Serre spectral sequence of Borel construction

Question

Let $G$ be finite $p$-group, $X$ be path-connected $G$-space, $E=EG\times_{G}X$ be the Borel construction and $BG$ be the classifying space of $G$. Consider Serre spectral sequence of the fibration $$X\to E\to BG$$ converges from $$E_2^{p,q}=H^p(G;H^q(X))$$ to $H^*_{G}(X)=H^*(E)$. The cohomology is taken $\mathbb{F}_p$ coefficient.

In Peter May's book (Equivariant homotopy and cohomology theory, chapter IV, section 2, pp.45, the second proof of Smith theorem), he said that when $G=\mathbb{Z}/p$ and $X$ is mod $p$ cohomology $n$-sphere, if the fixed points set $X^G$ is non-empty, then Serre spectral sequence collapses at $E_2$ because each fixed point gives a cross section.

My questions are:

1. How to prove Serre spectral sequence collapse in this case?

2. In general, $G=\mathbb{Z}/p$, $X$ is path-connected and $X^G$ is non-empty, whether Serre spectral sequence of above fibration collapses at $E_2$?

Thanks all your help.

Answer 1

It's terrible notation to use $p$ for both the prime in question and for the index in the spectral sequence, so I'll write the spectral sequence as $$E_2^{nm} = H^n(G;H^m(X)).$$ Since $X$ is a mod-$p$ homology sphere, for some $k$ we have that $H^m(X)$ is $\mathbb{F}_p$ for $m=0,k$ and is $0$ otherwise. The spectral sequence thus only has two nonzero rows. Saying that it degenerates is equivalent to saying that the differential connecting these two rows is $0$. The $m=0$ row is $H^n(G)$ on $E_2$, and when you go to $E_{\infty}$ you quotient it to get the image of the map from $H^n(G)$ to the nth cohomology group of the Borel construction. Since your fibration has a cross section, this map is split injective, so nothing gets quotiented and your differential must be $0$.

As you can see, this really has nothing to do with your setting and is just how the Serre spectral sequence of a fibration whose fiber is a homology sphere behaves. You can also see that there is no reason to expect degeneration if $X$ is anything else.