Tag: Heisenberg group

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

Conan1 Comment

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.

The Carnot-Carathéodory metric

Conan7 Comments

I remember looking at Gromov’s notes “Carnot-Carathéodory spaces seen from within” a long time ago and didn’t get anywhere. Recently I encountered it again through professor Guth. This time, with his effort in explaining, I did get some ideas. This thing is indeed pretty cool~ So I decided to write an elementary introduction about it here. We will construct a such metric in \mathbb{R}^3.

In general, if we have a Riemannian manifold, the Riemannian distance between two given points p, q is defined as

\inf_\Gamma(p,q) \int_0^1||\gamma'(t)||dt

where \Gamma is the collection of all differentiable curves \gamma connecting the two points.

However, if we have a lower dimensional sub-bundle E(M) of the tangent bundle (depending continuously on the base point). We may attempt to define the metric

d(p,q) = \inf_{\Gamma'} \int_0^1||\gamma'(t)||dt

where \Gamma' is the collection of curves connecting p, q with \gamma'(t) \in E(M) for all t. (i.e. we are only allowed to go along directions in the sub-bundle.

Now if we attempt to do this in \mathbb{R}^3, the first thing we may try is let the sub-bundle be the say, xy-plane at all points. It’s easy to realize that now we are ‘stuck’ in the same height: any two points with different z coordinate will have no curve connecting them (hence the distance is infinite). The resulting metric space is real number many discrete copies of \mathbb{R}^2. Of course that’s no longer homeomorphic to \mathbb{R}^3.

Hence for the metric to be finite, we have to require accessibility of the sub-bundle: Any point is connected to any other point by a curve with derivatives in the E(M).

For the metric to be equivalent to our original Riemannian metric (meaning generate the same topology), we need E(M) to be locally accessible: Any point less than \delta away from the original point p can be connected to p by a curve of length < \varepsilon going along E(M).

At the first glance the existence of a (non-trivial) such metric may not seem obvious. Let’s construct one on \mathbb{R}^3 that generates the same topology:

To start, we first identify our \mathbb{R}^3 with the 3 \times 3 real entry Heisenberg group H^3 (all 3 \times 3 upper triangular matrices with “1”s on the diagonal). i.e. we have homeomorphism

h(x,y,z) \mapsto \left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right)

Let g be a left-invariant metric on H_3.

In the Lie algebra T_e(H_3) (tangent space of the identity element), the elements X = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) , Y = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) and Z = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) form a basis.

At each point, we take the two dimensional sub-bundle E(H_3) of the tangent bundle generated by infinitesimal left translations by X, Y. Since the metric g is left invariant, we are free to restrict the metric to E(M) i.e. we have ||X_p|| = ||Y_p|| = 1 for each p \in M.

The interesting thing about H_3 is that all points are accessible from the origin via curves everywhere tangent to E(H_3). In other words, any points can be obtained by left translating any other point by multiples of elements X and Y.

The “unit grid” in \mathbb{R}^3 under this sub-Riemannian metric looks something like:

Since we have

\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x+1 & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) ,

the original x-direction stay the same, i.e. a bunch of horizontal lines connecting the original yz planes orthogonally.

However, if we look at a translation by Y, we have

\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x & z+x \\ 0 & 1 & y+1 \\ 0 & 0 & 1 \end{array} \right)

i.e. a unit length Y-vector not only add a 1 to the y-direction but also adds a height x to z, hence the grid of unit Y vectors in the above three yz planes look like:

We can now try to see the rough shape of balls by only allowing ourselves to go along the unit grid formed by X and Y lines constructed above. This corresponds to accessing all matrices with integer entry by words in X and Y.

The first question to ask is perhaps how to go from (0,0,0) to (0,0,1). –since going along the z axis is disabled. Observe that going through the following loop works:

We conclude that d_C((0,0,0), (0,0,1)) \leq 4 in fact up to a constant going along such loop gives the actual distance.

At this point one might feel that going along z axis in the C-C metric is always takes longer than the ordinary distance. Giving it a bit more thought, we will find this is NOT the case: Imagine what happens if we want to go from (0,0,0) to (0,0,10000)?

One way to do this is to go along X for 100 steps, then along Y for 100 steps (at this point each step in Y will raise 100 in z-coordinate, then Y^{-100} X^{-100}. This gives d_C((0,0,0), (0,0,10000)) \leq 400.

To illustrate, let’s first see the loop from (0,0,0) to (0,0,4):

The loop has length 8. (A lot shorter compare to length 4 for going 1 unit in z-direction)

i.e. for large Z, it’s much more efficient to travel in the C-C metric. d_C( (0,0,0), (0,0,N^2)) = 4N

In fact, we can see the ball of radius N is roughly an rectangle with dimension R \times R \times R^2 (meaning bounded from both inside and outside with a constant factor). Hence the volume of balls grow like R^4.

Balls are very “flat” when they are small and very “long” when they are large.