I encountered the statement that $H^*(G,\pi ^*_{S\rightarrow G}A)$ is a universal functor, when S is an open subgroup of $G$. Its statement is accompanied with the reasning that, as $S$ has finite index in $G$, $\pi ^*_{S\rightarrow G}A$ is injective whenever $A$ is. But how to derive this?
P.S. Some additional conditions were missing in this question. I apologize. Here $G$ is a profinite group, which acts on the $G$-module $A$. So $G$ is compact, which justifies the claim that $S$ has finite index in $G$.
Thanks for your attention.
When (G:S) is finite, why is every injective G-module, when considered as an S-module, still an injective S-module?
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group-cohomology
1 Answers
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The key reason it works is that in this case, $kG$ is (finitely generated) projective as a $kS$-module (i.e. $\pi^*_{S \to G} kG$ is finitely generated projective in your notation). Indeed $kG$ is a free $kS$-module with free generators a set of coset reps for $S$ in $G$. Here $k$ is whatever ring you are working over.
Injectivity of $\pi^*_{S \to G} A$ is equivalent to the vanishing of all groups $\operatorname{Ext}^n_{kS}(-, \pi^*_{S \to G} A)$ with $n>0$. But by the Eckmann-Shapiro Lemma,
$$\operatorname{Ext}^n_{kS}(M, \pi^*_{S \to G} A) \cong \operatorname{Ext}^n_{kG}( M \!\!\uparrow _S ^G, A) = 0 $$
because $A$ is injective.
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0So it avails of the theory of finite-dimensional representations to transform the injectivity of A to that of $\pi ^*_{S\rightarrow G}A$. Is this correct? Thanks for the answer. – 2012-12-06
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0No, finite-dimensionality isn't needed. – 2012-12-06
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0As far as I understand, it employs the finitely-generatedness, along with the projectivity of $kG$ to produce an isomorphism relating the extensions over $G$ and $S$. Is this correct, again? Still thanks for the explanation. – 2012-12-06
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0Maybe I'm missing something, but I don't actually think any finiteness hypotheses are needed. The ingredients of the proof are first that $kG$ is free over $kS$ and second Eckmann-Shapiro. But you find E-S stated in Benson - Representations and Cohomology I (or in Cartan-Eilenberg) with no assumptions on the rings or modules involved, and $kG$ is always free on a set of coset reps regardless of whether the index is finite. – 2012-12-06
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0So $\pi ^*_{S\rightarrow G}A$ is always injective when $A$ is, whether or not $kG$ is finitely-generated over $kS$? Because the Eckmann theorem, as stated in wiki avails of the finitely-generatedness of $kG$. Per chance I misunderstood something? I am only trying to understand the basic arguments of the cohomology theory, so might be led into the wrong paths. Thanks for the explanations. – 2012-12-08
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0The wiki article is confused. It says "$R[G]$ is finitely generated...so the previous applies" but "the previous" had no finiteness hypotheses. Probably the author isn't sure so is hedging their bets. You'll find it proved with no finiteness hypotheses for groups in Weibel's homological algebra book, or in Cartan-Eilenberg for arbitrary rings, or in Benson I. – 2012-12-08
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0You *do* need finiteness assumptions to prove that induction is the same as coinduction -- if I remember correctly -- or to get the nicest possible relationship between injective and projective modules. – 2012-12-08
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1You might like Kenneth Brown's notes http://www.math.cornell.edu/~kbrown/papers/cohomology_hangzhou.pdf on cohomology of groups – 2012-12-08
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0Thanks. I am at present reading the book by Cartan-Eilenberg. I hope I will proceed to that part soon. In addition, I am interested in your mast sentence, about the induction and co-induction, per chance you could explain or provide a reference? Thanks very much. Also thanks for the link! – 2012-12-08
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0You're welcome. I wrote some notes on induction/coinduction and related topics here https://sites.google.com/site/matthewtowers/home/induction_mark.pdf but they are in a slightly messy state. The Benson book I mentioned covers this, possibly the Brown notes but I haven't read them. – 2012-12-08
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0Sincere thanks here. – 2012-12-08