ProofComplete

Proof of the Hopf-Rinow Theorem

The Hopf-Rinow theorem connects metric completeness, geodesic completeness, and compactness properties of Riemannian manifolds. It is a cornerstone of global Riemannian geometry.


Statement

Theorem6.7Hopf-Rinow theorem

Let (M,g)(M, g) be a connected Riemannian manifold. The following are equivalent:

  1. (M,d)(M, d) is complete as a metric space (every Cauchy sequence converges).
  2. Every geodesic can be extended to all of R\mathbb{R} (geodesic completeness).
  3. For some point pMp \in M, the exponential map expp\exp_p is defined on all of TpMT_pM.
  4. Every closed bounded subset of MM is compact (Heine-Borel property).

Moreover, if any of these holds, then for any two points p,qMp, q \in M, there exists a minimizing geodesic from pp to qq (i.e., a geodesic whose length equals d(p,q)d(p,q)).


Proof

Proof

We prove the key implications.

(3) \Rightarrow (4) and the existence of minimizing geodesics. Fix pMp \in M and assume expp\exp_p is defined on all of TpMT_pM. Let qMq \in M with d(p,q)=rd(p,q) = r. We show there exists a minimizing geodesic from pp to qq.

Choose a small ϵ>0\epsilon > 0 such that expp\exp_p is a diffeomorphism on Bϵ(0)TpMB_\epsilon(0) \subset T_pM. On the geodesic sphere Sϵ=expp({v:v=ϵ})S_\epsilon = \exp_p(\{v : |v| = \epsilon\}), by compactness, there exists a point x0Sϵx_0 \in S_\epsilon minimizing d(x0,q)d(x_0, q).

Claim: d(p,q)=ϵ+d(x0,q)d(p, q) = \epsilon + d(x_0, q).

Any path from pp to qq must cross SϵS_\epsilon, and the shortest way to reach SϵS_\epsilon from pp is the radial geodesic of length ϵ\epsilon. So d(p,q)ϵ+minxSϵd(x,q)=ϵ+d(x0,q)d(p,q) \geq \epsilon + \min_{x \in S_\epsilon} d(x, q) = \epsilon + d(x_0, q). The reverse inequality follows from the triangle inequality.

Let γ\gamma be the unit-speed geodesic from pp through x0x_0. Define A={t[0,r]:d(p,γ(t))+d(γ(t),q)=r}A = \{t \in [0, r] : d(p, \gamma(t)) + d(\gamma(t), q) = r\}. Then ϵA\epsilon \in A and AA is closed. The key step: if t0At_0 \in A, repeat the argument at γ(t0)\gamma(t_0) to show t0+δAt_0 + \delta \in A for small δ>0\delta > 0. This shows A=[0,r]A = [0, r], so γ(r)\gamma(r) satisfies d(p,γ(r))=rd(p, \gamma(r)) = r and d(γ(r),q)=0d(\gamma(r), q) = 0, hence γ(r)=q\gamma(r) = q, and γ\gamma is minimizing.

For the Heine-Borel property: the closed ball Bˉr(p)={q:d(p,q)r}=expp({v:vr})\bar{B}_r(p) = \{q : d(p,q) \leq r\} = \exp_p(\{v : |v| \leq r\}) is the continuous image of a compact set, hence compact. Any closed bounded set is a closed subset of some Bˉr(p)\bar{B}_r(p), hence compact.

(4) \Rightarrow (1): Heine-Borel implies completeness in any metric space.

(1) \Rightarrow (2): If a geodesic γ\gamma cannot be extended past time TT, then γ(tn)\gamma(t_n) for tnTt_n \to T is Cauchy (since d(γ(s),γ(t))std(\gamma(s), \gamma(t)) \leq |s - t|), so it converges to a limit p0p_0. By the local existence of geodesics near p0p_0, we can extend γ\gamma past TT, contradiction.

(2) \Rightarrow (3): Immediate from the definition of the exponential map. \blacksquare


Applications

ExampleComplete manifolds
  • Rn\mathbb{R}^n, SnS^n, Hn\mathbb{H}^n, and any compact Riemannian manifold are complete.
  • An open ball B1(0)RnB_1(0) \subset \mathbb{R}^n with the Euclidean metric is not complete (geodesics reach the boundary in finite time).
  • Any closed submanifold of RN\mathbb{R}^N (with the induced metric) is complete.
RemarkCompleteness in physics

In general relativity, the analogue of geodesic completeness is crucial: a spacetime is geodesically incomplete if some freely falling observer reaches the "edge of spacetime" in finite proper time. The Penrose-Hawking singularity theorems show that, under natural energy conditions, spacetimes are generically geodesically incomplete, implying the existence of singularities (like black holes or the Big Bang).