Suppose that $E \to B$ is a fibration with fibers $F$. There is a long exact sequence in homotopy groups. In particular, we have boundary maps $\partial: \pi_n(B) \to \pi_{n-1} (F)$. Is there a good, geometric interpretation for this boundary map?

In particular, I want to understand the long exact sequence arising from $SO(2) \to SO(3) \to S^2$. How does an element $a \in \pi_2(S^2)$ induce a loop in $SO(2) \cong S^1$ via the boundary map. In particular, is it is visibly clear that the generator for $\pi_2(S^2)$ induces $[2] \in \pi_1(SO(1)) \cong \mathbb{Z}$?


Yes. Let's chase through the proof that this long exact sequence exists. First, the construction of the relative long exact sequence of a pair is reasonably elementary; and there is an isomorphism $p: \pi_n(E,F_b) \to \pi_n(B,b)$ given by projection commuting with the obvious map from $\pi_n(E,*)$ to both. And the boundary map is constructed very geometrically for the relative pair: you send a (homotopy class of) map $(D^n,S^{n-1}) \to (E,F_b)$ to the map $S^{n-1} \to F_b$. So the construction of $p^{-1}$ is the geometric content of the theorem.

Suppose you have a sphere in $B$. Think of the sphere as a homotopy $f_t: S^{n-1} \times [0,1] \to B$ with $f_0(x) = b_0$ and $f_1(x) = b$. Picking an arbitrary lift of $b_0$, applying the homotopy lifting property we get a disc in $E$ with boundary (corresponding to points of the form $(x,1)$) having having values in $F_b$. Another application of homotopy lifting shows this map is well-defined. That the two are mutually inverse is also simple.

For $S^1 \to S^3 \to S^2$, start with the identity map $S^2 \to S^2$. Thinking of $b_0$ as the north pole and $b$ as the south, and run the above procedure. In more generality, on the total space of the circle bundle with Euler number $n$, the construction will show that the boundary map in the corresponding long exact sequence $\pi_2(S^2) \to \pi_1(S^1)$ is multiplication by $n$. (As a hint, use that you can trivialize the bundles over the top and bottom hemispheres, while keeping in mind the data of the transition function over the center circle. It's easy to construct the lift over the trivial bundle.)

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  • $\begingroup$ Thanks, this was very helpful and I learned a lot form it! But how can I show that the circle bundle from the in my fibration has Euler number 2? $\endgroup$ – Lorenzo Najt Apr 24 '17 at 5:30
  • $\begingroup$ @AreaMan: Just to clarify, you see how to finish once you know the Euler number? As for calculating the Euler number, there are many equivalent ways to obtain it: "The homology of the total space is Z/|e|Z"; or the linking number of two fibers. I'll write up something more detailed to you later. $\endgroup$ – user98602 Apr 24 '17 at 14:05
  • $\begingroup$ I think the first two paragraphs are clear to me now. I need to think more about the third - I'll get back to you in a few days. $\endgroup$ – Lorenzo Najt Apr 25 '17 at 5:18

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