# Equivariant obstruction theory done wrong

Let $G$ be a finite group, and let $p:E\to B$ be a $G$-fibration with fibre $F$. The correct framework for studying equivariant sections of $p$ is Bredon cohomology. With the right definitions, most of the classical obstruction theory generalizes to this setting (even for $G$ compact Lie - details appear in a paper of Mukherjee-Mukherjee, for example).

One could also apply the Borel construction to $p$ to obtain a map $$1\times_Gp:EG\times_G E\to EG\times_G B,$$ where $EG\to BG$ is a numerable classifying bundle. (That this is a fibration was shown in the answers to the question Does the Borel functor take equivariant fibrations to fibrations?) The fibre is again $F$, and the obstructions live in the Borel equivariant cohomology groups $H^{n+1}(EG\times_GB;\pi_n(F))=:H^{n+1}_G(B;\pi_n(F))$. Clearly, equivariant sections of $p$ give sections of $1\times_G p$, but the converse is less clear (and probably false).

My questions is about whether anyone has studied the relationship between the obstruction theory of $1\times_G p$ and that of $p$? Perhaps in terms of the spectral sequence $$H^p(G;H^q(B;\pi_*(F)))\Rightarrow H^*_G(B;\pi_*(F))?$$

Any insight or references would be appreciated!

• Is it possible that it comes from the restriction map from the Bredon cohomology of $B$ to the Bredon cohomology of $EG \times B$, which can be identified (up to appropriate twisting) with the cohomology of the Borel construction? – Tyler Lawson Mar 2 '17 at 19:38
• @Tyler: That could well be. I was also wondering if difference cochains show up. For example, let $G=C_2$ generated by $\sigma$, then any (partial)section $s:B\to E$ can be "conjugated by $\sigma$" to produce another section $s^\sigma:B\to E$, and $s$ is equivariant iff $s=s^\sigma$. Now the difference cochain $\delta(s,s^\sigma)\in H^k(B,\pi_k(F))$ is defined, and probably is in $\ker(1+\sigma)$ so represents an element of $H^1(G;H^k(B,\pi_k(F))$. This is the kind of thing I was wondering about. It's all a bit reminiscent of the construction of Steenrod powers. – Mark Grant Mar 6 '17 at 9:34