What are the ingredients of the Sunada's theorem in this example?

Question

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?

Answer 1

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.