I'm interested in group cohomology calculations, specific ones in particular, so I can manage a better understanding of the theory.
The basic setup is one is given a group,
-modules (i.e. modules over
. And the starting point is calculating
, (this is before I look at
-equivariance), and I"m trying to figure it out for say a free
-module (from there I should be able to get all the finitely generated abelian groups by breaking the homs across direct sums), however the only thing I know for sure from the start is that
, which looks like
as an abelian group if I'm not mistaken. And if my group is free, I'm pretty sure the group ring is just
itself (as a free, abelian group) right? And if so, do I just calculate hom via the isomorphism type, or is there merit in explicitly understanding the generators? From there, how can one proceed to figure out the G-equivariant homomorphisms?
For any group
, the (integral) group ring
as an abelian group is the direct sum of
copies of your ring, not
To check equivariance, you only need to check the action of generators of
is cyclic, that is easy enough (and very concrete). I'm not sure what you mean by isomorphism type here.
I mean if say I don't have a free
module, it might help to know an explicit description of the modules
as quotients of some polynomial algebra and knowing a description of exacly what maps generate the abelian group hom-set. Or do I not need to be so specific in general? (i.e. if I know that say
is whatever then from abstract theory I just have that the abelian group homomorpisms look like n copies of
, but I'm wondering if--given a presentation of each module as, say, polynomial algebras knowing which are the generators--would that be helpful in understanding the cohomology, or generally do people only care about the isomorphism type in these kinds of calculations?) My backgroung with (co)homology is from topology where sometimes maps (like the Bott map) are useful to know explicitly is a reason I ask. In particular I know that the general theory can easily give me the isomorphism type of
, but I'm wondering if I need to know a better description than just that to get at the
-equivariant parts, i.e. in order to check the action (I know that checking on generators is enough).
OK, I see what you are trying to do.
Yes, you need to know more to get the
-equivariant part --- consider
with the usual
with the trivial
-action. These two give the same
), but usually different
. The important
-action which was thrown out in calculating
is definitely needed.
Of course, if you are calculating cohomology by taking free resolution of
(such as the bar resolution, which nobody use in practice to compute cohomology) then you just need
(together with the induced maps).