# Nonpositively curved implies conjugate-free

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## 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:

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

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:

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

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

## Application

This observation is crucial to the proof of the Cartan-Hadamard theorem.