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Aug 5, 2016 at 14:36 comment added Adel BETINA The number of generators is equal to the genus of $X_0(p)$.
Dec 7, 2013 at 2:42 comment added David E Speyer In principal, yes, Mumford's proof is constructive. And once you find any generators, arxiv.org/abs/1309.5243 will put them into a convenient normalized form. But I don't know how to get those generators in the first place.
Dec 7, 2013 at 1:48 comment added Hugo Chapdelaine Is it possible to give a finite list of explicit topological generators for $\Gamma$?
Dec 7, 2013 at 1:46 vote accept Hugo Chapdelaine
Dec 6, 2013 at 16:58 history edited David E Speyer CC BY-SA 3.0
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Dec 6, 2013 at 16:40 history answered David E Speyer CC BY-SA 3.0