Timeline for Branched coverings of Riemann surfaces with specified branch points.
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2010 at 16:35 | comment | added | H. Hasson | No problem, that's how we learn. The point is not that maps from pi_1 correspond to covers, but that finite quotients (by not nec. normal subgroups) of pi_1 correspond to finite covers. I'm mapping surjectively to a finite group, G; so I know this corresponds to a finite normal cover with deck transformations G (the quotient I'm thinking of is pi_1/kernel of this map). The pi_1 is of Sigma without the branch points. Once we get this cover, and algebraize it, there's only one way to fill in the points (this is true for curves; see 1.6 in Hartshorne). | |
| Apr 6, 2010 at 2:04 | comment | added | Ilya Grigoriev | This seems interesting, but I'm also confused by this answer. What exactly is the cover? I'm even confused by what is covered: is the mentioned $\Sigma$ the Riemann surface with points removed, or just the surface itself? I guess the reason I'm confused is that I'm used to covers corresponding to subgroups of $\pi_1$, but here you seem to say that the map from $\pi_1$ to somewhere correspond to covers. Also, at what point do you fill back in the branch points you removed? How do we know the resulting cover is finite? Thank you (if you have time to clarify), sorry if I'm just stupidly confused. | |
| Apr 2, 2010 at 4:06 | comment | added | H. Hasson | Yes... So for that case let's take the group to be non-abelian. Then it has a non-trivial commutator [x,y]. Let a_1 go to x, b_1 go to y, c_1 go to [x,y]^(-1), and the rest to 1. This is of course only possible for g>=1; but for g=0, as David mentioned, this is impossible. | |
| Apr 2, 2010 at 3:50 | comment | added | Dylan Thurston | Aren't you missing the case $r=1$? | |
| Apr 2, 2010 at 0:56 | history | answered | H. Hasson | CC BY-SA 2.5 |