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I need some sort of classification (up to isometry) of spherical polygons (i.e. polygons in $\mathbb{S}^2$ whose edges are given by geodesics) subject to the interior angles and the perimeter of the polygon itself. Most likely this question is already solved and can be found in some literature, but I couldn't find something in this direction. Does anyone know where I can find something related?

Most of the literature just deal with triangles but do not go further and study polygons with more than three vertices (not even a word is said about it). In case of triangles the situation is very clear! The interior angles determine a spherical triangle up to isometry. But what about more than three vertices?

Best regards

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  • $\begingroup$ In the case of more than three vertices you have more than $0$ degrees of freedom. What kind of classification are you looking for? E.g., what would the Euclidean counterpart be? $\endgroup$
    – Alex Degtyarev
    Commented Dec 13, 2014 at 11:50
  • $\begingroup$ You can triangulate any polygon with more than three vertices. This helps reduce some problems for general polygons (like the sum of angles as a function of area) to those with triangles, but it's hard to say more before you specify your question. $\endgroup$
    – Joonas Ilmavirta
    Commented Dec 13, 2014 at 12:27
  • $\begingroup$ Thank you for your comments, I will be more precise. First, it would help me to know if there are non isometric n-gons with $n\geq 4$ with same interior angles $\alpha_1,…,\alpha_n$ and same perimeter (i.e. sum of the side lengths) $a_1+…+a_n$. E.g. for n=4 I could construct a one parameter family of non isometric 4-gons with same interior angles, but which all have different perimeter. Are there two 4-gone which not only have the same interior angles but also the same perimeter? $\endgroup$
    – asd
    Commented Dec 13, 2014 at 13:42
  • $\begingroup$ @Alex: In Euclidean plane this problem (from the comment above) would mean to construct two n-gons with same interior angles, area and perimeter. This problem is easily solved. In spherical case this is less obvious to me. $\endgroup$
    – asd
    Commented Dec 13, 2014 at 13:51

1 Answer 1

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You should specify what you mean by a polygon: a broken line or a surface. If we are talking about a broken line, then some classification is given here:

MR1703691 Kapovich, Michael; Millson, John J. On the moduli space of a spherical polygonal linkage. Canad. Math. Bull. 42 (1999), no. 3, 307–320.

If we are talking about surfaces, this is a different problem which is much more complicated. When the angles are sufficiently small, so that the whole polygon is a subset of the sphere, a classification is obtained in

F. Luo and G. Tian, Liouville equation and spherical convex polytopes, Proc. AMS, 116 (1992) 4, 1119-1129.

If the interior angles can be arbitrarily large, there is no known classification, even for the case of quadrilaterals. Some partial results can be found in

Eremenko, Gabrielov, Tarasov, arXiv:1405.1738 Metrics with conic singularities and spherical polygons

We actually have a classification of quadrilaterals up to isometry but it is not ready for publication yet.

To answer your more specific question. A spherical polygon generally depends on 2n-3 real parameters: $n$ angles and $n-3$ additional parameters. For these additional parameters, one can take certain conformal moduli, for example accessory parameters in the differential equation which is associated to this polygon. So for $n\geq 4$, we have 5 parameters, but whether there exist two quadrilaterals with the same angles and perimeter we did not investigate. For $n\geq 5$ polygons with the same angles and perimeter certainly exist, by dimension count.

EDIT. That $n$-gons with the same angles and same perimeters exist, can be obtained from our other paper, arXiv:math/0405196 where polygons whose angles are integer multiples of $2\pi$ are classified.

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  • $\begingroup$ Thank you very much for your wonderful answer! The only open question is whether there exist two quadrilateral with same angle and perimeter. I have to think about this again. If I make any progress I will inform you here. If anyone has helpful ideas about this issue I would be glad to get help. $\endgroup$
    – asd
    Commented Dec 13, 2014 at 18:16
  • $\begingroup$ I am sure that the answer can be obtained from our classification, or even perhaps from our preprint arXiv:1409.1529. We just never thought about perimeter. Can you explain why do you need perimeter? $\endgroup$
    – Alexandre Eremenko
    Commented Dec 13, 2014 at 18:39

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