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'<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.

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Comments

  • gbbrian  On April 29, 2013 at 10:57 pm

    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}.

  • gbbrian  On April 29, 2013 at 11:00 pm

    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.

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