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Every vector field along a geodesic is a Jacobi field

Statement

False claim: every smooth vector field along every affinely parametrized geodesic in a Riemannian manifold is a Jacobi field.

Facts & Assumptions

Given: To refute this universal claim, it suffices to give one smooth vector field along one affinely parametrized geodesic whose Jacobi residual is nonzero.

[F1]

A smooth field $J$ along an affinely parametrized geodesic is Jacobi only if it satisfies $D_{t}^{2} J + R (J, \dot{γ}) \dot{γ} = 0.$ (Jacobi field)

[F2]

In coordinates, the Levi-Civita symbols are $Γ^{k}_{i j} = \frac{1}{2} \sum_{ℓ} g^{k ℓ} (\partial_{i} g_{j ℓ} + \partial_{j} g_{i ℓ} - \partial_{ℓ} g_{i j}) .$ (Christoffel formula for the levi civita connection)

[F3]

The curvature components are $R^{ℓ}_{k i j} = \partial_{i} Γ^{ℓ}_{j k} - \partial_{j} Γ^{ℓ}_{i k} + Γ^{m}_{j k} Γ^{ℓ}_{i m} - Γ^{m}_{i k} Γ^{ℓ}_{j m} .$ (Coordinate formula for the curvature tensor)

[F4]

In coordinates, a curve is a geodesic exactly when ${\ddot{x}}^{k} + Γ^{k}_{i j} (x) {\dot{x}}^{i} {\dot{x}}^{j} = 0.$ (Coordinate geodesic equation)

[F5]

Covariant differentiation along a curve is the pullback connection $D_{t} V = (γ^{*} \nabla)_{\partial / \partial t} V .$ (Covariant derivative along a curve)

[F6]

In a coordinate frame, the Christoffel symbols are the connection coefficients: $\nabla_{\partial_{i}} \partial_{j} = \sum_{k} Γ^{k}_{i j} \partial_{k} .$ (Christoffel symbols of an affine connection)

Refutation

1.1

Choose the Euclidean line, $M = R$ , $g = d x^{2}$ , $I = [0, 1]$ , $γ (t) = t$ , and $J (t) = t^{2} \partial_{x} ∣_{γ (t)}$ . [construct, given] The field is smooth up to both endpoints, and the interval is nondegenerate.

2.1

Verify $γ (t) = t$ is an affinely parametrized geodesic from the coordinate equations. [step 1.1, F2, F4, F6, algebra] Indeed $g_{11} = 1$ , so [F2] gives $Γ^{1}_{11} = 0$ . By [F6], $\nabla_{\partial_{x}} \partial_{x} = 0$ ; [F4] then reduces the geodesic equation for $γ (t) = t$ to $\ddot{γ} = 0$ . Thus $γ$ is a nonconstant, affinely parametrized geodesic.

2.2

Compute that the curvature component vanishes, so the Jacobi curvature term is zero. [step 1.1, F2, F3, algebra] There is only one curvature component. In [F3] its two derivative terms cancel because $i = j = 1$ , and its two product terms cancel for the same reason; indeed all Christoffel symbols are zero by [F2]. Hence $R^{1}_{111} = 0$ and $R (J, \dot{γ}) \dot{γ} = 0$ .

2.3

Define instead $\overset{ˉ}{J} (t) = t^{2} \partial_{x} ∣_{0}$ along the constant geodesic $\overset{ˉ}{γ} (t) = 0$ . [step 1.1, F1, F2, F3, F4, F5, F6, algebra] Here [F2]-[F3] again give $Γ = 0$ and $R = 0$ , [F4] says $\overset{ˉ}{γ}$ is geodesic, and [F5]-[F6] give $D_{t}^{2} \overset{ˉ}{J} = 2 \partial_{x} ∣_{0} \neq 0$ . Thus [F1] shows $\overset{ˉ}{J}$ is not Jacobi along $\overset{ˉ}{γ}$ .

3.1

Compute the nonzero Jacobi residual for $J$ by pullback covariant differentiation. [step 2.1, step 2.2, F1, F5, F6, algebra] From [F5] and [F6], $D_{t} \partial_{x} = 0$ along $γ$ . The product rule therefore gives $D_{t} J = 2 t \partial_{x}$ and $D_{t}^{2} J = 2 \partial_{x}$ . By step 2.2, $D_{t}^{2} J + R (J, \dot{γ}) \dot{γ} = 2 \partial_{x} \neq 0,$ so [F1] shows $J$ is not a Jacobi field.

4.1

The nonzero residual in step 3.1 refutes the universal claim. [step 1.1, step 2.3, step 3.1, F1, F5, given] In dimension zero every field along a geodesic is zero, and on the empty manifold there is no geodesic; neither case changes the one-dimensional counterexample. The zero field itself satisfies the Jacobi equation, but the specified field does not. The residual is nonzero throughout $[0, 1]$ , including its one-sided endpoint values. The examples are fixed data, require no choice, and make no if-and-only-if claim. ∎

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Sources

John M. Lee, Riemannian Manifolds: An Introduction to Curvature (1997) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (2025) (standard reference, not scraped)