TITLE: Why do manifolds with negative sectional curvature not have conjugate points?
QUESTION [8 upvotes]: 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

REPLY [4 votes]: 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.