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I came across the following problem:

Which compact surfaces admit a metric with negative sectional curvature ($K<0$)?

I know that we have a classification of the compact orientable surfaces by its genus. By Preismann's theorem, if in such a surface we have a metric with $K<0$ then its fundamental group must be non-abelian. Then the only possible orientable compact surfaces would be those with genus $g\geq 2$ (we can also prove that the sphere and the torus do not admit such a metric by Gauss-Bonnet theorem).

For these surfaces $\Sigma_g$ with $g\geq 2$, its universal cover is the hyperbolic plane, which we know that admits a metric with (constant) negative sectional curvature. I think we can pass this metric (how so?) to $\Sigma_g$ and so we would have that: the compact orientable surfaces that admits a metric with negative sectional/Gaussian curvature are those with genus $\geq 2$. Is it correct?

Let's move on to the non-orientable ones. We know that they are connected sums of $\mathbb{RP}^2$. For $\mathbb{RP}^2$ itself, by Preismann's theorem every non trivial abelian subgroup of the fundamental group should be isomorphic to $\mathbb{Z}$. Since $\pi_1(\mathbb{RP}^2)=\mathbb{Z}_2$, then $\mathbb{RP}^2$ does not admit a metric with $K<0$. What about the (non trivial) connected sums of $\mathbb{RP}^2$ ? I do not know what happens in these cases.

Thanks a lot for the help in advance!

Stefani Joanne
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  • There are many way to prove this, for instance, via Uniformization Theorem. – Moishe Kohan Mar 28 '25 at 13:30
  • @MoisheKohan, oh nice! So for the orientable ones the affirmation that I gave is correct, isn't it? And what about the non-orientable cases? – Stefani Joanne Mar 28 '25 at 13:34
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    It's the same story, just use the fact that anticonformal maps of the unit disk are also hyperbolic isometries. – Moishe Kohan Mar 28 '25 at 13:35
  • @MoisheKohan, I believe in you, but I confess that I do not know how to do it. My background in topology is probably not good enough to be confident enough to say that. – Stefani Joanne Mar 28 '25 at 13:38
  • Prove that every anticonformal map of the unit disk is a composition of a conformal map and complex conjugation (chances are, you already proved this in your complex analysis class). Here, conformal means a diffeomorphism which "preserves oriented angles" and anticonformal means "preserves angles and reverses orientation". Next, prove that complex conjugation is an isometry of the Poincare metric. – Moishe Kohan Mar 28 '25 at 13:49
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    Incidentally, a cleaner way to rule out existence of metrics (on surfaces) with definite curvature sign is via Gauss-Bonnet formula. – Moishe Kohan Mar 28 '25 at 14:12
  • @MoisheKohan, Can I think as follows: the orientable double cover of $#_k \mathbb{RP}^2$ is $\Sigma_g$ where $g = k-1$ (e.g.: for $\mathbb{RP}^2$ is $S^2$, for $\mathbb{RP}^2#\mathbb{RP}^2$ is the torus etc). Then by the "orientable case", there exists a metric with $K<0$ only for the connected sum of at least three $\mathbb{RP}^2$ (that is, $\mathbb{RP}^2#\mathbb{RP}^2#\mathbb{RP}^2$ and so on...) ? Is it correct? – Stefani Joanne Mar 28 '25 at 15:39
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    Yes, but, as I said, it's easier to avoid the separation in two cases and argue via Gauss-Bonnet which does not care about orientability. – Moishe Kohan Mar 28 '25 at 15:41
  • @MoisheKohan, but Gauss-Bonnet requires orientability, no? – Stefani Joanne Mar 28 '25 at 15:41
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    Not really, this was discussed in the past on MSE, say, here. – Moishe Kohan Mar 28 '25 at 15:44
  • Oh, nice! Btw, thanks for the discussion! It was very helpful. – Stefani Joanne Mar 28 '25 at 15:45

1 Answers1

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To convert comments to a proper answer:

  1. Gauss-Bonnet formula (which holds even in the nonorientable case) says that for a compact Riemannian surface with Gaussian curvature $K$, $$ 2\pi\chi(S)=\int_S K dA, $$ where $dA$ is understood as a density. Thus, if $K<0$ then $\chi(S)<0$. On the other hand, every Riemannian metric on a compact surface (regardless of orientability) is conformal to a metric of constant curvature (due to the Uniformization Theorem). Thus, if $\chi(S)<0$ then $S$ admits a metric of constant negative curvature. (There are also more explicit constructions using pairs of pants with geodesic boundary or fundamental polygons.)

Hence, a compact surface $S$ admits a metric of negative curvature if and only if $S$ admits a metric of constant negative curvature.

  1. Remarkably, the same conclusion holds for 3-dimensional manifolds, as a consequence of Perelman's Geometrization Theorem. However, for manifolds of dimension $\ge 4$ this conclusion is false. Moreover, for every $n\ge 4$ and $\epsilon>0$ there are $n$-dimensional compact manifolds $M$ such that $M$ admits a metric of sectional curvature varying in the interval $(-1, -1+\epsilon)$, but $M$ does not admit a metric of constant curvature. (This was proven by Gromov and Thurston.)
Moishe Kohan
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