# Why is the mapping class group of a surface with nonempty boundary torsion-free?

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 $$f^k = 1$$.
• 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? Mar 15 at 16:10
• 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 $f^{\circ n} = z + na_kz^k + \ldots$ Mar 16 at 9:01
• 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 $\infty$.) Mar 16 at 9:01
• 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 Mar 16 at 9:23