I know my question is very imprecise. I am trying to understand Tate-Farrell cohomology of the infinite Lie group $S^1$ (say, with coefficients in $\mathbb C$). I would expect that the answer is something like the space of Laurent polynomials $\mathbb C[t^{-1},t]$. Is there any geometric intuition for this? What would be the meaning of multiplication by $t$? What is the meaning of the completion $\mathbb C((t))$?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Loran should probably be Laurent :) $\endgroup$– Mariano Suárez-ÁlvarezNov 4, 2010 at 20:55
-
$\begingroup$ Are you talking about S^1 as a discrete group, a topological group or a Lie group? Rather, what are you more interested in: S^1 or Tate-Farrell cohomology? $\endgroup$– David Roberts ♦Nov 4, 2010 at 22:42
-
$\begingroup$ Lie group. More interested in S^1 :-) $\endgroup$– Roman FedorovNov 5, 2010 at 13:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Comment: Farrell-Tate cohomology as defined in Brown's book "Cohomology of Groups" requires the group to be of finite virtual cohomological dimension (i.e. the group has a finite index subgroup which has a finite projective resolution). But $S^1$ doesn't have finite virtual cohomological dimension because it has finite subgroups of arbitrary order.
There is a generalization of Farrell-Tate cohomology for arbitrary groups due to Benson-Carlson/Mislin, that is usually called "complete cohomoloy" or "complete Tate cohomology". I don't know if that cohomology has been computed for $S^1$ (most computations are done for discrete groups).
