Fiber bundle/ Hopf's fiber bundle

Question

The question:

A fiber bundle $(E,B=S^2,F=S^1)$ is given. Determine homotopy groups of $E$ in terms of homotopy groups of $S^2$.

Can I say-let $E=S^3$ and use Hopf's fiber bundle?

Answer 1

An essential tool for studying homotopy groups of bundles in general is the long exact sequence of a fibration, and bundles whose fibres are $S^1$ are made especially nice by the fact that $\pi_k(S^1) = 0$ for $k > 1$. In particular if $S^1 \to E \to B$ is a fibre bundle you immediately see that $\pi_k(E) \cong \pi_k(B)$ if $k > 2$, and the undetermined part of the sequence is

$$ 0 \to \pi_2(E) \stackrel{p_*}{\to} \pi_2(B) \stackrel{\partial}{\to} \pi_1(S^1) \stackrel{i_*}{\to} \pi_1(E) \to \pi_1(B) \to 0. $$

In the special case that $B = S^2$ it simplifies further to

$$ 0 \to \pi_2(E) \stackrel{p_*}{\to} \pi_2(S^2) \stackrel{\partial}{\to} \pi_1(S^1) \stackrel{i_*}{\to} \pi_1(E) \to 0 $$

and since $\pi_2(S^2) \cong \mathbb{Z} \cong \pi_1(S^1)$, exactness implies that the sequence is entirely determined by $\partial$. Either 1) $\partial \neq 0$ and is therefore injective so $\pi_2(E) = 0$ and $\pi_1(E) \cong \mathbb{Z}/im(\partial)$, or 2) $\partial = 0$ so both $p_*$ and $i_*$ are isomorphisms.

The Hopf fibration falls into Case 1 because $E=S^3$, and for this bundle $\partial$ is an isomorphism.

Case 2 includes the trivial bundle $S^1 \times S^2$.

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Without picking a specific bundle, it's impossible to say more at this point. However, if we consider the classification of principal $S^1$-bundles as homotopy classes of maps to a classifying space $$Prin_{S^1}(S^2) \cong [S^2, BS^1]\cong \pi_2(BS^1) \cong \mathbb{Z}$$ it's possible to describe $\partial$ in terms of a given classification map $c\colon S^2 \to BS^1$.

Exercise: If the bundle $E$ is classified by a map $c\colon S^2 \to BS^1$ so that the (chosen/fixed) isomorphism $\pi_2(BS^1)\cong \mathbb{Z}$ sends $[c]$ to the integer $d$, then $\partial\colon \pi_2(S^2) \to \pi_1(S^1)$ is multiplication by $\pm d$. In particular the trivial bundle is the only bundle (up to isomorphism) that falls into Case 2.

Hint: Use the fact that the classifying map induces a morphism of exact sequences between the l.e.s. of $E$ and the l.e.s. of the universal bundle $S^1 \to ES^1 \to BS^1$, which in particular induces isomorphisms between the homotopy groups of the fibres.

Note: I am making an apparent simplification by restricting to principal circle bundles to invoke the classification result, which are classified by maps to $BS^1$, as opposed to the more general case of "bundles whose fibers are $S^1$", which should really be classified by maps to $BHomeo(S^1)$. $Homeo(S^1)$ contains orientation-reversing homeomorphisms whereas $S^1\cong SO(2)$ does not, but it still strongly deformation retracts onto $O(2)$ (this result is non-trivial!). The functorial map $BS^1 \to BO(2)$ is a non-trivial double cover, so in particular $\pi_k(BO(2))\cong \pi_k(BS^1)$ for $k \geq 2$ and $\pi_1(BO(2))\cong \mathbb{Z}/2$. HOWEVER since $S^2$ is simply connected in fact $[S^2, BO(2)]\cong [S^2, BS^1]$ so in other words any bundle over $S^2$ with circle fibres admits a compatible principal bundle structure which is essentially unique.