Consider the function f on Sn which equals 1/n on all adjacent transpositions (i,i+1), where we let n+1=1, and 0 otherwise, and its Fourier transform hatf(rho) evaluated at the irreducible representations.
Recall the irreducible representations of Sn are indexed by the set of partitions of n. Partitions here are written as a finite non-increasing sequence of positive integers that add up to n.
When rho is the representation corresponding to the partition (n), the matrix hatf(rho) is simply the 1times1 matrix [1].
When rho is the representation corresponding to the partition (n−1,1), the resulting matrix hatf(rho) can be explicitly diagonalized, since it can be extended into a cyclic matrix on mathbbRn. The eigenvalues are simply cosfrac2pikn where k=1,ldotsn−1. Therefore the spectral gap for that matrix (the smallest gap between 1 and an eigenvalue not equal to 1) is simply
1−cosfrac2pin=1−cosfrac2(n−1)pin=frac2pi2n2+O(frac1n3)
the following questions are in increasing levels of difficulty and are interesting to Markov chain theorists:
- Is it true that all other eigenvalues of hatf(rho) for some irreducible representation rho are strictly less than 1−cosfrac2pin in absolute value?
Denote by elambda,j, j=1,ldots,dlambda the eigenvalues of hatf(rholambda), where rholambda is the representation associated with the partition lambda and dlambda is the dimension of that representation.
For any fixed kinmathbbN, is it true that (1−maxjelambda,j)le(n−lambda1)frac2pi2n2+O(frac1n3), for n−lambda1lek? Here lambda1 denotes the longest part of the partition lambda.
If lambda>lambda′ in the sense that one can move blocks in the Ferrers diagram of lambda′ in the up and right direction to obtain lambda, for instance (n−1,1)>(n−2,1,1), is it true that the spectral gap of hatf(rholambda) is smaller than that associated with lambda′?
Give an explicit formula for elambda,j. This is most likely not possible.
This question shows how hard it can be to diagonalize matrices and to understand the representation theory of Sn at a practical level.
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