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

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.

• 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. – Paul Oct 29 '18 at 14:39
• @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. – Ian Agol Oct 29 '18 at 19:33
• 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. – Paul Oct 29 '18 at 19:51
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 $$g$$.