# Uniqueness of Preferred Framing of a Solid Torus in $S^3$

One way to state my question tersely is: For a homeomorphism $f : S^1 \times \mathbb{D}^2 \rightarrow S^1 \times \mathbb{D}^2$, does $f|_{S^1 \times S^1}$ determine the isotopy class of $f$?

This is motivated by the question of why a "preferred framing" of a solid torus imbedded in $S^3$ is unique up to isotopy.

In what follows I will try to give the background for this question and how it reduces to the first question.

Let $V \subset S^3$ be a homeomorph of the solid torus ($S^1 \times \mathbb{D}^2$) continuously imbedded in the 3-sphere in such a way that the closure of $S^3 - V$ is a manifold (I'll call $X$)  with boundary equal to $\partial V$.  It can be seen from a Mayer-Vietoris sequence that there exists a simple closed curve $\lambda : S^1 \rightarrow \partial V$ which represents a generator of $H_1 (V)$ and is homologically trivial in $X$.  Call this a "preferred longitude for V". Moreover, for any other preferred longitude, $\lambda '$, we have $[\lambda]= \pm [\lambda ']$ in $H_1 (\partial V)$. In addition there exists a simple closed curve $\mu : S^1 \rightarrow \partial V$ which is homologically trivial in V but not in $\partial V$.  Call $\mu$ a "meridian of V".  It is not hard to see that, like preferred longitudes, meridians fall into only two homology classes in $H_1 (\partial V)$, depending on orientation.

Now a choice of meridian and preferred longitude determines the isotopy class of homeomorphisms $f': S^1 \times S^1 \rightarrow \partial V$ which have [$S^1 \times 1] \mapsto [\lambda]$ and $[1 \times S^1] \mapsto [\mu]$ under $H_1(f')$ (here let's say $S^1$ has the counterclockwise orientation).  That all such $f'$ are isotopic is not obvious, but not very difficult to prove.

Now because $1\times S^1$ is mapped to a meridian, any such $f'$ extends to a framing of $V$, i.e. a homeomorphism $f: S^1 \times \mathbb{D}^2 \rightarrow V$.  I would like to show that any two such $f$ are isotopic. I am allowed to assume that the two framings agree on the boundary because an isotopy of the boundary can be extended to an isotopy of the solid torus by means of a collar. This is why I believe an answer to the question at the top of the post would suffice.

Thank you for your time.

## 2 Answers

Consider two maps isotopic on the boundary, and quotient the meridian and longitude to a single point. This turns the solid torus into the ball. The question is then, if two maps from the ball to itself fixing a boundary point are isotopic on the boundary, are they isotopic on the ball? Since the ball is convex, I'm fairly certain this is true.

• Very interesting, Brian. Lets assume the map is the identity on the boundary and we want to show it is isotopic to the identity on the solid torus. In this case I agree that it descends to a homeomorphism of the closed ball which is the identity on the boundary sphere. I also agree that implies that this new map is isotopic to the identity on the ball (as i see it we can "push in the identity radially"). What isn't immediately obvious to me though is why this isotopy ascends back up to an isotopy of the torus. If this is a trivial point then I will just have to think on it a bit. Jul 20, 2013 at 22:46
• You're right, it's not completely obvious. But I feel that you could use universal properties of the quotient topology. Continuous maps out of a quotient are in 1-1 correspondence with continuous maps on the original space that are constant on the collapsed set, but that doesn't give you the pullback immediately... Jul 21, 2013 at 0:33

Elaboration of Brian's very clever solution:

As mentioned in the comments, we may assume one of the functions to be the identity and the other, $f$, to have $f|_{S^1 \times S^1} = id$. By means of a collar, we may then assume that $f$ is the identity in an open neighborhood of the $S^1 \times S^1$ (I believe this will resolve the issue I raised in my comment to Brian's answer). $f$ then descends to a homeomorphism, $f_1$ of the 3-ball which is the identity in a neighborhood of $S^2$. There exists an isotopy, $H_1$, from $f_1$ to the identity which is stable in an open neighborhood of $S^2$. Now we may define a function $$H: \{S^1 \times \mathbb{D}^2 \} \times I \rightarrow S^1 \times \mathbb{D}^2, \qquad H(p,t) = \left\{ \begin{array}{lr} H_1 (p,t) & : p \notin \ast \times S^1 \cup S^1 \times \ast \\ p & : p \in \ast \times S^1 \cup S^1 \times \ast \end{array} \right.$$

$H$ is clearly continuous whenever $p$ is not in the collapsed set. But it is also just the projection function in a neighborhood of $\{ \ast \times S^1 \cup S^1 \times \ast \} \times I$.