Jacobi fields and geodesic separation

Try

Shoot nearby unit-speed geodesics from one point. On a plane their separation initially grows linearly. On a sphere they eventually meet again at the antipode. In a hyperbolic plane their separation grows faster than linearly. Jacobi fields turn this comparison into a differential equation.

A Jacobi field is the variation field produced by a smooth family of geodesics, or equivalently a solution of the resulting linearized geodesic equation along a fixed geodesic. The linearization describes infinitesimal separation, not the exact distance between two widely separated curves.

Predict

Predict why curvature should enter the derivative-commutation step. Covariant derivatives in different directions need not commute; their discrepancy is the curvature tensor.

Predict whether every point where global shortest paths stop being unique is caused by a Jacobi field vanishing. A flat cylinder has competing routes from topology without positive-curvature focusing, so cut points and conjugate points are different concepts.

Experiment

For a unit-speed geodesic and perpendicular initial variation with j(0)=0,j′(0)=1, compare constant-curvature solutions.

K	Scalar separation j(s)	First positive zero
0	s	None
a², a>0	sin(as)/a	π/a
−a²	sinh(as)/a	None

Differentiate each expression twice and check both initial conditions. The normalization j′(0)=1 fixes the scale of the infinitesimal opening angle. It does not claim that arbitrary finite geodesic spacing equals j exactly.

Understand

Let F(s,ε) be a variation by geodesics, T=F_s and J=F_ε. Each radial curve satisfies D_sT=0. Differentiate this identity in ε and commute derivatives using our convention R(X,Y)Z=∇_X∇_YZ−∇_Y∇_XZ−∇_[X,Y]Z. Since the parameter fields commute and the connection is torsion-free, D_εT=D_sJ. The result is D_s²J+R(J,T)T=0.

Along a unit-speed surface geodesic choose a parallel unit field E perpendicular to T. For a normal variation J=jE, the second covariant derivative is j″E. The surface curvature identity gives R(E,T)T=KE. Therefore the Jacobi equation reduces to j″+Kj=0, with K evaluated along the geodesic.

The table solves this scalar ODE when K is constant. Positive curvature can force a second zero; in the round sphere this corresponds to the failure of the exponential map’s differential at the antipode. Nonpositive constant curvature produces no such zero in these radial examples.

Apply

A conjugate point is detected by a nonzero Jacobi field vanishing at both endpoints, equivalently a singularity of the exponential differential along the appropriate initial vector. This is a local variational phenomenon. A cut point concerns where a geodesic ceases to be globally minimizing, and can occur because another distinct minimizing route arrives first.

On a flat cylinder, points half a circumference apart have two equal shortest routes around the cylinder. Yet K=0 and the normal Jacobi solution s has no positive zero. This separates topological competition from curvature focusing.

Repair “Negative curvature makes all distances increase exponentially.” The table concerns an infinitesimal normal variation in constant curvature. A general statement about arbitrary pairs, variable curvature or nonunit parameters needs additional hypotheses and comparison theorems.

Continue

State the reference geodesic, initial variation and curvature convention before interpreting a Jacobi calculation. Continue to completeness and minimizing geodesics to examine global existence and shortest-path questions that local ODE theory alone does not settle.