# Are there "principal" bundles $S^1 \to S^3 \to S^2$ other then Hopf's? (They would be necessarily not locally trivial)

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.

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

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 class in $$\pi_3(S^2)$$ 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 $$z\in S^1= \{z\in \mathbb{C}, |z|=1\}$$.

• 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? Aug 23, 2016 at 20:31
• This is a really beautiful answer, but I still must object to the misuse of the word 'bundle', which to me means locally trivial. Aug 24, 2016 at 7:37
• @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 $S^2$. Aug 24, 2016 at 13:48
• 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? Aug 24, 2016 at 15:29
• 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. Aug 25, 2016 at 0:37