Take the 2-minute tour ×

Why in proof of proposition 6 of http://arxiv.org/abs/0909.1665, they claim that if a embedded surfaces $\Sigma^2 \subset (M^3,g)$ is homeomorphic to $\mathbb{RP}^2$, where $M$ is compact manifold, then, by uniformization theorem, there exist diffeomorphism $\phi: \mathbb{RP}^2 \rightarrow \Sigma$ such that $\phi^{*}g$ is conformal to standard metric on $\mathbb{RP}^2$?

Can someone help me?

share|improve this question
Because there is a unique conformal structure on the projective plane. That's a consequence of the uniformization theorem. –  Mariano Suárez-Alvarez Dec 20 '12 at 5:59
add comment

1 Answer 1

I think that in the definition of class $\mathscr F$ "embedded" means "smoothly embedded", not just topologically. Otherwise they would not be talking about Gaussian curvature, etc of an arbitrary surface $\Sigma\in\mathscr F$. So, the surface $\Sigma$ carries a Riemannian metric and is homeomorphic to $\mathbb RP^2$. What does it mean to uniformize $\Sigma$?

Uniformization theorem for surfaces comes in several flavors. Sometimes it's understood just as the existence of a metric of constant curvature on topological surfaces. Other times, it's about biholomorphic equivalence of complex 1-manifolds (Riemann surfaces). Yet another version relates the constant curvature metric to a pre-existing Riemannian metric: namely, they are related by a conformal diffeomorphism such as $\phi$ above. Many sources focus on the orientable case because they care about complex structures. But non-orientable compact surfaces such as $\Sigma$ can be uniformized too. I think the book Teichmüller Theory by Hubbard covers uniformization from the geometric point of view that is most relevant here.

share|improve this answer
add comment

Your Answer


