No this is not a medical procedure.

This week, postdoctoral candidate Fiona Burnell came to Oxford to interview for the prestigious All-Souls fellowship. (The last physicist to win this fellowship was Austen Lamacraft, who apparently reads this blog, at least once in a while). I’ve been working with Fiona on and off for much of the last year and I always learn something new from talking to her. While she was here this week, I finally grokked the mathematical idea of Dehn surgery – and actually it is very simple, I had just never seen it explained clearly anywhere before (although in retrospect, the Wikipedia entry on Dehn Surgery is actually not too bad).

First, let us think about a two dimensional surface, like the surface of a sphere. Cut two holes (disks) out of the sphere and throw them away. The boundary of the cutting region is two circles. Now, if I want to patch up these holes to get a smooth surface again, I need to find something that also has a boundary with two circles – and there are now two options. (1) The set of two disks has a boundary which is two circles (and if I sew them back in, I just get the original sphere). OR, (2) a hollow tube (circle cross interval, for those keeping track) also has a boundary which is two circles. If I instead sew on the hollow tube, connecting the two holes in the sphere, I end up getting a sphere with a “handle” attached, which is topologically a torus. This is a simple example of what mathematicians call surgery (for obvious reasons).

OK, now it gets a bit harder. Consider a smooth three-dimensional manifold, like the space around us (mathematicians like to add a “point at infinity” to change R3 into S3 but this is just to assure that we have a closed topology rather than an open topology). Draw some closed circle (usually called an S1) in this space. Then take a tubular neighborhood of this circle, so you have a solid torus (S1 cross disk. Disk is usually called D2 so this is S1 cross D2). Now remove this solid torus from the manifold. The boundary of what was removed is the torus surface (S1 cross S1) so if we want to sew something back in here, we need to find something whose boundary is S1 cross S1. There are two options (1) S1 cross D2. In this case we are sewing back in exactly the solid torus that we removed and we get back the original manifold. OR (2) D2 cross S1 also has the same boundary, and we can sew this back in instead along the same surface and get a new manifold! The key here is to realize that for a torus surface, the two S1’s (longitude and meridian) are on equal footing, but when we think about a solid torus, one S1 has been filled in and the other has not. Choosing a torus in a three manifold and switching which S1 is filled in is a surgery on the 3-manifold, known as Dehn surgery. Don’t try too hard to imagine it in detail, because you can’t put the result easily into our three dimensional space, and that tends to make your head explode if you think too hard about it.

Now to make life more interesting still, the torus we do surgery on may not be trivially embedded into our three manifold. In other words, when we started by drawing a closed circle in the manifold, it need not be a simple open circle, but could be knotted circle, like a trefoil. Further, we can give the original circle a “framing” meaning that the torus is embedded with a twist around its meridian before we do the surgery.

A theorem that strikes me as rather remarkable is that any closed three manifold can be obtained, starting from S3 (or any other simple three manifold) and doing successive Dehn surgeries of this type. This is known as the Lickorish-Wallace theorem and dates from the early 1960s. A relatively simple proof is provided by Rourke

J. London Math. Soc. (2) 31 (1985), no. 2, 373–376. (Sadly this probably requires a library subscription to have access – Project Euclid, has not yet put this on its list of freely available math journals)

“relatively simple” means that it is simple enough that more or less I can follow it and it doesn’t make my head explode. I found it rather entertaining that Witten, in his famous Field's medal winning paper on the Jones polynomial, refers to this theorem as “a not too deep result”. I suppose if you are Ed Witten, or a topologist, maybe that is true.

So you can describe any three manifold as successive Dehn surgeries starting with any other three manifold. However, it is possible to have more than one Dehn surgery give you the same result. Nonetheless, it is known exactly which combination of Dehn surgeries are equivalent to which other combination. The identification is given by some very very simple rules known as Kirby Calculus.

Added: Throughout, I'm assuming an orientable manifold.

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## 2 comments:

My rather naive understanding is that embedding does not affect the fundamental nature of the object. I would even say that it's not theembedding that's been putting S1 x D2 back into S1 x S1. Silly verbal exercises aside, I wanted to ask, whether the embedding really adds any serious complications (to the proof of the above theorem, say) beyond a few technical details, -- does it serve the same purpose as an atlas on a differentiable manifold -- change the charts and all you've done is skinned the same old cat in a different fashion. I could have, of course, mixed my tangerines with my quinces just now, but if you set me straight, I'd really appreciate it.

Stated another way, a torus is a torus no matter how you slice it, although in light of your post this statement is no longer on terra firma.

And finally, I wonder whether any topologists have been sued for malpractice after it was discovered that they'd left an empty S2 inside of the T3 after a Dehn surgery.

Back to tending my shrubberies.

I like the joke in the last line -- I'll probably use it.

If you start with S3 (R3 closed up at infinity) and you perform Dehn surgury on the standard torus (0-framed, or trivially framed) you actually get S2 cross S1 not the original S3. This is not too hard to argue. When you remove the solid torus from S3, you are left with something that has a handle -- ie, you can draw noncontractable loops in the remaining part of the manifold. Now when you replace the solid torus in the opposite way, you did not replace it in such a way that you can contract this loop. So you still have a noncontractable loop indicating a handle -- this is the S1 in the final result.

To see this more clearly, keep track of what sort of lines within the boundary of the torus can be contracted to zero. In the torus surface, S1 x S1 you can't contract either the meridian or the longitude. If you fill in one of the S1s to get D2 x S1 you can only contract one of the S1s.

Takes a bit of thought, but I'm pretty sure you get S2 x S1.

Of course you can also do Dehn surgury on a nontrivial torus (like a torus tied in a trefoil) or on a torus with nontrivial framing (twists around its own meridian). These give other results.

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