On page 201 of Farb and Margalit's Primer on Mapping Class Groups, they explain why the mapping class group $\mathrm{Mod}(S)$ is torsion-free when $\partial S \neq \varnothing$. Here is my understanding of the argument:

Let $S$ be a surface with a hyperbolic metric and let $\phi\colon S \to S$ be an isometry fixing $\partial S$ pointwise. Then $\phi$ fixes a frame (a basis for the tangent space) at every point of $\partial S$. Since isometries of surfaces are determined by their action on a frame, we must have that $\phi = \mathrm{id}$.

The Nielsen realization theorem (Theorem 7.1) states that for every order $k < \infty$ element $f \in \mathrm{Mod}(S)$, there is an isometry $\phi \in \mathrm{Homeo}^+(S)$ of order $k$ representing $f$. However, there is no guarantee that $\phi$ will fix the boundary, we only know that $\phi$ is in the free homotopy class $f$. Up to here I understand all the points that have been made.

What I don't understand is how they go from this, and the fact that Dehn twists about boundary components have infinite order, to conclude that $\mathrm{Mod}(S)$ is torsion-free whenever $\partial S \neq 0$. Is the point that if $f$ is a torsion element, then the isometry representative $\phi$ given by the NIelsen realization theorem must fix the boundary pointwise and therefore be the identity? I don't know how you would show this. I also don't understand why Dehn twists entered the argument.


I think the reason that Dehn twists enter is that we can take the differential of a (orientation-preserving) diffeomorphism $f$ of $S_g$ that fixes a chosen basepoint $\ast \in S_g$ at this point, and will get a map $d \colon \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$. The fiber of this fibration is then $\text{Diff}_{\partial}(S_{g,1})$. When you study the effect of the resulting fiber sequence $\text{Diff}_{\partial}(S_{g,1}) \to \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$ on homotopy groups, you obtain a short exact sequence $1 \to \mathbf{Z} \to \Gamma_{g,1} \to \Gamma_g^1 \to 1$ where $\Gamma_g^1$ stands for the mapping class group of a once-punctured surface of genus $g$, and the map $\mathbf{Z} \to \Gamma_{g,1}$ is given by Dehn twisting around the boundary curve.

Ok, so far so good, but now let us prove that $\Gamma_{g,1}$ is torsion-free, by contradiction: suppose $f \in \Gamma_{g,1}$ has finite order $k > 1$. Then (as you remarked), we can find $\phi \in \text{Diff}(S_g,\ast)$ of order $k$ representing $f$. We can also find a compatible metric on $S_g$. Then $d(f)$ actually lands in $SO(2) \subset \text{GL}_2^+(\mathbb R)$ and is given by a rotation, say around angle $\theta$. Then $k\theta = \ell2\pi$ for some non-zero integer $\ell$, and you can convince yourself that $f^k$ represents a non-trivial Dehn twist in $\Gamma_{g,1}$, contradicting $f^k = 1$.

  • $\begingroup$ Surely $k \theta$ is an integer multiple of $2 \pi$, not necessarily $2\pi$ itself, and you get the Dehn twist if the integer multiple is nonzero? $\endgroup$
    – Will Sawin
    Mar 15 at 16:10
  • $\begingroup$ Yes, of course. Thanks for pointing at this; I have corrected for it. $\endgroup$ Mar 15 at 16:11
  • 1
    $\begingroup$ For completeness let me add an argument for why the rotation angle $\theta$ can not vanish. Consider the set of formal power series $ \{ a_1 z + a_2 z^2 + a_3z^3 + \ldots \mid a_i \in \mathbf C \text{ for all } i,\,\, a_1 \neq 0\}$, which form a group under composition. It is easy to see that any nontrivial element with $a_1=1$ has infinite order in this group. Indeed, if $f = z + a_kz^k + \ldots$ with $a_k\neq 0$, then $f^{\circ n} = z + na_kz^k + \ldots $ $\endgroup$ Mar 16 at 9:01
  • 1
    $\begingroup$ It follows in particular that any conformal map $F$ from a Riemann surface to itself which fixes both a point and the tangent space at that point, must be the identity or of infinite order: consider the germ of $F$ at the fixed point. (The infinite order case occurs only on the Riemann sphere, in which case $z \mapsto z+1$ fixes both the point $\infty$ and the tangent space at $\infty$.) $\endgroup$ Mar 16 at 9:01
  • $\begingroup$ By thickening up the tangent space, there is an injective map $\text{Diff}(M,T_xM) \to \text{Homeo}(M,D)$. So this statement can alternatively be deduced from the more general fact that $\text{Homeo}(M,D)$ is torsion-free that I once learned here: mathoverflow.net/questions/266311/is-homeom-dn-torsion-free $\endgroup$ Mar 16 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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