For a complete non-compact Riemannian manifold with sectional curvature positive, it is diffeomorphic to $\mathbb{R}^n$ by known result. Choose a point $p$ on the manifold, is it possible that the distance function $d(p, \cdot)$ has a sequence of critical points going to the infinity of the manifold?

  • $\begingroup$ Sectional curvature is positive or nonpositive? $\endgroup$ – Mahdi Aug 6 '17 at 11:12
  • $\begingroup$ strictly positive sectional curvature. $\endgroup$ – mmaatthh Aug 6 '17 at 11:45

Yes, it is a classical result.

Let $q_1,\dots,q_n$ be a sequence of critical points such that $$|q_{n+1}-p|\ge 2\cdot |q_n-p|.$$ By Toponogov's comparison, $$\measuredangle [p\,^{q_i}_{q_j}]\ge \tfrac\pi3.$$ Hence we get a bound on $n$.

  • $\begingroup$ What about $3$-dim complete Riemannian manifold with $Rc> 0$? Schoen-Yau had proved it is diffeomorphic to $\mathbb{R}^3$, is it also true that all critical points with respect to a fixed point $p$ is in a compact set?@Anton Petrunin $\endgroup$ – mmaatthh Aug 7 '17 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.