Timeline for What is the minimal genus of a surface acted on by the symmetric group $S_n$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 30 at 9:48 | vote | accept | André Henriques | ||
| Jul 23 at 14:12 | comment | added | YCor | About existence: $\pi_1(\Sigma_2)$ ($\Sigma_g$ closed surface of genus $g$) has a $S_n$ quotient, which yields a Galois covering of genus $1+n!$ with free action of $S_n$. | |
| Jul 23 at 13:06 | comment | added | André Henriques | @DavidESpeyer Your last comment is a complete answer to my question. | |
| Jul 23 at 12:06 | comment | added | André Henriques | @DavidESpeyer My intention was to allow anti-holomorphic maps, but I realise that I might have been ambiguous. | |
| Jul 23 at 12:02 | comment | added | David E Speyer | I just realized -- do you mean a holomorphic action, or do you also allow anti-holomorphic maps? Because I think that Condor's paper, which I cited below, should show that $S_n$ acts on a surface with $n! = 168 (g-1)$, if you allow anti-holomorphic maps. | |
| Jul 23 at 11:03 | comment | added | user491858 | The case of $g=0$ and $g=1$ you can do by hand (and don't include $S_n$ for $n \ge 5$), but for higher genus you can't do better asymptotically than $O(|G|) = O(n!)$ (with an obvious explicit constant) by the Hurwitz bound. (compare mathoverflow.net/questions/179785/…). So the only question is whether you care about the constant. | |
| Jul 23 at 11:03 | answer | added | Sam Nead | timeline score: 11 | |
| Jul 23 at 10:41 | comment | added | HenrikRüping | But on the other hand $S_4$ already acts faithfully on $S^2$, by extending the canonical action on an tetrahedron, i.e. we get for an larger $n$ an even smaller genus. | |
| Jul 23 at 10:37 | comment | added | HenrikRüping | For example, I believe that $S_3$ acts faithfully on $T^2$. Think of $T^2$ as a regular hexagon where opposing edges are identified in an orientation preserving way. Then picking every second vertex gives a regular triangle. The canonical $S_3$ action can (also rather canonicallly) be extended to the entire $T^2$. Of course we have a global fixed point in the middle, but faithful only means that no group element acts as the identity. | |
| Jul 23 at 10:29 | history | edited | YCor |
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| Jul 23 at 10:16 | history | asked | André Henriques | CC BY-SA 4.0 |