Cup Product and Representability

[lavinia]

Out of curiosity, where does the more general statement with "for any G with torsion" come from? Is it true that if H<G, then the EM-space for G has the EM-space for H as a subcomplex?

I am not sure how the proof goes. For finite cyclic groups I think that H^2(Zn:Z2) generates cohomology classes in arbitrarily large dimensions. It follows immediately from this that the same is true for finite abelian groups.

I don't see how to generalize this. One idea might be to identify the second Z/2 cohomology of a group,G, with group extensions

0 -> Z/2 -> H -> G -> 0

and try to show that if the group,G, has torsion then there is always an extension that is not split and which restricts to a non-split extension over one of its finite cyclic subgroups.

There must be a general construction for a K(pi,1) as a CW complex that has cell in unbounded dimensions when the group has torsion. I will try to find one. I know for Z/2 it is the infinite dimensional real projective space which I think has a cell in every dimension.

A K(pi,1) is the universal classifying space for vector bundles whose structure group is the group,pi, with the discrete topology. These are so called flat bundles. Maybe one could show that if pi has torsion then in any dimension there is a manifold with a flat vector bundle over it whose structure group is,pi, and whose Euler class is not zero.

 

[Jamma]

There must be a general construction for a K(pi,1) as a CW complex that has cell in unbounded dimensions when the group has torsion. I will try to find one. I know for Z/2 it is the infinite dimensional real projective space which I think has a cell in every dimension.

Well, just finding cells in unbounded dimensions isn't enough, because the CW complex may be homotopic to one which doesn't. It would be sufficient however to find the homology of the space and show that has unboundedly many non-trivial groups, although it is obviously infeasible to compute the homology of arbitrary K(G,1)'s where G has torsion.

The paper goes through the construction for the EM1-spaces for the cyclic groups at the end- they are the infinite lens spaces and indeed their homology is non zero in infinitely many dimensions. It seems to go from this to arbitrary groups quite suddenly- I don't know if this is intentional to imply that not much more work needs to be done, or it is simply mentioning the generalisation. However, it seems that the K(Zn,1)'s having to be infinite dimensional should somehow imply the same for any K(G,1) for which G has torsion.

Your group extension idea is good though- I'm sure more can be done with that.

 

[Jamma]

Ahha, I seem to remember something along the lines of this:

If we have a group extension such as:

0 -> H -> G -> G/H -> 0

Then this leads canonically to a fibre bundle:

K(H,1) -> K(G,1) -> K(G/H)

I'm not 100% sure on this, will check up on it when I get time. I don't even know if this helps- perhaps there is some sort of spectral sequence argument, although it would seem that we should know something about K(G/H,1).

 

[Jamma]

Or perhaps I'm over-complicating things in the above- another way of defining group cohomology is to take the simplicial cohomology of the classifying space (which is a K(Z,1) in the discrete case). So maybe there is an argument in here with exact sequences.

 

[lavinia]

Or perhaps I'm over-complicating things in the above- another way of defining group cohomology is to take the simplicial cohomology of the classifying space (which is a K(Z,1) in the discrete case). So maybe there is an argument in here with exact sequences.

I think this line of thought would work. I would look for how one constructs the universal classifying space for flat bundles with a given structure group.