Parallel transport and holonomy

Try

Carry a tangent arrow around a latitude on the unit sphere while requiring its covariant derivative to vanish. After returning to the starting point, the arrow can have rotated relative to its original direction. Parallel transport is locally “no intrinsic turning,” but its result can depend on the path.

For a curve γ(t), a tangent field V is parallel when D_tV=0. In coordinates this is a linear ODE for its components. Smooth coefficients and an initial tangent vector give a unique transported field along the chosen interval.

Predict

Predict whether a parallel field keeps its length. Metric compatibility makes the answer yes. Predict whether its ambient Cartesian coordinates remain constant. On a curved surface a constant ambient vector may not remain tangent, so that is generally impossible.

Predict whether a closed loop must return every arrow unchanged. That would impose a strong flatness/path condition, not merely the definition of parallelism.

Experiment

Use the standard oriented sphere frame eθ,eφ away from the poles. Its connection form is ω=cos θ dφ.

Latitude colatitude θ	Frame angle change after a positive turn	Equivalent rotation modulo 2π
π/2	0	0
π/3	−π	π
π/6	−√3π	Reduce modulo 2π

The angle is measured relative to the returning latitude frame. Do not compare unnormalized coordinate basis vectors: eφ has been divided by the metric length of the angular coordinate vector.

Understand

For parallel V and W along the same curve, metric compatibility gives d⟨V,W⟩/dt=⟨D_tV,W⟩+⟨V,D_tW⟩=0. Thus lengths and mutual angles are preserved. In coordinates, differentiating V=V^jX_j and projecting tangentially gives dV^k/dt+Γ^k_(ij)u′iV^j=0. This linear equation is the computational transport rule.

On the unit sphere, take eθ=(cos θ cos φ,cos θ sin φ,−sin θ) and eφ=(−sin φ,cos φ,0). Tangent-projecting their derivatives gives ∇eθ=ωeφ and ∇eφ=−ωeθ with ω=cos θ dφ.

Write a unit field V=cos α eθ+sin α eφ. Differentiation shows its covariant derivative is (α′+ω(γ′)) times the perpendicular unit tangent field. Parallelism therefore requires α′=−ω(γ′). Around a latitude, integrate with θ fixed to obtain Δα=−2πcos θ. The result is a rotation understood modulo 2π.

Apply

Repair “Changing components means the field is not parallel.” Even in the plane, a constant Cartesian vector has varying polar components. Those component derivatives cancel the connection correction. Test the full covariant derivative.

For optional depth, holonomy is the linear isometry of the starting tangent plane obtained by parallel transport around a loop. Small-loop holonomy is controlled by curvature; a later moving-frame calculation will relate its angle to an integral of K. The latitude result already supplies an exact example but does not prove the general theorem.

Path dependence is also global. A flat geometry can have nontrivial topology, and statements about path independence need a domain and homotopy hypothesis. We do not infer global triviality from a single zero-curvature chart.

Continue

Record the loop, frame, initial vector and modulo-angle convention in a transport calculation. Continue to geodesic equations to apply the same derivative to the velocity of the path itself.