Notes of Alain Valette’s Rennes lecture nr 2

{L^2}-Betti numbers, continued

Recall Kyed-Petersen-Vaes’ definition as von Neumann dimension of {H^n(G,L^2(G))}.

1. How to compute them

Type I groups have the property that we have a Plancherel measure {\mu} on the unitary dual {\hat{G}}. A Fourier type formula gives the semi-finite trace

\displaystyle  \begin{array}{rcl}  \mathrm{Trace}(x^*x)= \end{array}

Theorem 1 (Petersen-Valette) For type I groups,

\displaystyle  \begin{array}{rcl}  \beta_{(2)}^n(G)=\int_{\hat{G}}\mathrm{dim}_{\mathbb{C}}\underline{H}^n(G,\mathcal{H}_\omega)\,d\mu(\omega). \end{array}

The Plancherel measure may have atoms, corresponding to square integrable representations, also known as discrete series representations. For those, one can use ordinary cohomology (no need to take reduced cohomology).

Corollary. If {G} has a normal amenable non compact subgroup, reduced cohomology vanishes a.e. on {\hat{G}}.

Beware : we have no examples where non discrete series contribute non trivially to the integral.

2. Applications

2.1. Semi-simple Lie groups

See Borel. {L^2}-Betti numbers vanish unless {\mathrm{dim}(G)=\mathrm{dim}(K)}. In this case, only the middle dimensional Betti number is non zero (and it is indeed non zero, an integer given as a sum of dimensions of discrete series representations).

In the case of {Sl(2,{\mathbb R})}, there is a factor of {2\pi} arising from

2.2. Simple {p}-adic groups

{L^2}-Betti numbers vanish except for the top dimensional one, which is equal to the dimension of the Steinberg representation. See Dymara-Januszkiewicz.

2.3. Groups acting on trees

Let {G=Aut(T)}, {T} a simplicial tree. Nebbia 2012 shows that there is a unique representation with non vanishing 1-cohomology, which is square-integrable.

Is {G} of type I ? Known for biregular trees (Olshanskii 1975).

3. Alternative approach to {\beta_{(2)}^1(G)}

With Thibault Pillon and Bachir Bekka.

3.1. Affine actions

1-cohomology classifies affine actions with a fixed linear part. Every affine action defines a cocycle and thus a cohomology class. Two affine actions are conjugate iff the cohomology classes are the same.

Say an affine action is irreducible is no invariant convex sets. Such exist iff {G} has not (T), Shalom 2000. Schur’s Lemma holds : Action is irreducible iff the commutant consists of translations (the translations by vectors invariant under the linear part). Therefore, for abelian groups, irreducible actions are actions by translations, the set of which having dense linear span.

The same holds for nilpotent groups (whereas irreducible linear representations of nilpotent groups can be non trivial). Restrictions of irreducible affine isometric actions to cocompact lattices stay irreducible, Neretin 1997 (fails for linear representations, see Cowling-Steger for simple Lie groups).

Beware of your linear intuition.

Example. Let {{\mathbb R}^2} act by translations on {{\mathbb R}}, either by the first coordinate, or by the second coordinate. The direct sum of these two representation is the usual action of {{\mathbb R}^2} on itself, which is irreducible.

3.2. Irreducible realizability

When is a given linear unitary representation the linear part of some irreducible affine isometric action ? For ICC groups, the first {L^2}-Betti number gives the answer

Theorem 2 Assume {G} is finitely generated and ICC. A Hilbert {G}-module {\mathcal{H}} is the linear part of some irreducible affine isometric action iff its von Neumann dimension is {\leq \beta_{(2)}^1(G)}.

Example. Free group, we have an explicit construction of the affine action.

The amenable case is pretty different. According to Andreas Thom, for amenable groups, isometric actions admit closed invariant subspaces of arbitrarily low dimension.


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