Conjugate points

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In differential geometry, conjugate points or focal points^[1] are, roughly, points that can almost be joined by a 1-parameter family of geodesics. For example, on a sphere, the north-pole and south-pole are connected by any meridian. Another viewpoint is that conjugate points tell when the geodesics fail to be length-minimizing. All geodesics are locally length-minimizing, but not globally. For example on a sphere, any geodesic passing through the north-pole can be extended to reach the south-pole, and hence any geodesic segment connecting the poles is not (uniquely) globally length minimizing. This tells us that any pair of antipodal points on the standard 2-sphere are conjugate points.^[2]

Definition[]

Suppose p and q are points on a Riemannian manifold, and ${\displaystyle \gamma }$ is a geodesic that connects p and q. Then p and q are conjugate points along ${\displaystyle \gamma }$ if there exists a non-zero Jacobi field along ${\displaystyle \gamma }$ that vanishes at p and q.

Recall that any Jacobi field can be written as the derivative of a geodesic variation (see the article on Jacobi fields). Therefore, if p and q are conjugate along ${\displaystyle \gamma }$, one can construct a family of geodesics that start at p and almost end at q. In particular, if ${\displaystyle \gamma _{s}(t)}$ is the family of geodesics whose derivative in s at ${\displaystyle s=0}$ generates the Jacobi field J, then the end point of the variation, namely ${\displaystyle \gamma _{s}(1)}$, is the point q only up to first order in s. Therefore, if two points are conjugate, it is not necessary that there exist two distinct geodesics joining them.

Examples[]

• On the sphere ${\displaystyle S^{2}}$, antipodal points are conjugate.
• On ${\displaystyle \mathbb {R} ^{n}}$, there are no conjugate points.
• On Riemannian manifolds with non-positive sectional curvature, there are no conjugate points.

See also[]

• Cut locus
• Jacobi field

References[]