Topological entropy of the geodesic flow for manifold with non-negative constant sectional curvature and norms of Jacobi fields

Question

I am trying to see why it's true that if we have a compact Riemannian manifold with non-negative constant sectional curvature $K$ then it's geodesic flow $\phi_t:TM\rightarrow TM$ has topology entropy $0$, i.e, $h_{top}(\phi)=0$.

Now to do this I am trying to see why we have that it's Lyapunov exponents are identically zero and the using Ruelle's Inequality I will get the desired result. So basically we need to compute $\limsup_{n\rightarrow \pm \infty}\frac{1}{n}\log \langle \langle d_{\theta}\phi_n(\xi),d_{\theta}\phi_n(\xi)\rangle\rangle= \limsup_{n\rightarrow \pm \infty}\frac{1}{n}\log (\langle J_{\xi}(n), J_\xi(n)\rangle +\langle \dot J_{\xi}(n),\dot J_{\xi}(n)\rangle$ and check that this is zero , where $J_{\xi}(t)$ is a jacobi field.

Now my initial idea for this would be to decompose the jacobi field into the tangential and normal jacobi fields and do the computations for this fields. Now for manifold of constant non-negative sectional curvature we know that if $J$ is a jacobi field with $J(0)=0$ then $J(t)=as_c(t)E(t)$ where $s_c(t)=R\sin(\frac{t}{R})$ if $K>0$ and $s_c(t)=t$ if $K=0$ and $E$ is any parallel unit normal vector field along $\gamma$. Now for these it's easy to see that the previous limit will be going to zero. But how can I deal with the case that $J(0)\neq 0$? I know that we have general expressions for the jacobi field but I don't know if there is anything that can help me here .

Basically my question is what happens to the norm of a normal jacobi field $J$ such that $J(0)\neq 0$, is it possible to know anything about this ?

Any help regarding this is appreciated.

Answer 1

First of all, I am sorry I could not find a complete reference for this fact: I think with some effort one can find something similar in Petersen, Riemannian geometry, Lee, Riemannian geometry, an introduction to curvature or Gallot, Hulin Lafontaine, Riemannian Geometry. But I could recover the result that follows ; there may be some typo. This is quite long but really is elementary.

Suppose $J$ is a Jacobi field along a geodesic $\gamma : I \to (M,g)$. We fixe some notations: we denote by $\sec$ the sectionnal curvature of $(M,g)$ and by $R_{\gamma}$ the tensor along $\gamma$ defined by $R_{\gamma}X = R(\gamma',X)\gamma'$. It is equivalent to say $g\left(R_{\gamma}X,X \right) = {\|X\|_g}^2\sec(\gamma',X)$. Recall the Jacobi equation is $$ J'' +R_\gamma J = 0 $$ The function $t \in I \mapsto \|J(t)\|^2$ is smooth, and we have \begin{align} \dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2g\left(J',J\right)\\ &= 2g\left(J'',J \right) + 2g\left(J',J'\right) \\ &=-2g\left(R_\gamma J, J\right) + 2\|J'\|^2 \end{align}

Moreover, the left hand side can be computed another way for $t$ such that $J(t) \neq 0$: \begin{align} \dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2\|J\|\|J\|' \\ &= 2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right) \end{align} and this tells us that for $t$ such that $J(t) \neq 0$, then $$ 2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right) = -2g\left(R_\gamma J, J\right) + 2\|J'\|^2 $$ from which we deduce $$ \left\|J \right\|'' = -\frac{g\left(R_{\gamma}J,J \right)}{\|J\|} = -\sec(\gamma',J)\|J\| $$ Suppose $M$ has positive sectionnal curvature bounded from below along $\gamma$, say $\sec \geqslant \kappa^2 >0$, and define on $I$ $f(t) = \|J(0)\|\cos(\kappa t) + \|J(0)\|'\frac{\sin(\kappa t)}{\kappa}$. It is solution to the second order ODE $y'' = -\kappa^2 y$.

Now appears the trick: consider $g(t) = f(t) \|J(t)\|' - f'(t)\|J(t)\|$. Then on an interval containing $0$ on which $f(t) \geqslant 0$ and $J(t) \neq 0$ we have \begin{align} g' &= f\left(\|J\|'' + \kappa^2 \|J\| \right) \\ &= f \left(\|J\|'' + \sec(\gamma',J)\|J\| - \sec(\gamma',J)\|J\| +\kappa^2\|J\| \right) \\ &= f \left(\kappa^2 - \sec(\gamma',J) \right)\|J\| \geqslant 0 \end{align} Hence, $g$ is non-decreasing and $g(0) = 0$. This shows that $$ \forall t \text{ as above},~ g(t) \geqslant g(0) = 0 $$ which turns out to be $$ \forall t \text{ as above},~ \frac{\|J\|'}{\|J\|} \leqslant \frac{f'}{f} $$ integrating gives $$ \ln \frac{\|J\|}{\|J(0)\|} \leqslant \ln \frac{f}{f(0)} $$ and as $f(0) = \|J(0)\|$, we can deduce that for all $t$ such that all the above works $$ 0 \leqslant \|J(t)\| \leqslant f(t) $$ Now, you can deduce vanishing properties of $J$ thanks to this inequality.

Also, the exact same study in case the sectionnal curvature is bounded from above $\sec \leqslant -\kappa^2 <0$ shows that $$ \forall t,~ \|J(t)\| \geqslant \|J(0)\|\cosh (\kappa t) + \|J(0)\|' \frac{\sinh(\kappa t)}{\kappa} $$

Comment: if $J(0) = 0$ one can adapt this proof to fix this. Also, $\|J\|$ may not be differentiable at $0$ if $J(0)=0$, but if the curvature is of constant sign, then above calculations imply that $\|J\|$ is concave or convex, thus differentiable from the right at zero and we are done.

Another comment: if $\kappa \geqslant 0$ without having a lower bound, we cannot say much, because the euclidean case shows that $J$ may not vanish for positive $t$.

Answer 2

There is also another way to do this , that is to check that it's true for the case where $K=0,1$ and then argue by multiplying the metric by a constant we can get the other cases since $h_{top}(cg)=\frac{h_{top}(g)}{\sqrt{c}}$.

For the case where $K=0$ it will follow since the terms are linear , for the case where $K=1$, one can check that the geodesic flow preserves the sasaki-metric if and only if $M$ has constant sectional curvature equal to $1$.