# Is $\mathrm{Diff}_0(S_g)$ torsion-free?

Let $S_g$ be a closed oriented smooth surface of genus $g>1$, and let us consider $\text{Diff}_0(S_g)$, the identity component of the diffeomorphism group of orientation preserving diffeomorphisms of $S_g$.

Is this a torsion-free group?

Sorry if this question is too elementary for experts in low-dimensional topology, but even after searching in the literature quite thourougly, I could not find any reference addressing the question.

• What if you take a surface $S_g \subset \mathbb{R}^3$ which is symmetric with respect to the $z$-axis, and consider a rotation of $180$ degrees with respect to this axis? This should be an orientation-preserving diffeomorphism of order $2$, right? – Francesco Polizzi Sep 16 '16 at 14:54
• @FrancescoPolizzi: The question is whether this rotation is isotopic to the identity. My feeling (and at the moment it is just a feeling) is that if you had a torsion element in $Diff_0(S_g)$ you could use it to produce a smooth effective action of a circle on $S_g$, and that such an action may not exist. – Mark Grant Sep 16 '16 at 15:00
• Yes, this group is torsion free. The same for homeomorphisms. Hint: Lift to a periodic homeomorphism of the hyperbolic plane and extend to a homeomorphism of the sphere fixing an open disk pointwise. – Misha Sep 16 '16 at 15:07
• @JensReinhold: That's right: The lifting always exists and it always extends to a homeomorphism of the closure of the Poincare disk. But since the action on $\pi_1$ is trivial, the extended action is by the identity on the boundary circle. Now you use either Smith' theory or a theorem by Newmann (?) to conclude that the periodic homeomorphism is the identity map. – Misha Sep 17 '16 at 2:49
• @HJRW: I just did. – Misha Sep 17 '16 at 15:54

Here is a proof that $Homeo_0(S)$ is torsion-free for every compact hyperbolic surface $S$. With more analytic assumptions on homeomorphisms one can get the same conclusion for noncompact hyperbolic surfaces.

1. Every element $f\in Homeo_0(S)$ has nonzero Lefschetz number (here we use hyperbolicity of $S$) and, thus, has a fixed point in $S$. I will identify $\pi_1(S,x)$ with the group $\Gamma$ of covering transformations of the universal cover $H^2\to S$, where $H^2$ is the hyperbolic plane and $\Gamma$ acts isometrically.

2. Every periodic element $f\in Homeo_0(S)$ lifts to a periodic homeomorphism $F$ of the hyperbolic plane $H^2$ which commutes with the group of covering transformations $\Gamma\cong \pi_1(S)$. (You have to a bit careful which lift to choose, it will be the one which fixes a lift of $x$ used to identify $\pi_1(S,x)$ and $\Gamma$.) The existence of a fixed point is critical here since for the torus $T^2$ there are nontrivial periodic elements of $Homeo_0(T^2)$ and they cannot lift to periodic homeomorphisms of the Euclidean plane.

3. Next, $F$ extends to a homeomorphism of the closed disk compactification of $H^2$ which is the identity on the boundary circle. (Here we use that $F$ commutes with every element of $G$.) Now, extend $F$ by the identity to the rest of the 2-dimensional sphere. (I identify $H^2$ with the Poincare disk) The result is a periodic homeomorphism $h: S^2\to S^2$ whose fixed point set has nonempty interior. There is an old (pre 1940) theorem in surface topology that such $h$ has to be the identity. I forgot who proved it. But a bit more modern proof is via Smith Theory. Use a suitable power of $h$ to reduce the question to the case where $h^p=id$, where $p$ is prime. Then Smith proved that the fixed point set of $h$ is a $Z_p$-homology manifold and $Z_p$-homology sphere of some dimension; the result holds in all dimensions, not just 2. (This is covered in Bredon's "Compact Transformation Groups", I think.) Since the fixed point set of $h$ has nonempty interior and is a $Z_p$-homology manifold, it has to be the entire $S^2$-sphere.

• Newman, "A theorem on periodic transformations of spaces" in 1930 (which was then generalized by Smith). – Chris Gerig Sep 18 '16 at 7:09

This theorem was proved by Hurwitz in the 19th century, who in fact showed the stronger theorem (also mentioned in Danny Ruberman's answer) that any finite-order diffeomorphism of a surface of genus at least $2$ acts nontrivially on homology. This is proved in the same paper where Hurwitz proved the more famous Riemann-Hurwitz formula. I give an exposition of what is essentially Hurwitz's original proof in my note "The action on homology of finite groups of automorphisms of surfaces and graphs", which is available on my webpage here. This proof is not as efficient as the proof using the Lefshetz fixed-point theorem that Danny mentions, but it is very direct and elementary (basically only using the definition of homology).

There's another proof of this that is well worth knowing, using the Lefschetz fixed point theorem (for the surface, not the hyperbolic plane as above). It's apparently due to Serre, and is nicely explained in Farb-Margalit's Primer on Mapping Class Groups. The proof yields a stronger fact, namely that a finite order diffeomorphism of a surface of genus at least $2$ must act non-trivially on the first homology. Since elements of $\text{Diff}_0(S_g)$ act as the identity, the result follows.

One of the reasons that I like this proof is that it generalizes to (many) Kaehler surfaces, if one assumes that the action is holomorphic. (See for instance the book Compact Complex Surfaces, by Barth-Peters-Hulek-Van de Ven) It is an interesting question as to whether this works for smooth (ie not necessarily holomorphic) actions.

Just to expand on Danny and Andy's answers, in fact every finite-order diffeomorphism acts non-trivially on the mod-$m$ homology for any $m\geq 3$. This is how you prove that the level-$m$ congruence'' subgroup of the mapping class group is torsion-free for $m\geq 3$. (This is not true for the level-2 congruence subgroup, since it contains all the hyperelliptic involutions.)

• I'm not precisely sure which proof that $\text{Mod}_g(m)$ is torsion-free you're thinking of. The only proof I know of this first proves that every finite-order element acts nontrivially on $H_1(\Sigma_g;\mathbb{Q})$ (this is what my proof and Danny's proof show). This implies that Torelli is torsion-free. You then prove by a purely algebraic argument that for $m \geq 3$ the level $m$ subgroup of of the symplectic group is torsion-free (this is just taking powers of matrices). These two facts imply that $\text{Mod}_g(m)$ is torsion-free. – Andy Putman Oct 7 '16 at 21:04
• But maybe I'm just mis-reading you in saying that you can prove directly that every finite-order diffeomorphism acts nontrivially on mod-$m$ homology and then deduce that $\text{Mod}_g(m)$ is torsion-free. – Andy Putman Oct 7 '16 at 21:06
• I wasn't thinking about the proof, but I believe you have it right. – Tom Church Oct 7 '16 at 21:25