Orbifolds with positive curvature and maximal diameter are investigated in this article, by J. Borzellino. Theorem 1 of the article states:

Let $\mathcal{O}$ be a complete $n$-dimensional Riemannian orbifold with Ricci curvature satisfying $\mathrm{Ric}_\mathcal{O}\geq n-1$ and diameter $\mathrm{diam}(\mathcal{O})=\pi$. Then $\mathcal{O}$ is a good orbifold.

Recall that a Riemannian orbifold is good when it is a quotient orbifold of a Riemannian manifold by an effective, (proper) discontinuous, isometric action. Also, $\mathrm{Ric}_\mathcal{O}\geq n-1$ means that the local charts $\tilde{U}\to U\cong \tilde{U}/\Gamma$ satisfy $\mathrm{Ric}_{\tilde{U}}\geq n-1$, and the diameter of $\mathcal{O}$ is the diameter of its underlying metric space (with the metric induced by the Riemannian structure).

The article aims for an analog of Cheng's maximal diameter sphere theorem, claiming (Theorem 2) that $\mathcal{O}$ must then be a quotient of the sphere by a finite group of $\mathrm{O}(n+1)$.

My question is: why isn't the $\mathbb{Z}_p$-teardrop orbifold, i.e., the $2$-orbifold with underlying space $\mathbb{S}^2$ and with a single cone singularity of order $p$, a counterexample to Theorem 1?

The proof of Theorem 1 is based on volume comparison and, sincerely, I don't quite understand it, mostly because the notation does not distinguish between $\mathcal{O}$ and its underlying space (that I will denote by $|\mathcal{O}|$). Am I missing something or the correct interpretation of the conclusion in Theorem 1 is actually: "$|\mathcal{O}|$ can be realized as the underlying space of a good Riemannian orbifold"?

  • 1
    $\begingroup$ I think the short answer to your question is that the teardrop orbifold with a metric with $Ric\geq 1$ must have diameter $< \pi$. If you look at the proof of Theorem 1 (starting on p. 50), it shows that two points $x, y$ with distance $d(x,y)=\pi$ are orbifold points of the same order quotient. But the teardrop orbifold has only one orbifold point. I'm not sure if there's a better explanation than a description of the proof... The extension of the Bishop comparison theorem to orbifolds is almost obvious if one understands the proof. $\endgroup$ – Ian Agol Feb 2 '18 at 6:06

Your Answer

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

Browse other questions tagged or ask your own question.