Let $S_{g,1}$ be the surface of genus $g \geq 1$ and $1$ boundary component. Let $Mod(S_{g,1})$ be the mapping class group in which we allow isotopies to rotate the action on the boundary (equivalently think of it as the mapping class group of the once-punctured surface of genus $g$).

Is every element of $Mod(S_{g,1})$ a composition of right-handed Dehn twists?

Note that this is true for $S_{g,0}$ as stated in page 124 of A primer on Mapping Class Groups by Farb and Margalit under the name of "a strange fact".

Edit: I am going to comment ThiKu's answer to avoid further confusion.

In A. Wand: Factorisation of Surface Diffeomorphisms and in Baker, Etnyre and Van Horn-Morris: Cabling, Contact Structures and and Mapping Class Monoids the authors, independently, provide with examples of diffeomorphisms in $Veer(\Sigma_{2,1},\partial \Sigma_{2,1})$ which are not in $Dehn^+(\Sigma_{2,1}, \partial \Sigma_{2,1})$. That is, right-veering diffeomorphisms which are not a product of right-handed Dehn twists. However, the mapping class group in which these results hold is $Mod( \Sigma_{2,1}, \partial \Sigma_{2,1})$, that is, the mapping class group of automorphisms fixing the boundary and isotopies fixing the boundary as well. This, a priori, does not yield counter-examples to my question (unless it does together with some other result that I do not know).

Observe that for all $g \geq 1$ there is a central extension

$$1 \to \mathbb{Z} \to Mod(\Sigma_{g,1}, \partial \Sigma_{g,1}) \to Mod( \Sigma_{g,1}) \to 1 $$

which is not split in general.


No. A. Wand: Factorisation of Surface Diffeomorphisms discusses conditions when a mapping class is a product of right-handed Dehn twists. Theorem 5.2 of that paper gives an explicit counterexample to your question.

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  • $\begingroup$ I deleted my previous comment because now I think these are not counter examples. Note that I asked for the mapping class group free on the boundary. It could happen that the image of Wands examples in $Mod(S_{g,1})$ are a composition of right handed Dehn twists. And that, by taking this composition in $Mod(S_{g,1}, \partial S_{g,1})$ you get the original automorphism plus some right handed Dehn twists around the boundary parallel curve. $\endgroup$ – Paul Oct 29 '18 at 14:39
  • $\begingroup$ Actually the identity is a (non-empty) composition of right Dehn twists in the mapping class group free on the boundary but not in the mapping class group relative on the boundary $\endgroup$ – Paul Oct 29 '18 at 14:45
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    $\begingroup$ @Paul: I think you've answered you're own question. If the identity is a product of right Dehn twists, then the whole mapping class group is. Take the product of right twists, set it $=1$, and move one twist to the right hand side of the equation. This expresses the inverse of right-hand twist (a left-hand twist) as a product of right-hand twists. Since the set of all Dehn twists generate, you can generate the mapping class group with right-hand twists. This is stated in the "strange fact" on p. 124 of Farb-Margalit. $\endgroup$ – Ian Agol Oct 29 '18 at 19:33
  • $\begingroup$ I had in mind a particular example for g=1 that's why I did not answer... but now I think that, since all genus and boundary components combinations appear as milnor fibers of brieskorn-pham singularities on $C^2$, the result is true in greater generality. I am going to think a little bit about it and if it is not true, at least I will say that it is true for some genus. $\endgroup$ – Paul Oct 29 '18 at 19:51

Yes. As pointed out by Ian Agol, I actually answered my question.

I observed that the identity is a non-empty composition of right handed Dehn twists in $Mod(S_{g,1})$. A priori this is not trivial. I was thinking about monodromies on Brieskorn-Pham singularities $(x^p+y^q)$ which are freely periodic and a composition of right-handed Dehn twists (by morsifying the singularity). One can easily see that this solves the problem in the cases that I was asking originally since all surfaces $S_{g,1}$ appear as Milnor fibers of such singularities for all $g$.

EDIT: I removed the last part of the answer since it was not true in that generality and this answers completely what I wanted.

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