# sectional curvature with distance between two points

I am wondering if the following statement is true or not:

Let $M$ be a complete Riemannian manifold with sectional curvature $K$. Let $\tilde{M}$ be the simply connected complete Riemannian manifold of constant curvature $\kappa$ with $K\geq \kappa$ and $\dim M = \dim \tilde{M} = n$. Let $p,q \in M$ and $\gamma$ a shortest geodesic from $p$ to $q$ with $L(\gamma) = r$ i.e. dist$(p,q) = r$. Does it hold that for arbitrary $\tilde{p} \in \tilde{M}$, there exists $\tilde{q} \in \tilde{M}$ with dist$( \tilde{p} , \tilde{q} ) \geq r$? thanks a lot

• No. $M= S^2$ with induced metric from $\mathbb R^2$ and $\bar{M}= D^2$ with Poincare metric. Is there any other relation between $M,\bar{M}$ other than dimension? – Anubhav Mukherjee Sep 9 '16 at 15:23

It seems to me that the statement is true. Indeed, if $\kappa \leq 0$ then for any $r > 0$ and any point $\tilde{p} \in \tilde{M}$, where $\tilde{M}$ is the simply connected and complete space form $\tilde{M}$ of constant sectional curvature $\kappa$, there is $\tilde{q} \in \tilde{M}$ such that $dist(\tilde{p},\tilde{q}) \geq r$. So we can assume $\kappa > 0$. In this case the diameter of $M$ is bounded by $\pi K^{-1/2}$ as follows from the classica Bonnet-Myers Theorem. On the other hand the space form of constant sectional curvatue $\kappa$ has diameter $\pi \kappa^{-1/2}$. So $$r \leq \pi K^{-1/2} \leq \pi \kappa^{-1/2} = dist(\tilde{p},\tilde{q}) \, .$$ Since space forms are homogeneous spaces the point $\tilde{p}$ is arbitrary. This answer positively the OP question.