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The Helly theorem in the Euclidean plane asserts that if $S_1, \dots, S_n$ are $n \ge 3$ convex subsets such that $S_i \cap S_j \cap S_k \ne \emptyset$ for all distinct triples $i,j,k$, then the total intersection $\bigcap_{i = 1}^n S_i$ is also nonempty.

I'm wondering if the same theorem is true in the hyperbolic plane (for concreteness, let's assume the Poincaré disk model). My understanding is that if the analogue of Radon's theorem is true in this setting, then Helly follows axiomatically.

Radon's theorem in the Euclidean plane asserts that given any four points $x_1, \dots, x_4$, there is a partition into two nonempty subsets such that the convex hulls intersect. The proof I know uses the affine structure on the Euclidean plane and so doesn't seem to port directly into hyperbolic space. On the other hand, I can verify that the Radon property holds for all the collections of four points I've looked at...

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2 Answers 2

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I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and vice versa, since the model maps convex sets to convex sets.

EDIT This has (almost) absolutely nothing to do with the above, but the topological version of Helly's theorem goes back to at least Debrunner (and it is a Monthly paper, so is human-readable), no need to allude to Farb's paper.

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    $\begingroup$ The topological version is proved basically by just abstracting Helly's original proof, so in spirit it should probably be attributed to Helly. I referenced Farb's paper for three reasons: 1. because it has a nice bibliography of related stuff, which might be useful, and 2. it focuses on things like CAT(0)-spaces, which I suspect are of interest to the OP, and 3. to amuse the OP by pointing out that he should have learned about this in grad school! $\endgroup$ Commented Feb 2, 2018 at 22:57
  • $\begingroup$ @AndyPutman Actually, Mike Davis was attributing the toplogical version to Hsiang (not sure which one, perhaps both) in the mid-sixties, and that was presumably stronger than Debrunner's result. Apparently, back when Mike was a grad student everyone did know this, but no more, alas. $\endgroup$
    – Igor Rivin
    Commented Feb 2, 2018 at 23:10
  • $\begingroup$ Kids these days... $\endgroup$ Commented Feb 4, 2018 at 3:34
  • $\begingroup$ @AndyPutman Good think the Old Skool types are around to keep them honest. $\endgroup$
    – Igor Rivin
    Commented Feb 4, 2018 at 4:05
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The original proof of Helly's theorem was topological and only uses basic homological properties of convex sets. It generalizes to all sorts of contexts, including the one you are interested in. Here is a general statement of what it can do. A homology cell is a topological space whose reduced singular homology is the same as that of a point (this implies in particular that it is nonempty).

Theorem: Let $X$ be a normal topological space such that for some $n \geq 1$, every open set $Y \subset X$ satisfies $H_q(Y)=0$ for $q \geq n$. Let $X_1,\ldots,X_k$ be a collection of closed homology cells in $X$. Assume that the intersection of any $r$ of the $X_i$ is nonempty for all $r \leq n+1$ and is a homology cell for $r \leq n$. Then the intersection of all the $X_i$ is a homology cell (and in particular is nonempty).

A discussion of this with references is in Section 3 of

B. Farb, Group actions and Helly's theorem, Adv. Math. 222 (2009), no. 5, 1574–1588.

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    $\begingroup$ An amusing personal note for Nick: Benson gave a colloquium about this paper when I was an undergraduate, and it was one of the main factors that led me to go to Chicago to work with him. But he didn't actually get around to writing the paper until after I got my PhD! $\endgroup$ Commented Feb 2, 2018 at 20:10

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