Let $G$ be a finitely generated discrete group that satisfies [Kazhdan's property T](http://en.wikipedia.org/wiki/Kazhdan%27s_property_%28T%29). Rapinchuk [has proved](http://www.ams.org/journals/proc/1999-127-05/S0002-9939-99-04696-1/home.html?pagingLink=%3Ca+href%3D%22%2Fepubsearch%2Fservlet%2FPubSearch%3Fco1%3Dand%26co2%3Dand%26co3%3Dand%26endmo%3D00%26f1%3Dmsc%26f2%3Dtitle%26f3%3Danywhere%26f4%3Dauthor%26pubname%3Dall%26sendit22%3DSearch%26sperpage%3D30%26ssort%3Dd%26startmo%3D00%26v4%3DRapinchuk%26startRec%3D1%22%3E) 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](https://mathoverflow.net/questions/88677/which-unitary-permutation-representations-are-isolated).



On infinite-dimensional unitary representations of Kazhdan groups
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>


