Lemma 92.5.3. In Example 92.5.1 let $\rho _0 : \Gamma \to \text{GL}_ k(V)$ be a finite dimensional representation. Then

$T\mathcal{D}\! \mathit{ef}_{V, \rho _0} = \mathop{\mathrm{Ext}}\nolimits ^1_{k[\Gamma ]}(V, V) = H^1(\Gamma , \text{End}_ k(V)) \quad \text{and}\quad \text{Inf}(\mathcal{D}\! \mathit{ef}_{V, \rho _0}) = H^0(\Gamma , \text{End}_ k(V))$

Thus $\text{Inf}(\mathcal{D}\! \mathit{ef}_{V, \rho _0})$ is always finite dimensional and $T\mathcal{D}\! \mathit{ef}_{V, \rho _0}$ is finite dimensional if $\Gamma$ is finitely generated.

Proof. We first deal with the infinitesimal automorphisms. Let $M = V \otimes _ k k[\epsilon ]$ with induced action $\rho _0' : \Gamma \to \text{GL}_ n(M)$. Then an infinitesimal automorphism, i.e., an element of $\text{Inf}(\mathcal{D}\! \mathit{ef}_{V, \rho _0})$, is given by an automorphism $\gamma = \text{id} + \epsilon \psi : M \to M$ as in the proof of Lemma 92.4.3, where moreover $\psi$ has to commute with the action of $\Gamma$ (given by $\rho _0$). Thus we see that

$\text{Inf}(\mathcal{D}\! \mathit{ef}_{V, \rho _0}) = H^0(\Gamma , \text{End}_ k(V))$

as predicted in the lemma.

Next, let $(k[\epsilon ], M, \rho )$ be an object of $\mathcal{F}$ over $k[\epsilon ]$ and let $\alpha : M \to V$ be a $\Gamma$-equivariant map inducing an isomorphism $M/\epsilon M \to V$. Since $M$ is free as a $k[\epsilon ]$-module we obtain an extension of $\Gamma$-modules

$0 \to V \to M \xrightarrow {\alpha } V \to 0$

We omit the detailed construction of the map on the left. Conversely, if we have an extension of $\Gamma$-modules as above, then we can use this to make a $k[\epsilon ]$-module structure on $M$ and get an object of $\mathcal{F}(k[\epsilon ])$ together with a map $\alpha$ as above. It follows that

$T\mathcal{D}\! \mathit{ef}_{V, \rho _0} = \mathop{\mathrm{Ext}}\nolimits ^1_{k[\Gamma ]}(V, V)$

as predicted in the lemma. This is equal to $H^1(\Gamma , \text{End}_ k(V))$ by Étale Cohomology, Lemma 59.57.4.

The statement on dimensions follows from Étale Cohomology, Lemma 59.57.5. $\square$

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