Surgery on the unknot - a $D^2$ bundle over $S^2$?

Question

Let $K$ be the unknot in $S^3$ and let $N(K)$ be a tubular neighborhood of $K$. Suppose we take a homeomorphism $h : S^1 \times D^2 \to N(K)$ where $S^1 \times \{ 0 \} \mapsto K$ and $S^1 \times \{ 1 \}$ maps to a longitude in $\partial N(K)$ that has linking number $-p$ with $K$ (when both are given parallel orientations). Using $h$ we can then attach $D^2 \times D^2$ to $S^3 = \partial D^4$ and the result will be a space I will denote $E_p$ which has as its boundary the lens space $L(p,1)$.

I was told that $E_p$ is a locally trivial $D^2$ bundle over $S^2$ - what is the map $E_p \to S^2$ and why is it such a fiber bundle? Does this map then restrict to a $S^1$ bundle over $S^2$ on $L(p,1)$? For $p=0$ why is this the trivial bundle?

Answer 1

Every $D^2$-bundle over the sphere $S^2$ can be constructed using the following construction. Fix $p \in \mathbb{Z}$, decompose $S^2=A \cup B$ as the union of two disks intersecting in their common boundary (north hemisphere and south hemisphere), form the products $A \times D^2 $ and $B \times D^2$, and glue them together using the identification $$\partial A \times D^2\to \partial B \times D^2: \ \ \ (\theta, v)\mapsto (\theta, \gamma(\theta) \cdot v) \ ,$$ where $\gamma$ denotes any loop such that $[\gamma]=p$ in $\pi_1(SO(2))= \mathbb{Z}$. As outcome of this operation you get a space $E_p$ that constitute the total space of a $D^2$-bundle over $S^2$: the bundle projection $E_p \to S^2$ is just the map induced on the quotient by the projections $A \times D^2 \to A $ and $B \times D^2 \to B $.

This is the same as gluing the 2-handle $h = A \times D^2$ to $D^4= B \times D^2$ along the knot $K= \partial B \times 0 $ with framing $$f(\theta)= \gamma(\theta) \cdot 1 \in \theta \times \partial D^2,\ \ \ \ \ \ \theta \in \partial B.$$ Notice that $K \subset \partial D^4= (\partial B \times D^2) \cup (B \times \partial D^2) $ is the unknot since it bounds a disk in $S^3 = \partial D^4$ (namely $D=(\partial B \times [0,1]) \cup (B \times 1)$). It is easy to see (exercise!) that $lk(K, f)=p$ and consequently that $E_p$ is the result of a $p$-framed 2-handle attachment along the unknot.

Notice that the $D^2$-bundle projection $E_p \to S^2$ restricts to an $S^1$-bundle projection $\partial E_p \to S^2$. As consequence we obtain that $L(p,1)= \partial E_p$ is the total space of an $S^1$-bundle over $S^2$.