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Jacobi fields in constant sectional curvature

Example

Assume exactly ACω through the declared constant-curvature interface. Let (M,g) be a finite-dimensional Riemannian manifold without boundary of constant sectional curvature K∈R, let I⊆R be a nondegenerate interval containing 0, and let γ:I→M be a unit-speed affinely parametrized geodesic. Put T=γ˙ and define

sK(t)={sin⁡(K t)/K,K>0,t,K=0,sinh⁡(−K t)/−K,K<0.

Then the normal Jacobi fields J with J(0)=0 are exactly J(t)=sK(t)E(t), where E is a unique parallel normal field along γ. The tangential Jacobi fields are exactly J(t)=(at+b)γ˙(t),a,b∈R. No completeness is assumed. If an endpoint of I is included, derivatives there are one-sided.

Facts & Assumptions

Given: The constant sectional curvature K, a nondegenerate interval I containing 0, and a supplied unit-speed affine geodesic γ on I.

[A1]

ACω is countable choice. Its only role here is inherited through the constant-sectional-curvature and curvature-tensor interfaces; the coordinate computations and the unique extensions of supplied vectors use no further choice, and no full AC is assumed (The Axiom of Countable Choice (ACω), Constant sectional curvature and space form, Curvature tensor of constant sectional curvature).

[F1]

Constant sectional curvature K means that every tangent two-plane has sectional curvature K. In dimensions zero and one this predicate is vacuous for every K (Constant sectional curvature and space form).

[F2]

Under this hypothesis, the curvature operator is R(X,Y)Z=K(g(Y,Z)X−g(X,Z)Y) (Curvature tensor of constant sectional curvature).

[F3]

A Jacobi field is a smooth field satisfying Dt2J+R(J,T)T=0 throughout the interval (Jacobi field).

[F4]

The affine geodesic equation gives DtT=0, and unit speed gives g(T,T)=1 (Geodesic of an affine connection, given).

[F5]

Levi-Civita metric compatibility, expressed in a local frame with metric matrix H and connection matrix B, gives H′=BTH+HB. The along-curve frame formula is Dt(eu)=e(u′+Bu), and covariant differentiation also satisfies Dt(fV)=f′V+fDtV. Therefore, for fields U,V along γ, (g(U,V))′=g(DtU,V)+g(U,DtV). (Levi civita connection, Metric compatible connection on a riemannian vector bundle, Local frame formula for covariant differentiation along a curve, Covariant derivative along a curve)

[F6]

Every supplied initial vector along γ has a unique parallel extension to all of I; a field is parallel when DtE=0 (Existence and uniqueness of parallel sections, Parallel section along a curve).

[F7]
[F8]

Initial value and covariant derivative at 0 determine a unique Jacobi field along all of I, including when 0 is an included endpoint (Existence and uniqueness of jacobi fields from initial data).

[F9]

Since γ has unit speed it is nonconstant, and for every smooth f:I→R, fT is Jacobi exactly when f(t)=at+b for constants a,b (Tangential jacobi fields are affine multiples of the velocity).

Verification

technique · parallel transport reduces the normal Jacobi equation to a scalar initial-value equation
1.1givenalgebra

Direct differentiation in the three sign cases gives sK′′+KsK=0, sK(0)=0, and sK′(0)=1 on the full interval, regardless of later zeros of sK.

1.2F9given

Unit speed makes γ nonconstant, so [F9] proves both directions of the tangential classification: a tangential Jacobi field has affine coefficient and every affine coefficient gives a Jacobi field.

2.1F2F3F4F5F6step 1.1

For any parallel normal E, the product rule gives Dt(sKE)=sK′E and Dt2(sKE)=sK′′E; [F2] and [F4] give R(sKE,T)T=K(g(T,T)sKE−g(sKE,T)T)=KsKE, so [F3] and step 1.1 make it Jacobi, normal, and zero at 0.

3.1F4F5F6F7F8step 1.1step 2.1

If J is normal Jacobi with J(0)=0, differentiating g(J,T)=0 at 0 using [F5] and [F4] shows W=DtJ(0)⊥T(0); extend W uniquely to a parallel E by [F6]. Then h=g(E,T) has zero derivative by [F5], is zero at 0, and so vanishes on I by [F7]. Since [F6] gives E(0)=W, steps 1.1 and 2.1 give (sKE)(0)=0 and Dt(sKE)(0)=sK′(0)E(0)=W. Jacobi initial-data uniqueness [F8] now gives J=sKE on all of I, and [F6] makes E unique.

4.1A1F1F3F4F6F8F9step 1.1step 3.1step 1.2∎

A supplied geodesic rules out empty M; dimension zero has no unit-speed geodesic, and in dimension one the normal subspace is zero, so the normal case is J=E=0 while the tangential result remains valid. [F1] records the vacuous low-dimensional curvature convention. The interval is nondegenerate; included endpoints use one-sided derivatives; W=0 gives E=J=0; the zero tangential field has a=b=0; and constant geodesics are excluded by unit speed. The three signs of K are covered in step 1.1. Exactly the inherited ACω of [A1] is assumed, and no choice beyond unique extension of a supplied vector is made.

Source notes

Lee, Riemannian Manifolds, Lemma 10.8 and proof, printed pp.179–180 / PDF labels P195–196, lines 6974–7020, derives the normal equation from the constant-curvature tensor formula and solves the scalar equation. His proof uses the dimension of the normal Jacobi space to conclude exhaustiveness; the argument above instead matches the initial derivative and uses the library's Jacobi initial-value uniqueness theorem. Datar, Lectures on Riemannian Geometry, Proposition 24.1.1 and proof, printed pp.174–175 / PDF labels P181–182, lines 9810–9848, gives the same normal-field formula for a complete space form. Completeness is part of Datar's setting and is not used here. The tangential branch is supplied by the separately cited library proposition.

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Sources