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 <J,J> 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.