On the eigenvalues of certain Cayley graphs and arrangement graphs

Chen, Bai Fan; Ghorbani, Ebrahim; Wong, Kok Bin

doi:10.48550/arxiv.1310.8078

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“…Proof. The eigenvalues of A(4, 3) and A(5, 3) are determined by a computer; these are {−3 [1] , −2 [6] , −1 [3] , 0 [4] , 1 [3] , 2 [6] , 3 [1] } and {−3 [14] , −2 [5] , −1 [12] , 1 [14] , 2 [6] , 4 [8] , 6 [1] }, respectively, which agree with the assertion. (Note that letting n = 4 in the assertion, the sum of the multiplicities of −3 and n − 7 equals 1.)…”

Section: Eigenvalues Of A(n 3) and A(n 4)supporting

confidence: 56%

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Cyclic Decomposition of $k$-Permutations and Eigenvalues of the Arrangement Graphs

Chen¹,

Ghorbani

Wong

2013

Electron. J. Combin.

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The (n, k)-arrangement graph A(n, k) is a graph with all the k-permutations of an n-element set as vertices where two k-permutations are adjacent if they agree in exactly k − 1 positions. We introduce a cyclic decomposition for k-permutations and show that this gives rise to a very fine equitable partition of A(n, k). This equitable partition can be employed to compute the complete set of eigenvalues (of the adjacency matrix) of A(n, k). Consequently, we determine the eigenvalues of A(n, k) for small values of k. Finally, we show that any eigenvalue of the Johnson graph J(n, k) is an eigenvalue of A(n, k) and that −k is the smallest eigenvalue of A(n, k) with multiplicity O(n k ) for fixed k.We then define the cycle type for a k-permutation π to be the list of integers consisting of the lengths of the cycles and the paths appeared in the decomposition of π. This gives rise to the 'cycle-type partition' of V (n, k) in which each class consists of all elements of V (n, k) sharing the same cycle type. The reason for studying the cycle-type partition of k-permutations will become clear below.The (n, k)-arrangement graph A(n, k) has V (n, k) as its vertices, and two k-permutations π = (u 1 , . . . , u k ) and ρ = (v 1 , . . . , v k ) are adjacent if they agree in exactly k−1 positions, i.e. if for exactly one i 0 , u i 0 = v i 0 and for all i = i 0 , u i = v i . The family of arrangement graphs was first introduced in [10] as an interconnection network model for parallel computation. In the interconnection network model, each processor has its own memory unit and communicates with the other processors through a topological network, i.e. a graph.For this purpose, the arrangement graphs possess many nice properties such as having small diameter, a hierarchical structure, vertex and edge symmetry, simple shortest path routing, high connectivity, etc. Many properties of arrangement graphs have been studied by a number of authors, see, e.g. [4,5,6,7,8,19,20].Another family of graphs with the same nature as the arrangement graphs are the derangement graphs.The n-derangement graph is a graph whose vertices correspond to all the permutations of [n] where two permutations are adjacent if they differ in all n positions. It is known that the eigenvalues of the derangement graph are integers (see [1,11,15,17]). For other properties of the eigenvalues of the derangement graph, we refer the reader to [13,14,18].As an application of the cycle-type partition of V (n, k), we consider the problem of determining the eigenvalues of the arrangment graphs. It turns out that the cycle-type partition of V (n, k) is indeed an equitable partition of the graph A(n, k). Normally, the eigenvalues of equitable partitions of a graph give a subset of the set of eigenvalues of the graph. However, in view of a result of Godsil and McKay [12], the cycle-type partition of V (n, k) is fine enough to give the complete set of eigenvalues as well as their multiplicities. Consequently, we will be able to determine the eigenvalues of A(n, k) for small values ...

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Section: Eigenvalues Of A(n 3) and A(n 4)supporting

confidence: 56%

“…For any integer k, there is an integer n 0 such that for all n ≥ n 0 , −k is the only negative eigenvalue of A(n, k). 1 Recently, Conjecture 2 was settled affirmatively in [3].…”

Section: The Smallest Eigenvaluementioning

confidence: 98%