Mathematics.

curvature

Jacobi Fields

Differential Geometry75 minDifficulty8 out of 10

Overview

A Jacobi field is a vector field along a geodesic that arises as the variational vector field of a one-parameter family of geodesics. Jacobi fields satisfy a second-order linear ODE — the Jacobi equation — that involves the Riemann curvature tensor. They describe how nearby geodesics spread apart or converge, making them the fundamental tool for understanding conjugate points, the exponential map's critical locus, and comparison theorems in Riemannian geometry.

Intuition

Imagine a family of geodesics fanning out from a single point, like great circles leaving the north pole of a sphere. The Jacobi field measures how fast these geodesics are moving apart from one another. On a positively curved surface (like a sphere), nearby geodesics converge and can even refocus — this is the conjugate point. On a negatively curved space, they diverge faster than on flat space. Curvature directly controls the spreading via the Jacobi equation.

Formal Definition

Definition

Let (M, g) be a Riemannian manifold and γ: I → M a geodesic. A Jacobi field is a vector field J along γ satisfying the Jacobi equation.

D2Jdt2+R(J,γ˙)γ˙=0\frac{D^2 J}{dt^2} + R(J, \dot{\gamma})\dot{\gamma} = 0

J is a Jacobi field if it satisfies this second-order linear ODE along γ

Jacobi equation
R(J,γ˙)γ˙=RijklJjγ˙kγ˙liR(J,\dot{\gamma})\dot{\gamma} = R^i{}_{jkl}\, J^j \, \dot{\gamma}^k \, \dot{\gamma}^l \, \partial_i
Curvature term in local coordinates
d2Jidt2+Rijklγ˙jJkγ˙l=0\frac{d^2 J^i}{dt^2} + R^i{}_{jkl}\, \dot{\gamma}^j \, J^k \, \dot{\gamma}^l = 0
Jacobi equation in coordinates (for the Levi-Civita connection)

Theorems

Theorem 1: Jacobi Fields from Geodesic Variations
If Γ(t,s) is a smooth variation of geodesics with Γ(t,0)=γ(t), then J(t)=sΓ(t,0) is a Jacobi field along γ.\text{If } \Gamma(t,s) \text{ is a smooth variation of geodesics with } \Gamma(t,0) = \gamma(t), \text{ then } J(t) = \partial_s \Gamma(t,0) \text{ is a Jacobi field along } \gamma.
Theorem 2: Dimension of Jacobi Fields
The space of Jacobi fields along a geodesic γ:[a,b]M is a 2n-dimensional vector space (n=dimM).\text{The space of Jacobi fields along a geodesic } \gamma: [a,b] \to M \text{ is a } 2n\text{-dimensional vector space (}n = \dim M\text{).}
Theorem 3: Conjugate Points
A point γ(t0) is conjugate to γ(0) along γ if there exists a non-zero Jacobi field J with J(0)=0 and J(t0)=0.\text{A point } \gamma(t_0) \text{ is conjugate to } \gamma(0) \text{ along } \gamma \text{ if there exists a non-zero Jacobi field } J \text{ with } J(0) = 0 \text{ and } J(t_0) = 0.

Worked Examples

  1. 1

    Take S² with the round metric. Let γ(t) = (sin t, 0, cos t) be the unit-speed great circle (a geodesic). The sectional curvature is K = 1 everywhere.

    K=1K = 1
  2. 2

    A Jacobi field J perpendicular to γ̇ satisfies the simplified Jacobi equation in dimension 2. Along γ, the curvature term gives R(J, γ̇)γ̇ = K · g(γ̇, γ̇) · J = J (for |γ̇| = 1 and K = 1).

    D2Jdt2+J=0\frac{D^2 J}{dt^2} + J = 0
  3. 3

    This scalar ODE has general solution:

    J(t)=(Asint+Bcost)N(t)J(t) = (A \sin t + B \cos t)\, N(t)
  4. 4

    where N(t) is the unit normal to γ. Initial conditions J(0) = 0 require B = 0, giving J(t) = A sin(t) N(t).

    J(t)=Asin(t)N(t)J(t) = A\sin(t)\, N(t)
  5. 5

    J vanishes again when sin(t) = 0, i.e., t = π. So the first conjugate point is at distance π — the antipodal point on S².

    J(π)=0    t0=π is the first conjugate pointJ(\pi) = 0 \implies t_0 = \pi \text{ is the first conjugate point}

✓ Answer

On S², Jacobi fields along a great circle are J(t) = A sin(t) N(t) (normal direction). The first conjugate point occurs at t = π (the antipodal point), consistent with positive curvature K = 1 causing geodesics to refocus.

Practice Problems

Mediumfree response

In flat Euclidean space ℝⁿ, what are the Jacobi fields along a straight-line geodesic γ(t) = p + tv?

Hardproof writing

Prove that if J is a Jacobi field along a geodesic γ and J is everywhere tangent to γ, then J(t) = (at + b) γ̇(t) for constants a, b.

Quiz

What does a Jacobi field along a geodesic γ measure?
On the unit sphere S², the first conjugate point along a great circle to the starting point is at geodesic distance:
The Jacobi equation along a geodesic γ is:

Summary

  • Jacobi fields J along a geodesic γ satisfy D²J/dt² + R(J, γ̇)γ̇ = 0, a linear second-order ODE involving curvature.
  • They arise as variational fields of geodesic variations: J = ∂_s Γ(t,0) for a family Γ of geodesics.
  • On a space of constant curvature K, |J(t)| = |J'(0)| · s_K(t) with s_K = sin/t/sinh scaled by K.
  • Conjugate points are zeros of non-trivial Jacobi fields with J(0) = 0; positive curvature causes them, negative curvature prevents them.
  • The space of Jacobi fields along γ is 2n-dimensional; tangential Jacobi fields are J = (at+b)γ̇.