8
$\begingroup$

Let $G$ be the mapping class group of a surface of genus $g > 1$. Is it known for which positive integer $k$ one can find a subgroup $H$ of $G$ generated by a finite number of Dehn twists and a Dehn twist $\tau \in G$ which is not in $H$ but such that $\tau^k$ is in $H$, $k$ being minimal for this property ?

Using the chain relation, one can construct examples with $k = 2$, but I cannot find any examples with higher powers.

EDIT: I realized I should put a little motivation to this question. So let us take $U_d$ the space of smooth degree $d$ planar complex curves. There is a monodromy map $\pi_1(U_d)\rightarrow G$, and the image of this map is generated by Dehn twists obtained for example by taking a generic pencil of curves. Now I can construct a loop in $U_d$ whose monodromy is a power of some Dehn twist and I am wondering if there should be a reason coming from the structure of $G$ for the Dehn twist itself to be in the image of the monodromy.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Edited:

The paper Dehn twists have roots proves that Dehn twists have roots, naturally. The limits on that construction can be found in the paper Roots of Dehn twists; they bound the degree of the root (linearly, as I recall) in terms of the complexity of the surface.

Now, if I understand correctly, you are asking for ways to realize $\tau_\alpha^p$, a power of a Dehn twist about $\alpha$, as a product of other twists. If we found a root of $\tau_\alpha^p$, which was not itself a twist, then that would answer your question. But I think that the main idea behind the above papers will work. For: any root of $\tau_\alpha^p$ has the same canonical reduction system, namely $\alpha$. So the root of $\tau_\alpha^p$ stabilizes $\alpha$. You can now build a periodic map in the complement of $\alpha$ that does a fractional twist about $\alpha$: for example $p/q$ fraction of a right twist. Take the $q$-power of this get the desired $\tau_\alpha^p$.

Note that it is far from clear that these constructions are the only way to get a subgroup $H$ of the type you desire.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks a lot for your answer! Those are very interesting results. However, I am not sure they answer my question as I would like $\tau$ to be a Dehn twist itself. In the papers you mention only $\tau^k$ is a Dehn twist. $\endgroup$
    – Rémi Cr.
    Commented Jan 20, 2016 at 8:41
  • $\begingroup$ Bibliographical note: Monden independently discovered results similar to McCullough-Rajeevsarathy in arxiv.org/abs/0911.5079 $\endgroup$
    – Daniel Moskovich
    Commented Jan 20, 2016 at 9:46
  • $\begingroup$ Dear Rémi - Ah, yes - I see your point. I'll edit my answer. But the papers I referenced are still what you want - those techniques will give you examples of the kind you want. $\endgroup$
    – Sam Nead
    Commented Jan 20, 2016 at 20:06
  • $\begingroup$ Oh, I see now! Thanks a lot, I will think about it more. $\endgroup$
    – Rémi Cr.
    Commented Jan 21, 2016 at 12:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .