Why do manifolds with negative sectional curvature not have conjugate points?

Question

I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.

many regards, and thanks in advance, Leo

Answer 1

Let $\gamma:[0,\epsilon]\rightarrow M$ be a geodesic curve, $\epsilon>0$ and $J$ be a Jacobi vector field along $\gamma$ with $J(0)=J(\epsilon)=0$.

I remind you that :

1. $J$ being a Jacobi vector field means that $\dfrac{D^2}{dt^2}J+R(\gamma',J)\gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.
2. $\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right] \kappa(X,Y)=\langle R(X,Y)X,Y\rangle$ for any $X,Y\in TM$ with $\kappa$ the sectional curvature.

Now look at the map $\phi:t\mapsto \|J(t)\|^2$. Its second derivative is easy to compute and using the two remarks you get : $$\begin{align} \phi''(t) & = \langle \dfrac{D^2}{dt^2}J,J\rangle + \langle \dfrac{D}{dt}J,\dfrac{D}{dt}J\rangle\\ & = -\langle R(\gamma',J)\gamma',J\rangle + \|\dfrac{D}{dt}J\|^2 \\ & = -\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right]\kappa(\gamma',J) + \|\dfrac{D}{dt}J\|^2.\end{align}$$

In particular, if all sectional curvatures are non-positive then $\phi''\geq 0$. So $\phi$ is a convex map and since $\phi(0)=\phi(\epsilon)=0$, $\phi(t)=0$ for any $t\in [0,\epsilon]$. It follows that $J$ has to be trivial.

So we can't have conjugate points in $M$.

Answer 2

This is very late, but to provide an alternative proof:

The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $\mathbb{R}^m$.

Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $p\in M$, and that $\gamma:\left[0,T\right]\to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $\gamma$ with $J(0)=0$ and $||J'(0)||>0$.

Let $\tilde\gamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $\mathbb{R}^m$. Let $\tilde{J}$ be the Jacobian vector field along $\tilde\gamma$ satisfying $\tilde{J}(0)=0$, $\tilde{J}'(0)=(<\gamma'(0),J'(0)>,(||J'(0)||^2-<\gamma'(0),J'(0)>^2)^\frac{1}{2},0,0,...,0)$. Then $J(0)=0=\tilde{J}(0)$, $||\tilde{J}'(0)||=||J'(0)||$, and $<\tilde{\gamma}'(0),\tilde{J}'(0)>=<\gamma(0),J'(0)>$.

For all $t\in\left[0,T\right]$, $max(K(\gamma'(t),X))\leq0=min(\tilde{K}(\tilde{\gamma}'(t),\tilde{X}))$ and $dim(\mathbb{R}^m)\leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||\geq||\tilde{J}(t)||$ for all $t\in\left[0,T\right]$. In particular, since $\mathbb{R}^m$ has no conjugate points, $\tilde{J}'(0)\not=0$ guarantees that $||\tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $\gamma$.

Since $p$ and $\gamma$ were arbitrary, $M$ has no conjugate points.