# On infinite-dimensional unitary representations of Kazhdan groups

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.

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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

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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