Timeline for Is $\mathrm{Diff}_0(S_g)$ torsion-free?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2016 at 21:25 | comment | added | Tom Church | I wasn't thinking about the proof, but I believe you have it right. | |
| Oct 7, 2016 at 21:06 | comment | added | Andy Putman | 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. | |
| Oct 7, 2016 at 21:04 | comment | added | Andy Putman | 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. | |
| Oct 7, 2016 at 19:55 | history | answered | Tom Church | CC BY-SA 3.0 |