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:

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.

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.

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.