Mathematics.

global geometry

Complete Riemannian Manifolds

Differential Geometry65 minDifficulty8 out of 10

You should know: geodesics riemannian, riemannian metric

Overview

A Riemannian manifold (M, g) is complete if every geodesic can be extended to all time — equivalently, by the Hopf-Rinow theorem, if it is complete as a metric space (every Cauchy sequence converges). Complete manifolds are the natural setting for global differential geometry: they are the spaces where geodesics connect any two points. The Hopf-Rinow theorem equates five different notions of completeness, and the theorem of Hadamard-Cartan shows that nonpositive curvature implies simple topology.

Intuition

Think of a manifold as a surface you can walk on. Completeness means the surface has no 'edges' or 'holes' you can fall off — you can always keep walking in any direction forever. A punctured plane ℝ² \ {0} is not complete: a geodesic heading toward the origin 'falls off' at finite time. The unit sphere S² IS complete: great circles go on forever (wrapping around). Completeness is the right condition for global geometry to work properly.

Formal Definition

Definition

Let (M, g) be a Riemannian manifold. The Riemannian distance is d(p,q) = inf{L(γ) : γ a piecewise smooth curve from p to q}. (M,g) is complete in several equivalent senses:

Geodesically complete: every geodesic γ:[0,a)→M extends to γ:R→M\text{Geodesically complete: every geodesic } \gamma: [0,a) \to M \text{ extends to } \gamma: \mathbb{R} \to M
Geodesic completeness
Metrically complete: (M,d) is a complete metric space\text{Metrically complete: } (M, d) \text{ is a complete metric space}
Metric completeness (Cauchy sequences converge)
exp⁡p:TpM→M is defined on all of TpM for some (hence every) p∈M\exp_p: T_pM \to M \text{ is defined on all of } T_pM \text{ for some (hence every) } p \in M
Exponential map completeness

Theorems

Theorem 1: Hopf-Rinow Theorem
For a connected Riemannian manifold (M,g), the following are equivalent:(1)  M is geodesically complete.(2)  M is complete as a metric space.(3)  exp⁡p is defined on all TpM for some p.(4)  Every closed bounded subset of M is compact.Moreover, any of these implies: any two points are joined by a length-minimizing geodesic.\text{For a connected Riemannian manifold } (M,g), \text{ the following are equivalent:}\newline (1)\; M \text{ is geodesically complete.}\newline (2)\; M \text{ is complete as a metric space.}\newline (3)\; \exp_p \text{ is defined on all } T_pM \text{ for some } p.\newline (4)\; \text{Every closed bounded subset of } M \text{ is compact.}\newline \text{Moreover, any of these implies: any two points are joined by a length-minimizing geodesic.}
Theorem 2: Hadamard-Cartan Theorem
If (M,g) is a complete simply-connected Riemannian manifold with K≤0 everywhere, then exp⁡p:TpM→M is a diffeomorphism for every p, so M≅Rn diffeomorphically.\text{If } (M,g) \text{ is a complete simply-connected Riemannian manifold with } K \leq 0 \text{ everywhere, then } \exp_p: T_pM \to M \text{ is a diffeomorphism for every } p, \text{ so } M \cong \mathbb{R}^n \text{ diffeomorphically.}
Theorem 3: Bonnet-Myers Theorem
If (M,g) is a complete Riemannian manifold with Ricci curvature Ric≥(n−1)K>0, then diam(M)≤π/K and M is compact.\text{If } (M,g) \text{ is a complete Riemannian manifold with Ricci curvature } \mathrm{Ric} \geq (n-1)K > 0, \text{ then } \mathrm{diam}(M) \leq \pi/\sqrt{K} \text{ and } M \text{ is compact.}

Worked Examples

  1. 1

    In the Euclidean metric, geodesics are straight lines. Consider the straight line γ(t) = (1−t, 0) for t ∈ [0,1).

    γ(t)=(1−t, 0),t∈[0,1)\gamma(t) = (1-t,\, 0), \quad t \in [0,1)
  2. 2

    γ is a geodesic in ℝ² \ {0} for t ∈ [0,1) since (1−t, 0) ≠ (0,0) for t < 1. But as t → 1, γ(t) → (0,0) which is not in M.

    lim⁡t→1γ(t)=(0,0)∉M\lim_{t \to 1} \gamma(t) = (0,0) \notin M
  3. 3

    The geodesic cannot be extended beyond t = 1 within M. Hence M is not geodesically complete.

    γ cannot extend to t=1 in M=R2∖{0}\gamma \text{ cannot extend to } t = 1 \text{ in } M = \mathbb{R}^2 \setminus \{0\}
  4. 4

    Equivalently: the Cauchy sequence γ(1 − 1/n) = (1/n, 0) converges in ℝ² to (0,0), which is not in M. So M is not metrically complete (Hopf-Rinow equivalence).

    (1n,0)→(0,0)∉M\left(\frac{1}{n}, 0\right) \to (0,0) \notin M

✓ Answer

ℝ² \ {0} is not geodesically complete: the straight-line geodesic toward the origin reaches the missing point in finite time t = 1 and cannot be extended. By Hopf-Rinow, this equivalently shows it is not a complete metric space.

Practice Problems

Mediumfree response

State the Hopf-Rinow theorem and give one example of a manifold that is complete and one that is not.

Hardfree response

The Bonnet-Myers theorem says that if Ric ≥ (n−1)K > 0 then M is compact with diam(M) ≤ π/√K. What does this imply for the fundamental group π₁(M)?

Quiz

A Riemannian manifold is complete if and only if (Hopf-Rinow):
The Hadamard-Cartan theorem applies to complete simply-connected manifolds with:
The Bonnet-Myers theorem concludes that a complete manifold with Ric > 0 is:

Summary

  • A Riemannian manifold is complete if geodesics extend to all of ℝ; Hopf-Rinow shows this is equivalent to metric completeness and compactness of closed bounded sets.
  • Hopf-Rinow also guarantees: any two points in a complete manifold are joined by a minimizing geodesic.
  • Hadamard-Cartan: complete + simply connected + K ≤ 0 → M ≅ ℝⁿ (diffeomorphically); exp_p is a global diffeomorphism.
  • Bonnet-Myers: complete + Ric ≥ (n−1)K > 0 → M compact with diam ≤ π/√K; π₁(M) finite.
  • ℝ² \ {0} is a canonical non-complete example; compact manifolds are always complete.