(co)homology of cyclic groups

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.