3
$\begingroup$

It is well-known that in the picture below we have $t_d=(t_at_b)^6$ as elements in the mapping class group of a two-holed torus, ($t_\gamma$ represents positive Dehn twist about the curve $\gamma$). Is it possible to get another factorization of $t_d$ in terms of the curves $a$,$b$,$c$ and $e$ below?

enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ Does "factorization" mean "expression for" or does it mean "root of"? $\endgroup$
    – Sam Nead
    Feb 27, 2016 at 11:01

1 Answer 1

Reset to default
2
$\begingroup$

Let's define the mapping class group to be homeomorphisms of the surface, up to isotopy. Thus a Dehn twist about a boundary component is isotopic to the identity map. Now the twist about $d$ can be expressed in terms of those about $c$, $e$, and $a$, using the lantern relation.

Improve this answer
$\endgroup$
4
  • $\begingroup$ How does the boundary parallel dehn twists vanish in your setting? I meant a relation other than the one coming from lantern. for intance for $e$ we have:$e=b^{-1}a^{-2}cbc^{-1}a^2b$. I'm looking for sth similar for the curve $d$. $\endgroup$
    – nikita
    Feb 27, 2016 at 14:35
  • $\begingroup$ 1) They are isotopic to the identity. 2) I don't see why you are ruling out the lantern relation. Perhaps you could add some details to your question, so I can understand what you are asking? 3) The curves $a$, $b$, $c$, and $e$ are all non-separating, so yes, the Dehn twists on those curves are conjugate to each other. The curve $d$ is separating, so it cannot be conjugate to any of the others. $\endgroup$
    – Sam Nead
    Feb 27, 2016 at 15:30
  • $\begingroup$ this is the problem I am thinking about: the problem is comparing two contact structures on the Brieskorn manifold (2,3,11). One induced by the Milnor fiber and the other one by a plumbing tree. According to a result of Ghiginni-Shoenberger:arxiv.org/abs/math/0201099 the tight contact structures on brieskorn manifold (2,3,11) are realized by two possible Legendrian realizations of the -3-knot at the tail of the plumbing. I want to check that open book from the Milnor fiber with monodromy=(ab)^11 is equivalent to the open book obtained by "rolling up" the plumbing diagram. $\endgroup$
    – nikita
    Feb 27, 2016 at 16:44
  • $\begingroup$ Here the rolling up method is explained: arxiv.org/abs/math/0607344. The problem is I could not get the from $\psi=(ab)^{11}=(ab)^5 d$ to the open book obtained by rolling up the diagram, which does not involve the curve $d$. $\endgroup$
    – nikita
    Feb 27, 2016 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.