Length of the transversal for surfaces with cusps

Question

In Peter Buser's Geometry and Spectra of Compact Riemann Surfaces he shows that the length of the transverse curve to a geodesic in a pants decomposition on a compact hyperbolic surface has length a convex function of the twist parameter about that geodesic (Proposition 3.3.11).

Does a similar result hold in the case of surfaces with cusps?

In this case, Buser's argumentation breaks down as many of the lengths of the curves he examines become infinite and his geometric picture is no longer useful. In fact, As there are no geodesics inside of a cusp, you cannot even define the transversal for surfaces with cusps in the same way as for surfaces without. Any suggestions would be helpful.

Answer 1

I do not know about arguments in Buser's book, but the proofs of convexity by Kerchkoff here and Wolpert here are completely general and have nothing to do with existence or nonexistence of cusps. Concerning your question about transversality: A (possibly nonsimple) smooth curve $\alpha$ is transversal to a simple curve $\beta$ in a surface $S$ iff the map $\alpha: {\mathbb R}\to S$ is transversal to the submanifold $\beta$ in the sense of differential topology; in your special case, geodesics are transversal iff their images are distinct. Note that the intersection, of course, could be empty. Then convexity holds for the trivial reason that a constant function is always convex.