Let $S$ be a surface of negative Euler characteristic (for simplicity let's assume $S$ to be closed), and let $\mathcal{M}(S)$ denote the moduli space of hyperbolic surfaces homeomorphic to $S$.

Given $X \in \mathcal{M}(S)$, let $L(X)= l_{1} \leq l_{2} \leq l_{3} \leq...$ denote the increasing sequence of lengths of all closed geodesics on $X$. This is the so-called "unmarked length spectrum". It is known that $L(X)$ does not determine $X$ up to isometry; indeed, if $\mathcal{V} \subset \mathcal{M}(S)$ denotes the set of all surfaces $X$ for which there exists $Z \neq X$ with $L(X) = L(Z)$ (in this case $X$ and $Z$ are called $\textit{iso-spectral}$), then it is a theorem of Wolpert that $\mathcal{V}$ is a real-analytic subvariety of $\mathcal{M}(S)$ with positive dimension. On the other hand, $\mathcal{V}$ also has positive codimension, so "generically" $L(X)$ does determine the metric.

McKean proved that any subset $W \subset \mathcal{M}(S)$ consisting of pairwise iso-spectral surfaces must be finite, and Buser gave an explicit upper bound on $|W|$ which depends only on the genus of $S$.

Let $d_{T}( , )$ denote the projection of the Teichmuller metric to $\mathcal{M}(S)$, and similarly $d_{WP}( , )$ for the Weil-Petersson metric.

$\textbf{Question 1}:$ Let $D_{T}(S):= \sup \left\{ d_{T}(X,Y) : \hspace{2 mm} X,Y \in \mathcal{M}(S), X \hspace{2 mm} \mbox{iso-spectral to} \hspace{2 mm} Y \right\}$, and define $D_{WP}(S)$ similarly. Is $D_{T} < \infty$? How about $D_{WP}$? (of course if $D_{T}$ is finite then so is $D_{WP}$).

$\textbf{Question 2:}$ Let $LS(X)$ denote the subsequence of $L(X)$ consisting only of lengths corresponding to $\textbf{simple}$ closed geodesics. Is it the case that $LS(X)= LS(Y)$ implies $L(X)= L(Y)$?

$\textbf{Regarding Question 1:}$ I suspect that the answer is "no" for both $D_{T}$ and $D_{WP}$, but I don't have an argument in mind, nor am I very familiar with what is known about iso-spectrality in general. I'll remark that there exists fairly general/algebraic methods for constructing iso-spectral surfaces with arbitrarily large genus, such as the following due to Sunada:

Let $\phi: \pi_{1}(X) \rightarrow G$ be a surjective homomorphism on to a finite group $G$, and let $H_{1}, H_{2}$ be $\textit{almost conjugate}$ subgroups of $G$, meaning that

$$ | [g] \cap H_{1}| = | [g] \cap H_{2} |, \forall g \in G, $$

where $[g]$ denotes the conjugacy class of $g$. Then the covering spaces of $X$ associated to the subgroups $\phi^{-1}(H_{1})$ and $\phi^{-1}(H_{2})$ will be iso-spectral. Perhaps, one can construct iso-spectral surfaces that are very far apart in $\mathcal{M}(S)$ by carefully choosing $G$, $\phi$ and $X$.

$\textbf{Regarding question $2$:}$ It is known that the converse is false. That is, $L(X)= L(Y)$ does not imply that $LS(X)= LS(Y)$ (see the following preprint of Maungchang: http://arxiv.org/abs/1111.4317). The proof uses Sunada's construction described above.

Thank you very much for reading.

  • 1
    $\begingroup$ I think you are correct, regarding $D_T$ -- Sunada's construction will give a negative answer to Question 1. Fix $S$ the closed surface of genus two. Fix data $G$, $\phi$, $H_1$ and $H_2$ as required by Sunada. Vary $X \in M(S)$ as follows. Fix a pants decomposition of $S$. For any $n > 0$... $\endgroup$
    – Sam Nead
    Apr 5, 2015 at 12:50
  • $\begingroup$ let $X_n$ to be a hyperbolic genus two surface where all cuffs have length $1/n$. Let $Y_n$ and $Z_n$ be the two covers. These will also have many short curves, but in different (non-homeomorphic) configurations. Thus $d_T(Y_n,Z_n)$ goes to infinity with $n$. --- You might be able to shorten the proof by factoring through the map to $F_2$ (free group of rank two) that kills the conjugacy classes of the cuffs. Then $Y_n$ and $Z_n$ come to us with distinct pants decompositions. $\endgroup$
    – Sam Nead
    Apr 5, 2015 at 12:51

2 Answers 2


As a partial answer to question one: The diameter of $M(S)$ with respect to the Weil--Petersson metric is finite. So $D_{WP}$ is finite.

  • $\begingroup$ Ha, of course. Thank you. Thanks also for the comments- your suggestion makes perfect sense $\endgroup$ Apr 5, 2015 at 13:35

For $D_T:$ if you have a bound on $\lambda_1,$ then, by Cheeger's inequality you have a (lower) bound on the shortest geodesic. An estimate then follows from the known estimate on the diameter of the $\epsilon$-thick part of moduli space. (due to, who else, Kasra Rafi).

EDIT 1 As pointed out by Ian in the comments, that is not quite true ($\lambda_1$ gives a bound on the shortest separating multicurve, not the shortest curve).

EDIT 2 Re Question 2: This is true the punctured torus, as shown (but not written down in any detail) by McShane and Rivin around 20 years ago in our CRAS note. If you write down the detailed proof, mankind will thank you (as usual, for the punctured torus, this comes down to continued fractions: you can back out the marking from the lengths).

  • $\begingroup$ A lower bound on $\lambda_1$ doesn't necessarily give a lower bound on the shortest geodesic, if the geodesic is non-separating (maybe if $\lambda_1 >1/4$). $\endgroup$
    – Ian Agol
    Apr 5, 2015 at 18:38
  • $\begingroup$ @IanAgol good point. In any case, this kind of reasoning seems to indicate that there is no upper bound as the first eigenvalue collapses to zero... $\endgroup$
    – Igor Rivin
    Apr 5, 2015 at 19:46

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