Hopf fibration from Hopf link: change to a different ambient space or a different type of link

Question

Hopf fibration from Hopf link

The Hopf fibration is a continuous function from the 3-sphere (a three-dimensional surface in four-dimensional Euclidean space) into the more familiar 2-sphere, with the property that the inverse image of each point on the 2-sphere is a circle. Thus, these images decompose the 3-sphere into a continuous family of circles, and each two distinct circles form a Hopf link. This was Hopf's motivation for studying the Hopf link: because each two fibers are linked, the Hopf fibration is a nontrivial fibration. This example began the study of homotopy groups of spheres.

My question

since the Hopf fibration of $S^1 \hookrightarrow S^3 \rightarrow S^2$ demands that any two $S^1$ fiber attached to any two pointson the base $S^2$ must be Hopf link,

1. do we need to specify which "SPACE" do we consider the two $S^1$ fibers are Hopf linked? Is that space a Euclidean $\mathbf{R}^3$, or a 3-sphere $S^3$? or can we change that space to something more exotic space? (like $\mathbf{RP}^3$ or $S^2 \times S^1$ or Lens space $L(p,q)$?

2. Could we change to a different ambient space or a different type of link? such as any link from this: http://katlas.org/wiki/The_Thistlethwaite_Link_Table

3. If so, could we classify the fibration $S^1 \rightarrow E \rightarrow S^2$ for the total space $E$ via how any $S^1$ fiber is linked with another fiber? What will be the classification? An integer $\mathbf{Z}$ class?

Answer 1

I'm not sure I understand question 1 -- the total space of the Hopf fibration is $S^3$, so you consider the fibers in $S^3$. If you're asking about the definition of a Hopf link, let's say it is a link in the ball $B^3$ that when included in $S^3$ is isotopic to two fibers of the Hopf fibration; then a Hopf link in a 3-manifold $M$ is defined to be the image of this link via an embedding of the ball in $M$. If you're asking about whether you could study fibrations $S^1\hookrightarrow E\to S^2$ for other spaces $E$, then sure, but those wouldn't be called Hopf fibrations.

Let's focus on question 3, considering only the fact it's a fibration by circles over $S^2$. Every fibration $S^1\hookrightarrow E\to S^2$ is a Seifert fiber space with no exceptional fibers and orbit space $S^2$. In the language of Hatcher's 3-manifolds book, $E$ is isomorphic to a model Seifert fiber space $M(0,0;n)$ for some $n\in\mathbb{Z}$. Up to orientation-preserving diffeomorphism, these fiberings are all distinct, and orientation-reversing diffeomorphism gives $M(0,0;n)\approx M(0,0;-n)$. For example,

• $M(0,0;0)$ is $S^1\times S^2$,
• $M(0,0;\pm 1)$ are the two mirror images of the Hopf fibrations, and
• $M(0,0;\pm 2)$ are fibrations of $\mathbb{R}\mathrm{P}^3$.

In general, there is a description of $M(0,0;n)$ as $n$-surgery on an unknot, which is the Lens space $L(n,1)$. This is the classification. First homology is $H_1(M(0,0;n))=\mathbb{Z}/n\mathbb{Z}$ so that's enough to classify them up to diffeomorphism.

Except when $n=\pm 1$, the fibers are homotopically nontrivial. There is a sense in which you could say pairs of fibers have linking number $1/n$ using rational linking numbers, but in any case pairs of fibers when $n\neq \pm 1$ do not form a Hopf link, and furthermore they are not 2-component links from the Thistlethwaite Link Table in your question.

More general than a fibration by circles is a foliation by circles. In 1, simplifying things a bit, Epstein showed that every smooth foliation by circles of a compact orientable smooth three-manifold (possibly with boundary) is a Seifert fiber space. These are also all classified. There are some examples of Seifert fibrations of $S^3$ in this answer.

1 Epstein, D. B. A., Periodic flows on three-manifolds, Ann. Math. (2) 95, 66-82 (1972). ZBL0231.58009.