22
$\begingroup$

Suppose a finite group $G$ acts piecewise linearly on a polyhedron $X$. Then there are two kinds of equivariant cohomology (or homology).

$\bullet$ With coefficients in a $\Bbb Z G$-module $M$. A reference is K. Brown's "Cohomology of groups". Namely, $H^\ast_G(X;M)=H^\ast(Hom_{\Bbb Z G}(C_\ast,M))$, where $C_\ast$ is the simplicial chain complex of a suitable invariant triangulation of $X$. Note that $H^\ast_G(X;M)\simeq H^\ast_G(X\times EG;M)$. Generalized cohomology (including stable cohomotopy) of $X$ with coefficients in a module can be defined via generalized cohomology of $X\times_G EG$ with local coefficients, which in turn is defined in the last chapter of the Buoncristiano-Rourke-Sanderson book.

$\bullet$ Representation-graded. A reference is "Equivariant homotopy and cohomology theory" by J. P. May et al. (but to simplify matters, I'm thinking of trivial coefficients, as in Segal and Kosniowski rather than coefficients in a Mackey functor). Instead of defining $H^*_G$ let me only say that there is a Hurewicz homomorphism $\omega^*_G(X)\to H^*_G(X)$, and the equivariant stable cohomotopy $\omega_G^{V-W}(X)$ can be defined as the equivariant homotopy set $[S^{W+U}\wedge X,S^{V+U}]_G$, where $U$ is a sufficiently large (with respect to the partial order by inclusion) finite-dimensional $\Bbb RG$-submodule of $\Bbb R G\oplus\Bbb RG\oplus\dots$, and the "representation sphere" $S^U$ is the one-point compactification of $U$ (viewed as a Euclidean space with an action of $G$).

What relations are there between these two kinds of (e.g. ordinary) equivariant cohomology? More specifically,

is cohomology with coefficients in a $\Bbb ZG$-module a special case of representation-graded cohomology?

A related (equivalent?) question is whether cohomology with coefficients in a (nontrivial!) module is representable in some reasonable sense.

For instance, by reducing both sides to non-equivariant cohomology, I know that $H^m_{\Bbb Z/2}(X;I^{\otimes m})\simeq H_{\Bbb Z/2}^{mT}(X_+)$, where $\Bbb Z/2$ acts freely on $X$, $X_+=X\sqcup$(basepoint), $I$ is the augmentation ideal of $\Bbb Z[\Bbb Z/2]$ (so $I^{\otimes m}$ is $\Bbb Z$ if $m$ is even and $I$ if $m$ is odd) and $T$ is the augmentation ideal of $\Bbb R[\Bbb Z/2]$ (so $mT$ is $\Bbb R^m$ with the antipodal involution). Something of this kind can also be done for stable cohomotopy. However this all depends on a twisted Thom isomorphism which I don't know for more general modules.

$\endgroup$
2
  • $\begingroup$ Another reference for generalized cohomology with local coefficients is K. Brown's paper, ams.org/journals/tran/1973-186-00/S0002-9947-1973-0341469-9 $\endgroup$ Oct 26, 2011 at 18:58
  • $\begingroup$ When the action is free, $H^n_G(X;M)\simeq [X,K(M,n)]_G$; the relative case looks a bit more complicated, $H^n_G(X,Y;M)\simeq [(X,Y),(K(M,n)\times BG,BG)]_G$ (see VI.3.11 in the Goerss-Jardine book and projecteuclid.org/euclid.ijm/1256052280). On the other hand, for every $G$-spectrum $E$, $h^{V-W}_G(X):=[S^W\wedge X,S^V\wedge E]_G$ is an equivariant cohomology theory. So we want to find for an $M$ e.g. a $V$ such that $H^{dim V+1}_G(S^V\wedge B|M|)\simeq M$, where $|M|$ is $M$ with the trivial action of $G$. This is indeed possible for $G=\Bbb Z/2$ but not in general. $\endgroup$ Nov 18, 2011 at 0:36

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.