Consider configurations consisting of 4 distinct circles on the sphere. Two configurations are equivalent if they can be mapped onto each other by a homeomorphism of the sphere. How to enumerate/classify such configurations?

Equivalent problem: classify the arrangements of 4 hyperbolic planes in the hyperbolic space, up to homeomorphisms of the space.

Before voting to close this question as trivial, you may look at the classification of generic configurations which we obtained by brute force:

configurations without disjoint pairs of circles

configurations with at least one disjoint pair of circles

Each region bounded by more than 3 sides is labeled by the number of its boundary sides. This is used to show that all configurations are non-equivalent.

Questions: Is this new? Is there a scientific method to obtain this? Is there any structure on these 35 configurations?

There is a large research area about hyperplane arrangements in a Euclidean space. How about hyperbolic space? There is also a large body of research on hyperbolic tetrahedra. But it is always assumed that the tetrahedron is compact (or has only vertices at infinity).

We encountered this question in our studies of the Heun and Painlevé VI equations with real coefficients. (See Appendix II in the linked paper). Projective monodromy groups associated to these equations are generated by 4 reflections in circles.

EDIT. It seems that the problem is of purely topological nature: for any collection of Jordan curves on the sphere, such that each pair intersects transversally at at most two points, there exists an equivalent configuration of circles: A. Bobenko, B. Springborn, Variational principles for circle patterns and Koebe's theorem. Trans. Amer. Math. Soc. 356 (2004), no. 2, 659–689.

EDIT2. The previous remark is incorrect (thanks to Ivan Izmestiev vor his comment). A counterexample with 5 curves is contained in this paper: MR3216670 Kang, Ross J.; Müller, Tobias Arrangements of pseudocircles and circles, Discrete Comput. Geom. 51 (2014), no. 4.

  • 1
    The paper "Connected sum at infinity and Cantrell-Stallings hyperplane unknotting" by Jack S. Calcut, Henry C. King, and Laurent C. Siebenmann seems relevant, see 9.2, 9.3 in – Igor Belegradek Jun 11 at 17:28
  • 1
    Sphere arrangements (in the spirit of classical works on hyperplane arrangements) were studied in "Arrangements of spheres and projective spaces" by Deshpande, Rocky Mountain J. Math, 2016. – Ivan Izmestiev Jun 11 at 20:14
  • 3
    I suspect that your last edit (topological nature of the problem) is false. In the similar setting of arrangements of lines and pseudolines there exist so called nonstretchable arrangements of pseudolines. (Even if you require that at most two lines pass through each point.) An example with 9 lines can be "doubled" and an equator can be added so that one gets a "nonstretchable" arrangement of 10 Jordan curves on the sphere. – Ivan Izmestiev Jun 12 at 16:20
  • 1
    Hi Alex, I asked a very similar question to yours, back in February.… – Ryan Budney Jun 13 at 2:02
  • 1
    If I'm reading the review of the Kang and Muller paper correctly, the five curve counterexample is due to J. Linhart and R. Ortner [Beiträge Algebra Geom. 46 (2005), no. 2, 351–356; MR2196921]. What Kang and Muller do is prove the Linhart-Ortner conjecture that this counterexample is minimal. – Gerry Myerson Jun 13 at 2:47

Not an answer; don't have enough rep to comment. You may be able to impose some structure by using the operation of crossing/uncrossing two circles which are otherwise adjacent (That is, they currently do not cross and crossing them such that there is a new region with two boundaries, or conversely removing that region).

For example, $X$ is only adjacent to $W$, but $W$ is adjacent to $S$, $V$, $X$, and $X'$. This would break it into families - $T$ cannot be made from $X$.

I'm not sure whether this structure is interesting, but it was a thought I had upon seeing your classification. I can't provide much more of the structure, as it's rather difficult to work out by hand.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.