## Meyer’s Thm

Thm:(Meyers) Suppose ${M}$ is a ${n-}$dimensional complete manifold and

$\displaystyle \text{Ric}_M\geq (n-1)K>0$

Then ${\text{diam}M<\frac{\pi}{\sqrt{K}}}$ hence ${M}$ is compact and ${\pi_1(M)}$ is finite.

This theorem is proved by the following lemma.

Lemma: Suppose ${\text{Ric}_M\geq (n-1)K>0}$, then every geodesic which is longer than ${\displaystyle \frac{\pi}{\sqrt{K}}}$ on ${M}$ must have conjugate points hence not a minimal geodesic.

Proof: Suppose ${\gamma(t):[0,l]\rightarrow M}$ is a mininal normal geodesic with length ${\displaystyle l>\frac{\pi}{\sqrt{K}}}$, ${t}$ is the arc length parameter. Then there exists ${\displaystyle K'=\left(\frac{\pi}{l}\right)^2}$ such that ${K'.

For ${T_{\gamma(0)}M}$, there exists an orthonormal basis ${\{\frac{\partial }{\partial t}, V_i, i=1,\cdots,n-1\}}$. Let ${V_i(t)}$ be the parallel displacement of ${V_i}$ along ${\gamma}$, ${T}$ is the one of ${\frac{\partial }{\partial t}}$.

Define ${W_i=V_i(t)\sin \sqrt{K'}t }$. Then the second variation of ${\gamma}$ along ${W_i}$ is

$\displaystyle I(W_i,W_i)=\int_{0}^l\langle\nabla_TW_i, \nabla_TW_i \rangle+R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l-\langle W_i, \nabla_T\nabla_TW_i\rangle+ R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l K'\sin^2\sqrt{K'}t+\sin^2\sqrt{K'}tR(T,V_i,T,V_i)dt$

So

$\displaystyle \sum_1^{n-1} I(W_i,W_i)=\int_0^l[(n-1)K'-\text{Ric}_M(V_i)]\sin^2\sqrt{K'}tdt<0$

So there exists a ${I(W_i,W_i)<0}$, for such ${i}$, this means ${\gamma}$ is locally maximal. This is contradiction, because we choose ${\gamma}$ is a minimal geodesic.

There is an easier way to prove the Meyers’ theorem. The mean curvature $H_r$ of the geodesic sphere with radius $r$ is alway less than or equal to that of spheres with the same radius in $S^n$ of constant curvature K, which is easily to be seen as $(n-1)\cos( \sqrt{K}r)/ \sin(\sqrt{K}r)$. It goes to $-\infty$ as $r\to \pi/\sqrt{K}$, so the diameter of $(M,g)$ must be less than $\pi/\sqrt{K}$.
Also when the diameter is equal to $\pi/\sqrt{K}$, 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.