Myers's theorem

For a connected Riemannian manifold of dimension $n\ge 2$, assume geodesic completeness and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$. The Riemannian distance then has diameter at most $\pi /\sqrt{\kappa }$; the manifold is compact and its fundamental group is finite. The dimension restriction excludes the vacuous one-dimensional Ricci bound.

The graph $z=x^2+y^2$ in Euclidean three-space has the displayed induced Riemannian metric and positive Gaussian curvature. It is smooth at its vertex; the polar-coordinate singularity is only a coordinate artifact. The metric dominates the Euclidean plane metric, so upward stability of Riemannian completeness proves completeness. It has infinite diameter. In dimension two Ricci curvature equals Gaussian curvature times the metric, so this is a counterexample to compactness under merely pointwise positive Ricci curvature. Its curvature tends to zero at infinity.

If a connected complete Riemannian manifold has $\mathrm{Ric}\ge (n-1)\kappa g$ with $\kappa >0$, its universal cover with the lifted metric is complete and satisfies the same bound. The Bonnet-Myers theorem makes that cover compact. A covering fiber is closed and discrete and hence finite; the number of its elements equals the order of the fundamental group. Applying the diameter theorem only to the base would not prove this conclusion: the positive lower bound must also be used on the cover.

This metric pulls back the round spherical metric by $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$. It has unit Gaussian curvature and Ricci curvature equal to its metric, yet the meridian reaches a missing boundary in finite time. Every curve between $(0,0)$ and $(0,L)$ has length at least $|L|/\sqrt{2}$, so its diameter is infinite. It shows that the positive Ricci bound alone does not replace geodesic completeness in the Bonnet-Myers theorem.