Good afternoon everyone !

I have the following question of Riemannian geometry :

Let $M$ be a smooth closed orientable manifold of dimension at least $3$, and let $\mathcal{T} = \{ $ smooth Riemannian metric on $M$ with sectional curvature pinched between $-1- \epsilon$ and $-1$ $\}$ where $\epsilon$ is an arbitrary positive number. Assume that $\mathcal{T}$ is non-empty.

Let $\gamma $ be a simple closed curve in $M$. It is classical that for every negatively curved metric $g$ there is a unique closed curve in the free homotopy class of $\gamma$ that is length minimizing. Note $L_g(\gamma)$ the length of such a curve.

1) Is it known whether $ \inf_{g \in \mathcal{T} }{L_g(\gamma)}$ is positive or zero ?

2) Is it known whether $ \sup_{g \in \mathcal{T} }{L_g(\gamma)}$ is finite or infinite ?

3) Can one say more in specific cases, say when $M$ is hyperbolic ?

Obviously in dimension 2 Fenchel Nielsen coordinates show that $ \inf_{g \in \mathcal{T} }{L_g(\gamma)}$ is zero and $ \sup_{g \in \mathcal{T} }{L_g(\gamma)}$ is infinite. Nonetheless, the lack of topological symmetries for higher-dimensional hyperbolic manifolds make me hope that the opposite might be true.

Thanks for your attention !


We have $0<\inf_{g\in \mathcal{T}} L_g(\gamma) \leq \sup_{g\in \mathcal{T}} L_g(\gamma) <\infty$. In fact, there should be a universal bound on the ratio $\sup_{g\in \mathcal{T}} L_g(\gamma)/ \inf_{g\in \mathcal{T}} L_g(\gamma)$ for all $\gamma \in \pi_1 M$.

This follows from a theorem of Belegradek, who proves that the class of such metrics (actually, with just a fixed fundamental group $\pi$) is precompact in the Lipschitz topology. In particular, all such metrics are uniformly bi-Lipschitz, and thus one has the bound on the ratio between maximal and minimal lengths, as well as absolute bounds.

Quoting from the paper: Recall that the class of all compact Riemannian manifolds of a given dimension has the so-called Lipschitz topology, namely, two manifolds $M$ and $N$ are said to be $\epsilon$-close if there exists a diffeomorphism $f : M → N$ such that both $f$ and $f^{−1}$ are $e^\epsilon$-Lipschitz. A class of manifolds is called precompact if for any positive $\epsilon$, every sequence of manifolds in the class has a subsequence whose members are mutually $\epsilon$-close.

Now, suppose there is no upper bound $C$ so that any two metric $g,h\in \mathcal{T}$ are $C$-close. Take sequences $g_i,h_i\in \mathcal{T}$, such that $g_i$ and $h_i$ are not $N_i$-close, for a sequence $N_i\to \infty$. Passing to subsequences, we may assume that $\{g_i\}$ are mutually $\delta$-close for any $\delta>0$, and similarly for $\{h_i\}$. But $g_i$ is $\delta$-close to $g_1$, which is $C$-close to $h_1$ for some $C$ (since these are metrics on the same manifold), and which is $\delta$-close to $h_i$ for all $i$. Thus, $g_i$ is $2\delta+C$-close to $h_i$ for all $i$, a contradiction.

So we see that any two metrics in $\mathcal{T}$ are $C$-close for some $C$. This implies that $\sup_{g\in \mathcal{T}} L_g(\gamma)/ \inf_{g\in \mathcal{T}} L_g(\gamma)\leq e^C$ for all $\gamma \in \pi_1 M$. Moreover, comparing to any fixed metric in $\mathcal{T}$, we see that $0<\inf_{g\in \mathcal{T}} L_g(\gamma) \leq \sup_{g\in \mathcal{T}} L_g(\gamma) <\infty$.

  • $\begingroup$ Thank you Ian, Belgradek's paper answer the very first question I had in mind, namely how different can two negatively curved metric be on the same underlying manifold. $\endgroup$ – Selim G Apr 25 '15 at 10:18
  • $\begingroup$ Dear Ian, I checked in detail this morning your argument. There is one thing which is not clear to me(in the last part). Two metric on $M$ are said to be $C$-close if there is a diffeomorphism $f$ of $M$ such that both $f$ and $f^{-1}$ are $e^C$-Lipschitz, but to get the bound on $L_g(\gamma)$ one would would have to be sure that $f$ preserves the class of $\gamma$. We might need here something like the finiteness of the mapping class group of $M$ to conclude right ? $\endgroup$ – Selim G Apr 27 '15 at 16:32
  • $\begingroup$ @SelimG: Yes, you're essentially correct. In fact, for a fixed $M$, there are boundedly many curves of bounded length. So a subsequence will preserve the homotopy class of $\gamma$. I should fix the answer to reflect this. $\endgroup$ – Ian Agol Apr 27 '15 at 20:47
  • $\begingroup$ Sorry Ian, maybe I'm being thick, but I don't understand why you can assume $g_i$'s and $h_i$'s are $\delta$-close among themselves. Could you elaborate ? $\endgroup$ – BS. Aug 10 '15 at 21:19
  • $\begingroup$ By the precompactness theorem - see the third paragraph. $\endgroup$ – Ian Agol Aug 13 '15 at 5:48

A very partial answer: by the results of Richard Bamler, bounds on Ricci curvature (which is less than what you are requiring here) and diameter (more than you are requiring), implies that the metric is Lipschitz close to an Einstein metric. In three dimension, this means that the length of the homotopy class is close to that of the hyperbolic manifold. In higher dimension, if your manifold is (topologically) hyperbolic, then the Einstein metric is actually hyperbolic (Leung, 1997), so same as above.

PS Of course, if you have a diameter bound, then your lengths are bounded below in the two-dimensional case (by the collar lemma, if you like).


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