In Witten's “QFT and Jones Polynomials” paper, page 383, it states that: "It is a not too deep result that every 3-manifold can be obtained from or reduced to $S^3$ (or any other desired 3-manifold) by repeated surgeries on knots.

What are the methods to show this? In the simplest intuitive level?

I understand this may be a relevant post, but I hope there are better illuminations. Thanks.

p.s. I am a QM/QFT theorist trying to understand the topology better.

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    $\begingroup$ In response to the (currently) three votes to close: Yes, this is a very standard fact in 3-manifold topology, but it is certainly not obvious to someone outside the field. I think it is a perfectly fine MO question. (In case it is relevant, I myself am a low-dimensional topologist.) $\endgroup$ – Kevin Walker Mar 21 '15 at 22:17
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    $\begingroup$ I thought about voting to close, but I then realized that this is really a reference request. As such it is appropriate for MO... $\endgroup$ – Sam Nead Mar 21 '15 at 23:30

This is a result of Lickorish, in his paper "A representation of orientable combinatorial 3-manifolds". The paper is only eleven pages, and is very readable. In his proof, Lickorish rediscovers some ideas first investigated by Max Dehn 40 years earlier.

Lickorish's theorem was unexpected at the time; I don't think that Witten's remark is really justified...

Here is a very sketchy overview of Lickorish's proof.

The first step is to note that every closed, connected, orientable three-manifold $M$ has a Heegaard splitting. That is, you can get $M$ by gluing together a pair of handlebodies. (That this is true for combinatorial manifolds is an easy exercise. That all three-manifolds are combinatorial is a very difficult theorem.)

The second step is to prove that any orientation-preserving homeomorphism of a closed surface is isotopic to a product of Dehn twists. This requires proving that you can get from any (non-separating) curve to any other by a chain of curves, with each pair meeting exactly once.

The third step is to relate Dehn twists to a special case of Dehn surgeries - this is probably the most "three-dimensional" part of the proof.

Since the three-sphere has a Heegaard splitting in every genus, the theorem will follow.

Other references include Rolfsen's book (I think) and the book by Prasolov and Sossinsky (definitely).

  • $\begingroup$ Why do you attribute this theorem to Lickorish, instead of Dehn? $\endgroup$ – Igor Rivin Mar 22 '15 at 0:28
  • $\begingroup$ The best source to read the proof is probably Lickorish's textbook "Knot theory" which has a Chapter on mapping class groups with the proof that the mapping class group is generated by Dehn twists. (Which was the new result that Lickorish proved at the time.) $\endgroup$ – ThiKu Mar 22 '15 at 2:19
  • $\begingroup$ It is chapter 12 in scribd.com/mobile/doc/183097631 $\endgroup$ – ThiKu Mar 22 '15 at 2:25
  • $\begingroup$ @Igor - I know of no evidence that Dehn realized the connection between Dehn twists on a surface and integral Dehn surgeries along knots in three-manifolds. Stillwell asserts that Dehn did not have this idea - see the first paragraph of his introduction to his translation of Dehn's paper "Die Gruppe der Abbildungsklassen". $\endgroup$ – Sam Nead Mar 22 '15 at 13:34
  • $\begingroup$ @Igor - Dehn did prove that Dehn twists generate the mapping class group - but that is just step two of the outline given above. $\endgroup$ – Sam Nead Mar 22 '15 at 13:35

The question of which closed, orientable 3-manifolds can be obtained as surgery along a link in $S^3$ appears at the very end of Bing's paper Necessary and Sufficient Conditions that a 3-Manifold Be $S^3$.

The affirmative answer to Bing's question is generally attributed to Lickorish and Wallace. Both worked independently and also used different methods. A treatment of Lickorish's work is given in Sam Nead's answer above, so this answer is meant to highlight Andrew Wallace's paper, Modifications and Cobounding Manifolds.

Relevant to this discussion, Section 5 of that paper culminates in Theorem 6. Wallace's perspective is to consider cobordisms between two manifolds. Restricting to the case of 3-manifolds he is able to show that any closed orientable 3-manifold can be obtained from repeated used of a specific type of cobordism that can be understood as a Dehn surgery on one component of a link in $S^3$.


This is really an extended comment. Firstly, as pointed out by Neil Hoffman, the result was proved by Wallace two years before Lickorish, and Lickorish was well aware of this (Wallace's paper is cited in Lickorish's). I really can't understand why Lickorish's name is attached to this result at all. Lickorish's reasonable claim was that his argument was more elementary than Wallace's. Indeed, this is true, since Lickorish's entire paper is eight pages (Wallace's is 28), BUT Lickorish (as suggested in my comment re Sam Nead's answer) does NOT prove that a finite number of Dehn twists suffice (that result is proved in a 1964 paper - two years after, and if you look at the math review, the reviewer is not so sure that it is really proved). Of course, had Lickorish been aware of Dehn's work, his 1962 paper would be down to around two pages. So, another (somewhat complicated) example of Arnold's principle at work.

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    $\begingroup$ I think Lickorish's attachment to the theorem might come from Rolfsen's "Knots and Links." Chapter 9.I is titled "The fundamental theorem of Lickorish and Wallace" and discusses this very topic. $\endgroup$ – Neil Hoffman Mar 23 '15 at 3:00

Math Reviews says of the paper "A type of homology-preserving surgery on 3-manifolds", J. London Math. Soc. 17 (1978), 183–185 by Norris Weaver:

"Let $M$ be a 3-manifold and let $T \subset {\rm int}\,M$ be a tame submanifold homeomorphic to $S^1\times D^2$. Let $g: \partial T \to \partial T$ be an orientation-preserving homeomorphism, and let $n$ be an integer. Glue $T$ or $M - {\rm int}\, T$ by $g^n$ to obtain a manifold $M_1$. Call this construction an $n$th power surgery. The following result is obtained. Theorem: If $n\equiv 0 \, ({\rm mod}\,6)$ and $0\leq n\leq 36$, then $H^1(M_1:\mathbb{Z}_n) \cong H^1(M:\mathbb{Z}_n)$. If $n \not\equiv 0\,({\rm mod}\,6)$ then every closed, orientable 3-manifold can be obtained from $S^3$ by successively performing a finite number of $n$th power surgeries."

  • $\begingroup$ A relative of yours? $\endgroup$ – Igor Rivin Mar 23 '15 at 2:30
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    $\begingroup$ Yes, this was my father. $\endgroup$ – Nik Weaver Mar 23 '15 at 3:49

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