# Can lens spaces be realized by surgery along torus links?

As we know, the Lens space $L(p,q)$ is the quotient of $S^3$ by a $\mathbb{Z}_p$ action: $(z_1,z_2) \rightarrow (e^{2\pi i/p}z_1,e^{2\pi iq/p}z_2)$.

It seems that the Lens space $L(1,0)$, a.k.a $S^3$, could be realized by surgery along the Hopf link in $S^2\times S^1$. The Hopf link could be represented by the torus link $T(2,2)$.

So I was wondering if the Lens space $L(p,1)$ could be obtained by surgery along the torus link $T(2,2p)$ in $S^2\times S^1$?

• By a torus link in $S^2 \times S^1$, do you mean a link sitting on a Heegaard surface? – magicker72 Sep 27 '16 at 0:41
• Why are you focusing on a link? $S^3$ is surgery on $\{*\}\times S^1 \subset S^2 \times S^1$, and every $L(p, 1)$ is surgery on the same knot. – magicker72 Sep 27 '16 at 0:46
• Exactly, a link on a Heegaard surface. By Lickorish’s theorem, any three-manifold $M$ can be obtained by surgery on a link $\mathbb{L}$ in $S^3$, thus in $S^2\times S^1$. While the Lens space $L(p,1)$ is homeomorphic to $S^3/\mathbb{Z}_p$. Thus I wish to set up a connection between $L(p,1)$ and $T(2,2p)$. @magicker72 – Franklin Wu Sep 27 '16 at 5:01
• @magicker72: I think that you only get $S^3$ by doing (integral) surgeries along $\{*\}\times S^1$. In the 4-dimensional cobordism picture, you are attaching a 2-handle that cancels the 1-handle. However, if I'm not mistaken, the answer to the original question is yes; in fact, $L(p,1)$ is an integral surgery along any $T(2,2q)$, and this is a good exercise with handle diagrams. – Marco Golla Sep 27 '16 at 7:55
• @MarcoGolla Thanks for answering. What I don't understand is an integral surgery along any $T(2,2q)$, even if $q\neq p$. – Franklin Wu Sep 27 '16 at 16:23

## 1 Answer

OP told me about his question a few months ago—now might be as good a time as any to attempt an answer.

Disclaimer: I know second to nothing about knots, surgery, or anything in geometric topology at large.

I think that any lens space can be obtained from a torus link $T(2, 2p)$ in $S^2 \times S^1$. So, the answer to your question is yes. There are several ways to see it. As attached, surgery diagrams work. $0$-surgery on the unknot gives $S^2 \times S^1$. A torus link $T(2, 2p)$ there can be modified into $T(2, 0)$ by twisting along the unknot with coefficient $0$. Then we have a simpler surgery diagram which consists of the connected sum of two Hopf links, where the new coefficients are $x$, $y$, and $0$. It is easy to modify it to the final form by using the Kirby calculus. For example, $x$ must be an integer, but $y$ can be any rational number. Hence any lens space can appear.