I've been scouring google and asking friend about something I was certain must be absolutely the easiest thing to people who do homological algebra, and none seem to know the answer to this, so if it's something really easy, I apologize.

The setup is relatively simple, I have a profinite group, $G$, and a sequence of $G$-modules--that is to say $\mathbb{Z}[G]$ modules--

$$1\to X_0\to X_1\to X_2\to 1$$

Given a trivial $G$ module, $R$ which is injective as an abelian group, I'd like to analyze the Galois cohomology groups, $H^*(G, R\otimes_{\mathbb{Z}} X_i)$ given knowledge of $H^*(G, X_i)$. If it helps at all, one can assume I'm talking about $R=\mathbb{R}$, as even that case would be of great use. As far as I can tell, with a trivial $G$ action, $R$ is a flat $G$ module, so the snake lemma still applies to the modified sequence. The main problem I have is finding any result which indicates how the new cohomology groups relate to the old ones.

Furthermore, if I have a $\mathbb{Q}[G]$ module, $V$--that is, a vector space with a continuous $G$ action--how can I compute $H^*(G, \mathbb{F}\widehat{\otimes}_{\mathbb{Q}}V)$ where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and the hat means I intend to complete my vector space $\mathbb{F}\otimes V$ with respect to some topology (my case is a norm topology if that helps at all). Is it just the same as $H^*(G,\mathbb{F}\otimes_{\mathbb{Q}}V)$--i.e. does completion affect anything? (I'm inclined to think not, as the $\mathbb{F}$ are endowed with a trivial $G$ action, but I'm not enough of an expert to be confident of that.) If so, is then $H^*(G, \mathbb{F}\otimes_{\mathbb{Q}}V)$ the same as $H^*(G, V)$ or a simple operation away from $H^*(G,V)$?

Edit: The answer posted by user 43326 makes me realize that I must have gotten the wrong tensor, it should be over $\mathbb{Q}$ (or even $\mathbb{Z}$) and not, $\mathbb{Q}[G]$ since the latter trvializes the $G$ action, which should not happen.

  • 2
    $\begingroup$ With trivial G action, R can't be flat. $\endgroup$
    – user43326
    Commented Jan 30, 2014 at 9:51
  • $\begingroup$ On the second paragraph, the completion of finitely generated modules is exact, but $F$ is not finitely generated over $R$ or $C$, so probably the completion messes things up. $\endgroup$
    – user43326
    Commented Jan 30, 2014 at 9:55
  • $\begingroup$ Since $\mathbb{F}$ is free over $\mathbb{Q}$, the tensoring is exact, so we have $H^*(G,{\mathbb F} \otimes _{Q} V)\cong H^*(G,V)\otimes _{Q} {\mathbb F}$. The completion can still mess things up, do we know, for example, if the topology is Hausdorff? $\endgroup$
    – user43326
    Commented Jan 30, 2014 at 17:18
  • $\begingroup$ Yes, the space $V$ is metric. $\endgroup$ Commented Jan 30, 2014 at 20:48
  • $\begingroup$ OK, so the completion won't do something too weird. Is V compact? Another important point: how do you topologize ${\mathbb F}\otimes _QV$? $\endgroup$
    – user43326
    Commented Jan 31, 2014 at 7:25

1 Answer 1


Since you seem to let $G$ act trivially on ${\mathbb F}$, the module ${\mathbb F}\otimes _{Q[G]}V$ is isomorphic to ${\mathbb F}\otimes _{\mathbb Q}{\mathbb Q}\otimes_{Q[G]}V$ where $G$ acts trivially on ${\mathbb Q}$. But ${\mathbb Q}\otimes_{Q[G]}V$ is just the coinvariant $V/G$, on which $G$ acts no longer. So you get $$H^*(G,{\mathbb F} \otimes _{Q[G]} V)\cong H^*(G,{\mathbb F})\otimes _{\mathbb Q}V/G.$$

Added after the edition of the original post.

Since $R$ is flat over $Z$, $\otimes _ZR$ is exact in the category of abelian groups. The forgetful functor from the category of $Z[G]$-modules to abelian groups is exact, which means that taking the (co)homology of (co)chain complexes is same in two categories. So we have $$H^*(G,X\otimes _Z R)\cong H^*(G,X)\otimes _Z R.$$

  • $\begingroup$ I must have erred in making the tensor over $\mathbb{Q}[G]$ then, certainly the action is not trivial on $V$. I'll edit my original post accordingly, and thanks for pointing this out. $\endgroup$ Commented Jan 30, 2014 at 17:07

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