Question

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.

Answer 1

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.

Answer 2

I think the answer to this pathological question is "no". Here's why. Let the pro-$p$-group $G$ be the $p$-adic integers. This embeds into an algebraic closure of the $p$-adic numbers, which is isomorphic as an abstract field to an algebraic closure $\overline{\mathbb{Q}}_\ell$ of the $\ell$-adic numbers, because any algebraically closed fields of characteristic zero and cardinality that of the reals must be isomorphic (assuming the axiom of choice). Now consider the "normalised" trace map from $\overline{\mathbb{Q}}_\ell$ to ${\mathbb{Q}}_\ell$ defined on a degree $d$ extension $K/\mathbb{Q}_\ell$ as $1/d$ times the usual trace map. Putting all these maps together, we get a stupid discontinuous non-trivial (because it sends 1 to 1) group homomorphism from $G=\mathbb{Z}_p$ to $\mathbb{Q}_\ell$ and now letting $\mathbb{Q}_\ell$ act on a 2-dimensional vector space by letting $a$ act via the matrix $(1\ a;0\ 1)$ gives a counterexample.