Let $l$, $p$ be primes. Is it true that the functor of taking invariants under pro-$p$-group $P$ of finite-dimensional $\mathbb Q\_l$-vector spaces ($l\neq p$) is an exact functor?


Thanks!


NOTE 1: I am not assuming that the action is discrete.


NOTE 2: I am assuming the action of the group on the vector space is continuous.


ANSWER: The answer is yes. The proof is as follows. First, note that if $V$ is a continuous
$\mathbb Q\_l[P]$-module, then there exists a $\mathbb Z\_l[P]$-lattice $L\subset V$ (use the compactness of $P$). Then $L = \varprojlim L/l^nL$ and $H^1\_{cont}(P,L) = \varprojlim\_n H^1\_{cont}(P,L/l^nL) = 0$ [N, Prop. 2.3.5]. Since $H^1(P,V) = H^1(P,L)\otimes\_{\mathbb Z\_l}\mathbb Q\_l$ [N, Prop. 2.3.10], we conclude.


[N] Neukirch, Schmidt, Wingberg, Cohomology of Number Fields.



Taking invariants under pro-p-group is exact?
Yes, I think taking invariants is exact in your context. 


**Proof:** Let be $G$ a pro-p group and $0 \to A \to B \to C\to 0$ a short exact sequence (s.e.s) of finite dim. $\mathbb{Q}\_l$-vector spaces on which $G$ acts continuously and $\mathbb{Q}\_l$-linearly. There is a long exact sequence (l.e.s.) [N, 2.3.2]


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


of $\mathbb{Q}\_l$-vector spaces ($H^1\_c$ benotes continuous cohomology). Hence it's enough to show $H^1\_c(G,A)=0$. 


Write $A= \mathbb{Q}\_l^n$. By the OP's comment above, there is a $G$-submodule $T = \mathbb{Z}\_l^n$ of $A$ and $G$ acts continously and $\mathbb{Z}\_l$-linearly. With $W := A/T = (\mathbb{Q}\_l/\mathbb{Z}\_l)^n$ we have the s.e.s. of $\mathbb{Z}\_l$-modules $$0 \to T \to A \to W\to 0.\tag{1}$$  

Since $W$ is a discrete $G$-module, $H^i\_c(G,W)=H^i(G,W)=0$ $(i>0)$ by the discrete case below and the l.e.s. of $(1)$ yields the surjection of $\mathbb{Z}\_l$-modules 


$$H^1\_c(G,T) \twoheadrightarrow H^1\_c(G,A).\tag{2}$$ 


[N,2.3.9] states: 



> 
> Assume that the cohomology groups of $G$ with coefficients in finite $l$-primary modules are finite. Then $H^i\_c(G,T)$ is a finitely generated $\mathbb{Z}\_l$-module for all $i$. 
> 
> 
> 


By the discrete case below we can apply this theorem and find that $H^1\_c(G,T)$ is a f.g. 
$\mathbb{Z}\_l$-module. Therefore, by $(2)$, the $\mathbb{Q}\_l$-vector space $H^1\_c(G,A)$ is f.g. as $\mathbb{Z}\_l$-module what is only possible if $H^1\_c(G,A)=0$. q.e.d.




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**Discrete case:** Let $G$ be a pro-p group and $A$ a discrete $G$-module such that $A \to A, x \mapsto px$ is an automorphism. Then $H^i(G,A)=0$ for $i>0$. In particular, for each short exact sequence of discrete $G$-modules 
$0 \to A \to B \to C\to 0$, the induced sequence $$0 \to A^G \to B^G \to C^G\to 0$$ is exact. 


**Proof:** By the long exact cohomology sequence [RZ, 6.6.1] the latter follows from $H^1(G,A)=0$. Let $i>0$ and $x\in H^i(G,A)$. Since $H^i(G,A)= \varinjlim\_U H^i(G/U,A^U )$ [RZ, 6.5.6], there is an open normal subgroup $U\le G$ and $y \in H^i(G/U,A^U)$ such that $x=\text{inf}(y)$. Since $G/U$ is a finite p-group, $y$ and hence $x$ is annulated by a power of p. But multiplication with p is an automorphism on $H^i(G,A)$. Thus $x=0$ and $H^i(G,A)=0$ follow. q.e.d. 




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[N]$\;\;$ Neukirch, et. al.: Cohomology of Number Fields. 


[RZ] Ribes, Zalesskii: Profinite Groups, 2nd Edition. 


