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