# Unbounded curvature implies infinite diameter on complete metric spaces

I recently asked this question Unbounded sectional curvature implies infinite diameter?. I would like now to ask something similar, but in another context.

Suppose you have a complete metric space $$(M,d)$$. Assume that the curvature of $$M$$ nas no upper bound. Can one concludes that the diameter of $$M$$ is infinite?

If it helps, one can assume that $$(M,d)$$ is the limit of a sequence of compact manifolds.

EDIT

Sorry for being vague on the first time. I am assuming I have a length space $$(M,d)$$ which is actually complete. It is obtained as the Gromov-Hausdorff limit of a sequence of compact Riemannian manifolds. So my question is: if the Alexandrov curvature of $$(M,d)$$ is not bounded from above, is the diameter of $$(M,d)$$ infinite?

• Your condition on curvature is irrelevant. If you have a complete non-compact surface of finite diameter, you can always make a sequence of very little holes in it and glue in spheres of very large curvature. So the question is equivalent to: "does there exist a complete Riemannian manifold of finite diameter?" – Alexandre Eremenko Dec 20 '19 at 14:56
• Could you say exactly what you mean by unbounded curvature of a metric space? For instance, do you mean it doesn't satisfy the $CAT(\kappa)$ condition for any $\kappa$? – HJRW Dec 20 '19 at 15:01
• The curvature of a (complete) metric space has no, at least obvious, meaning. Within certain classes, you'll find Alexandrov spaces as limits of Riemannian manifolds where certain curvature bounds make sense, but you need to formulate a more precise question to get a reasonable answer. – Andy Sanders Dec 20 '19 at 15:02
• If the space (with a Riemannian metric) is complete but not compact, it must have infinite diameter. – Alexandre Eremenko Dec 20 '19 at 15:05
• I have updated the question. – L.F. Cavenaghi Dec 20 '19 at 18:14

Here is counter example. Consider the surface of the unit cube in $$\mathbb{R}^3$$. This is compact and has infinite curvature at the corners. One can also construct a sequence of smooth compact manifolds that converge to this surface, some care needs to be taken of what kind of convergence you want.
Edit: Thanks to the helpful discussion in the comments. As a result we can say that the surface of the unit cube does have infinite curvature in the sense that it is not a $$CAT(k)$$ space for any finite $$k$$.
• This isn't true of the only definition I know of curvature of a metric space -- viz, satisfying the $CAT(\kappa)$ condition for some $\kappa$. The surface of the unit cube is a $CAT(1)$ space. (EDIT: On reflection, it can be rescaled to be $CAT(1)$. So it's $CAT(\kappa)$ for some $\kappa>0$, but I'm too lazy to work out the exact value.) – HJRW Dec 20 '19 at 15:16
• @HJRW No, the surface of a cube (with Euclidean metric on faces and with the geodesic metric) is not CAT$(\kappa)$ for any $\kappa<+\infty$. Indeed, every tangent cone $\lim(X,x_0,nd)$ of a $CAT(\kappa)$ space $X$ for some $\kappa<+\infty$, is $CAT(0)$ (indeed the rescale $(X,nd)$ is $CAT(\kappa/n^b)$ for some $b\in\{1/2,1,2\}$, I'm not sure which one but it doesn't matter). But the tangent cone at the singular point looks like the corner of an infinite cube, and this clearly not CAT($0$). – YCor Dec 20 '19 at 15:53
• @HJRW, curvature conditions on metric spaces is a vast subject (as you likely know): $CAT(\kappa)$, Alexandrov spaces and $\delta$-hyperbolicity are just some of the most well studied. I'm not trying to be a grouch, but I think discussing this question, absent a philosophical safari, is kind of pointless until the OP tells us what he actually means. – Andy Sanders Dec 20 '19 at 15:54