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 $M$.