Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S_g$ be a closed surface of genus $g\geq 2$. Given $r \in \mathbb{N}$, what is the number of elements of order $r$ in the mapping class group? Is it finite or infinite? If it is infinite is there any way to generate such a class? If it is finite is there an upper bound on the number? Any reference of link will be extremely helpful. Thanks in advance.

share|improve this question

migrated from mathoverflow.net Feb 6 '14 at 22:16

This question came from our site for professional mathematicians.

The number should always be $\infty$ or $0$. Think about a 180-degree rotation: this provides one order-2 element. But all of its conjugates (i.e. mapping classes which look in some coordinates like a 180-degree rotation) will also have order 2. Except for the hyperelliptic involution in genus 2, I believe every finite-order element of the mapping class group should have infinitely many conjugates (hence the claim, since there are other involutions in genus 2). Depending on your application, a better question might be "How many conjugacy classes are there of elements of order $r$?" –  Tom Church Feb 6 '14 at 15:10
Centralizers in these groups are almost always (infinite) cyclic, no? That should immediately give infinitely many conjugates. –  Steve D Feb 6 '14 at 17:27
I think you should revise your question to ask for conjugacy classes of finite-order elements, of which there are finitely many, and they are enumerable. Otherwise, I would vote to close the question or shift it over to math.stackexchange –  Agol Feb 6 '14 at 17:28
Please see, for example, "A primer on mapping class groups", by Farb and Margalit. I've voted to close, by moving the question. –  Sam Nead Feb 6 '14 at 17:49

Your Answer


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