Bonnet-Myers Theorem

The Bonnet-Myers theorem shows that a complete Riemannian manifold with positive Ricci curvature lower bound must be compact with bounded diameter. It is one of the fundamental results connecting curvature to global topology.

Statement

Theorem8.6Bonnet-Myers theorem

Let $(M^n, g)$ be a complete Riemannian manifold with Ricci curvature satisfying $\mathrm{Ric} \geq (n-1)\kappa g$ for some constant $\kappa > 0$ . Then:

The diameter of $M$ is bounded: $\mathrm{diam}(M) \leq \frac{\pi}{\sqrt{\kappa}}$ .
$M$ is compact.
The fundamental group $\pi_1(M)$ is finite.

Proof

Step 1: Diameter bound. Let $p, q \in M$ with $d(p,q) = L$ . By Hopf-Rinow (completeness), there exists a minimizing unit-speed geodesic $\gamma: [0, L] \to M$ from $p$ to $q$ .

Choose parallel orthonormal fields $E_2(t), \ldots, E_n(t)$ along $\gamma$ perpendicular to $\gamma'(t)$ . Consider the variation vector fields $V_i(t) = \sin\left(\frac{\pi t}{L}\right) E_i(t)$ for $i = 2, \ldots, n$ . Since $\gamma$ is minimizing, the second variation of energy is non-negative:

\frac{d^2 E}{ds^2}\Big|_{s=0} \geq 0 \quad \text{for any fixed-endpoint variation.}

Computing the second variation for $V_i$ :

I(V_i, V_i) = \int_0^L \left(|\nabla_{\gamma'} V_i|^2 - g(R(V_i, \gamma')\gamma', V_i)\right) dt.

We have $\nabla_{\gamma'} V_i = \frac{\pi}{L}\cos\frac{\pi t}{L} E_i$ (since $E_i$ is parallel), so $|\nabla_{\gamma'} V_i|^2 = \frac{\pi^2}{L^2}\cos^2\frac{\pi t}{L}$ .

Summing over $i = 2, \ldots, n$ :

\sum_{i=2}^n I(V_i, V_i) = \int_0^L \sin^2\frac{\pi t}{L}\left(\frac{(n-1)\pi^2}{L^2}\frac{\cos^2\frac{\pi t}{L}}{\sin^2\frac{\pi t}{L}} - \mathrm{Ric}(\gamma', \gamma')\right) dt.

Wait -- more carefully: $\sum_i I(V_i, V_i) = \int_0^L \left(\frac{(n-1)\pi^2}{L^2}\cos^2\frac{\pi t}{L} - \sin^2\frac{\pi t}{L}\mathrm{Ric}(\gamma', \gamma')\right) dt$ .

Using $\mathrm{Ric}(\gamma', \gamma') \geq (n-1)\kappa$ :

\sum_i I(V_i, V_i) \leq (n-1)\int_0^L \left(\frac{\pi^2}{L^2}\cos^2\frac{\pi t}{L} - \kappa\sin^2\frac{\pi t}{L}\right) dt = \frac{(n-1)}{2}\left(\frac{\pi^2}{L} - \kappa L\right).

If $L > \pi/\sqrt{\kappa}$ , then $\sum_i I(V_i, V_i) < 0$ , so at least one $I(V_i, V_i) < 0$ , contradicting the minimality of $\gamma$ . Hence $L \leq \pi/\sqrt{\kappa}$ .

Step 2: Compactness. The diameter bound implies $M = \bar{B}_{\pi/\sqrt{\kappa}}(p)$ is a closed bounded set. By Hopf-Rinow (completeness implies Heine-Borel), this is compact.

Step 3: Finite fundamental group. The universal cover $\tilde{M}$ inherits the Riemannian metric with the same Ricci bound. By Steps 1-2, $\tilde{M}$ is compact. The fundamental group $\pi_1(M)$ acts on $\tilde{M}$ by deck transformations (isometries). Since $\tilde{M}$ is compact, the fibers of $\tilde{M} \to M$ are finite, so $\pi_1(M)$ is finite. $\blacksquare$

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Applications

ExampleDiameter bounds

$S^n$ with $K = 1$ : $\mathrm{Ric} = (n-1)g$ , so $\kappa = 1$ and $\mathrm{diam} \leq \pi$ . Equality holds.
$S^n(r)$ with $K = 1/r^2$ : $\mathrm{Ric} = \frac{n-1}{r^2}g$ , so $\mathrm{diam} \leq \pi r$ . Again equality holds.
Any Einstein manifold with $\mathrm{Ric} = \lambda g$ , $\lambda > 0$ , has $\mathrm{diam} \leq \pi\sqrt{\frac{n-1}{\lambda}}$ and is compact.

RemarkRigidity: Cheng's theorem

Cheng's maximal diameter theorem states that if $\mathrm{Ric} \geq (n-1)\kappa > 0$ and $\mathrm{diam}(M) = \pi/\sqrt{\kappa}$ (equality), then $M$ is isometric to the sphere $S^n(1/\sqrt{\kappa})$ . This is a rigidity result: the sphere is the only manifold achieving the maximal diameter.