MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S_{g,1}$ be a orientable compact surface of genus $g$ with one boundary component and $\Gamma_{g,1}$ the mapping class group. By $F_n$ I denote the free group on $n$ generators.

One obtains a representation $\rho: \Gamma_{g,1} \rightarrow Aut(F_{2g})$.

What is the kernel of $\rho$?

share|cite|improve this question
To be clear, you are placing a base point on the boundary of $S_{g,1}$. Otherwise, you only get a representation into $\mathrm{Out}(F)$. – HJRW Jan 30 '12 at 13:30
@HW and lsw: Could you clarify what is $\rho$ and why it depends on where the basepoint is (as long as it is fixed)? – Mark Sapir Jan 30 '12 at 15:07
@HW: Thank you for making this precise. @Mark Sapir: You have to consider the induced action on the fundamental group of $S_{g,1}$. By fixing a base-point there is no $Inn(\pi_1)$-action. – lsw Jan 30 '12 at 15:16
@Isw: Why is the kernel non-trivial? – Mark Sapir Jan 30 '12 at 16:24
@Mark Sapir: I don't know. Why is it trivial? – lsw Jan 30 '12 at 16:39
up vote 5 down vote accepted

The representation is faithful, since a mapping class is determined by its action on the fundamental group of the surface. A surface is a $K(\pi,1)$, so given any element $Aut(S_{g,1})$, one obtains a (pointed) map $\varphi:S_{g,1}\to S_{g,1}$ which is unique up to homotopy. Now one needs to know that two homotopic homeomorphisms of a surface are isotopic, which is classic (at least one may find this in a paper of Waldhausen). In fact, one may identify the image in $Aut(F_{2g})$ as the subgroup preserving the peripheral element. Also, note that everything should be fixing a basepoint in the boundary, as in HW's comment.

share|cite|improve this answer
Thank you, this is a very nice answer. – lsw Jan 30 '12 at 19:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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