Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finitely generated discrete group that satisfies Kazhdan's property T. Rapinchuk has proved that every unitary representation $\rho$ of $G$ on some finite-dimensional Hilbert space is locally rigid: this means that if another representation is sufficiently closed to $\rho$ (on a fixed finite set of generators) then it must be conjugated to $\rho$. The proof is nice and goes as follows:

  1. Consider the adjoint action $\rho^{\rm ad}$ on the Lie algebra $gl_n$. As proved by André Weil, if $H^1(G, \rho^{\rm ad})=0$ then the representation $\rho$ is locally rigid.

  2. Since $G$ is Kazhdan, every action on a Hilbert space has a fixed point. Using cocycles, this is equivalent to say that $H^1(G, \eta)=0$ for any unitary representation $\eta$ on any Hilbert space.

  3. If $H^1(G, \rho^{\rm ad})\neq 0$ we get a contradiction, because it is easy to find an invariant positive-definite scalar product on $gl_n$ which turns the adjoint action $\rho'$ into a unitary action.

Points 1 and 3 heavily rely on the fact that the representation $\rho$ is finite-dimensional, where point 2 does not. I am interested in infinite-dimensional Hilbert spaces:

Can a Kazhdan group $G$ have a non locally rigid unitary representation into some infinite-dimensional Hilbert space?

The question might depend on which notion of local rigidity one uses in infinite-dimension, i.e. on which topology (or distance) one puts on the set of all bounded (actually, unitary) operators. I don't know if there is a standard accepted notion. I suspect that there should be plenty of non-rigid representations but I don't know any example.

A related question on non-rigidity in infinite dimension is here.

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

The situation in infinite dimensions is different for Kazhdan groups.

If $\Gamma$ contains a non-abelian free group, then the left-regular representation $\lambda \colon \Gamma \to U(\ell^2 \Gamma)$ admits a deformation $\lambda_t$ (for $t \in [0,1]$ say), such that $\lambda_t$ is a unitary representation, $$\sup_{g \in \Gamma} \|\lambda_t(g) - \lambda_s(g)\| \leq |s-t|,$$ $\lambda_0=\lambda$ and $\lambda_t$ not equivalent to $\lambda$ for $t \neq 0$. This is a very strong condition on a deformation; typically one does not require a uniform deformation and usually also using the strong operator topology.

An explicit construction of such a deformation for the free group itself goes back to Pytlik-Swarc in

T. Pytlik and R. Swarc, An analytic family of uniformly bounded representations of tree groups, Acta Math. 157(3-4):289-309, 1986.

The idea is then to induce this deformation to $\Gamma$ and check that the continuity is preserved (this is easy because of the strength of the assumption) and that the resulting representations remain inequivalent.

A similar behaviour is conjectured to hold for all non-amenable groups. These ideas have appeared in

Marc Burger,Narutaka Ozawa, and Andreas Thom, On Ulam stability, to appear in Israel J. Math., http://arxiv.org/abs/1010.0565

share|improve this answer
Thank you very much, I'll read the paper. Do you use Mackey induction to construct a representation for the whole of $\Gamma$ starting from the Pytlik-Swarc representation on the non-abelian free subgroup? –  Bruno Martelli Jun 22 '12 at 11:17
Yes, precisely. –  Andreas Thom Jun 23 '12 at 9:23
@ Andreas: Can't you simply assume that $\Gamma$ contains an element of infinite order, and induce from characters of that infinite cyclic subgroup? –  Alain Valette Aug 11 '12 at 8:35
If you are interested in the strong topology: let $G$ be a simple Lie group with property (T) (say $G=SL(n,\mathbb{R}),n\geq 3$), and $\Gamma$ a lattice in $G$. Then $G$ has "continuous" series of irreducible unitary representations (e.g. the principal series), which are continuous in the strong topology, and it is a famous result by Cowling and Steger that, in restriction to $\Gamma$, those remain irreducible and pairwise inequivalent. –  Alain Valette Aug 11 '12 at 8:39
Alain, you are right. What I wrote is relevant for another notion of rigidity; where one is interested in uniform estimates. –  Andreas Thom Aug 11 '12 at 8:39
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.