While reading a proof of a theorem about Reshetikhin Turaev topological quantum field theory, I encountered the following problem.

Suppose we have several unlinked unknots $K_i$, $i=1, \dots, g$ in $x$-$z$ plane of $\mathbb{R}^3$. Assume each unknot interests with $x$-axis at two points.

We complete the $x$-$y$ plane to obtain $S^2=\mathbb{R}^2 \cup \{\infty\}\subset S^3$

Consider the regular neighborhood of $S^2\cup \{K_i\}_{i=1}^g$ in $S^3$. Let us call it $N$.

Then we do a Dehn surgery along unknots $K_i$ in $N$. (Assuming the framings of knots are zero?)

What I want to prove is that the resulting 3-manifold is homeomorphic to a cylinder over a surface $S$ with genus $g$, namely $S\times [0, 1]$.

I have no idea how to calculate this surgery to get the result. I have studied the basic Dehn surgery theory.

Any help is appreciated.

  • $\begingroup$ It is Dehn surgery that I have in mind. $\endgroup$ – Snow Jun 18 '14 at 1:29
  • $\begingroup$ What do you mean by a cylinder with genus g surface? The product of this surface with the real line? $\endgroup$ – Moishe Kohan Jun 18 '14 at 2:43
  • $\begingroup$ By a cylinder over a surface $S$, I mean $S\times [0, 1]$. $\endgroup$ – Snow Jun 18 '14 at 3:55
  • $\begingroup$ @user126154 I assume that framings of unknots are zero, so I think it is 0-surgery. Is it still not enough to specify a surgery? $\endgroup$ – Snow Jun 18 '14 at 10:27
  • $\begingroup$ @user126154 Could you explain how to get the cylinder if meridians and longitudes are exchanged? $\endgroup$ – Snow Jun 18 '14 at 12:45

Disclaimer: this kind of proofs are extremely difficult to describe without the help of pictures. So I'll try to give the idea, and I do not claim that the folloing is a complete formal proof.

Now, consider for simplicity the case $g=1$. You have a solid torus $S$ that cut a horizontal thik plane $P$ (the regular neighborhood of the plane).

$P$ is naturally fibered by vertical segments so that it is a product of an interval by a plane.

The boundary $T$ of $S$ intersect $P$ in two annuli, which are fibered by vertical segments. $P$ has the product structure od $D^2\times S^1$ and the annuli are parallel to $\partial D^2$.

Now, perform a Dehn surgery, along the core curve of $S$, in such a way that you exchange the meridian of $S$ with its longitude. That is to say remove a torus of the form $D^2\times S^1$ and glue it back by itendifying $\partial D^2$ with $S^1$.

From the point of view of the torus $S$, this is equivalent to switch the annuli so that they become parallel to $S^1$ in the structure of $D^2\times S^1$. So the annuli now no longer bound disks.

Now, you can give the new torus the structure of $[0,1]\times[0,1]\times S^1$ where the annuli are exactly $A_0=[0,1]\times\{0\}\times S^1$ and $A_1=[0,1]\times\{1\}\times S^1$ and the fibration of $A_i$ is by segments of the form $[0,1]\times\{i\}\times\{\theta\}$ for $i=0,1$.

It is then clear that such a product fibration extend to the whole torus by

$[0,1]\times\{t\}\times\{\theta\}$ for $t\in[0,1]$.

This product structure is in fact an extension of the product structure of the thick plane and gives you the required product structure.

EDIT: An alternative way to see it is the following. Again, let $g=1$ for simplicity. The space $N$ is obtained by attaching an annulus (the spehere minus two disks) to a solid torus $D^2\times S^1$, and then by thikening a little the annulus.

The circle where the annulus is attached are boundary of disks-fibers of $D^2\times S^1$.

If you think a little, you see that performing the Dehn surgery that switches the meridian with longitude is equivalent to attach the annulus to the solid torus in a different way, namely along circles parallel to the $S^1$ component of $D^2\times S^1$. That is to say, you are attaching a cylinder (which is homeomorphic to an anulus) to a donut, in such a way that the "hole" of the donut corresponds to the "hole" of the cylinder. After thikening, this is equivalent to identify the two boundary components of the cylinder (and then thikening). The result is therefore $T^2\times S^1$.

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  • $\begingroup$ Thank you for the answer. Could you explain why we can ignore a part of the original solid torus which is not intersecting with the thicken plane? $\endgroup$ – Snow Aug 23 '14 at 2:24

It's possible to visualise the surgery. This really needs some pictures to illustrate, but I'll do my best without. I'll describe the case $g=1$. In this case, your manifold is obtained by gluing an $S^2$ with 2 disks removed to a solid torus, $T=S^1 \times D^2$. The $S^2$ minus 2 disks is homeomorphic to a fattened annulus, $S^1 \times [0,1] \times [0,1]$. And your manifold is obtained by gluing this fattened annulus along the two annuli contained in the boundary specified by $S^1 \times \{0,1\}\times [0,1]$. Since the surgery coefficient is $0$, the $S^1$ factor of each boundary component we are gluing along attaches to a longitude of the solid torus $T$. Such a construction can be seen to give a $S^1 \times S^1 \times [0,1]$.

The case $g>1$ can be build up inductively from this case.

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