Surface automorphisms and conformal automorphisms

Question

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.

Answer 1

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.

Answer 2

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.