Chapter 18 Polynomial Schemes

Wednesday, March 3, 1993

Lemma 18.1 Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ denote the symmetric scheme with associated matrices $A_0, A_1, \ldots, A_D$. Then the following are equivalent.

$(i)$ The graph $\Gamma = (X, R_1)$ is distance-regular, and $R_0, \ldots, R_D$ are labelled so that

\[R_i = \{xy\mid \partial(x,y) = i\}.\]

$(ii)$ There exists $f_i\in \mathbb{C}[\lambda]$, $\deg f_i = i$ such that $f_i(A_1) = A_i$ for all $i$ with $0\leq i\leq D$.

$(iii)$ The parameter $p^h_{ij}$

\[\begin{cases} = 0 & \text{if one of $h, i, j$ is larger than the sum of the other two,}\\ \neq 0 & \text{if one of $h,i,j$ is equal to the sum of the other two.}\end{cases}\]

Proof.

$(i)\Rightarrow (ii)$: Lemma 16.1.

$(ii)\Rightarrow (iii)$: Define

\[k_i \equiv p^0_{ii} = |\{z\mid z\in X, \; \partial(x,z) = i, \; ((x,z)\in R_i)\}|\] for any $x\in X$. Then $k_i \neq 0$ $(0\leq i\leq D)$, $k_0 = 1$.

(By symmetricity, $(x,y)\in R_i$ if and only if $(y,x)\in R_i$.)

Claim. \[\begin{align} k_hp^h_{ij} & = k_ip^i_{hj} = k_jp^j_{ih}\\ & = |X|^{-1}|\{xyz\in X^3\mid \partial(x,y) = h, \partial(x,z) = i, \partial(y,z) = j\}|. \end{align}\] Pf. The number of $xyz\in X^3$, $\partial(x,y) = h, \partial(x,z) = i, \partial(y,z) = j$ is equal to \[|X|k_hp^h_{ij} = |X|k_ip^i_{hj} = k_jp^j_{ih}.\]

In particular, \[p^h_{ij} = 0 \leftrightarrow p^i_{hj} =0 \leftrightarrow p^j_{ih} = 0.\] Hence, it suffices to show \[\begin{cases} p^h_{ij} = 0 & \text{if }\; h > i+j\\ p^h_{ij} \neq 0 & \text{if }\; h = i+j. \end{cases}\]

Fix $i,j$. Without loss of generality, we may assume that $i+j\leq D$ as trivial otherwise. \[f_i(A)f_j(A) = A_iA_j = \sum_{\ell = 0}^Dp^{\ell}_{ij}A_\ell = \sum_{\ell=0}^Dp^\ell_{ij}f_\ell(A).\] \[\begin{align} i + j & = \deg \mathrm{LHS}\\ & = \deg \mathrm{RHS}\\ & = \max\{\ell\mid p^\ell_{ij}\neq 0\}. \end{align}\]

$(iii)\Rightarrow (i)$

Let $A = A_1$, and consider a graph $\Gamma$ with adjacency matrix $A$. \[\begin{align} AA_j & = \sum_{h}p^h_{1j}A_h\\ & = p^{j+1}_{1j} A_{j+1} + p^j_{1j}A_j + p^{j-1}_{1j}A_{j-1}. \end{align}\]

Then, $p^{j+1}_{1j} \neq 0 \neq p^{j-1}_{1j}$.

Fix a vertex $x\in X$, and set $R_i(x) = \{y\mid (x,y)\in R_i\}$.

Then each $y\in R_i(x)$ is adjacent in $\Gamma$ to exactly \[\begin{align} p^i_{1,i+1} & \neq 0 \quad \text{vertices in }\; R_{i+1}(x),\\ p^i_{1i} & \qquad \text{vertices in }\; R_{i}(x),\\ p^i_{1,i-1} & \neq 0 \quad \text{vertices in }\; R_{i-1}(x). \end{align}\] Hence, by induction, \[\begin{align} R_i(x) & = \{y\mid \partial(x,y) = i \text{ in }\Gamma\} && (0\leq i\leq D), \end{align}\] and $\Gamma$ is distance regular.