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Let $G$ be an algebraic group with Lie algebra $\mathfrak g$ and let $\Gamma$ be any finitely generated (discrete) group. One can consider the representation variety $\mathfrak R=\mathrm{Hom}(\Gamma,G)$. If one considers the group scheme $\mathfrak G$ associated to $G$, there is a nice argument that shows that infinitesimal deformations of a homomorphism $\phi\in\mathfrak R(\mathbb R)$ are the (Zariski) tangent space to $\mathfrak R$ at $\phi$ given by the kernel of the map $$ \mathfrak R(\eta)\colon\mathfrak R(\mathbb R[\epsilon])\to \mathfrak R(\mathbb R) $$ where $\mathbb R[\epsilon]$ is the algebra of dual numbers and $\eta$ is the augmentation $\eta(\epsilon)=0$.

One checks that $$\mathfrak R(\mathbb R[\epsilon])=\mathrm{Hom}(\Gamma,\mathfrak G(\mathbb R[\epsilon]))=\mathrm{Hom}(\Gamma,\mathfrak g\rtimes G) $$ and $T_\phi(\mathfrak R)$ is the fibre of $\mathfrak R(\eta)$ over $\phi$, i.e. homomorphisms into $\mathfrak g\rtimes G$ that project onto $\phi$, which must be of the form $\gamma\mapsto(\tau(\gamma),\phi(\gamma))$ for some function $\tau\colon\Gamma\to\mathfrak g$. Finally, the requirement that this homomorphism be indeed a homomorphism implies that $\tau$ must be a $1$-cocycle, i.e. $\tau\in Z^1(\Gamma,\mathfrak g)$, where $\Gamma$ acts on $\mathfrak g$ via $\phi\colon\Gamma\to G$.

My question is, whether the above is also true for Lie groups, which in general may not be algebraic groups.

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In the case of finitely presented groups the proof could be found in Raghunathan's book "Discrete subgroups of Lie groups". The computation itself goes back to A.Weil's paper "Remarks on cohomology of groups." All you need is real-analytic structure on the Lie group. –  Misha Jul 21 '12 at 4:14

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This fact holds for general Lie groups $G$ (which I will equip with a real-analytic structure) and finitely-generated groups $\Gamma$. It is explained in detail in Raghunathan's book "Discrete subgroups of Lie groups", sections 6.1-6.9), but goes back to Andre Weil's paper "Remarks on cohomology of groups", Annals of Math, 1964. Here is an outline:

Let $x_1,...,x_n$ be generators of $\Gamma$ and $R_i, i\in I$, be the relators (this set need not be finite). Each relator $R_i$ (regarded as a word in the letters $x_k^{\pm 1}$) defines an analytic map $f_i: G^n\to G$. Thus, $Hom(\Gamma,G)$ is identified naturally with the real-analytic variety $$ \{(g_1,...,g_n)\in G^n: f_i(g_1,...,g_n)=1, i\in I\}, $$ $$ \rho\mapsto (\rho(x_1),...,\rho(x_n)). $$ Now, you say that for $\rho\in Hom(\Gamma,G)$, the tangent space $T_{\rho}(\Gamma,G)$ is the intersection of kernels of the derivatives $df_i$ at the point $\rho$. In the case when $G$ is algebraic, this is the Zariski tangent space, otherwise, this is the tangent space $m_\rho/m^2_\rho$ of $Hom(\Gamma,G)$ regarded as an analytic variety (that need not be reduced). Weil and Raghunathan also explain why this is natural in the context of group homomorphisms. Then, they prove that $T_{\rho}(\Gamma,G)\cong Z^1(\Gamma, Ad\circ \rho)$, see 6.9 in Raghunathan's book (he is only interested in local rigidity, but his proof is general). In particular, if $H^1(\Gamma, Ad\circ \rho)=0$, then $\rho$ is locally rigid.

The advantage in the algebraic groups case is that you do not need to go through group presentations to see all this, maybe there is a similarly clean proof in the real analytic case.

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A quick small answer: for reductive real Lie groups, there are the "rigidity theorems" of Mostow, Margulis, and others. Basically, the point is that for real rank above 1, "lattice subgroups" admit no deformations whatsoever (and are essentially "arithmetic"). These groups are typically either algebraic or isogenous to an algebraic group (the latter illustrated significantly by "metaplectic" two-fold covers of symplectic groups Sp(n,R).

I do not know about nilpotent and solvable Lie groups, which I do suspect are less algebraic (already illustrated by various incarnations of three-dimensional Heisenberg groups, within my ken).

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Thanks. The point of the above reasoning with tangent spaces is that $Z^1(\Gamma,\mathfrak g)$ is general non-trivial, but for Weil rigidity et al., $B^1$, which coincides with the space of inifinitesimal deformations induced by the conjugation action, equals $Z^1$, so that we don't obtain any interesting deformation. I am obviously only interested in the case, where a lattice can be deformed in an interesting way. For example, the (standard incarnation) of the Heisenberg group does admit deformations, and I presume other, less algebraic incarnations of it will admit deformations, too. –  Earthliŋ Jul 21 '12 at 0:51
    
Aha! Ok, but/and unfortunately I don't know anything about these! Barely aware... :) –  paul garrett Jul 21 '12 at 1:31
    
Would you mind expanding on what other incarnations of the Heisenberg group you are thinking of? –  Earthliŋ Jul 21 '12 at 1:35
    
Oh, sorry to be opaque: just the obvious, to my mind, namely, the upper-triangular unipotent real matrices, versus those modulo the matrices with 1's on the diagonal, and integers (!) in the upper right corner. That is, the commutator is converted from $\mathbb R$ to $\mathbb R/\mathbb Z$. From a physical/practical viewpoint (all the better), the distinction between the two is unsurprising, obvious, ... but/and, abstractly, it does require a little something to "abstractly" [sic] distinguish these. I should emphasize that I've not thought about this much... Said almost more than I know. :) –  paul garrett Jul 21 '12 at 1:51

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