Geodesic equations and exponential maps
Derive zero covariant acceleration, solve the local initial-value problem, and distinguish the exponential map's local invertibility from global uniqueness.
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The question for today
What you’ll understand
- Write the affinely parameterized geodesic equation.
- Define the exponential map with its initial-velocity convention.
- Identify why a geodesic can continue without remaining globally minimizing.
A straight route on a manifold is defined through its own metric rather than an ambient picture. Geodesics have zero covariant acceleration, which becomes a second-order differential equation in coordinates. Initial position and velocity determine a local solution. Gathering those solutions into the exponential map creates useful normal coordinates, but does not promise that all geodesic routes are shortest or that every endpoint has a unique initial velocity.
Zero intrinsic acceleration is an ODE
An affinely parameterized geodesic satisfies . Expanding the covariant derivative gives the coordinate equation below. Smooth Christoffel coefficients turn this into a smooth first-order system for position and velocity, so standard ODE existence and uniqueness applies locally to every initial tangent vector. The parameter is part of the equation: affine changes of time preserve geodesic form, whereas an arbitrary nonlinear reparameterization can introduce tangential acceleration.
Metric compatibility makes the squared speed constant along a geodesic. The first variation of energy shows that fixed-endpoint critical curves are precisely these affinely parameterized geodesics. Being a critical point does not itself mean a global minimum. Sufficiently short geodesic segments minimize distance locally, but continuation past a cut point can lose minimality even though the same smooth geodesic equation continues to hold. Global route selection therefore requires information beyond local ODE uniqueness.
Initial velocities become local coordinates
For , let be the geodesic with initial position p and initial velocity v. When it exists up to time one, define . Its domain contains an open neighborhood of the zero vector and is star shaped under scaling toward zero. The zero vector maps to p. Smooth dependence on ODE initial data makes the exponential smooth, and its derivative at zero is the identity after identifying the relevant tangent spaces naturally.
The inverse function theorem therefore provides a neighborhood on which the exponential is a diffeomorphism, giving normal coordinates near p. Radial lines in this tangent-space neighborhood map to geodesics. The radius of useful injectivity can be limited by singular derivatives or competing routes. Completeness will make the exponential defined on the entire tangent space, but it will not in general make it globally injective. A sphere is the clearest example of these distinct statements.
A geodesic can travel beyond a shortest route
On the unit sphere, geodesics with nonzero speed are great circles. Every initial direction of length π reaches the antipodal point, so the exponential at p is far from injective there. Great circles can then continue indefinitely and return to previously visited points. This does not violate the initial-value uniqueness theorem, which fixes the entire initial velocity rather than only the endpoint. Different initial velocities can lead to the same endpoint.
The geodesic equation also must be interpreted on the correct manifold. A circle drawn in the Euclidean plane has nonzero intrinsic acceleration as a plane curve. The same circular motion, regarded as motion on the one-dimensional circle with its induced metric, is a geodesic when its angular speed is constant. Tangent spaces and the connection have changed. An extrinsic picture of a bent curve is therefore insufficient to decide whether its intrinsic covariant acceleration vanishes.
Work it through
Two spherical velocities with one endpoint
At the north pole p=(0,0,1) of the unit sphere, compare initial tangent vectors and . Find their exponential endpoints and route lengths.
The sphere formula with r=π/2 gives cosine zero and sine one, so expₚ(v)=(1,0,0). The geodesic has constant speed π/2 during the unit time interval, hence length π/2.
For w, cosine of 5π/2 is again zero and sine is one. Its unit tangent direction is the same, so expₚ(w) also equals (1,0,0), while its route length is 5π/2.
The shorter arc realizes the distance π/2; the longer route loops once around the great circle before arriving. Both solve the geodesic equation. Equal endpoints do not imply equal initial velocities or equal route lengths.
The exponential map has the same endpoint for these two velocities, although only the shorter route is minimizing.
The living notebook
Explore the model
Interactive lab · vectors
A circle endpoint does not remember winding
Pause and Reset, then set angle = 180° and magnitude = 1. Predict the vector tip is (−1,0). Reset and set both angle = −180° and magnitude = 1; predict the same tip while the signed angular parameters differ by a full turn. Interpret this as the unit-circle endpoint map exp(iθ), whose two half-circle geodesic routes differ. The instrument plots a planar vector; it does not solve a sphere's geodesic ODE or show a tangent-space exponential map on a general manifold.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forLocal uniqueness from initial data and global nonuniqueness of routes to an endpoint can coexist.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the Euclidean plane with polar metric , test the radial curve r=t, θ=0 for t>0 and the circle r=1, θ=t. Which is a plane geodesic?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The radial equation is r¨−rθ̇²=0 and the angular equation is θ¨+2ṙθ̇/r=0. Evaluate both, rather than looking only at how the curve is drawn.
Compare with the solution
For the radial curve, ṙ=1, r¨=0, and both angular derivatives vanish, so both equations hold. For the circle, ṙ=r¨=0 and θ̇=1, so the radial left side equals −1 and the geodesic equation fails. The circle is not a geodesic of the plane, though constant-speed motion is a geodesic of the circle considered as its own one-dimensional Riemannian manifold. This distinction depends on the intrinsic connection used in each problem.
Keep the idea
Take it with you
- Geodesics solve zero covariant acceleration with an affine parameter.
- The exponential map is locally invertible at zero by its identity derivative.
- Complete geodesics can revisit points and cease to minimize distance globally.
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