Counting spanning trees of the hypercube and its $q$-analogs by explicit block diagonalization

Mathematics – Combinatorics

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Scientific paper

The number of spanning trees of a graph $G$ is called the {\em complexity} of $G$ and is denoted $c(G)$. Let C(n) denote the {\em (binary) hypercube} of dimension $n$. A classical result in enumerative combinatorics (based on explicit diagonalization) states that $c(C(n)) = \prod_{k=2}^n (2k)^{n\choose k}$. In this paper we use the explicit block diagonalization methodology to derive formulas for the complexity of two $q$-analogs of C(n), the {\em nonbinary hypercube} $\Cq(n)$, defined for $q\geq 2$, and the {\em vector space analog of the hypercube} $\Cfq(n)$, defined for prime powers $q$. We consider the nonbinary and vector space analogs of the Boolean algebra. We show the existence, in both cases, of a graded Jordan basis (with respect to the up operator) that is orthogonal (with respect to the standard inner product) and we write down explicit formulas for the ratio of the lengths of the successive vectors in the Jordan chains (i.e., the singular values). With respect to (the normalizations of) these bases the Laplacians of $\Cq(n)$ and $\Cfq(n)$ block diagonalize, with quadratically many distinct blocks in the nonbinary case and linearly many distinct blocks in the vector space case, and with each block an explicitly written down real, symmetric, tridiagonal matrix of known multiplicity and size at most $n+1$. In the nonbinary case we further determine the eigenvalues of the blocks, by explicitly writing out the eigenvectors, yielding an explicit formula for $c(\Cq(n))$ (this proof yields new information even in the binary case). In the vector space case we have been unable to determine the eigenvalues of the blocks but we give a useful formula for $c(\Cfq(n))$ involving "small" determinants (of size at most $n$).

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