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5 votes

Equivariant complex $K$-theory of a real representation sphere

Here is one possible approach, which is specific to the adjoint representation. Let $X$ denote $U(n)$, regarded as a $U(n)$-space by the rule $g.x=gxg^{-1}$. Put $X_k=\{x\in X:\text{rank}(x-1)\leq k …
Neil Strickland's user avatar
7 votes

$E^G_\ast(E)$ tensored with the rationals

For any finite abelian $G$ and $H\leq G$ we have a geometric fixed-point functor $\phi^H\colon\text{Sp}_G\to\text{Sp}$ which preserves smash products and sends the equivariant sphere $S^0_G$ to $S^0$. …
Neil Strickland's user avatar
1 vote

Do Mackey (co)homology functors factor through derived categories? References with details?

The book "Equivariant Homotopy and Cohomology Theory" by May et al probably counts as the standard reference. I don't think that it answers your last question, however. That is probably best address …
Neil Strickland's user avatar
7 votes
Accepted

Equivariant cohomology ring is an integer domain

It is very rare for these rings to be integral domains. To see this, put $$ f_V(t)=\sum_kc_k(V)t^{\dim(V)-k} \in H^*(BG)[t]. $$ (All cohomology here has rational coefficients.) It is then standard …
Neil Strickland's user avatar
5 votes
Accepted

$RO(G)$-Graded Cohomology Theories

There is no relationship in general. If $E$ is a module over the $G$-equivariant $MU$ spectrum then $E_G^{n+V}(\text{pt})$ is naturally isomorphic to $E^{n+\text{dim}_{\mathbb{R}}(V)}(\text{pt})$ for …
Neil Strickland's user avatar
3 votes

Equivariant cohomology and complex non-degenerate bilinear forms

I'll assume that you mean the Borel cohomology $H^*(EG\times_GM)$. If so, the answer is $$ \mathbb{Z}[\![c_1,\dotsc,c_n]\!]\otimes\Lambda^*(a_1,\dotsc,a_n), $$ where $|c_k|=2k$ and $|a_k|=2k-1$. To …
Neil Strickland's user avatar
11 votes

Reference request: Equivariant Topology

\bib{MR1413302}{book}{ author={May, J. P.}, title={Equivariant homotopy and cohomology theory}, series={CBMS Regional Conference Series in Mathematics}, volume={91}, note={With contribu …
Neil Strickland's user avatar