(compact, non-empty boundary )Surface Geodesics on Hyperbolic Geometry

I have a basic knowledge of hyperbolic geometry . I am trying to understand the meaning of "a compact surface S with non-empty boundary(which is neither a disk nor an annulus )with a complete hyperbolic metric with geodesic boundary" .

I know about the upper-half plane model with the metric $$ds^2 =\frac {dx^2+dy^2}{y^2}$$

in which the geodesics are half-circles with the endpoints in the Real line and vertical half-lines with an endpoint on the Real line. But I don't have a clear idea about the issue of geodesics in a surface with a hyperbolic metric. I imagine we may use charts to pullback the metric locally from $\mathbb R^2$ and then somehow patch it up with partitions of unity (which should have no problem working because of the compactness of S ). I guess completeness then follows from compactness of S. Could someone please give me some intro. comments and refs? Thanks.

• How much Riemannian geometry do you know? Do you know the definition of a Riemannian manifold with boundary? What a totally geodesic submanifold is? – Moishe Kohan Oct 11 '14 at 2:53
• @studiosus: Yes, I do know the basics of Riemannian manifold in general, but not of one with boundary. – Hyper Oct 11 '14 at 2:54
• You have a long way to go. The simplest example is called a pair of pants, one way to make things simple is to insist that the three boundary curves be geodesics in the original surface en.wikipedia.org/wiki/Pair_of_pants_%28mathematics%29 in which case the original surface cannot be a sphere or torus. – Will Jagy Oct 11 '14 at 4:01
• Well, but how do we just insist that the boundary curves are geodesics? My main question is : how do we define the metric on a surface with boundary so that the boundary curves are geodesics in that metric? – Hyper Oct 19 '14 at 0:46

Let's start with generalities. Suppose that $M$ is a smooth manifold with boundary. In order to define the notion of a Riemannian metric on $M$, embed $M$ as a codimension 0 submanifold in a smooth manifold $M'$ without boundary (this is always possible). Now, a Riemannian metric $g$ on $M$ is the restriction of a Riemannian metric $g'$ on $M'$. The manifold $(M,g)$ has totally geodesic boundary iff the boundary of $M$ is totally geodesic in $(M',g')$.

One say that $(M,g)$ has negative curvature iff the curvature of $(M',g')$ is negative at each point of $M$. Same for "constant curvature".

Now, in dimension 2, you can use Gauss-Bonnet to show that if $(M,g)$ is compact negatively curved with geodesic boundary, then $\chi(M)<0$. This rules out disks and annuli. Conversely, given a compact surface $M$ with $\chi(M)<0$, $M$ admits a metric of constant negative curvature with geodesic boundary. You can prove it by gluing "pairs of pants" with geodesic boundary and equal boundary lengths.

Here are two books as references:

1. W.P.Thurston "Three-dimensional Geometry and Topology", section 4.6.

2. W.Abikoff "Real-analytic theory of Teichmuller space", chapter 2, section 3.

• Thanks, but what I am ultimately looking for is how to choose the metric so that it is of negative (sectional) curvature and so that the boundary curves (working with a surface) are geodesics. Maybe you answered this? – Hyper Oct 19 '14 at 0:44
• @Hyper: Yes, I did. – Moishe Kohan Oct 19 '14 at 1:41
• Would you please elaborate? – Hyper Oct 19 '14 at 1:44
• I guess we need to use the fact that we are gluing pairs of pants with geodesic boundary. If so, could you give a ref. for this, please? – Hyper Oct 19 '14 at 3:25