On the Eigenvalues of Simplicial Rook Graphs

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On the Eigenvalues of Jeremy L. Martin (University of Kansas) Jennifer D. Wagner (Washburn University) AMS Southeastern Sectional Meeting Tulane University October 13 14, 2012

Let d, n N, and let n d 1 denote the dilated simplex {v = (v 1,..., v d ) R d : d v i = n}. i=1 The simplicial rook graph SR(d, n) is the graph with vertices V (d, n) = n d 1 N d with two vertices adjacent iff they differ in exactly two coordinates.

030 120 021 210 111 012 300 201 102 003

030 120 021 210 111 012 300 201 102 003

V (d, n) = v = ( ) n+d 1 d 1 SR(d, n) is regular of degree δ = (d 1)n Eigenspaces of adjacency matrix A and Laplacian matrix L are the same because AX = λx LX = (δ λ)x Independence number α(sr(d, n)) = maximum number of nonattacking rooks on a simplicial chessboard α(sr(3, n)) = (2n + 3)/3 [Nivasch Lev 2005; Blackburn Paterson Stinson 2011]

The Adjacency and Laplacian Matrices Adjacency matrix of a graph G: A = A(G) = matrix with rows and columns indexed by V (G) with 1s for edges, 0s for non-edges Laplacian matrix of G: L = D A, where D = diagonal matrix of vertex degrees A acts on the vector space RV by Av = w neighbors w of v Eigenvalues of A, L = connectivity, spanning trees,... G regular = eigenspaces of A, L are the same

The Spectrum of A(3, n) Theorem (JLM/JDW, 2012) The eigenvalues of A(3, n) = A(SR(3, n)) are as follows: n = 2m + 1 odd n = 2m even Eigenvalue Multiplicity Eigenvalue Multiplicity ( 3 2m ) ( 2 3 2m 1 ) 2 2,..., m 3 3 2,..., m 4 3 m 1 2 m 3 2 m,..., n 2 3 m 1,..., n 2 3 2n 1 2n 1 Method of proof: Construct explicit eigenvectors.

Counting Spanning Trees Corollary The number of spanning trees of SR(3, n) is 32(2n + 3) (n 1 2 ) 2n+2 a 3 a=n+2 if n is odd, 3(n + 1) 2 (n + 2)(3n + 5) 3 32(2n + 3) (n 1 2 ) 2n+2 a 3 a=n+2 if n is even. 3(n + 1)(n + 2) 2 (3n + 4) 3

in Arbitrary Dimension Conjecture The graph SR(d, n) is integral for all d and n. Partial results for least eigenvalue λ and corresp. eigenspace W : Eigenvectors come from lattice permutohedra. If n ( ( d 2), then λ = d ) ( 2 and dim W = n (d 1)(d 2)/2 ) d 1. Note that dim W lim n V (d, n) = 1. If n < ( d 2), then the least eigenvalue appears to be n, and dim W is the Mahonian number M(d, n) of permutations in S d with exactly n inversions.

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3.

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3.

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3.

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3.

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3. 0 1 1 1 0 1 0 1 1 0

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3. 0 1 1 1 0 1 0 1 1 0

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3. 0 1 1 1 0 1 0 1 1 0

Hexagon Vectors in V (3, n) For each internal vertex v V (3, n) (i.e., v i > 0 for all i), the signed characteristic vector of the hexagon centered at v is an eigenvector with eigenvalue 3. 0 1 1 1 0 1 0 1 1 0

Hexagon Vectors in V (3, n) Number of possible centers for a hexagon vector = number of interior vertices of n d 1 = ( ) v 1. 2 The hexagon vectors are all linearly independent. The other ( ) ( v+2 2 v 2 ) 2 = 3v eigenvectors have explicit formulas in terms of characteristic vectors of lattice lines.

Permutohedron Vectors in G(d, n) Definition Let p Z d (if d is odd) or (Z + 1 2 )d (if d is even). The lattice permutohedron centered at p is Per(p) = {p + σ(w): σ S d } where S d is the symmetric group and ( 1 d 3 d w =,,..., 2 2 d 3, 2 ) d 1. 2 Most eigenvectors of SR(d, n) are signed characteristic vectors H p of lattice permutohedra inscribed in the simplex n d 1. [SHOW THE NIFTY SAGE PICTURE]

Permutohedron Eigenvectors Each H p is an eigenvalue of A(d, n) with eigenvalue ( ) d 2 The H p are linearly independent. Permutohedron vectors account for most eigenvectors: #{p: Per(p) V (d, n)} V (d, n) = ( n ( d 1 2 ) d 1 ( n+d 1 d 1 ) ) 1 as n.

The Case n < ( ) d 2 When n < ( d 2), the simplex n d 1 is too small to contain any lattice permutohedra.

The Case n < ( ) d 2 When n < ( d 2), the simplex n d 1 is too small to contain any lattice permutohedra. On the other hand, characteristic vectors of partial permutohedra Per(p) n d 1 are eigenvectors with eigenvalue n.

The Case n < ( ) d 2 When n < ( d 2), the simplex n d 1 is too small to contain any lattice permutohedra. On the other hand, characteristic vectors of partial permutohedra Per(p) n d 1 are eigenvectors with eigenvalue n. Number of partial permutohedra = Mahonian number M(d, n) = number of permutations in S d with n inversions = coefficient of q n in (1 + q)(1 + q + q 2 ) (1 + q + + q d 1 )

The Case n < ( ) d 2 When n < ( d 2), the simplex n d 1 is too small to contain any lattice permutohedra. On the other hand, characteristic vectors of partial permutohedra Per(p) n d 1 are eigenvectors with eigenvalue n. Number of partial permutohedra = Mahonian number M(d, n) = number of permutations in S d with n inversions = coefficient of q n in (1 + q)(1 + q + q 2 ) (1 + q + + q d 1 ) Construction uses (ordinary, non-simplicial) rook theory!

The Case n < ( ) d 2 Permutation π S d with n inversions inversion word (a 1,..., a d ), where a i = #{j [d] : π i > π j } (note that a i = n) Rook placement σ on skyline Ferrers board (a 1,..., a d ) lattice point x(σ) = (a i + i σ i ) n d 1 Eigenvector X π = σ ε(σ)x(σ) Proof that X π is an eigenvector: sign-reversing involution moving rooks around

Open Problems (The big one.) Prove that A(d, n) (equivalently, L(d, n)) has integral spectrum for all d, n. (Verified for lots of d, n.) The induced subgraphs SR(d, n) V (d,n) Per(p) also appear to be Laplacian integral for all d, n, p. (Verified for d 6.) Is A(d, n) determined up to isomorphism by its spectrum? (We don t know.)

Acknowledgements Thanks to... The MathOverflow crowd (in particular Cristi Stoica and Noam Elkies) Sage (sagemath.org) Mahir, Michael and Jeff for organizing You for listening! Preprint: arxiv:1209.3493