The paper "Cohomology of p-adic analytic groups" by Symonds and Weigel is considered one of the main references for continuous cohomology of profinite groups. There is a passage I do not understand: the context is the following.

We consider $G$ to be a pro-$p$ group Poincaré dual: this is defined in §4.4 to be a group of finite cohomological dimension $n$ such that $\mathbb{Z}_p$ (considered as a trivial module) admits a projective resolution of finite length. Moreover we have $H^k(G; \mathbb{Z}_p [\![G]\!] )=0$ for $k \neq n$ and $H^k(G; \mathbb{Z}_p [\![G]\!] )=\mathbb{Z}_p$.

The first two conditions are only for finiteness and well behavior of the associated cohomology groups, the key property is the third which ensures $H^*(G;-)$ and $H_{n-*}(G;{-})$ are related by an appropriate duality.

In fact we can show the existence of a right $\mathbb{Z}_p [\![G]\!] $-module $D_p(G)$ whose underlying abelian group is $\mathbb{Z}_p$ and such that there are natural bjiections \begin{equation} H^*(G;M)\cong H_{n-*}(G; D_p(G)\otimes M) \end{equation} for $M$ in an appropriate category of left $\mathbb{Z}_p [\![G]\!] $-modules.

My question regards Corollary 4.5.2 of this paper: it states that for such $G$ and any module $M$ which is isomorphic to $\mathbb{Z}_p$ as abelian group then the following two conditions are equivalent

  1. $H^n(G;M)\neq 0$
  2. $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M \cong \mathbb{Z}_p$ as $\mathbb{Z}_p [\![G]\!]$-modules.

I do not see why 1 should imply 2: the duality implies $H^n(G;M)\cong H_{0}(G; D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)\neq 0$ and we know $H_0(G;N)\cong N_G=N/_{ \overline{\langle g.n-n \mathrel: g \in G, \ n \in N \rangle}}$. It seems to state that $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G \neq 0$ implies $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G=\mathbb{Z}_p$ and by duality the top cohomolgy $H^n(G;M)$ can only be $0$ or $\mathbb{Z}_p$.

But this should be false: a priori we could have that $\overline{\langle g.x-x \mathrel: g \in G, \ x \in D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M}\rangle$ is a proper subgroup of $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M$ (which is just $\mathbb{Z}_p$ as abelian group).

Also it is easy to provide a counterexample to the statement that $H^*(G;M)$ is either $0$ or $\mathbb{Z}_p$. Take $G=1+p\mathbb{Z}_p$ and $M=\mathbb{Z}_p$ with the action by multiplication, we can compute $H^1(G;M)=\mathbb{Z}/p$ and $H^k(G;M)=0$ for $k \neq 1$.


1 Answer 1


I privately asked Symonds about the matter and he seems to agree with me that the corollary in this form is wrong. My counterexample should be valid.

The point is that this corollary 4.5.2 is used in corollary 5.2.5 of the cited paper to prove that the right action on the dualizing module of a Poincarè group is given by the determinant of the adjoint action on $L(G)$, the logarithm algebra associated to $G$. In the proof we actually show that that for a precise module $M$ we have $H^n(G;M)\otimes \mathbb{Q}_p \neq 0$ and we use the implication $1)\Rightarrow 2)$ to conclude.

Since we show the rationalization is not trivial, modifying $1)$ to ask $H^n(G;M)$ to be torsion-free or $\mathbb{Z}_p$ does not break the proof of corollary 5.2.5. I am not aware of any instance in the literature where this corollary 4.5.2 is used in this wrong form, if you know such a case maybe reporting it here could be useful to the discussion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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