I am looking at my algebraic topology notes right now, and I am looking at our definition for the Serre Spectral Sequence and it requires that the action of the fundamental group of the base space of a fibration $F\to E\to B$ be trivial on all homology groups of the fiber. Then we can construct the SSS etc. etc. What exactly does it mean for the action of the fundamental group to be trivial (rather, what is this action at all, how is it defined)? Intuitively, does this mean that if we take the fiber at a point, take its homology, and then follow the that homology along a loop, we come back to the same homology?
Any guidance would be much appreciated. I am currently trying to use the SSS to find the cohomology groups of path loop spaces of even dimensional spheres, and we have an example using the odd dimensional spheres, so I imagine I could just assume all the requirements are met to use the SSS, but I'd like to know what exactly is going on here.
Thanks!
Best Answer
let $p:\ E\rightarrow B$ be the fibration, and let $B$ be path-connected; then all fibers $p^{-1}(b)$ are homotopy equivalent. Furthermore, a path $f:\ [0,1]\rightarrow B$ defines a homotopy class $f_*$ of homotopy equivalence between $p^{-1}(f(0))$ and $p^{-1}(f(1))$. Even better, this only depends on the homotopy class of $f$, relative its endpoints. So specializing to the fundamental group, based at $b\in B$, we have a group homomorphism from $\pi_1(B,b)$ to $S$, where $S$ is the homotopy classes of homotopy equivalences between $p^{-1}(b)$ and $p^{-1}(b)$. If we let $F=p^{-1}(b)$, then each of these homotopy equivalences induces an automorphism of $H_n(F)$ for every $n$; that is, we get a group homomorphism from $\pi_1(B,b)$ to $Aut(H_n(F))$ for every $n$.
This is all covered well in chapters 6 and 9-10 of Kirk and Davis's "Lecture Notes in Algebraic Topology".