Hopf–Rinow and geodesic completeness
Match metric and geodesic completeness under the correct manifold hypotheses, and separate existence of minimizing routes from completeness.
On this page
The question for today
What you’ll understand
- State the connected boundaryless finite-dimensional Hopf–Rinow theorem.
- Detect missing finite-distance endpoints in an incomplete metric.
- Compare two metrics on the same open interval using an explicit isometry.
Local geodesic existence does not prevent a route from reaching a missing endpoint in finite time. Metric completeness expresses the same concern through Cauchy sequences. Hopf–Rinow connects these analytic and geometric notions on finite-dimensional Riemannian manifolds and adds compactness of bounded regions. Its hypotheses matter, and its minimizing-geodesic consequence should not be reversed into an equivalent condition without checking the actual theorem.
Three equivalent forms of completeness
Let M be a connected smooth finite-dimensional Riemannian manifold without boundary, equipped with its intrinsic Riemannian distance. Hopf–Rinow says metric completeness is equivalent to geodesic completeness and to compactness of every closed bounded subset. Here metric completeness means every Cauchy sequence converges to a point of M. Geodesic completeness means every affinely parameterized geodesic extends for all real parameter values, not merely that one selected path can be continued.
Under these equivalent conditions, any pair of points is joined by a distance-minimizing geodesic, and each exponential map is defined on the full tangent space. Connectedness ensures finite distances between all pairs; if the manifold is disconnected, these statements are interpreted separately on components. The no-boundary assumption avoids geodesics simply hitting a physical boundary. The finite-dimensional Riemannian setting also supplies local compactness; a general complete metric space or infinite-dimensional Hilbert space need not have compact closed bounded balls.
A missing endpoint can still allow many shortest routes
The open interval (0,1) with the ordinary metric is incomplete: the sequence 1/n, starting with n≥2, is Cauchy but converges only to the missing endpoint zero in the ambient real line. The geodesic starting at one half with unit positive velocity reaches the other missing endpoint after time one half and cannot extend there as a curve in M. Both failures describe the same finite-distance escape.
Nevertheless, every pair of points inside this interval is joined by its straight segment, which realizes their distance. Thus the existence of a minimizing geodesic for every pair is a consequence of completeness, but by itself is not sufficient for completeness. Likewise, boundedness alone does not imply compactness: the entire interval is bounded and closed relative to itself, yet not compact. Statements about closedness must specify the ambient space; an excluded Euclidean boundary point is not restored by using relative topology.
Completeness depends on the metric
Keep the same smooth interval but replace its metric by the one displayed above. The coordinate u is now a global isometry to the entire Euclidean line. Approaching either missing coordinate endpoint requires infinite metric distance, so no geodesic can escape there in finite affine time. The same underlying smooth manifold can therefore carry an incomplete metric and a complete one. Completeness is not determined just by whether a chosen coordinate description looks bounded or has excluded endpoints.
The compactness conclusion is useful because minimizing sequences of routes or points can now be localized in compact metric balls. It does not make the manifold itself compact unless it is also bounded in intrinsic distance. Euclidean space is complete and unbounded. A round sphere is complete and bounded, hence compact. These examples separate local existence, global extendibility, boundedness, compactness, and existence of shortest routes rather than treating all five words as interchangeable.
Work it through
A bounded coordinate interval with infinite ends
On (0,1), use . Compute the distance from x=1/4 to x=3/4, and decide whether approaching x=1 takes finite or infinite distance.
In the isometric u-coordinate, x=1/4 gives u=−log 3 and x=3/4 gives u=log 3. Their distance is therefore 2log 3, approximately 2.1972.
As x tends to one, log(x/(1−x)) tends to positive infinity. Thus the distance from any fixed interior point to those approaching coordinates diverges, rather than ending at a finite-distance missing point.
The image is all of the complete real line, so this metric is complete and Hopf–Rinow applies. A unit-speed geodesic can be written x(t)=1/(1+exp(−(t+u₀))); it approaches a coordinate endpoint only as affine time tends to infinity.
The interior distance is 2log 3; both coordinate endpoints lie infinitely far away in this complete metric.
The living notebook
Explore the model
Interactive lab · optimization
Convergence in an ambient space is not completeness of a subset
Pause and Reset, then set step η = 0.5 and initial x₀ = 2. Predict four Steps give x₄=0.125, then check. Regard these positive positions as points of the Euclidean interval (0,3): their limiting point zero is excluded. Reset and set both η = 0.5 and x₀ = 1 to predict x₄=0.0625 and the same excluded limit. The instrument runs a quadratic descent recurrence, not a geodesic equation; it illustrates a Cauchy sequence whose ambient limit is missing from the chosen manifold.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. 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 forA convergent numerical sequence in ℝ can fail to converge inside an open subset equipped with its inherited metric.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the punctured Euclidean plane, consider p=(−1,0) and q=(1,0). Show their intrinsic distance is two but no curve attains it. Use a small semicircular detour and explain the completeness failure.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A route can follow the x-axis to −ε, use a semicircle of radius ε around the origin, then follow the axis from +ε. Equality in the Euclidean length bound requires the straight segment.
Compare with the solution
Every connecting curve has Euclidean length at least two. The described detour has length , so the infimum is two as ε tends to zero. Attaining length two would require the straight segment between p and q, which passes through the removed origin and is inadmissible. The punctured plane is incomplete because sequences approaching the origin can be Cauchy without a limit in the manifold. Its geodesics can also run into that missing point in finite time.
Keep the idea
Take it with you
- Hopf–Rinow equates metric completeness, geodesic completeness, and properness in its stated setting.
- Minimizing routes for all pairs alone do not guarantee completeness.
- Changing the Riemannian metric can move a coordinate endpoint from finite to infinite distance.
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