Obtaining the three torus via Dehn surgery

Question

It is a well known theorem from the '60 (Lickorish-Wallace) that any closed orientable three dimensional smooth manifold can be obtained performing a sequence of integral Dehn surgeries along knots in $\mathbb{S}^3$. The most common examples found in any book are $\mathbb{S}^3, \mathbb{S}^2\times \mathbb{S}^1 $ and the Lens spaces $L(p,q)$. Curiosly, I can't find how to get the three torus $\mathbb{T}^3 = \mathbb{S}^1\times\mathbb{S}^1 \times \mathbb{S}^1$ which is a quite ubiquitous $3$-manifold in Geometry.

How can I obtain $\mathbb{T}^3$ via (rational and integral) Dehn surgery from $\mathbb{S}^3$?

Answer 1

May be the following explanation, following Ethan Dlugie, gives a visualized picture.

To show that 0-surgery on three components of the Borromean link yields $S^1\times S^1\times S^1,$ what is required to be done is showing that 0-surgery on the two components link in $S^1\times D^2$ yields $(T^2-int(D^2))\times S^1,$ such that the boundary of each slice $(T^2-int(D^2))\times\{ t\}$ is a meridian on the boundary of the solid torus.

Construct the sections as follows, select a meridional disk to make it intersect the link at two points. Consider the case when the meridional disk doesn't intersect the other component first. Assume a tubular neighborhood $N(L)$ of the link is fixed. At the beginning, the section is a surgery on the meridional disk, disjoint from $N(L)$. As the meridional disk moves to right, the radius of the tube becomes smaller, when it touches $N(L)$, they intersect at an annulus. When the section moves further to right, it intersects $N(L)$ at annulus, till the tube moves to the other side outside $N(L)$. As shown below, three states during the movement of the meridional section are exhibited.

Now consider when the meridional disk passes to the other component. The radius of the tube becomes bigger till it intersects the section from the other side. The intersection situation is described below, a vertical tube intersects a horizontal tube at a curved rectangle, combined together to be a punctured torus. After passing this, the section jumps to the other component of the link.

This indicates that the solid torus becomes $(T^2-int(D^2))\times S^1$ after surgery, combining the 0-surgery outside, doing 0-surgery on three components Borromean link yields $T^2\times S^1=S^1\times S^1\times S^1.$

Answer 2

According to @MikeMiller, Exercise 5.4.3(c) of Gompf and Stipsicz identifies this as the $0$-framed surgery on the Borromean rings.

Answer 3

Here's a way not to solve this. I don't know if that's something StackExchange goes for, but I feel like it's still instructive.

How to recognize $T^3$ as a manifold? The most obvious thing in my view is that it's a $T^2$ bundle over $S^1$ with trivial monodromy. So I was hoping to find a genus 1 fiber surface for the Borromean rings complement with trivial monodromy that gives 0-slopes on each boundary component. Then the filled manifold would fill in that fiber surface with three disks to get a torus, and again the filled manifold would be fibered with trivial monodromy.

So here are two diagrams for the Borromean rings. I like the second one better because it makes it easier to draw a Seifert surface.

If I want a connected Seifert surface that gives 0-slopes on each boundary component, probably just start with the the obvious three disks and resolve their singularities in some way, right? Here's what that looks like:

This is certainly a genus 1 Seifert surface, as can be verified with a little Euler characteristic calculation. But it's not a fiber surface, and least not with the trivial monodromy. (I verified this with the criterion given in this paper by S. Baader and C. Graf. Specifically their Cor 2.5.) Even worse, this surface doesn't give the 0-framing on the 'clasp' components!

Then my friend pointed out that the Borromean rings are hyperbolic, hence they can't be fibered with trivial monodromy (else they would have e.g. essential tori)! So alas, this approach seems to just not work. I'll give another answer with the real solution once I find it.