5
$\begingroup$

This is to some extent a continuation of an earlier question of mine. Now that I'm all cleared up on what it means for a finite group to have periodic cohomology, I have another question; first I will put up some background info. Sorry about the long read.

I am working on the exercise at the end of Chapter 8, Section 4 in Serre's Local Fields (p.134), which asks the reader to generalize the Herbrand quotient $h(A)$ of a $G$-module $A$ to the case when $G$ is any finite group with periodic cohomology, not necessarily a finite cyclic group. If $q\in\mathbb{N}$ is the period of the cohomology of $G$ - i.e., $q$ is the smallest natural number such that there exists a $g\in \widehat{H}{}^q(G,\mathbb{Z})$ for which $$\widehat{H}{}^n(G,A)\,\,\,\,\xrightarrow{\,\,\,\,\,\,g\,\,\,{\small \smile}\,\,\,\cdot\,\,\,\,\,\,}\,\,\,\,\widehat{H}{}^{n+q}(G,A)$$ is an isomorphism for all $n\in\mathbb{Z}$ (see my previous question for what's going on here) - then I've defined the generalized Herbrand quotient of a $G$-module $A$ to be $$h(A)=\prod_{n=0}^{q-1}|\widehat{H}{}^n(G,A)|^{(-1)^n}$$ As it turns out, the period $q$ must be even, so this definition is "balanaced", i.e. a generalized Herbrand quotient always looks like $$h(A)=\frac{|\widehat{H}{}^0(G,A)|\cdot|\widehat{H}{}^2(G,A)|\cdots|\widehat{H}{}^{q-2}(G,A)|}{|\widehat{H}{}^1(G,A)|\cdot|\widehat{H}{}^3(G,A)|\cdots|\widehat{H}{}^{q-1}(G,A)|}.$$ I've already answered the first part of the exercise, which is to prove that if $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of $G$-modules ($G$ being any finite group with periodic cohomology), and at least two of $h(A)$, $h(B)$, and $h(C)$ are defined, then the third is as well and $h(B)=h(A)\cdot h(C)$. So I feel pretty confident that this definition is correct. My current question concerns the other part of the exercise, which is to show that if $A$ is a finite $G$-module, then $h(A)=1$.


Hypothesis: if $G$ is a finite group with periodic cohomology, and $A$ is a finite $G$-module, then $$|\widehat{H}{}^{-1}(G,A)|=|\widehat{H}{}^{0}(G,A)|.$$ Because $|{}_NA|\cdot|NA|=|A|$, this is eqiuvalent to $$|I_GA|=[A:A^G].$$

I know that this is true when $G$ is a finite cyclic group (which always have periodic cohomology); in fact this is exactly how they proved the statement $h(A)=1$ for the standard Herbrand quotient, because for a finite cyclic group the period of the cohomology is always $q=2$, so $$h(A)=\frac{|\widehat{H}{}^0(G,A)|}{|\widehat{H}{}^1(G,A)|}=\frac{|\widehat{H}{}^0(G,A)|}{|\widehat{H}{}^{-1}(G,A)|}=1.$$ My motivation for this hypothesis is that it is in fact equivalent to the statement that $$|\widehat{H}{}^{n}(G,A)|=|\widehat{H}{}^{m}(G,A)|$$ for any $n,m\in\mathbb{Z}$, which would clearly imply the desired $h(A)=1$ for the generalized Herbrand quotient. It's equivalent to this stronger statement by the use of dimension shifting; if it's true that $|\widehat{H}{}^{-1}(G,A)|=|\widehat{H}{}^{0}(G,A)|$ for any finite $G$-module $A$, then given a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where $B$ is finite and co-induced (for example, $B=\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[G],A)\cong$ $\bigoplus_{G}A$), then $C$ is a finite $G$-module for which $$\widehat{H}{}^{n}(G,A)\cong\widehat{H}{}^{n-1}(G,C)\text{ for all }n\in\mathbb{Z},$$ hence
$$|\widehat{H}{}^{0}(G,A)|=|\widehat{H}{}^{-1}(G,C)|=|\widehat{H}{}^{0}(G,C)|=|\widehat{H}{}^{1}(G,A)|.$$ There is a dual process that works in the other direction with induced modules (in fact induced = co-induced since $G$ is finite). Repeating as necessary we can get that $|\widehat{H}{}^{n}(G,A)|=|\widehat{H}{}^{m}(G,A)|$ for any $n,m\in\mathbb{Z}$.


My question is whether we actually need the assumption that $G$ have periodic cohomology for this to be true. I strongly suspect that we do, but I'm not fluent enough in group cohomology yet to be very good at constructing examples or counter-examples to anything. So,

Question: Does there exist a finite group $G$ and finite $G$-module $A$ such that $$|\widehat{H}{}^{-1}(G,A)|\neq|\widehat{H}{}^0(G,A)|?$$

It may be useful to know that TFAE for a finite group $G$:

  • $G$ has periodic cohomology.

  • Every abelian subgroup of $G$ is cyclic.

  • Every $p$-subgroup of $G$ is either cyclic or is a generalized quaternion group.

  • Every Sylow subgroup of $G$ is either cyclic or is a generalized quaternion group.

So if you can construct a counterexample using a group with periodic cohomology, thereby showing my hypothesis to be incorrect, that would be especially useful!


Postscript: Of course, I would really appreciate any help on this question, regardless of whether it's along the lines of my current approach above. If you know the answer, please only post the smallest hints you can conceivably make.

My initial idea was

Hey, $A$ is a finite abelian group! Write it as a direct sum of cyclic $p$-groups $\bigoplus_i\mathbb{Z}/p_i^{k_i}\mathbb{Z}$, use the fact that cohomology factors over finite direct sums to get $$\widehat{H}{}^{n}(G,A)\cong\bigoplus_i \widehat{H}{}^{n}(G,\mathbb{Z}/p_i^{k_i}\mathbb{Z})\text{ for all }n\in\mathbb{Z}$$ and hence $$h(A)=\prod_i h(\mathbb{Z}/p_i^{k_i}\mathbb{Z}),$$ then use the previous part of the problem (the generalized Herbrand quotient is multiplicative on exact sequences) on the exact sequence $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/p_i^{k_i}\mathbb{Z}\rightarrow0$$ to get that $h(\mathbb{Z})=h(\mathbb{Z})\cdot h(\mathbb{Z}/p_i^{k_i}\mathbb{Z})$, hence $h(\mathbb{Z}/p_i^{k_i}\mathbb{Z})=1$ for all $i$, so we're done!

What I then realized was that this really only works if the action of $G$ on $A$ is trivial; just because $A$ can be decomposed nicely as an abelian group doesn't mean that it decomposes as a $G$-module. Similarly, there may very well not be an exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/p_i^{k_i}\mathbb{Z}\rightarrow0$ as $G$-modules, because $\mathbb{Z}$ can only be acted on trivially or by negation, while $\mathbb{Z}/p_i^{k_i}\mathbb{Z}$ will have lots of automorphisms.

Currently, I think this might be salvageable by breaking up the cohomology groups $\widehat{H}{}^n(G,A)$ into their $p$-primary components, or in other words proving that $h(A)=1$ by proving for each $p$ dividing $|G|$ that the number of $p$'s in the numerator and denominator are the same, but I don't know how I might go about doing that, which is why I'm trying another approach for now.

If you've read this far, thanks for your interest!

  • 1
    You expect me to read this book?2011-06-14
  • 1
    @jspecter: It's all relevant to the question. I set my notation; I explained where my question is coming from; I asked my question; and I mentioned other thoughts that I had about the question both to demonstrate that I've thought about it on my own, and also on the off-chance that someone sees how to improve them. I agree, it is long. If I had to cut something, I guess it would be the notation, mainly because anyone who knows how to help me would be familiar with it already.2011-06-14
  • 1
    I know. I'm just jibing you. I'm not done reading this yet but it might be helpful to you to read the chapter in Lang's "Algebra" on the Euler Characteristic of which the Hereband quotient (including your generalizations) is a special case. It's very straight forward. But if you don't feel like that, the main fact to take away from it is the Euler characteristic (Hereband quotient) commutes with taking homology. Using this fact, showing $h(A) = 1$ becomes trivial.2011-06-14
  • 0
    @jspecter: Ah, ok :) Thanks for your suggestion! I'd looked a little bit into the Euler characteristic of a complex (since Serre mentions it after proving that $h(A)=1$ for the usual Herbrand quotient), but I didn't understand his comment because I could only find references that dealt with the case that almost all cohomology vanished, which periodic cohomology would not satisfy. But the theorem in Lang (Theorem XX.3.1) makes the periodic case clear; obviously this is what Serre was getting at.2011-06-14
  • 0
    Our original complex is just $K^i=\text{Hom}_G(P_i,A)$ (where $P^\bullet\rightarrow\mathbb{Z}\rightarrow0$ is a projective resolution of $\mathbb{Z}$), we take our Euler-Poincare map to be the rank of an abelian group (taken "multiplicatively"), every $K^i$ is finite because $P_i$ is finitely generated and $A$ is finite, so that by the theorem the Euler characteristic is also 1. Or were we supposed to use somewhere that every $\phi(\widehat{H}{}^n(G,A))=1$ because they're also all finite?2011-06-14
  • 0
    @Zev: you can use `\quad`, or even `\qquad` instead of many `\,`'s :)2011-06-14
  • 0
    @Mariano: Ah, I'd forgotten about those. Thanks for the tip!2011-06-14
  • 3
    Dear Zev, Here is one comment on your proof strategy, in case you don't know this: if $A$ is a finite abelian group, then the decomposition of $A$ as a direct sum of Sylow subgroups is *functorial*. Namely, if we let $p$ be a prime, and let $A[p^{\infty}]$ denote the $p$-power torsion subgroup of $A$ (which *is* a subgroup, because $A$ is abelian), then the natural map $\oplus_p A[p^{\infty}] \to A$ is an isomorphism. Furthermore, since the formation of $A[p^{\infty}]$ is functorial, if $A$ is a $G$-module, then $A[p^{\infty}]$ will be a $G$-submodule.2011-06-15
  • 2
    If you look at the $p$-adic filtration on each $A[p^{\infty}]$, you furthermore obtain a filtration by $G$-submodules each quotient of which is killed by $p$. Thus many questions about group cohomology with finite coefficients can be reduced to the case when the module $A$ is $p$-torsion for some prime $p$. This is not quite as simple as the case of $\mathbb Z/p$ that you wanted to consider, but is not too far from it, and is the basis for many arguments in the cohomological proofs of local and global class field theory (which I guess is what you're working towards). Regards,2011-06-15
  • 0
    @Matt: That's really neat! It reminds me of some post on MO that talked about how, in some class of argument in algebraic geometry, there is a standard process to reduce the question from any ring to a noetherian one, to a local noetherian one, to an artinian one, to a field (or something like that).2011-06-16
  • 0
    @jspecter: Ok, I obviously wasn't thinking very clearly when I posted my second comment above; we don't want the rank of an abelian group, we want its cardinality! Sorry to ask for another hint, but I'm having a bit of trouble figuring out how to implement the fact that the Euler characteristic commutes with taking homology, because I can't think of a complex that is going to give $A$'s cohomology, that also obviously has Euler characteristic 1. I can use the standard resolution of $\mathbb{Z}$ and $\text{Hom}$ it with $A$ (as I mentioned in my comment),2011-06-16
  • 0
    producing a complex of finite abelian groups whose homology is what we want, but the sizes of the $K^i$ are increasing, if I'm not mistaken. For the cyclic case, Serre constructs a complex $K(A)$ all of whose entries are $A$, and therefore will obviously have Euler characteristic 1, but the construction depended on $G$ being cyclic because it needed the map $D$ (multiplication by $s-1$ where $s$ is a generator of $G$).2011-06-16
  • 0
    I suspect that whatever it is I'm still missing, is related to the reason Serre alters the definition of $K(A)$ on p.133 to be graded by $\mathbb{Z}/2\mathbb{Z}$. Perhaps I am fuzzy about the definition of a graded complex, but what does that do that the original $K(A)$ can't? Does he mean that he's forming a closed complex of length 2 (i.e. with both entries being $A$, and a single $N$ map and a single $D$ map), so that he can apply the result from Lang? (The way it's stated in Lang, it seems like it wouldn't quite apply to $K(A)$ because it's not a closed complex).2011-06-16
  • 0
    Thanks again for your help!2011-06-16
  • 0
    @MattE: I wonder, could you point me to a "standard" example where such an argument is done out in full detail? After thinking some more about it, I am quite sure that the strategy described in your comment applies to my problem, so we are reduced to the case of proving that $h(A)=1$ when $A$ is a finite $p$-torsion $G$-module. I wonder if there are further reductions that can be made by splitting up $G$ into its Sylow subgroups (especially because the fact that $G$ has periodic cohomology implies that all its $p$-Sylow subgroups, except possibly for $p=2$, are cyclic,2011-10-19
  • 0
    which is the case for which we already know it is true). My first thought was that perhaps only the $p$-Sylow subgroups of $G$ would have non-trivial action on a finite $p$-torsion $G$-module $A$, but that is false. And so far, the only relation I could find with Sylow subgroups ([Thm 2.1 and Cor 2.2 in Lang's book](http://i.stack.imgur.com/kgVSQ.png)) would only produce an injection in general, which need not preserve the size of the cohomology groups and therefore need not preserve the (generalized) Herbrand quotient. As always, I greatly appreciate your insight and help.2011-10-19
  • 0
    @zev: Dear Zev, Have you read the proof of the Tate--Nakayama theorem that underlies the cohomological proofs of class field theory? This involves these sorts of reductions. Another example is the proof of Tate local duality (although that is a little more involved, I think, because it incorporates the cohomological results from local class field theory). I think you can find the proof in Serre's {\em Galois cohomology} book. Regards,2011-10-19
  • 0
    @MattE: No, I have not yet, but I will take a look in *Local Fields* and *Galois Cohomology* for it. Thanks again for your help.2011-10-23

0 Answers 0