A geodesic on a Riemannian manifold locally minimizes length. Does a geodesic always globally minimize length between its endpoints?
AYes — geodesics are always the shortest paths
BNo — geodesics are only locally length-minimizing; beyond the cut point, shorter paths may exist
CNo — geodesics maximize length, not minimize it
DGeodesics minimize length if and only if the manifold has non-negative curvature
Geodesics are locally length-minimizing: any sufficiently short segment is the shortest path between its endpoints. But globally, a geodesic may cease to minimize. On the sphere, a great-circle arc shorter than a half-circle is minimizing, but past the antipodal point, there are shorter paths going the other way. The cut point is where a geodesic first fails to minimize. Between a point and its cut point, the geodesic is the unique shortest path. Beyond it, shorter paths exist.
Question 2 Multiple Choice
The geodesic equation in coordinates is d²γᵏ/dt² + Γᵏᵢⱼ (dγⁱ/dt)(dγʲ/dt) = 0. This is a second-order ODE, so geodesics are determined by...
AA starting point p only
BA starting point p and initial velocity v ∈ TpM
CTwo distinct points p, q on the manifold
DA starting point p, initial velocity v, and the curvature at p
As a second-order ODE, the geodesic equation has a unique solution given initial position γ(0) = p and initial velocity γ'(0) = v. This is analogous to Newton's second law: the trajectory of a particle is determined by its initial position and velocity. By contrast, two points do NOT uniquely determine a geodesic — on a sphere, there are infinitely many great circles through two non-antipodal points (well, exactly one great circle, but antipodal points have infinitely many). The curvature enters through the Christoffel symbols, not as separate initial data.
Question 3 Short Answer
On a Riemannian manifold, geodesics are both locally length-minimizing curves and curves that parallel-transport their own velocity vector. Why are these two characterizations equivalent?
Think about your answer, then reveal below.
Model answer: The length-minimizing characterization comes from the calculus of variations: critical points of the length (or energy) functional satisfy the Euler-Lagrange equation, which turns out to be exactly the geodesic equation ∇_{γ'}γ' = 0. The parallel-transport characterization says the velocity is 'constant' along the curve — the curve has zero acceleration. Both descriptions yield the same ODE. The connection is that a curve with constant-speed and zero acceleration is precisely one that cannot be shortened by small perturbations.
Technically, geodesics are critical points of the energy functional E(γ) = ½∫g(γ',γ')dt, not always minima (they could be saddle points). The equivalence between ∇_{γ'}γ' = 0 and the Euler-Lagrange equation for E is a direct computation in local coordinates. The energy functional is preferred over the length functional because it gives a nicer (non-degenerate) second variation and its critical points are automatically constant-speed.
Question 4 True / False
All geodesics on a compact Riemannian manifold are defined for all time (complete).
TTrue
FFalse
Answer: True
This is the Hopf-Rinow theorem: a Riemannian manifold is geodesically complete (all geodesics extend to all time) if and only if it is complete as a metric space. Compact manifolds are complete, so they are geodesically complete. On non-compact manifolds, geodesics can fail to be complete — for instance, on the punctured plane ℝ² {0}, geodesics aimed at the origin cannot be extended past the missing point.