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I recall what this theorem says: Let M be a Riemannian manifold upon which a finite group G acts by isometries; let H and K be subgroups of G that act freely. Suppose that H and K are almost conjugate i.e., there is a bijection f : H → K carrying every element h of H to an element f (h) of K that is conjugate in G to h. Then the quotient manifolds M1 = H/M and M2 = K/M are isospectral. I found an example in which we isospectral manifolds: http://www.geom.uiuc.edu/docs/research/drums/planar/node2.html#SECTION00020000000000000000 It seems that M is given in Fig 2 and that the quotient manifolds are in Fig 3. What is G and its subgroups H and K? thanks. Xsnl wrote that the Buser propellers are not of sunada type and it is true. But if you read https://arxiv.org/pdf/2008.12498.pdf you can see that Berard gave a proof of Sunada theorem which is still true without the free action condition. It uses orbifolds but we still have G ans the subgroups H and K. What are they here?

Naima
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  • Buser propeller examples are not of Sunada type; to begin with, a quotient by a free group action has a nontrivial fundamental group, while Buser propellers are (topologically) just disks. – xsnl Nov 29 '23 at 18:37
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    One can get a paving of the plane with the left propeller (or the right one). this paving has symmetries by 60 degrees rotations around points of a network. they generate a group G. are H and K (subgroups) what we are looking for? – Naima Nov 30 '23 at 08:57
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    Benjamin Linowitz wrote that " Sunada’s method is extremely versatile and is responsible for the majority of the known examples of isospectral but not isometric Riemannianmanifolds." This is why it would be worthwhile to see how it works on a given example. – Naima Dec 02 '23 at 21:19

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Buser, Conway and Doyle give in their article https://arxiv.org/abs/1005.1839 the ingredients that we are looking for. There is a set called the Fano projective plane which contains 7 points and 7 "lines". We can put them (with their number) on a triangle with its medians. "0" is at its barycenter. Choose one side and label its middle with "1" Then complete the labeling of the 5 other poinrs to get the sequence: 1 5 2 3 4 6 around the triangle. G is the group of the collineations of the points: https://en.wikipedia.org/wiki/Collineation

If with a given k you map each point with number n to the point with number n+k module 7 you have a collineation in G. Conway gives 3 such collineations

a = (0 1)(2 5) b = (0 2)(4 3) c = (0 4)(1 6) They gernerate a first subgroup of G. He then uses the duality between points and lines to get a second subgroup genrated by

a' = (0 4)(2 3) b' = (0 1)(4 6) c' = (0 2)(1 5) b' = (0 1)(4 6) I recall that the Sunada theorem is still true without the free action condition.

Naima
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