MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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.

share|cite|improve this question
If $V=\mathbb{Q}_l^n$ is such a vector space. Can we assume that $\mathbb{Z}_l^n$ is a $G$-submodule of $V$ ? – Ralph Feb 20 '13 at 20:14
Yes, it is easy to see that there always is a $G$-invariant $\mathbb Z_l$-lattice. – Nicolás Feb 20 '13 at 21:46
Nice. I found a proof (haven't seen your proof while writing down mine) that is based on [N, 2.3.9] in place of [N, 2.3.5]. But my approach is more complicated. – Ralph Feb 21 '13 at 0:15
Perhaps one should note in addition that $\varinjlim_n H^1_{cont}(P,L/l^nL)=0$ because $L/l^nL$ is discrete and hence $H^1_{cont}(P,L/l^nL)=H^1(P,L/l^nL)=0$ as explained in the 2nd part of my answer (this is certainly superfluous for specialists, but might be helpful for others). – Ralph Feb 21 '13 at 0:23
This question is getting more ridiculous by the minute. You said you were not assuming the action was discrete. But now you say you are assuming it's continuous. But if $G$ is $p$-adic and we have a map $G\to GL(n,\mathbb{Q}_\ell)$ which is continuous then the image lands in a compact, which after conjugation can be $GL(n,\mathbb{Z}_\ell)$, and this has a finite index subgroup which is pro-$\ell$. This proves that any continuous action is discrete! – user30035 Feb 21 '13 at 9:17
up vote 6 down vote accepted

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.

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.

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

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

share|cite|improve this answer
Note that I am not assuming that the modules are discrete. – Nicolás Feb 20 '13 at 18:38
Yes it is also in Rubin's Euler systems in appendix B. – Arijit Feb 21 '13 at 16:55

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.

share|cite|improve this answer
Note that I am assuming that the action is continuous (I have amended the statement in my question). – Nicolás Feb 20 '13 at 21:46
But you said you were not assuming the action was discrete!! These are equivalent in this setting! – user30035 Feb 21 '13 at 9:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.