Formal self-duality and numerical self-duality for symmetric association schemes

Abstract: Let ${\mathcal X} = (X, {R_i}{i=0}^d)$ denote a symmetric association scheme. Fix an ordering ${E_i}{i=0}^d$ of the primitive idempotents of $\mathcal{X}$, and let $P$ (resp.\ $Q$) denote the corresponding first eigenmatrix (resp.\ second eigenmatrix) of $\mathcal X$. The scheme $\mathcal X$ is said to be formally self-dual (with respect to the ordering ${E_i}{i=0}^d$) whenever $P=Q$. We define $\mathcal X$ to be numerically self-dual (with respect to the ordering ${E_i}{i=0}^d$) whenever the intersection numbers and Krein parameters satisfy $p^h_{i,j} =q^h_{i,j}$ for $0 \leq h,i,j \leq d$. It is known that with respect to the ordering ${E_i}{i=0}^d$, formal self-duality implies numerical self-duality. This raises the following question: is it possible that with respect to the ordering ${E_i}{i=0}^d$, $\mathcal X$ is numerically self-dual but not formally self-dual? This is possible as we will show. We display an example of a symmetric association scheme and an ordering the primitive idempotents with respect to which the scheme is numerically self-dual but not formally self-dual. We have the following additional results about self-duality. Assume that $\mathcal X$ is $P$-polynomial. We show that the following are equivalent: (i) $\mathcal X$ is formally self-dual with respect to the ordering ${E_i}{i=0}^d$; (ii) $\mathcal X$ is numerically self-dual with respect to the ordering ${E_i}{i=0}^d$. Assume that the ordering ${E_i}{i=0}^d$ is $Q$-polynomial. We show that the following are equivalent: (i) $\mathcal X$ is formally self-dual with respect to the ordering ${E_i}{i=0}^d$; (ii) $\mathcal X$ is numerically self-dual with respect to the ordering ${E_i}_{i=0}^d$.