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I have a question about Comparison Geometry:

I have a Riemannian Manifold $(X,g)$, complete and simply connected, with sectional curvature upper bounded by a positive constant $k>0$, so I can compare $(X,g)$ with the sphere of ray $\frac{1}{\sqrt{k}}$ and constant curvature $k$, say $S_k$.

I want to show that $r_{cut}(x)\geqslant \frac{\pi}{\sqrt{k}}$ for every $x\in X$ in order to prove that diam$X$ $\geqslant$ diam$S_k$ (it would be enough to show that it's true for only one point and only one direction) and I'm trying to do it for absurd: if I suppose that for a point $x\in X$ and for a direction $u$ (i.e. $u\in T_x X$ and $|u|=1$) is $r_{cut}(x,u)<\frac{\pi}{\sqrt{k}}$ I know by some theorems that, if $y$ is the cut-point of $x$ along $u$, there are two different geodesics $\gamma_1$ and $\gamma_2$ joining $x$ and $y$, and by the simply connection hypothesis I can say that the two geodesics are homotopic relative to $\{0,1\}$. And then .... ? what can I say more?

Maybe there is another way for prove that diam$X$ $\geqslant$ diam$S_k$ ...

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I don't think what you want to show is true, at least not without assuming also that sectional curvature is everywhere positive (that makes it Klingerberg's theorem, whose proof is not exactly trivial). In the where the sectional curvature is allowed to be negative, I think you can get a counterexample to the cut locus estimate by looking at a dumbbell shaped surface (something that looks roughly like two spheres joined by a skinny cylinder). – Willie Wong Jan 11 '13 at 11:46
I haven't worked it out completely, but I suspect you will have better luck considering directly the diameter instead of the cut-locus. Let $p,q\in X$ realise the diameter, you should be able to apply Rauch's comparison theorem to derive a contradiction. – Willie Wong Jan 11 '13 at 12:16
Thanks for the advices, I'm going to try in another way, without speaking of cut-locus. I've looked for in some books of Riemannian Geometry but the are few things about sectional curvature upper bounded by a POSITIVE constant! – AX.J Jan 11 '13 at 12:28
This is false, see counter examples in my answer: – studiosus Apr 24 '14 at 21:18

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