Given a closed orientable surface $S$ and a topological automorphism $\sigma$ of $S$, it is not in general possible to find a conformal structure $\Sigma$ on $S$ so that $\sigma$ is isotopic to a conformal automorphism of the Riemann surface $(S,\Sigma)$. For example by the theorem of Hurwitz that the conformal automorphism group is finite, while $\sigma$ on the other hand may be of infinite order in the mapping class group. But by a theorem of Colin Maclachlan in "Modulus space is simply connected", Proc. Amer. Math. Soc. 29 (1971), 85–86, every surface automorphism is isotopic to the composition of finitely many conformal automorphisms (for varying complex structures on $S$). For being isotopic to a conformal automorphism is equivalent to being isotopic to an topological automorphism of finite order (one direction by Hurwitz, the other by averaging a metric). Maclachlan proved that the mapping class group is generated by elements of finite order.

I am interested in the minimal number $m(\sigma)$ of conformal structures required, especially for the torus, where the mapping classes have a nice explicit description. Unfortunately when I tried to use this explicit description, it translated into some obscure number theory with a Diophantine flavor. I could not even show that for the torus in general arbitrarily many conformal structures would be needed, i.e. that $m(\sigma)$ is unbounded for $T^2$. This is my question. An upper bound for $m(\sigma)$ for $T^2$ in terms of the explicit description of $\sigma$ by a 2 by 2 integer matrix would also be interesting. Perhaps the higher genus case could be worth looking at after the torus case. I got stuck on $T^2$ and gave up quite a long time ago. But now that Math Overflow is here, I can ask this as a question.

CLARIFICATION: For the case of a topological torus I am concretely asking how many conformal automorphisms (relative to various complex structures on the topological torus) I need to compose together to have enough freedom to represent a topological automorphism up to isotopy. I tend to agree with Sam Nead that for every positive integer $K$ there will be some topological automorphism $\sigma$ that will require at least $K$ conformal automorphisms in order to be so represented. But I don't know how to prove this, though Sam Nead's comment on proving it seems reasonable.

  • 1
    $\begingroup$ @Marius - Your question is not clearly stated, as evidenced by the fact that Andy and Scott are telling you that various mapping class groups are generated by torsion while I have read something very different (probably due to you mentioning "Diophantine"). Would you care to clarify your question? $\endgroup$
    – Sam Nead
    Feb 24, 2011 at 17:38
  • $\begingroup$ So, in conclusion, I think the question is `Do the torsion elements of SL(2,Z) boundedly generate?' Quite a lot of work has been done on bounded generation, I believe. $\endgroup$
    – HJRW
    Feb 24, 2011 at 19:09
  • $\begingroup$ The answer is "no". This follows from work of Bestvina and Fujiwara. See below. $\endgroup$
    – Sam Nead
    Feb 24, 2011 at 19:30
  • $\begingroup$ Your question is also posed in the final paragraph of Brendle and Farb's paper "Every mapping class group is generated by 6 involutions" and is explicitly answered in papers of Korkmaz and also Kotschick -- see the second footnote of [BF]. $\endgroup$
    – Sam Nead
    Feb 24, 2011 at 19:55

2 Answers 2


Complete edit, after talking to a colleague -

Suppose that $G$ is the mapping class group of a surface $S$. Then you are asking:

Is there a number $K$, depending only on $S$, with the following property? For every $\sigma \in G$ there are torsion elements $\tau_i \in G$ so that $\sigma = \Pi_{i = 1}^K \tau_i$.

As Henry Wilton puts it - you are asking if the mapping class group is boundedly generated by torsion. The answer is "No". This follows from a paper of Bestvina and Fujiwara "Bounded cohomology of subgroups of mapping class groups". They show that the group $G$ admits unbounded quasi-homomorphisms. See the first five pages of their paper.

Edit - to give a few details. A $D$-quasi-homomorphism is a map $\phi$ from $G$ to the reals so that for all $g,f \in G$ we have $|\phi(gf) - \phi(g) - \phi(f)| < D$. It is an exercise to show that if $g$ is torsion then $|\phi(g)| < D$. Thus, if $G$ was boundedly generated, say with constant $K$, then we would have, for all $g \in G$, that $|\phi(g)| < 2KD$. This is a contradiction.

  • $\begingroup$ Yes, this is the tangible version of my question for the case of torus. I agree with you that the answer is likely to be "no". $\endgroup$ Feb 24, 2011 at 17:47
  • $\begingroup$ Remark - Marius is referring to the previous version of this answer. $\endgroup$
    – Sam Nead
    Feb 25, 2011 at 20:21

It was proven in

J. MacCarthy and A. Papadopoulos. Involutions in surface mapping class groups. Enseign. Math. 33 (1987), 275–290.

that the mapping class group is generated by the conjugates of a single involution (for $g$ large, I think at least $2$). Call this involution $\tau$ and let $S$ be the Riemann surface that $\tau$ acts on. Then the conjugate $\tau^f$ by $f$ in the mapping class group acts on the Riemann surface $f_{\ast} (S)$, which is isomorphic to $S$. In other words, up to isomorphism you only need one conformal structure.

  • $\begingroup$ @Andy - I think you misunderstood my question. Unfortunately I did not formulate the question precisely enough, so that there are now several different answers. It looks like the problem with your answer, from my point of view, is that the conjugating map $f$ may not be of finite order. If it were, your answer probably would give $m(\sigma) \leq 3$, though I am not quite sure of this. $\endgroup$ Feb 24, 2011 at 17:46
  • $\begingroup$ @Sam -- Scott and I gave answers to a reasonable interpretation of the question. It isn't fair to downvote us for not reading the OP's mind correctly. $\endgroup$ Feb 24, 2011 at 17:52
  • $\begingroup$ @Andy - I've removed the downvote and I apologize for doing so. A vote, up or down, is a single bit and thus is a limited form of communication. Let me instead ask you directly: why is your answer on-point, given Marius' edit of his post and then his comment just above? $\endgroup$
    – Sam Nead
    Feb 25, 2011 at 20:20
  • $\begingroup$ @Marius : It doesn't matter whether $f$ is of finite order or not. If $S$ is a Riemann surface and $\phi : S' \rightarrow S$ is a diffeomorphism, then we can pull back the Riemann surface structure on $S$ to get a Riemann surface structure on $S'$. With this Riemann surface structure, $\phi : S' \rightarrow S$ will be a biholomorphism. Moreover, if $\tau : S \rightarrow S$ is a biholomorphic map, then $\tau \circ \phi \circ \tau^{-1} : S' \rightarrow S'$ will be a biholomorphic map. I'm just applying this fact when $S' = S$. $\endgroup$ Feb 25, 2011 at 20:34
  • $\begingroup$ (continued) The following might make this less confusing. In my answer, $f : S \rightarrow S$ is a biholomorphic map between two different Riemann surfaces on the same topological surface $S$ (they are isomorphic Riemann surfaces, but the surfaces are "wearing" their Riemann surfaces structures on different ways). This is exactly what happens when the whole mapping class group acts on Teichmuller space. Two points in the same mapping class group orbit on Teichmuller space are isomorphic Riemann surfaces, but they are not the same Riemann surface! $\endgroup$ Feb 25, 2011 at 20:37

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.