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.



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


