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One of the foundational results about mapping class groups of surfaces is that they are generated by Dehn twists. A mapping class is a connected component in a space of diffeomorphisms, so another way of stating this is "any diffeomorphism of a surface is generated by diffeomorphisms each of which is supported in an annulus". This was proven by Dehn, and independently in a stronger form by Lickorish.

I'm teaching a summer reading course, and I am toying with the idea of presenting a proof to this statement. But the proof I know is a bit involved- you use the Birman exact sequence to relate the mapping class group of a surface Σ to the mapping class group of $\Sigma-D^2$ (not so trivial), then you use the fact that the complex of curves on a surface is connected (also non-trivial), and finally that for two non-disjoint connected curves α and β there exists a product of Dehn twists T such that $T(\alpha)=\beta$. This proof looks too involved to present properly in a single lecture.

80 or so years after Dehn's proof, and 47 years after Lickorish's:

Do you know an elegant proof that the mapping class group of a surface is generated by Dehn twists?
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    $\begingroup$ You can avoid using the Birman exact sequence by proving that the arc complex (consisting of isotopy classes of arcs joining two fixed points on two boundary components) is connected -- the base case would then be a once-punctured torus. However, I rather doubt that you are going to find a proof that is significantly simpler than the one you describe, which strikes me as a pretty elegant proof. $\endgroup$
    – Andy Putman
    Commented May 19, 2011 at 15:35
  • $\begingroup$ By the way, the proof you describe above is related to Lickorish's proof, but is not identical to it. Indeed, Lickorish's paper predates both the Birman exact sequence (which was Birman's thesis and was published in 1969) and the curve complex (which was first defined by W. Harvey in the late '70's). I believe that the proof you describe is due to Ivanov in his 1998 survey on mapping class groups -- he describes it as a simplification of an argument of Birman. $\endgroup$
    – Andy Putman
    Commented May 19, 2011 at 21:10
  • $\begingroup$ That's interesting! I learnt this proof from Massuyeau's survey, and it's really the only one I have gone through in any detail. But I hope there are other ways- maybe something slicker and easier to teach. I'd feel enriched by any fundamentally different proof, in fact- important facts should have many findamentally different proofs. $\endgroup$
    – Daniel Moskovich
    Commented May 20, 2011 at 0:54
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    $\begingroup$ @Daniel : I don't know, how many proofs do you know that $SL_n(\mathbb{Z})$ is generated by elementary matrices? I think of the two facts as being analogous. Anyway, I'll give an answer with some extended comments below. $\endgroup$
    – Andy Putman
    Commented May 20, 2011 at 5:45
  • $\begingroup$ You need to say "orientation preserving" at the right moment... $\endgroup$
    – Sam Nead
    Commented Sep 17, 2013 at 4:17

2 Answers 2

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I'm pretty sure there doesn't exist a "slicker" proof of this fact in the literature. The proof you describe exists in many forms starting with Dehn and Lickorish -- as I said in a comment, the particular arrangement of it you gave (making use of the complex of curves) is basically due to Ivanov. The only fundamentally different approach I know of is in the paper

MR1425631 (98c:20061) McCool, James(3-TRNT) Generating the mapping class group (an algebraic approach). (English summary) Publ. Mat. 40 (1996), no. 2, 457–468.

This proof is purely algebraic (and rather more complicated than the topological proof). It is based on McCool's 1975 proof that the mapping class group is finitely presentable. By the way, McCool is often written out of the history of the subject, but his work predates Hatcher-Thurston and is the first paper in the literature that explicitly proves that the mapping class group is finitely presentable (though the algebraic geometers proved equivalent facts in the early '60's).

I tend to view the fact that the mapping class is generated by Dehn twists as equivalent to the fact that the complex of curves is connected. For this, there exist several alternate proofs. Many of these proofs also give higher connectivity; I'll try to indicate that as I go.

  1. There is the combinatorial approach taken in many sources (eg Farb-Margalit's "Primer on Mapping Class Groups"). I don't know who to attribute this to. You could probably prove that the curve complex is simply-connected by this method, but I doubt you could prove higher connectivity.

  2. You could use Teichmuller theory -- this is how Harer originally proved that the curve complex is highly connected.

  3. There is a Morse-theoretic proof due to Ivanov (it is described in his survey "Mapping Class Groups"). It is an adaption of the Cerf theory approach to proving the mapping class group is finitely presentable which is due to Hatcher and Thurston. In his paper "Complexes of curves and Teichmüller modular groups", Ivanov uses this approach to prove that the complex of curves is highly connected.

  4. You could easily deduce that the curve complex is connected from Hatcher's slick proof that the arc complex is contractible, which is located in his paper "On triangulations of surfaces". I don't think they have this written up, but Hatcher and Vogtmann have a proof that the curve complex is highly connected with this as the starting point.

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  • $\begingroup$ As far as how to present it in a single lecture, it's perfectly doable if you don't insist on giving every detail. Just give the big picture (assuming the connectivity of the complex of curves and the Birman exact sequence). This could easily be done in 30 minutes if the students already know what a Dehn twist is or an hour if they don't. If you have time remaining, sketch one of the two assumed facts. $\endgroup$
    – Andy Putman
    Commented May 20, 2011 at 5:40
  • $\begingroup$ @Andy: I believe the standard combinatorial proof of connectivity is due to ... Lickorish. Of course, as you said in a previous comment, his work predates Harvey's definition of the curve complex. But it's abundantly clear in hindsight that connectivity of the curve complex is exactly what he's proving. An analogous phenomenon: Schreier's normal form for 3-braids (published in 1924) sorts conjugacy classes into 5 families. In hindsight, it's clear that one family is reducible, three are periodic, and one is pseudo-Anosov. But this predates Thurston's work by a half-century! $\endgroup$
    – Dave Futer
    Commented May 24, 2011 at 21:05
  • $\begingroup$ @Dave : I've not read Dehn's original proof that the mapping class group is finitely generated in detail, but my impression is that the germ of the idea that the curve complex is connected is contained in there as well. In any case, I definitely agree that it's lurking in the background of Lickorish's paper. $\endgroup$
    – Andy Putman
    Commented May 24, 2011 at 21:57
  • $\begingroup$ @AndyPutman: Could you give a reference where the algebraic geometers proved the finite-presentability of mapping class group? Thanks. $\endgroup$
    – QcH
    Commented Jan 26, 2014 at 17:28
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    $\begingroup$ @QcH : Notice that I said they proved "equivalent" results; they did not explicitly prove that the mapping class group is finitely presentable. Here's what I meant. Ignoring stacky issues, the mapping class group is the fundamental group of the moduli space of curves. Quasiprojective varieties are homotopy equivalent to compact CW-complexes, and in particular have finitely presentable fundamental groups. The desired result thus follows from the fact that the moduli space of curves is quasiprojective, which was originally proven by Baily in 1960. $\endgroup$
    – Andy Putman
    Commented Jan 26, 2014 at 21:10
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Here is what springs to mind: break the proof into two lectures. The first lecture proves that Dehn twists generate the pure spherical braid group. The second lecture gives enough of Lickorish's proof to show that you can send any oriented non-separating curve to any other such. Finally induct on genus. The first lecture gives the base case. For the induction step appeal to Alexander's trick.

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