Posted by: matheuscmss | April 29, 2013

## IPAM’s program “Interactions between Analysis and Geometry” and John Pardon’s talk on Hilbert-Smith conjecture for 3-manifolds

Two weeks ago, I was in Los Angeles to attend Workshop II: Dynamics of Groups and Rational Maps of the IPAM program Interactions between Analysis and Geometry.

The workshop was very interesting in several aspects. First, the topics of the talks concerned different research specialities (as you can see from the schedule here), so that it was an excellent opportunity to learn about advances in other related areas. Secondly, the schedule gave sufficient free time so that we could talk to each other. Also, I was happy to meet new people that I knew previously only through their work (e.g., Alex Kontorovich and John Pardon).

In particular, we had two free afternoons on Wednesday and Friday, and I certainly enjoyed both of them: on Wednesday Alex Eskin drove me to the beach and we spent a significant part of the afternoon talking to each other there, and on Friday I went to Getty Center with Sasha Bufetov, Ursula Hamenstadt, Pat Hooper, John Pardon, Federico Rodriguez-Hertz, John Smillie, and Anton Zorich, where, besides classical painters like Monet, Renoir, etc., I saw

As usual, the talks were very nice (and they will be available at IPAM website here in a near future), and hence I decided to transcript in this post my notes of one of the talks, namely, John Pardon’s talk on his solution of Hilbert-Smith conjecture for 3-manifolds. Of course, the eventual mistakes in what follows are my entire responsibility.

1. Statement of Hilbert-Smith conjecture

In this section, we will quickly review some of the history behind the Hilbert-Smith conjecture. For a more serious reading, we recommend consulting Terence Tao’s notes on Hilbert’s 5th problem (as well as his post here on Hilbert-Smith conjecture).

The 5th problem in the famous list of Hilbert’s problems (stated in 1900) is the following conjecture.

Conjecture (Hilbert’s 5th problem). Let ${G}$ be a locally Euclidean topological group. Then, ${G}$ has a (unique) Lie group structure.

After the works of Gleason (in 1951-1952), Yamabe (in 1953) and Montgomery-Zippin (in 1952), we know that the answer to Hilbert’s 5th problem is yes.

An important step towards the solution of Hilbert’s 5th problem is the following theorem of Gleason and Yamabe:

Theorem 1 (Gleason-Yamabe) Let ${G}$ be locally compact group. Let ${U\subset G}$ be an open set containing the identity element ${e\in G}$. Then, there exists ${K\subset U}$ compact and an open subgroup ${G'\subset G}$ such that ${G'/K}$ is a Lie group.

Remark 1 Another way of stating this theorem is: we have an exact sequence

$\displaystyle 1\rightarrow \lim\limits_{\leftarrow}\textrm{Lie}\rightarrow G\rightarrow \textrm{discrete}\rightarrow 1$

where ${\lim\limits_{\leftarrow}\textrm{Lie}}$ is the inverse limit of the family of Lie groups ${G^0/K}$ obtained by shrinking ${U}$ towards ${e}$ and “discrete” stands for a discrete group space (cf. Terence Tao’s comment below).

An important corollary of Gleason-Yamabe theorem is:

Corollary 2 ${G}$ is NSS (no small subgroups) if and only if ${G}$ is a Lie group.

In other words, this corollary provides a criterion to recognize Lie groups (and thus it explains the interest of Gleason-Yamabe theorem to Hilbert’s 5th problem). Namely, if ${G}$ has no small subgroups (i.e., there exists a neighborhood of the identity element ${e\in G}$ containing no non-trivial subgroup of ${G}$), then ${G}$ is a Lie group.

As it turns out, Hilbert-Smith conjecture is a generalization of Hilbert’s 5th problem where one asks whether Lie groups are the sole (locally compact) groups to act faithfully on manifolds:

Conjecture (Hilbert-Smith). If a (locally compact) group ${G}$ acts faithfully on a manifold ${M}$ (i.e., we have an injective continuous homomorphism ${G\hookrightarrow \textrm{Homeo}(M)}$), then ${G}$ is a Lie group.

It is known (see, e.g., this post of Terence Tao for further explanations) that the Hilbert-Smith is equivalent to the following “${p}$-adic version”:

Conjecture (Hilbert-Smith for ${p}$-adic actions). There is no injective continuous homomorphism ${\mathbb{Z}_p\hookrightarrow\textrm{Homeo}(M)}$ where ${\mathbb{Z}_p}$ is the group of p-adic integers.

It is not hard to convince oneself that the p-adic case is important for Hilbert-Smith conjecture: let ${f}$ generate ${\mathbb{Z}_p}$ and, by abusing notation, denote by ${f:M\rightarrow M}$ the corresponding homeomorphism; then, the sequence ${f, f^p, f^{p^2}, \dots}$ converges to the identity map ${id}$. Of course, this occurs because ${p^k\cdot\mathbb{Z}_p\subset \mathbb{Z}_p}$ (for ${k\in\mathbb{N}}$) are small subgroups of ${\mathbb{Z}_p}$ (a non-Lie group!).

In summary, the philosophy behind the Hilbert-Smith conjecture is that a compact group acting non-trivially on ${M}$ can not act very close to ${id}$.

As it is nicely explained in this post of Terence Tao, the reduction of Hilbert-Smith conjecture to the p-adic case uses the following result of M. Newman in 1931:

Theorem 3 (Newman) Let ${U\subset\mathbb{R}^n}$ be an open set containing the unit (Euclidean) closed ball ${B(1)}$. Suppose that ${U}$ has a ${\mathbb{Z}/p\mathbb{Z}}$-action whose orbits have diameter bounded by ${1/10}$, i.e., we have an homeomorphism ${T:U\rightarrow U}$ of period ${p}$ (that is, ${T^p=id}$) such that all ${T}$-orbits have diameter ${\leq1/10}$. Then, the action is trivial.

Proof: A rough sketch of proof goes like this.

Assume that the action is not trivial and consider the map ${F:U\rightarrow\mathbb{R}^n}$, ${F(x):=\frac{1}{p}\sum\limits_{a\in\mathbb{Z}/p\mathbb{Z}}T^a(x)}$.

From the fact that ${T}$-orbits have diameter ${\leq 1/10}$, one can check that ${F\simeq id}$ (${F}$ is homotopic to the identity map) and thus its degree ${deg(F)}$ is ${1}$.

On the other hand, ${F}$ factors through a map ${\overline{F}:U/(\mathbb{Z}/p\mathbb{Z})\rightarrow\mathbb{R}^n}$ via the natural projection map ${\pi:U\rightarrow U/(\mathbb{Z}/p\mathbb{Z})}$. Since the projection ${\pi}$ has degree ${p}$, it follows that the degree of ${F}$ is ${p\cdot deg(\overline{F})}$, a multiple of ${p\geq 2}$.

This contradiction “proves” the theorem. $\Box$

Using this type of argument, one can also show that:

Theorem 4 Let ${M}$ be a manifold with a metric. Then, there exists ${\varepsilon>0}$ such that, if ${G}$ is a compact Lie group acting on ${M}$ with orbits of diameter ${\leq\varepsilon}$, then the action is trivial.

After this short discussion of the reduction of Hilbert-Smith conjecture to the p-adic case, let us close this section by pointing out that the general case of Hilbert-Smith conjecture is open. Nevertheless, it was known to be true for low-dimensional manifolds, namely, Montgomery-Zippin showed in 1955 that the conjecture is true for ${1}$ and ${2}$ dimensional manifolds.

In next section, we will discuss the case of ${3}$-manifolds (after Pardon).

2. Hilbert-Smith conjecture in dimension 3

For the sake of this exposition, let ${M^3}$ be a connected, orientable, irreducible ${3}$-manifold with ${H_2(M)=\mathbb{Z}}$ and exactly two ends, e.g., ${M^3=\Sigma_g\times\mathbb{R}}$ where ${\Sigma_g}$ is a genus ${g\geq 1}$ surface.

Using the orientation, it makes sense to call one end of ${M}$ the “${+}$ end” and the other end of ${M}$ the “${-}$ end”.

The basic idea of Pardon to show the Hilbert-Smith conjecture in dimension ${3}$ is to reduce it to a ${2}$-dimensional problem (that one can handle using our knowledge of the mapping class group of surfaces). In this direction, let ${S(M)}$ be the set of surfaces ${F^2\subset M^3}$ such that ${F^2}$ is incompressible (i.e., ${\pi_1(F)}$ injects into ${\pi_1(M)}$) and separates the – end from the + end (i.e, ${[F]}$ generates ${H_2(M)}$) modulo isotopies (or, equivalently, homotopies).

Definition 5 Given two surfaces ${F, G}$, we say that ${[F]\leq [G]}$ if and only if there are surfaces ${F', G'}$ isotopic to ${F,G}$ (resp.) such that ${F'}$ is contained in the “- end” ${M-G'}$, that is, ${F'}$ is to the “left” of ${G'}$. For example, when ${M=\Sigma_g\times\mathbb{R}}$, the surface ${F=\Sigma_g\times\{0\}}$ is to the left of ${G=\Sigma_g\times\{1\}}$.

The first important fact about ${\leq}$ is:

Lemma 6 ${(S(M),\leq)}$ is a partially ordered set.

The second (crucial) fact about ${\leq}$ is the following lemma suggested by Ian Agol to John Pardon:

Lemma 7 (Agol) ${(S(M),\leq)}$ is a lattice, i.e., for all ${F,G\in S(M)}$, the set

$\displaystyle X(F,G)=\{H\in S(M): [F]\leq [H], [G]\leq [H]\}$

has a least element.

Proof: A rough sketch of proof goes as follows. By looking at the figure below

one sees that ${H_0=\partial((M-F)_+\cap (M-G)_-)}$ is a natural choice (where ${(M-F)_+}$, resp. ${(M-G)_-}$, is the “+ end”, resp. “- end” of ${M-F}$, resp. ${M-G}$).

However, this might not be a good choice because the intersection between ${F}$ and ${G}$ might be “artificially complicated” like in the figure below:

Here, J. Pardon overcomes this difficulty by using the following result of M. Freedman, J. Hass and P. Scott saying that if the representatives ${F}$ and ${G}$ of ${[F]}$ and ${[G]}$ minimize area, then the intersection between ${F}$ and ${G}$ is “minimal”:

Theorem 8 (Freedman-Hass-Scott) Let ${F^2, G^2\subset M^3}$ be incompressible. Assume that ${F}$ and ${G}$ are area-minimizing representatives of their homology classes. If ${F}$ and ${G}$ can be isotoped to be disjoint, then ${F}$ and ${G}$ are already disjoint unless they coincide.

Using this result, Pardon shows that ${H_0=\partial((M-\overline{F})_+\cap (M-\overline{G})_-)}$ is a least element of ${X(F,G)}$ (where ${\overline{F}}$ and ${\overline{G}}$ are area-minimizing representatives of ${[F]}$ and ${[G]}$). $\Box$

Remark 2 It is implicit in Pardon’s arguments above that the topological and PL (piecewise linear) categories coincide for ${3}$-dimensional manifolds (that is, a topological ${3}$-manifold can be triangulated), a profound theorem of E. Moise. Of course, this result doesn’t extend to higher dimensions and this partly explains why Pardon’s arguments really are “${3}$-dimensional”.

At this point, we are ready to give a sketch of proof of Pardon’s theorem:

Theorem 9 (Pardon) There is no injective continuous homomorphism

$\displaystyle \mathbb{Z}_p\hookrightarrow \textrm{Homeo}(M^3)$

Proof: Suppose by contradiction that there exists a ${\mathbb{Z}_p}$-action on ${M^3}$. Up to replacing ${\mathbb{Z}_p}$ by ${p^k\cdot\mathbb{Z}_p}$ for some large ${k\in\mathbb{N}}$, we can assume that this action is very close to the identity.

Let ${K_0}$ be a handlebody of genus 2 and denote by ${K=\mathbb{Z}_pK_0}$ its orbit under the ${\mathbb{Z}_p}$-action. Consider now ${L_0}$ a small arc connecting two boundary points of ${K_0}$, denote by ${L=\mathbb{Z}_p L_0}$ its orbit under the ${\mathbb{Z}_p}$-action, and define ${Z:=K\cup L}$:

Since ${\mathbb{Z}_p}$ acts very close to the identity:

• (1) ${Z}$ looks like a handlebody of genus 2 in a coarse scale.

On the other hand, Pardon shows that:

• (2) ${\mathbb{Z}_p\hookrightarrow \check{H}^1(Z)}$ is non-trivial (here, ${\check{H}^1}$ stands for Cech cohomology).

Now, let ${N_{\varepsilon}(Z)}$ be the ${\varepsilon}$-neighborhood of ${Z}$ (for some ${\mathbb{Z}_p}$-invariant metric) with ${\varepsilon>0}$ very small, and define ${U=N_{\varepsilon}(Z)-Z}$.

By definition, ${\mathbb{Z}_p}$ acts on the (invariant) ${3}$-manifold ${U}$, and, a fortiori, on the set ${S(U)}$ of incompressible separating surfaces on ${U}$ modulo isotopies.

Since ${S(U)}$ is a lattice (cf. Lemma 7) and a least element of ${S(U)}$ is fixed up to isotopy by the ${\mathbb{Z}_p}$ action, we get an action

$\displaystyle \mathbb{Z}_p\rightarrow MCG(F)$

of ${\mathbb{Z}_p}$ on the mapping-class group ${MCG(F)}$ of ${F}$.

At this point, we get a contradiction as follows. By item (1), if we look at the projections to ${F}$ of the curves ${\alpha_1,\alpha_2,\beta_1,\beta_2}$ shown in the figure below

we see that ${H_1(F,\mathbb{Z})}$ contains a submodule fixed by ${\mathbb{Z}_p}$ where the intersection is

$\displaystyle \left(\begin{array}{cccc} 0&-1&0&0\\1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0\end{array}\right)=\left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)^{\oplus 2} \ \ \ \ \ (1)$

However, by item (2), the action of ${\mathbb{Z}_p}$ on ${H_1(F,\mathbb{Z})}$ (via ${\mathbb{Z}_p\rightarrow MCG(F)}$) is non-trivial.

Using these informations, one can deduce the existence of a cyclic subgroup ${\mathbb{Z}/p\mathbb{Z}}$ of ${MCG(F)}$ (essentially the image of ${\mathbb{Z}_p}$ under the map ${\mathbb{Z}_p\rightarrow MCG(F)}$) such that the module ${H_1(F)^{\mathbb{Z}/p\mathbb{Z}}}$ annihilated by ${\mathbb{Z}/p\mathbb{Z}\subset MCF(F)}$ has a submodule where the intersection form is given by equation (1).

But, Pardon proves that this is a contradiction as follows. Using Nielsen’s classification of cyclic subgroups of the mapping class group (saying that any ${\mathbb{Z}/p\mathbb{Z}\subset MCF(F)}$ is realized by a ${\mathbb{Z}/p\mathbb{Z}}$-action on ${F}$ by isometries in some metric), he shows that the intersection form on the module ${H_1(F)^{\mathbb{Z}/p\mathbb{Z}}}$ is:

• either ${\left(\begin{array}{cc} 0&-p\\p&0\end{array}\right)^{\oplus g-1}\oplus\left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)}$
• or ${\left(\begin{array}{cc} 0&-p\\p&0\end{array}\right)^{\oplus g}}$

where ${g}$ is the genus of ${F/(\mathbb{Z}/p\mathbb{Z})}$. Thus, there is no submodule of ${H_1(F)^{\mathbb{Z}/p\mathbb{Z}}}$ where the intersection form is given by equation (1) and this completes the sketch of proof of Hilbert-Smith conjecture for ${3}$-manifolds. $\Box$

## Responses

1. Nice post! A minor nitpick: Remark 1 is not quite correct as stated, because one cannot get the inverse limit of Lie groups to be normal in general, so the quotient is a discrete _space_ but not a discrete _group_. (That said, it took me a while to come up with a concrete counterexample: one such counterexample is the group G of 4 x 4 unipotent upper triangular matrices with entries in the p-adics ${\bf Q}_p$. Any open normal subgroup G’ of this totally disconnected group must then contain all the matrices in G that vanish on the diagonal above the main diagonal, and any map from G’ to a Lie group will then annihilate the centre (the matrices which vanish on the two diagonals above the main diagonal), rendering it impossible for G’ to be the inverse limit of Lie groups.)

• Thank you! I have changed Remark 1 accordingly and I added a reference to your comment.