Question
Suppose that $\Gamma < \text{SU}(n,1)$ is a cocompact lattice, and let $\rho \colon \Gamma \to G$ be a representation to a non-compact simple Lie group (most interesting case for me: $G = \text{SU}(p,p)$). Suppose that $\rho$ is Zariski-dense. Is it true that $\rho$ is not obstructed, that is, every infinitesimal deformation in $H^1(\Gamma, \text{Ad}(\rho))$ is tangent to an actual family $\rho_t \colon \Gamma \to G$ with $\rho_0 = \rho$?

Background and motivation
The question can be posed more generally for Kähler groups or for more general (cocompact) lattices $\Gamma$, but the hyperbolicity hypothesis could help. By Goldman-Millson theory, this would mean that the bracket map $$ H^1(\Gamma, \text{Ad}(\rho)) \times H^1(\Gamma, \text{Ad}(\rho)) \to H^2(\Gamma, \text{Ad}(\rho)) $$ vanishes identically, but I think there is no hope that $H^2$ vanishes itself (quite to the contrary, Carlson-Toledo conjecture expects $H^2(\Gamma, \mathbb{R}) \neq 0$ always, but this does not say much about the above map).

I ask this because the only two examples of rigid but not infinitesimally rigid representations I know of are the one constructed by Goldman and Millson, who considered the embedding $\text{SU}(n,1) \hookrightarrow \text{SU}(n+1,1)$ and the one constructed by Kim, Klingler and Pansu, that is $$ \text{SU}(n,1) \hookrightarrow \text{Sp}(n,1) \hookrightarrow \text{SU}(2n,2). $$ However, in both cases the representation comes from an infinitesimally rigid representation (in the first case: up to "trivial deformations") to a smaller subgroup of $\text{SU}(p,q)$. The $H^1$ in this bigger group turns out to be non-zero, and along such directions there are obstructions, but this does not happen if we restrict to the Zariski closures. Furthermore, in both cases the representation factors through the whole of $\text{SU}(n,1)$, an hypothesis I am not assuming.

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    This is a well known question in the case of Kahler groups, going back to at least 1990s. It is very much unclear what the difference the hyperbolicity assumption would make. – Misha Sep 10 '14 at 1:24
  • Thank you very much, at least now I know that it is not a triviality I missed! – Marco Spinaci Sep 10 '14 at 7:52
  • By the way, do you happen to know if that is true at least in the case of variations of Hodge structure? – Marco Spinaci Sep 10 '14 at 14:35
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    I think it is open even in the case of VHS. My personal guess is that there are examples of Zariski dense representations of Kahler groups at which representation varieties are singular. The fact that nobody found them yet reflects, I think, current scarcity of examples of Kahler groups and our lack of imagination. – Misha Sep 10 '14 at 16:38

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