5

I have a question about the surgery construction of lens spaces.

Let $T=S^1 \times D$ be a solid torus. Let $T'$ be another torus.

We fix a meridian $m$ and longitude $l$ of the torus. Then the lens space $L(p, q)$ is obtained by gluing two tori $T$ and $T'$ along a homeomorphism $f:\partial T \to \partial T'$ sending the meridian to $qm+pl$.

I also understand the surgery along the unknot with framing $p/q$ gives the lens space $L(p,q)$.

What I do not understand is that the surgery along a Hopf link with integer framings also give a lens space.

$S^3$ decomposes into two solid tori each of which contains the unknot. We do surgery along this unknot and glue these space back. I understand every steps but I cannot show that this space is a lens space.

I want a rigorous argument.

I found a similar question Here.

In the answer there, it is claimed

Now I can do the surgery along this Hopf link by doing it in $T_1$ and $T_2$ separately. In particular, give $K_1$ a $-p$ framing, and $K_2$ a $-q$ framing. After the surgeries, I get two spaces $S_1$ and $S_2$. But both of these are still just solid tori; in fact, if $A_i\in GL(2,\mathbb{Z})$ is the gluing map used for the surgery along $K_i$, then the map $A_i^{-1}:\ \partial T_i\rightarrow \partial S_i$ extends to a homeomorphism from $T_i$ to $S_i$.

What I do not understand well here are

  1. $S_i$ are still solid torus.
  2. If $A_i\in GL(2, \mathbb{Z})$ is the gluing map, then the map $A_i^{-1}:\ \partial T_i\rightarrow \partial S_i$ extends to a homeomorphism from $T_i$ to $S_i$.

Especially, if the second part is clear for me, I think I can prove my claim.

I appreciate any help.

  • I love the book :Lectures on the topology of 3-manifold, from N.Saveliev. It's very clear... – Léo Feb 25 '14 at 21:49

1 Answers1

5

Maybe not what you want but this may help:

Surgery on the Hopf link is the same as gluing two solid tori to the two sides of $S^1 \times S^1 \times [0,1]$ by the gluing maps determined by the framings. Let $T_1,T_2$ be solid tori where $T_1$ is glued to $S^1 \times S^1 \times \{0\}$ with framing $p$, and $T_2$ is glued to $S^1 \times S^1 \times \{1\}$ with framing $q$.

The diagonal circle $x \times x \times \{.5\} : x \in S^1$ is a meridian plus a longitude of $S^1 \times S^1 \times \{.5\}$. It is homotopic to $p$ longitudes of $T_1$ and $q$ longitudes of $T_2$ by the surgery framings.

So if we just contract the $S^1 \times S^1 \times [0,1]$ to $S^1 \times S^1 \times \{.5\}$, then we have $T_1$ glued to $T_2$ such that $p$ longitudes of $T_1$ is homotopic to $q$ longitudes of $T_2$, so it is the $p,q$ lens space.

Carl
  • 3,408
  • 1
  • 18
  • 34