# length comparison on negatively curved surfaces

Suppose $g_1$, and $g_2$ are two Riemannian metrics on a closed surface $S$, provided that the Gaussian curvature $K_{g_1}$ $<$ $K_{g_2}\leq -1$. Denote by $\mathcal{C}$ the set of free homotopy classes of closed curves in $S$. The marked length spectrum function is $l_{g_i}: \mathcal{C} → \mathbb{R}^{+}$ which assigns to the class $[\gamma]$ the length $l_{g_i}(\gamma)$ of the closed geodesic in $[\gamma]$.

My question is that given the curvature condition $K_{g_1}$ $<$ $K_{g_2}\leq -1$, can one conclude that $$l_{g1}([\gamma])\leq l_{g_2}([\gamma]),$$ for all $[\gamma]\in\mathcal {C}$? i.e. the length of a closed geodesic in $(S,g_2)$ is longer then the length of the corresponding closed geodesic in $(S,g_1)$.

p.s. the converse is not true. Thanks to the answer from @ Igor Rivin and @Anton Petrunin.

Thanks for the help.

• Is the curvature constant? – Igor Rivin Mar 10 '15 at 1:11
• No. It could be variable. – Nyima Kao Mar 10 '15 at 2:53

This is false even if $K_{g_1} = K_{g_2} \equiv -1,$ in case the two surfaces are not isometric.
• I see thank you. But what if $K_{g_1}<K_{g_2}$? – Nyima Kao Mar 10 '15 at 3:35
• @NyimaKao multiply $g_1$ by a constant slightly smaller than 1. – Anton Petrunin Mar 10 '15 at 3:40
• Yes @AntonPetrunin , thank you. Just curious, is the converse true? i.e. Does $K_{g_1}<K_{g_2}\leq -1$ imply $l_{g_1}([\gamma])\leq l_{g_2}([\gamma])$? – Nyima Kao Mar 10 '15 at 5:21