# Surgery of $S^3$

I have been troubled by this seemingly simple question recently. How do we easily visualize the statement:

Surgery of $S^3$ over a trivial unknot gives $S^1 \times S^2$?

All I can think of for the first step is to remove a point from $S^3$ so we can work on $R^3$, but then I have no idea how to go on.

As a further generalization, what is the result if we do surgery of $S^3$ over $n$ unlinked unknot?

I am an algebraist so I have no clue how people usually visualize these objects in more than 3 dimension... Any hints will be appreciated.

$S^3$ is a union of two solid tori, glued along their boundary, where a meridian is glued to the longtitude, and conversely. When you do the surgery, you get a solid torus doubled along the boundary. If you think of a solid torus as $D^2 \times S^1,$ the double is a double of $D^2$ times $S^1.$ The double of $D^2$ is $S^2,$ hence, when the smoke clears, you get $S^2 \times S^1.$

• This is some nice thinking, though I am not really convinced that $S^3$ is union of two solid tori... (my topology professor says it is a trivial result...) Also how can this argument be generalized to the $n$ unlinked unknot case?
– Ivan
Jan 18, 2016 at 5:42
• $S^3=\partial D^4 \cong \partial(D^2\times D^2)=D^2\times S^1\cup S^1\times D^2$. For the $n$ unlinked unknot case, write $S^3$ as a connected sum of $n$ copies of $S^3$, each with one unknot in it. Jan 18, 2016 at 9:58
• You can also use S^3 = {|w|^2 + |z|^2 = 2} = {|w|\leq 1, |w|^2 + |z|^2 = 2} \cup {|w|\geq 1, |w|^2 + |z|^2 = 2} in \C^2. (This amounts to the same thing.) Jan 18, 2016 at 21:37
• Here's a hands on way to see that $S^3$ is the union of two tori. First visualize a ball in $S^3$ which we think of as $\mathbb R^3$ union a point at $\infty$. The complement is also a ball. Now add drill a hole through the middle of the ball. You get a solid torus. This has the effect of adding a solid tube to the exterior ball, making it a solid torus as well. Sep 28, 2017 at 15:05

This is all written up very well in the book of Prasolov and Sossinsky.

Surgery in Dimension 3 is explained in Chapter V and VI.

Proposition 14.4. shows that surgery along the unknot yields $S^1\times S^2$ (as explained in the other answer). Figure 8.7 for example shows how to see $S^3$ as the union of two solid tori.

For the second question: Consider a link in $S^3$ that you can devide by an $S^2$ in two sublinks $L_1$ and $L_2$, and if you denote by $S^3(L)$ the result of surgery along $L$. Then you cann do the surgery in two disjoint balls, so the resulting manifold is the connected sum $S^3(L)=S^3(L_1)\#S^3(L_2)$. So in your special example the result is $n$-times the connected sum of $S^1\times S^2$ with itself, ie $\#_n S^1\times S^2$.

Another very nice book where all this is explained is the book of Rolfsen.

As the others have said, the key is seeing $S^3$ as the union of two solid tori with their axes forming the Hopf link. Here is how you can see this (hat tip to Milnor). Write $\newcommand{\bC}{\mathbb{C}}$

$$S^3=\bigl\{ (z_1,z_2)\in\bC^2;\;\;|z_1|^2+|z_2|^2=2\;\bigr\}.$$

The two solid tori are

$$T_k=\left\{ |z_1|^2\leq 1\;\right\}\subset S^3,\;\;k=1,2$$

The map

$$S^3\supset T_1\to S^1\times D^2,\;\;(z_1,z_2)\mapsto \left(\frac{z_2}{|z_2|},z_1\right)\in S^1\times D^2$$ is a diffeomorphism. The diffeomorphism $T_2\to S^1\times D^2$ is constructed similarly.

Note that the axis of $T_1$ is the circle $z_1=0, |z_2|=\sqrt{2}$ and the axis of $T_2$ is $z_2=0,|z_1|=\sqrt{2}$. Clearly the axis of $T_1$ is a meridian of the axis of $T_2$ and vice-versa.

Here's a more 4-dimensional proof, or rather a 4-dimensional interpretation of Igor Rivin's answer above.

To do surgery along a knot in $S^3$, one attaches a 4-dimensional 2-handle to $B^4$, and takes the boundary of the resulting 4-manifold (there is some ambiguity in the framing, but the OP seems not to care about it).

To see that $S^1\times S^2$ arises as surgery along the unknot in $S^3$, we can identify $B^4$ with $B^2\times B^2$; the unknot is $K = \{0\}\times S^1 \subset \partial B^2\times B^2$ (indeed, it bounds an obvious disc).

Attaching a 2-handle along $K$ means attaching another copy of $B^2\times B^2$ along $B^2\times S^1$; there is an obvious way of doing so, which is just doubling $B^2\times B^2$ along half of its boundary, thus obtaining $B^2\times S^2$, whose boundary is precisely $S^1\times S^1$.

Full disclosure: I've swept all issues about smoothing corners under the rug, but I hope that this at least gives some intuition.