Subject: Kitchloo's K theory question. From: "John Greenlees" Date: Fri, 16 Feb 2007 12:22:20 -0000 Dear Nitu, I think that what you mean by K(H) if H acts trivially on H is the inflation of the usual (non-equivariant) K theory spectrum. This means that the map i: K(Htriv)--->K(Hcomplete) is A(G)--->R(G) in even degrees, where A(G) is the Burnside ring and R(G) is the complex representation ring [I think this is clear for connective K theory in degree 0 and then follows by making the result periodic]. This certainly isn't split in general in the naive sense (eg if most of the subgroups of G are non-cyclic). On the other hand, K(Hcomplete) is `a split ring spectrum' in the sense that the map i is a non-equivariant equivalence. It may be that this is what you intended? Once we've got this straight, we could look at other examples of H, and to your second question. John