Geodesic curvature on hyperbolic manifold with boundary

Question

Let $\Sigma$ be a compact oriented surface of genus $1$ having a single boundary component (i.e. $T^2$ minus an open disk) and let $g$ be a Riemannian metric on $\Sigma$ with constant Gaussian curvature $K=-1$. Is it necessarily the case that $\int_{\partial \Sigma} k_gds \leq 0$? The Gauss-Bonnet theorem gives a lower bound $\int_{\partial \Sigma} k_gds = \text{Vol}(\Sigma)-2\pi \geq -2\pi $ but I am wondering about an upper bound.

Answer 1

Given a topological surface $S$ which is the torus with closed disk removed, there is are two types of complete hyperbolic structures $h$ on $S$:

1. The ones of finite (hyperbolic area) area. In this case, the only "end" of $(S,h)$ is a "cusp."

2. The ones of infinite area, where the the only "end" of $(S,h)$ is a funnel.

I will be using the 2nd type. The hyperbolic surface $(S,h)$ in this case has the following decomposition: There exists a unique simple closed geodesic $c$ on $(S,h)$ such that cutting $S$ open along $c$ results in a compact hyperbolic surface with geodesic boundary and an annulus $S$. Then annulus $S$ is foliated by simple loops $c_t$ equidistant from $c$ (the distance from each point of $c_t$ to $c$ equals $t>0$). These curves are convex in $(S,h)$ and have strictly positive curvature: If I cut $S$ open along $c_t$, the result is a compact surface $\Sigma_t$ with convex boundary and an annulus $A_t$.

Now, take one of the surfaces $\Sigma_t$ as your surface $\Sigma$.

To get a better idea about the nature of the curves $c_t$, lift the closed geodesic $c$ to a biinfinite geodesic $\tilde{c}$ in the hyperbolic plane (the universal cover of $(S,h)$). I will be using the upper half-plane model of the hyperbolic plane. Then one can arrange it so that $\tilde{c}$ is given by $$ \{(x,y): x> 0, y=0\}, $$ the positive $y$-semiaxis. Then each $c_t$ lifts to a straight line $y=m_t x, x>0$; WLOG, $m_t>0$ for all $t$.