对称结合方案的形式自对偶与数值自对偶
Formal self-duality and numerical self-duality for symmetric association schemes
AI总结：
该研究定义对称结合方案的数值自对偶概念，证明存在数值自对偶但非形式自对偶的实例，并证明在P-多项式或Q-多项式条件下二者等价。
摘要：
设 $\mathcal{X} = (X, \{R_i\}_{i=0}^d)$ 表示一个对称结合方案。固定 $\mathcal{X}$ 的本原幂等元的一个排序 $\{E_i\}_{i=0}^d$，令 $P$（相应地，$Q$）表示 $\mathcal{X}$ 对应的第一特征矩阵（相应地，第二特征矩阵）。当 $P=Q$ 时，称方案 $\mathcal{X}$ （关于排序 $\{E_i\}_{i=0}^d$）是形式自对偶的。我们定义：当相交数与Krein参数满足对所有 $0 \leq h,i,j \leq d$ 有 $p^h_{i,j} =q^h_{i,j}$ 时，$\mathcal{X}$ （关于排序 $\{E_i\}_{i=0}^d$）是数值自对偶的。已知关于排序 $\{E_i\}_{i=0}^d$，形式自对偶蕴含数值自对偶。这引出了如下问题：关于排序 $\{E_i\}_{i=0}^d$，是否存在 $\mathcal{X}$ 是数值自对偶但非形式自对偶的情况？我们将证明这种情况是存在的。我们给出了一个对称结合方案的例子，以及一个本原幂等元的排序，在该排序下该方案是数值自对偶的，但不是形式自对偶的。关于自对偶性，我们还有如下额外结果。假设 $\mathcal{X}$ 是 $P$-多项式的，我们证明以下两条等价：(i) $\mathcal{X}$ 关于排序 $\{E_i\}_{i=0}^d$ 是形式自对偶的；(ii) $\mathcal{X}$ 关于排序 $\{E_i\}_{i=0}^d$ 是数值自对偶的。假设排序 $\{E_i\}_{i=0}^d$ 是 $Q$-多项式的，我们证明以下两条等价：(i) $\mathcal{X}$ 关于排序 $\{E_i\}_{i=0}^d$ 是形式自对偶的；(ii) $\mathcal{X}$ 关于排序 $\{E_i\}_{i=0}^d$ 是数值自对偶的。
英文摘要：
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 $\{E_i\}_{i=0}^d$.
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