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Sep 11, 2016 at 18:54 comment added QGravity At least in this case, $\Gamma\subset{\bf {PSL}}(2,\mathbb{R})$, (${\bf (PSL)}(2,\mathbb{R}$) is the group of orientation preserving authomorphisms of upper half plane $\mathbb{H}$) and acts freely on ${\mathbb{H}}$.
Sep 11, 2016 at 17:47 comment added user1688 Well, no, it is not a free group. The first $2g-1$ generators generate a free subgroup, though.
Sep 11, 2016 at 16:44 comment added QGravity Thank you for your answer, so there are $2g-2$ independent generators.
Sep 11, 2016 at 11:01 history answered user1688 CC BY-SA 3.0