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.


1 Answer 1


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, 2021 at 16:10
  • $\begingroup$ Yes, of course. Thanks for pointing at this; I have corrected for it. $\endgroup$ Mar 15, 2021 at 16:11
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    $\begingroup$ 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, 2021 at 9:01
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    $\begingroup$ 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, 2021 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, 2021 at 9:23

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