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.


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?

  • $\begingroup$ 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?" $\endgroup$ Dec 20, 2019 at 14:56
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    $\begingroup$ 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 $\kappa$? $\endgroup$
    – HJRW
    Dec 20, 2019 at 15:01
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    $\begingroup$ 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. $\endgroup$ Dec 20, 2019 at 15:02
  • $\begingroup$ If the space (with a Riemannian metric) is complete but not compact, it must have infinite diameter. $\endgroup$ Dec 20, 2019 at 15:05
  • $\begingroup$ I have updated the question. $\endgroup$ Dec 20, 2019 at 18:14

1 Answer 1


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 $k$.

  • $\begingroup$ 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.) $\endgroup$
    – HJRW
    Dec 20, 2019 at 15:16
  • $\begingroup$ @HJRW I will have to look into this further, it's been a while that I worked in the area. My intuition was that you could take a sequence of smooth manifolds starting at the sphere converging to the cube surface and then the entire curvature would accumulate in the corners. $\endgroup$
    – quarague
    Dec 20, 2019 at 15:20
  • $\begingroup$ I can imagine a way to make that true doing differential geometry of manifolds with corners. But the question is about metric spaces. $\endgroup$
    – HJRW
    Dec 20, 2019 at 15:24
  • $\begingroup$ @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($0$). $\endgroup$
    – YCor
    Dec 20, 2019 at 15:53
  • $\begingroup$ @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. $\endgroup$ Dec 20, 2019 at 15:54

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