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It is well known that the only principal locally trivial fiber bundle $S^1 \to S^3 \to S^2$ is Hopf map $h$ (see, for example, [1]).

What if we drop the local triviality but mantain a "principality" hypothesis meaning that the fibers are given by an $S^1$ action? Are there well known families of such bundles $S^1 \to S^3 \to S^2$ whose homotopy class is not that of $h$?

In particular, are the homotopy class of the nontrivial multiples $k[h]$ of Hopf's map in $\pi_3(S^2)$, $k \not\in \{ \pm 1,0 \}$, given by such bundles?

Such bundles would be necessarily not locally trivial, but hopefully they can be fibrations.

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The 3-sphere has infinitely many Seifert fibrations with generic fiber a torus knot (including the unknot).

Seifert fibration of 3-sphere

For a $(p,q)$ torus knot, the Hopf invariant will be $pq$ (up to sign). To see this, note that the fibration has two exceptional fibers which form the Hopf link (these are the red line and yellow circle in the image). The generic fibers wrap $p$ times around one of these, and $q$ times around the other, so have linking number $pq$.

In fact, then, one can represent each homotopy of $\pi_3(S^2)$ by a Seifert fibration with a single exceptional fiber (so all fibers are unknots, generically a $(p,1)$ curve on the Clifford torus).

These are (non-locally trivial) ``principal bundles" in the sense that there is an action of $S^1$ whose orbits are the fibers of the fibration. If $S^3\subset \mathbb{C}^2$ as the unit sphere, then the action is $(z_1,z_2)\to (z^pz_1,z^qz_2)$, for $z\in S^1= \{z\in \mathbb{C}, |z|=1\}$.

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  • $\begingroup$ Fantastic @IanAgol ! This is was the kind of answer I was expecting, the picture was a nice bonus. Do you know a reference for these constructions? $\endgroup$ – Lucas Seco Aug 23 '16 at 20:31
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    $\begingroup$ This is a really beautiful answer, but I still must object to the misuse of the word 'bundle', which to me means locally trivial. $\endgroup$ – Mark Grant Aug 24 '16 at 7:37
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    $\begingroup$ @MarkGrant: agreed, but I would blame the poser of the question for using bundle improperly (and he makes it clear that the bundle need not be locally trivial). The fibration is a Seifert fibration, and can be regarded as a principal bundle over an orbifold; however, of course, the orbifold structure on the base is being forgotten to get $S^2$. $\endgroup$ – Ian Agol Aug 24 '16 at 13:48
  • $\begingroup$ Thanks again @IanAgol! Personally I feel that these examples should always be given when one remarks that the Hopf map generates $\pi_3(S^2)$, don't you think? $\endgroup$ – Lucas Seco Aug 24 '16 at 15:29
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    $\begingroup$ I think it should be pointed out that, although the fibers are given by an $S^1$ action, that action is not free on two of the fibers. On $z_1=0$ and $z_2=0$, the action has a finite (nontrivial) stabilizer. Basically, you are exploiting that $S^1$ is a finite cover of itself. $\endgroup$ – David E Speyer Aug 25 '16 at 0:37

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