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Nonpositive sectional curvature excludes conjugate points ((∣J∣2)′′≥0)

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Calculus of variations Second variation Conjugate point
2026-10-07  0 By others on same topic  0 Discussions Create my own version
For a Jacobi field along a geodesic with nonpositive sectional curvature, the curvature convention with positive round-sphere curvature gives
(∣J∣2)′′=2∣Dt​J∣2−2⟨R(J,γ˙​)γ˙​,J⟩≥0.
(1)
Thus ∣J∣2 is nonnegative and convex. If it vanishes at both ends of a segment, convexity forces it to vanish everywhere. No nonzero Jacobi field can vanish at both ends, so the segment has no conjugate points. Only curvature of planes containing the geodesic tangent is needed, and completeness is unnecessary.

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  • Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 14 / 2 / Solution

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