Notes of Romain Tessera’s Lille lecture nr 1

Hyperbolic Lie groups

1. Non vanishing of {\ell^p} cohomology

Theorem 1 Let {X} be a connected graph with bounded degree, which is hyperbolic. Assume that {X} contains en quasi-isometrically embedded 3-regular tree {T}. Then {\ell^p\bar{H}^1(X)\not=0} for {p} lage enough.

To simplify the proof, I will use an extra assumption: Given a base point {o\in X}, {\exists C} such that every point sits at distance at most {C} from a geodesic ray passing through {o}.

{\partial T} is a Cantor set embedded in {\partial X}. Let {O_1} and {O_2} be disjoint open sets that both intersect {\partial T} an together cover {\partial T}. Consider the function on {\partial X}

\displaystyle  \begin{array}{rcl}  F(x)=\max\{0,1-\frac{1}{t}d(x,O_1), \end{array}

where {t=d(O_1,O_2)>0}. Extend {F} radially to a function {f} on {X}. Then {f\in D^p} for {p} large enough. Indeed, {df(e)} decays exponentially in terms of the distance of edge {e} to {o}. Its restriction belongs to {D^p(T)}. It pairs non trivially with the {\ell^{p'}} cycle supported on {T} which takes value 1 on an edge, {1/2} on neighboring edges, {1/4} on the next layer, and so on. Therefore its reduced cohomology class does not vanish.

2. {L^p} cohomology of locally compact groups

Let {G} be locally compact, compactly generated. Fix a compact generating set. This turns {G} into a metric space.

Fact. There exists a connected bounded degree graph {X} which is quasiisometric to {G}.

Indeed, pick a maximal set of disjoint unit balls in {G}, these are the vertices, and put edges when centers are at distance {\leq 4}.


\displaystyle  \begin{array}{rcl}  D^p(G)=\{f:G\rightarrow{\mathbb R}\,;\,\forall g\in G,\,f-\rho(g)f\in L^p(G)\}. \end{array}


\displaystyle  \begin{array}{rcl}  \bar{H}^1_p(G)=D^p(G)/\overline{L^p(G)+{\mathbb R}}. \end{array}

Gromov showed that non elementary hyperbolic groups contain quasiisometrically embedded trees. The argument (ping-pong) extends to compactly generated groups.

Corollary 2 If {G} is hyperbolic as a metric space, then {\bar{H}^1_p(G)\not=0} for {p} large enough.

3. Hyperbolic Lie groups

Question. Which connected Lie groups are hyperbolic ?

Example 1 Heintze groups, the affine group of the line (it is Heintze), the affine group of {{\mathbb Q}_p}, {Sl(2,{\mathbb R})}, {Sl(2,{\mathbb Q}_p)}.

Note these groups either act transitively on negatively curved manifolds, or cocompactly on trees. We shall prove that this is general.

3.1. Reduction

Let {G} be a connected Lie group (over {{\mathbb R}} or {{\mathbb Q}_p}). There exists a cocompact closed subgroup {H<G} and a compact normal subgroup {W< H} such that {B=H/W} satisfies

\displaystyle  \begin{array}{rcl}  1\rightarrow U\rightarrow B\rightarrow {\mathbb Z}^d \rightarrow 1 \end{array}

where {U} is simply connected nilpotent.

Note that one really needs {{\mathbb Z}^d} and not {{\mathbb R}^d}.

3.2. Classification

Theorem 3 (Cornulier-Tessera) Let {G} be a connected Lie group (over {{\mathbb R}} or {{\mathbb Q}_p}). The following are equivalent.

  1. {G} is hyperbolic.
  2. {B} is Heintze in the sense that {B=U\times {\mathbb Z}} and the action of {{\mathbb Z}} on {U} is contracting.
  3. {\bar{H}^1_p(G)\not=0} for some {p>1}.
  4. {\bar{H}^1_p(G)\not=0} for {p} large enough.

We have only partial results for general compactly generated groups: we can treat (with Caprace and Monod) the case when {G} has an amenable cocompact subgroup.

Note that such a classification fails for discrete groups. For instance, {{\mathbb Z}^2\star{\mathbb Z}} has nonzero reduced {\ell^p} cohomology for all {p} (it has infinitely many ends) but it is not hyperbolic.


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
This entry was posted in Workshop lecture and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s