3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Is the curvature constant? $\endgroup$
    – Igor Rivin
    Mar 10, 2015 at 1:11
  • $\begingroup$ No. It could be variable. $\endgroup$
    – Nyima Kao
    Mar 10, 2015 at 2:53

1 Answer 1

7
$\begingroup$

This is false even if $K_{g_1} = K_{g_2} \equiv -1,$ in case the two surfaces are not isometric.

$\endgroup$
3
  • $\begingroup$ I see thank you. But what if $K_{g_1}<K_{g_2}$? $\endgroup$
    – Nyima Kao
    Mar 10, 2015 at 3:35
  • 2
    $\begingroup$ @NyimaKao multiply $g_1$ by a constant slightly smaller than 1. $\endgroup$ Mar 10, 2015 at 3:40
  • $\begingroup$ Yes @AntonPetrunin , thank you. Just curious, is the converse true? i.e. Does $K_{g_1}<K_{g_2}\leq -1$ imply $l_{g_1}([\gamma])\leq l_{g_2}([\gamma])$? $\endgroup$
    – Nyima Kao
    Mar 10, 2015 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.