By Philipp Lampe

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The number of such subsets is equal to c2n−2,p−1,q . But not every such subset A yields an admissible set A. The subsets A that do not yield admissible sets A are precisely the sets with 2n − 2 ∈ A . For those subsets we have 2n − 3 ∈ / A , so that A is the union of {2n − 2} and a subset A of {1, 2, . . , 2n − 4} with exactly p − 1 odd, exactly q − 1 even and no consecutive elements. Every such subset A yields an admissible subset A and the number of such subsets is equal to c2n−4,p−1,q−1 . Altogether we get c2n−2,p−1,q − c2n−4,p−1,q−1 sets A.

We deduce | AC | · | BD | ≤ | AB| · |CD | + | AD | · | BC |. e. the quotient of the numbers is a positive real number. The quotient is known as the the cross ratio. We conclude that we have equality especially in the case when A, B, C, D lie in this order on the circle, in which case the polar angles of the quotients ( A − B)/( A − D ) and (C − D )/( B − C ) are equal. Unfortunately, Euclidean geometry yields no direct model of cluster algebras. If the points A, B, C, D lie on a circle, then the distances also satisfy the relation | AB| · | BC | · | AC | + | BC | · |CD | · | BD = | AB| · | DA| · | BD | + | BC | · |CD | · | BD |.

Write X = ( x1 , x2 , . . , xn ) as a sequence of column vector. Suppose that i1 < i2 < . . < id−1 and j1 < j2 < . . < jd+1 are sequences as above. We want to show that d +1 ∑ (−1)r det(xi , xi , . . , xi − , x j ) det(x j , x j , . . , x j , . . , x j + ) = 0. 1 2 d 1 r 1 2 r d 1 r =1 We expand the determinant in the first factor of each summand along the last column according to Laplace’s rule. Denote the (d − 1) × (d − 1) matrix obtained from ( xi1 , xi2 , . . , xid−1 ) by deleting (s) the s-th row by Xi .