Which homotopy classes $S^3 \to S^2$ lift to embeddings $S^3 \to S^2 \times D^3$?

Question

The question is, for a smooth embedding

$$f : S^3 \to S^2 \times D^3$$

one can compose the map $f$ with projection $\pi : S^2 \times D^3 \to S^2$, giving the map $\pi \circ f : S^3 \to S^2$.

Which homotopy classes of maps are realizable?

The analogous question where you vary the dimension of the disc, i.e.

$$ f : S^3 \to S^2 \times D^k $$

has immediate answers when $k \neq 3$. When $k < 3$ the answer is only the constant map, and when $k > 3$ the answer is all homotopy classes, since $S^3$ lives in the boundary of $D^k$. The $k < 3$ argument is a basic cut-and-paste topology argument, to argue that the projection must be null-homotopic.

I suspect the answer to this question exists in the literature, as you can view this as the problem of if one can "link" a linearly-embedded $S^2$ with a non-linearly embedded $S^3$ in $S^5$. But I have looked through the old Haefliger-Zeeman literature on mixed-dimensional links, without much luck. A short summary of the theory is here, written-up by Skopenkov: http://www.map.mpim-bonn.mpg.de/High_codimension_links

A closely-related MO question is on the $k=2$ case, but where you let $f$ be an immersion. This was answered positively in the comments. Hopf fibration inside the retraction of R^4 minus line -> S^2?

edit: I suppose I could add I have a suspicion the answer is at least the subgroup of index two in $\pi_3 S^2 \simeq \mathbb Z$. At present I do not see how to obstruct a generator of $\pi_3 S^2$ being realizable.

Answer 1

This is not really an answer, but it provides some context. We can consider $S^3$ as the space of unit quaternions, $S^2$ as the subspace of unit pure-imaginary quaternions, and $D^3$ as the space of pure-imaginary quaternions of norm at most one. With these models, the generator of the group $\pi_3(S^2)$ is the map $\eta(u)=u\,i\,\overline{u}$. The obvious way to try to define a corresponding map $f\colon S^3\to S^2\times D^3$ is to put $f(u)=(\eta(u),\text{Im}(u))$. It is not hard to see that this is injective with a single exception, namely that $f(1)=f(-1)$. So the question is whether we can eliminate this failure of injectivity by modifying $f$.

Answer 2

Earlier I thought I had an argument that twice the Hopf map was realizable for an embedding $S^3 \to S^2 \times D^3$, but there was a mistake in my argument -- the map I suggested failed to be a smooth embedding.

Here is a more elementary observation on the problem.

Proposition: Given a smooth embedding $f : S^3 \to S^2 \times D^3$, then the projection map $\pi : S^2 \times D^3 \to S^2$ can not restrict to a locally-trivial fiber bundle on $f(S^3)$.

This proposition does not answer the question, but it provides some restrictions on answers.

The idea is to consider the simplest case. Assume $\pi$ restricts to a locally-trivial fiber bundle on $f(S^3)$, and that $f(S^3)$ intersects the fibers of $\pi$ in unknotted loops. You could cite Hatcher's work here and conclude they must be linearly embedded loops.

This implies the Hopf fibration is classified by a map $S^2 \to V_{3,2} / SO_2$. We know the Hopf fibration is classified by the generator of $\pi_2 (V_{4,2} / SO_2)$, and $\pi_2 (V_{3,2}/SO_2)$ is the index two subgroup.

I believe this argument extends to the case the fibers are non-linearly embedded -- in that case you can argue the Euler class must be zero. That said, there is probably a simpler way to rule out non-linearly embedded fibers.

Comment:

There is a map $f : S^3 \to S^2 \times D^3$ that is a 2-to-1 immersion, with the projection $S^3 \to S^2$ equal to the Hopf fibration. The idea is that $S^3$ modulo antipodal points has a canonical identification with the unit tangent bundle of $S^2$, which is a subset of $S^2 \times D^3$. I suppose using Neil's notation this would be the map $(\eta(u), \tau(u))$ where $\tau(u) = uj\overline{u}$.