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Is there a way to construct a compact Riemann Surface X of genus[topological] g when g is given.

flag	asked May 4 2010 at 23:07 priyanka 27

Do you care about complex structure? If not, take a 4g-gon, with edges identified (clockwise) as $a_1 * b_1 * a_1^{-1} b_1^{-1} * \cdots * a_g * b_g * a_g^{-1} b_g^{-1}$, which will give you a topological one... – Simon Rose May 4 2010 at 23:13

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What do you mean by construct? There are various ways either with explicit equations, or as quotients of the disk, or as ramified coverings of $\mathbb{P}^1$... Also, I don't think your question is suitable for MathOverflow, please read the FAQ. – Andrea Ferretti May 4 2010 at 23:14

@Simon: your construction will also give you a complex structure, provided your 4g-gon is part of a tiling of the hyperbolic plane (meaning the sides of the tiling are geodesics). – Andrea Ferretti May 4 2010 at 23:17

Oh, certainly, but that requires a small amount more work, and realistically should require that you talk about discrete subgroups of $PSL_2(\mathbb{Z})$, and I didn't really want to go there. – Simon Rose May 5 2010 at 3:26

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-1: You have asked a question sufficiently vaguely so that it admits many possible answers, each of which is very well known and well explained in standard texts. Thus you seem to be wasting people's time. – Pete L. Clark May 5 2010 at 9:14

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S. Carnahan · Answer 1 · 2010-05-05 00:26:06Z UTC

Choose $2g+2$ distinct complex numbers $z_i$, and take a double cover of $\mathbb{P}^1$ branched at these points. This is typically written as a plane curve with an affine patch defined by $y^2 = \prod (x-z_i)$. See the Wikipedia article (which could use some polish).

Topologically, you can picture a genus $g$ surface with the handles lined up in a row, and take a quotient by a 180 degree rotation along the axis of symmetry to get a sphere.

compact riemann surface of genus g

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