**Thm:(Meyers)** Suppose is a dimensional complete manifold and

Then hence is compact and is finite.

This theorem is proved by the following lemma.

**Lemma:** Suppose , then every geodesic which is longer than on must have conjugate points hence not a minimal geodesic.

**Proof:** Suppose is a mininal normal geodesic with length , is the arc length parameter. Then there exists such that .

For , there exists an orthonormal basis . Let be the parallel displacement of along , is the one of .

Define . Then the second variation of along is

So

So there exists a , for such , this means is locally maximal. This is contradiction, because we choose is a minimal geodesic.

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## Comments

There is an easier way to prove the Meyers’ theorem. The mean curvature of the geodesic sphere with radius is alway less than or equal to that of spheres with the same radius in of constant curvature K, which is easily to be seen as . It goes to as , so the diameter of must be less than .

Also when the diameter is equal to , then the manifold is isometric to the standard sphere, which is Cheng’s well-known maximal diameter theorem. Rong might ask you such a question.