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



(co)homology of cyclic groups

> 
> 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. 


