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Hello to all,

While sprucing up my knowledge of group (co)homology,I stumbled onto the following question: The first step you usually take to compute various (co)homologies is to construct the infamous "bar resolution" which resolves $\mathbb{Z}$ by free $\mathbb{Z}[G]$-modules (I'll assume everyone knows which one I mean).

Now, in the case of the (co)homology of cyclic groups, one creates a 2-periodic resolution by splicing together certain exact sequences involving the norm element of $\mathbb{Z}[G]$. I was wondering if it was possible to distill this 2-periodic resolution somehow out of the standard bar-resolution above in some natural way ? In the case of $\mathbb{Z}_2$, this is quite trivial, but the higher cases are a mystery to me!

Thank you and merry Fields day

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No one really computes cohomology using the bar resolution... It is nice to know that one has it available to prove theorems and such, but it is far too big to be of any practical use. I don't think one can «distill» in any sensible sense (that is, without knowing the answer beforehand) the periodic resolution from the bar resolution in the case of cyclic groups. If I recall correctly, the periodic one can be obtained as the Gruenberg resolution for the obvious presentation of cyclic groups (see, e.g., Hilton-Stammbach's book) –  Mariano Suárez-Alvarez Aug 19 '10 at 15:50
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I would recommend a change in title. There is something called Cyclic Homology, and What you mean is (Co)Homology of Cyclic groups. –  Sean Tilson Aug 19 '10 at 19:18
    
The likely map would go the other way: into the bar resolution. May uses this to some effect in the May ss computing the cohomology of the Steenrod algebra. ('some' means 'I don't remember exactly how useful this is, but there must have been some point to it.') –  Robert Bruner Aug 19 '10 at 22:50
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Louis asked: "I was wondering if it was possible to distill this 2-periodic resolution somehow out of the standard bar-resolution above in some natural way ?"

By a result of Benson and Carlson [Complexity and Multiple Complexes. Math. Zeit. 195(1987), 221-238, Theorem 4.4], for finite groups there is a general procedure that produces a resolution that is the tensor product of r periodic complexes where r is the rank of the group.

Given a projective resolution and a set of r cocycles that represents a homogeneous system of parameters of the integral cohomology ring, the construction of the periodic complexes is explicit and quite simple. If a cocycle has degree d than the corresponding periodic complex is d-periodic.

Now consider a finite cyclic group. Then r = 1. If you figure out a cocycle of the bar resolution that generates the second integral cohomology group and apply the construction of Benson-Carlson to this cocycle then you'll end up with the usual 2-periodic resolution.

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For this you have to know the cohomology, so it is not that useful in order to compute the cohomology :) –  Mariano Suárez-Alvarez Aug 20 '10 at 0:27
    
@Mariano: 1. The cohomology of a cyclic group is known anyway. 2. In general, to apply the cited construction, you primarily need (information about) parameters. That's no big deal: For example, you can pick them from Chern classes of a faithful unitary representation. In particular no cohomology compuations are needed to obtain their degrees. –  Ralph Aug 20 '10 at 4:00
    
(continuation) The vast cohomology computations of Carlson and Green-King (all 2-groups of order at most 128) rely on the computation of projective resolutions. In this point of view the Benson-Carlson construction is an effective way to compute a projective resolution of minimal complexity: Just compute a projective resolution up to the highest degree of the parameters and apply the construction. –  Ralph Aug 20 '10 at 4:01
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You can interpret cyclic cohomology as Ext groups using representations of the cyclic category. In this setting there will be a bar resolution (which no-one uses) and the resolution as a double complex. I accept Mariano's comment that this is not useful for computations but it does give some insight into cyclic cohomology.

For instance this point of view gives the relationship between cyclic cohomology and U(1)-equivariant cohomology.

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Bruce, the title notwithstanding, I think he means "cohomology of cyclic groups", not "cyclic cohomology of groups". –  Victor Protsak Aug 19 '10 at 18:41
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The answer to your question is simple. Choose a generator $\sigma$ of $G$. Then we have an exact sequence $$0 \to \mathbb{Z} \xrightarrow{m \mapsto m 1_G} \mathbb{Z}[G] \xrightarrow{\sigma-1} \mathbb{Z}[G] \xrightarrow{g \in G \mapsto 1} \mathbb{Z} \to 0.$$ Since $\mathbb{Z}[G]$ is a free (hence projective) $\mathbb{Z}[G]$ module, this means that we have isomorphisms $\mathrm{Ext}^n(\mathbb{Z},M) \cong \mathrm{Ext}^{n+1}(\mathrm{Im}(\sigma-1),M) \cong \mathrm{Ext}^{n+2}(\mathbb{Z},M)$ for $n \ge 1$, and since $\mathrm{Ext}(\mathbb{Z},M) \cong H^n(M)$, this gives us the desired periodicity.

This is related to the bar resolution in the sense that the bar resolution gives us group cohomology specifically because $\mathrm{Ext}^n(\mathbb{Z},M) \cong H^n(G,M)$. It follows that $\mathrm{Ext}$ can be computed by applying $\mathrm{Hom}(-,M)$ to a projective resolution of $\mathbb{Z}$, and the bar resolution is precisely such a resolution.

Note that by $\mathrm{Ext}$ we mean over the category $\mathbb{Z}[G]$-Mod.

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This is somewhat unrelated to Louis's question, no? –  Mariano Suárez-Alvarez Aug 20 '10 at 1:58
    
I agree that this seems to be answering a question other than the one written, and I'm not sure why the author thinks it helps here. –  Yemon Choi Aug 23 '10 at 9:44
    
Doesn't the exact sequence above involving $\mathbb{Z}$ give us an exact sequence of resolutions, one of which is the bar resolution? This, in turn, is what gives us the resolution which shows the group cohomology is $2$-periodic. –  David Corwin Aug 23 '10 at 14:50
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