, D) \leq \frac{D}{D-C} \kappa$! In fact, when we are allowing relatively large distorsion (compare to constant $C$) this theorem is sharp:

Theorem 2 (Mendel-Naor): There exists constant $C'$ and a family of metric spaces $\{ X_\alpha \ | \ \alpha \in \mathbb{R}^+ \}$ such that for any small $\varepsilon$, no subset of $X_\alpha$ with dimension $\geq (1-\varepsilon)\alpha$ embeds into Hilbert space with distorsion $\leq C'/\varepsilon$.

This construction makes use of our all-time favorite: expander graphs! (see pervious post for definitions)

So what if $D$ is small? Turns out, unlike in the linear case when $D<2$, there is no $\alpha(\kappa, D)$, in the paper they also produced spaces $X_\alpha$ for each $\alpha$ where the only subset that embeds with distorsion < $2$ are of Hausdorff dimension 0!

In his words, the proof of theorem 1 (i.e. the paper) takes five hours to explain, hence I will not do that. But among many neat things appeared in the proof, instead of embedding into the Hilbert space directly they embedded it into an ultrametric space with distorsion $D$, and then make use of the fact that any such space sits in the Hilbert space isometrically.

Definition: An ultrametric on space $X$ is a metric with the additional property that for any three points $x, y, z \in X$, we have the so-called strong triangle inequality, i.e.

$d(x, y) \leq \max \{ d(x, z), d(y,z) \}$

If one pause and think a bit, this is a quite weird condition: for example, all triangles in an ultrametric space are isosceles with the two same length edge both longer than the third edge; every point is the center of the ball, etc.

Exercise: Come up with an example of an ultrametric that’s not the discrete metric, on an infinite space.

Not so easy, hum? My guess: the first one you came up with is the space of all words in 2-element alphabet $\Sigma_2$, with the dictionary metric (two points are distance $2^{-n}$ apart where $n$ is the first digit they differ), right?

In fact in some sense all ultrametrics look like that. (i.e. they are represented by an infinite tree with weights assigned to vertices, in the above case a 2-regular tree with equal weights on same level) Topologically our $\Sigma_2$ is a Cantor set and sits isometrically in $l_2$. It’s not hard to see in fact all such space embeds isometrically in $l_2$.

I just want to say that this operation of first construct an ultrametric space according to our given metric space, embed, and then embed the ultrametric metric space into the Hilbert space somehow reminds me of when we want to construct a measure satisfying a certain property, we first construct it on a Cantor set, then divide up the space and use the values on parts of the Cantor set for parts of the space…On this vein the whole problem gets translated to producing a tree that gives the ultrametric and work with the tree (by no means simple, though).

Finally, let’s see a cute statement which can be proved by applying this result:

Urbanski’s conjecture: Any compact metric space of Hausdorff dimension > $n$ can be mapped surjectively [thanks to Tushar Das for pointing out the typo] onto the unit ball $B_1^n$ by a Lipschitz map.

(note strictly larger is important for our theorem to apply…since we need to subtract an $\varepsilon$) and hence another cute statement that’s still not known:

Question: Does any compact metric space with positive Hausdorff $n$-dimensional measure maps Lipschitzly onto $B_1^n$?