Nonpositively curved implies conjugate-free

Statement

Verbal statement

If the sectional curvature of a Riemannian manifold is everywhere nonpositive, then the Riemannian manifold is conjugate-free, viz it does not contain a pair of conjugate points.

Proof

Let $M$ be the Riemannian manifold, $\gamma$ a geodesic in $M$ and $J$ a Jacobi field about $\gamma$ that vanishes at the endpoints. We must show that $J$ is everywhere zero.

Consider the Jacobi equation:

${\frac {D^{2}J}{dt^{2}}}+R(J,V)V=0$

Taking the inner product of both these with $J$, the second term becomes the sectional curvature of the tangent plane spanned by $J$ and $V$, which is negative. Hence we conclude that:

$<{\frac {D^{2}J}{dt^{2}}},J>\geq 0$

This tells us that $<{\frac {DJ}{dt}},J>$ is nondecreasing, and hence, that $<J,J>$ is nondecreasing. Thus $J$ cannot be zero at both endpoints unless it is identically zero.