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Proof that higher genus surface admits a metric of negative Ricci scalar everywhere

+ 4 like - 0 dislike
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In the Green, Schwarz and Witten Superstring Theory textbook, the paragraph below equation 3.3.15 says,

For genus greater than one, it can be shown that the surface admits a metric of everywhere negative scalar curvature.

This statement is essential in the proof that $C_g$ for $g>1$ is always zero, but I cannot prove that. Could you point out how to prove this?

This post imported from StackExchange Physics at 2015-03-30 13:49 (UTC), posted by SE-user Han Yan
asked Dec 12, 2013 in Mathematics by Han Yan (110 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

This result follows from i) Uniformization theorem and ii) Gauss-Bonnet theorem in 2d.

According to the statement of uniformization theorem from this wiki page :

every connected Riemann surface X admits a unique complete 2-dimensional real Riemann metric with constant curvature −1, 0 or 1 inducing the same conformal structure

On the other hand, according to Gauss-Bonnet theorem, integral of the scalar curvature on a 2d surface is a positive multiple of the Euler characteristic ($\chi=2-2g$). Since the Euler characteristic is negative for $g>1$ so from the uniformization theorem it follows that -

any Riemann surface with genus $g>1$ admits a unique complete 2-dimensional real Riemann metric with constant curvature −1 inducing the same conformal structure


Note : I am not aware of any good reference where uniformization thereom is proved in the form stated above. However, I hope you can find a proof in some of the references mentioned in the corresponding wiki article.

This post imported from StackExchange Physics at 2015-03-30 13:49 (UTC), posted by SE-user user10001
answered Dec 12, 2013 by user10001 (635 points) [ no revision ]
Thank you very much, I think that's the answer. To add some materials here: For the proof of Uniformization theorem, see Uniformization of Riemann Surfaces . For further reading, see GTM 71, Riemann Surfaces, Chpter IV

This post imported from StackExchange Physics at 2015-03-30 13:49 (UTC), posted by SE-user Han Yan
+ 1 like - 0 dislike

Why not use explicit construction for such a surface?

From The Manifold Atlas:

Any hyperbolic metric on a closed, orientable surface $S_g$ of genus $g\ge 2$ is obtained by the following construction: choose a geodesic $4g$-gon in the hyperbolic plane ${\Bbb H}^2$ with area $4(g-1)\pi$. (This implies that the sum of interior angles is $2\pi$.) Then choose orientation-preserving isometries $I_1,J_1,\ldots,I_g,J_g$ which realise the gluing pattern of $S_g$: for $j=1,\ldots,g$ we require that $I_j$ maps $a_j$ to $\overline{a}_j$, $J_j$ maps $b_j$ to $\overline{b}_j$. Let $\Gamma\subset Isom^+\left({\Bbb H}^2\right)$ be the subgroup generated by $I_1,J_1,\ldots,I_g,J_g$. Then $\Gamma$ is a discrete subgroup of $Isom^+\left({\Bbb H}^2\right)$ and $\Gamma\backslash{\Bbb H}^2$ is a hyperbolic surface diffeomorphic to ${\Bbb H}^2$.

Here is an illustration for genus 2 surface glued from an octagon in hyperbolic plane (image taken from here):

image

This post imported from StackExchange Physics at 2015-03-30 13:49 (UTC), posted by SE-user user23660
answered Dec 12, 2013 by user23660 (40 points) [ no revision ]

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