# Infinitesimal rigidity vs. local rigidity

I am thinking about homomorphisms $\mathrm{Hom}(\Gamma,G)$, where $G$ is a Lie group and $\Gamma$ is a discrete, finitely generated subgroup.

This question talked about the difference of infinitesimal rigidity vs. local rigidity and the second answer shows that infinitesimal rigidity implies local rigidity. Are there examples (preferably in the context of discrete subgroups of Lie groups) such that $\mathrm{Hom}(\Gamma,G)$ is locally rigid, but not infinitesimally rigid?

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First, there are many examples of representations $\rho: \Gamma \to G$ which are locally but not infinitesimally rigid. The earliest example is due to Lubotzky and Magid, it is a reducible representation $\rho_0$ of a Euclidean Coxeter group of rank 3 (namely, $T(3,3,3)$, the affine Weyl group of type $A_2$) to $SO(3)$, see their book "Varieties of Representations of Finitely Generated Groups". In general, character schemes of Coxeter groups satisfy "universality" ("Murphy's law"): All singularities defined over the integers are realized as germs of character varieties. For instance, you could take the non-reduced scheme $x^{175}=0$ and realize it as a singularity of a character variety (to $PO(3)$) of an appropriate Coxeter group. The corresponding representation will have infinitesimal deformations of orders $1$ through $174$, but be locally rigid.

On the other hand, as far as I can tell, there are no current examples of the following: $\Gamma\subset G$ is a discrete subgroup of a Lie group $G$ and $\rho$ is the identity representation, which is locally but not infinitesimally rigid. (This is impossible for discrete f.g. discrete subgroups in $O(3,1)$, however.) It does not mean that such examples do not exist, I am pretty sure they do, just nobody bothered with making computations.

Here is one possible construction: Let $\Gamma\subset SU(n,1)$, $n\ge 1$, be a uniform lattice and let $\rho: \Gamma\to SU(n+1,1)$ be the natural representation. Suppose that $\Gamma$ has finite abelianization, while $H^1(\Gamma, {\mathbb C}^{n,1})\ne 0$, where $\Gamma$ acts on ${\mathbb C}^{n,1}$ through its embedding in $SU(n,1)$. Then $\rho$ would be locally but not infinitesimally rigid, see the paper of Goldman and Millson "Local rigidity of discrete groups acting on complex hyperbolic space". It is well-known how to construct $\Gamma$'s with both finite abelianization and with nonzero $H^1(\Gamma, {\mathbb C}^{n,1})$. However, having both properties simultaneously is tricky. Another possible candidate is to take the Coxeter example of Lubotzky and Magid. Then $\rho_0: \Gamma\to O(2)$ is the linear part of the action $\rho$ of $\Gamma=T(3,3,3)$ on ${\mathbb R}^2$ as a Euclidean lattice. Now, there is a decent chance that $\rho$ is locally but not infinitesimally rigid as a representation $\Gamma\to G=Isom({\mathbb R}^3)$.

Update: Here is an example of a locally rigid but not infinitesimally rigid discrete embedding. Consider $\Gamma=T(6,6,6)$, the Coxeter group whose coxeter graph is the triangle with three edge-labels 3. Let $\phi: \Gamma\to \Gamma_0=T(3,3,3)$ be the natural epimorphism to the affine Coxeter group of type $A_2$, which sends generators to generators. Let $\rho_0: \Gamma_0\to O(2)\subset SO(3)$ be the representation as above. Now, let $\rho=\rho_1\times\rho_2: \Gamma\to Isom({\mathbb H}^2)\times SO(3)$ be the product homomorphism, where $\rho_1$ is the discrete embedding of $\Gamma$ to the isometry group of the hyperbolic plane and $\rho_2=\rho_0\circ \phi$. Now, each factor of $\rho$ is rigid, hence $\rho$ is also rigid. The representation $\rho$ is a discrete embedding since $\rho_1$ is. However, $\rho$ is not infinitesimally rigid, since $\rho_2$ is not (since $\rho_0$ was not).

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Very helpful, as always. Thank you! Is there any geometric reason why local deformations of f.g. discrete subgroups in $\mathrm O(3,1)$ must be inifinitesimally rigid? Why is it just $\mathrm O(3,1)$? –  s.barmeier Jul 23 '12 at 8:12
What is wrong with the following argument: If the identity representation $\phi$ is locally rigid, then $\phi\in\mathrm{Hom}(\Gamma,G)$ has an open neighbourhood $G\cdot\phi$. But that means that the subspace of the tangent space $T_\phi(\mathrm{Hom}(\Gamma,G))=Z^1(\Gamma,\mathfrak g)$ corresponding to the conjugation action (which is equal to $B^1(\Gamma,\mathfrak g)$) has maximal dimension, whence $H^1=Z^1/B^1=0$ and $\phi$ is infinitesimally rigid. –  s.barmeier Jul 23 '12 at 8:33
@s.barmeier: 1. Smoothness of character varieties to $SO(3,1)$ at discrete embeddings is proven in my book "Hyperbolic manifolds and discrete groups". It uses heavily 3-dimensional topology and fails in hyperbolic 4-space (an example is given by Johnson and Millson). 2. You have to know a bit of algebraic geometry. Consider the non-reduced scheme $x^2=0$ in the affine line. It consists of a single isolated point, but its Zariski tangent space is 1-dimensional. –  Misha Jul 25 '12 at 6:58
And this phenomenon (points with 1-dimensional tangent spaces) also occurs for Lie groups (or algebraic groups)? –  s.barmeier Jul 25 '12 at 15:09
Yes, character schemes are universal, read my original answer. –  Misha Jul 25 '12 at 17:19