Lemma 93.5.3. In Example 93.5.1 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 93.4.3, 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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