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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!

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2 Answers 2

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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".

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  • $\begingroup$ Just to be clear, when you say $f$ defines a homotopy class $f_\ast$ of homotopy equivalences, you are saying that we have a class of homotopy equivalences between $p^{-1}(f(0))$ and $p^{-1}(f(1))$? That is, $f$ somehow restricts which homotopy equivalences we can have between those two fibers? $\endgroup$
    – Jonathan Beardsley
    Commented Apr 12, 2011 at 2:40
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    $\begingroup$ I don't know what you mean by "restricts". $f$ induces a homotopy equivalence between those two fibers, similar to the way a continuous map induces a homomorphism on fundamental groups. The explicit construction is thus (let's only worry about the fundamental group): Let $\alpha\in\pi_1(B,b)$, and choose a loop $g:\ [0,1]\rightarrow B$ representing $\alpha$. Define a map $G:\ F\times [0,1]\rightarrow B$ by $G(x,t)=g(t)$. Since $p:\ E\rightarrow B$ is a fibration, there is (many!) lifts of $G$; that is, maps $H:\ F\times [0,1]\rightarrow E$ such that $G = pH$... $\endgroup$
    – user641
    Commented Apr 12, 2011 at 2:49
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    $\begingroup$ (continued)... So the point is $H$ is not really well-defined, but it is well-defined up to homotopy; Furthermore, $H(\cdot,1):\ F\rightarrow E$ is really a map from $F$ to $F$. So starting with $\alpha$, we've gotten a map $H(\cdot, 1)$, well-defined up to homotopy, which is a homotopy equivalence of $F$ to itself. $\endgroup$
    – user641
    Commented Apr 12, 2011 at 2:52
  • $\begingroup$ Thanks so much for your help! This helps a lot. This uses that diagram which defines fibrations, I see now. I should probably become very familiar with that. $\endgroup$
    – Jonathan Beardsley
    Commented Apr 12, 2011 at 2:54
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No, it means that the homomorphism induced by following a loop is trivial. You will always get the same group back but a generator may get sent to its negative. Take a Möbius strip $\mathbb R \to M \to S^1 $ and mod out each fiber of the strip by the action of $\mathbb Z$. This is a twisted torus (might be the Klein bottle). Pick any point $p \in S^1$, and a generator $\omega$ of $H_1(\pi^{-1}(p))$. The loop around the base induces an action of $H_1(\pi^{-1}(p))$ that sends $\omega$ to $-\omega$.

Does anyone know what happens to the Serre Spectral Sequence when this is not trivial?

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  • $\begingroup$ Could you be more clear, for an absolute beginner? I.e. what do you mean the homomorphism induced by following a loop? Can this be stated algebraically, i.e. how does a loop induce a map on the base, and hence a homomorphism in homology? $\endgroup$
    – Jonathan Beardsley
    Commented Apr 12, 2011 at 2:30
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    $\begingroup$ In the case the action is non-trivial, you need to use local or twisted coefficients; That is, we have $E^2_{p,q}=H_p(B,H_q(F))$, but the coefficient group $H_q(F)$ will have the structure of a $\mathbb{Z}[\pi_1(B)]$-module. I found this much easier to understand after doing some group cohomology, where it is often the case the coefficient group carries some additional structure as a $\mathbb{Z}[G]$-module. $\endgroup$
    – user641
    Commented Apr 12, 2011 at 2:35

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