$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gl{Gl}$Let $\mathcal{D}=\mathbb{C}[[t\_{1},\dotsc , t\_{n}]]$ be a formal polydisc over $\mathbb{C}$, and $G$ be a finite group. On Lemma 7.8 of [Etingof and Ma - Lecture notes on Cherednik algebras](https://arxiv.org/abs/1001.0432) it is stated that every action $\rho : G\rightarrow \Aut(\mathcal{D})$ is linearizable. The proof given in the notes goes as follows:


We have a decomposition $G = \Gl\_{n}(\mathbb{C})\ltimes \Aut\_{U}(\mathcal{D})$, where $\Aut\_{U}(\mathcal{D})$ are the automorphisms such that their derivative at the origin is the identity.
We need to show that the image of $\rho$ can be conjugated into $\Gl\_{n}(\mathbb{C})$. The obstruction to this is in the cohomology group $H^{1}(G, \Aut\_{U}(\mathcal{D}))$, which is trivial as $\Aut\_{U}(\mathcal{D})$ is a pro-unipotent algebraic group over $\mathbb{C}$.


I have two questions regarding this proof:


1. This proof seems to be assuming that the group $\Aut\_{U}(\mathcal{D})$ is abelian. I would like to see why is this the case.
2. Why does the fact that $\Aut\_{U}(\mathcal{D})$ is a pro-unipotent algebraic group over $\mathbb{C}$ imply the vanishing of $H^{1}(G, \Aut\_{U}(\mathcal{D}))$?



On the linearizability of the action of a finite group on a formal polydisc
For $1 \to A \to B \to C \to 1$ an exact sequence of (not necessarily abelian) algebraic groups, we have a long exact sequence


$$ H^0(G, A) \to H^0(G, B) \to H^0(G, C) \to H^1(G, A) \to H^1(G, B) \to H^1(G, C) $$


It follows that if $H^1(G,A) = H^1(G, C) =0$ then $H^1(G,B)=0$.


We can now prove that for finite $G$, and $U$ a unipotent algebraic group $H^1(G,U)=0$, by induction on $\dim U$. There is always a surjection from $U$ to a vector space $V$, and the kernel $K$ is unipotent of smaller dimension. Now $H^1(G,K) =0$ by the induction hypothesis, and $H^1(G, V)=0$ by representation theory of finite groups over characteristic zero fields, so $H^1(G,U)=0$, completing the induction step.


Now for finite $G$ and pro-unipotent $U$, $H^1(G,U)=0$ by a limiting argument.




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A limiting argument, probably not the best way:


Write $U$ as an inverse limit of unipotent groups $U\_n$. We want to show our cocycle in $C^1( G, U) $ is the coboundary of an element of $U$. We know for each $n$ there exists an element in $U\_n$ whose coboundary is the projection of our cocycle to $C^1(G, U\_n)$. In fact, the set of such elements of $U\_n$ form an algebraic variety, a coset of an algebraic subgroup.


The image of this variety in $U\_1$ is a coset of a subgroup in $U\_1$, thus a closed subvariety. As $n$ grows, these subvarieties form a descending chain, which stabilizes at some $n$. Pick a point $x\_1$ in the stabilization. Then for each $n$ the intersection of the subvariety with the inverse image of $x\_1$ is nonempty, and still a coset of a closed subgroup.


Look at the images of these intersections in $U\_2$. Choose a point $x\_2$ that's in all of them. Then repeat...


Then $(x\_1,x\_2,\dots)$ gives the desired point of $U$.


