Chapter 24 Balanced Conditions, I

Wednesday, March 23, 1993

No Class on Friday (another conference).

Proof (Proof of Lemma 23.1 continued). Let $E_j$ be a primitive idempotent, $H = E_jV$ and \[\rho: X\to H \quad (x\mapsto E_j\hat{x}).\]

$(v)$ Every representation $(\rho, H)$ of $Y$ is equivalent to a representation of above type, for some $j$ $(0\leq j\leq D)$ and $j$ is unique.

Let $E:=(\langle \rho(x), \rho(y))_{x,y\in X}$.

By $\mathrm{R2}$, \[E = \sum_{i = 0}^D \sigma_i A_i, \quad \text{some}\; \sigma_0, \sigma_1, \ldots, \sigma_D\in \mathbb{C}.\] Hence, $E$ belongs to the Bose-Mesner algebra $M$ of $Y$.

We want to show that $E$ is a scalar multiple of a primitive idempotent.

Fix $x\in X$ and fix $i$ $(0\leq i\leq D)$.

By $\mathrm{R3}$, \[\begin{equation} \sum_{y\in X, (y,x)\in R_i}\rho(y) = \alpha \rho(x), \quad \text{some }\; \alpha\in \mathbb{C}. \tag{24.1} \end{equation}\] So, \[k_i\overline{\sigma_i} = \left\langle \sum_{y\in X, (y,x)\in R_i}\rho(y),\rho(x)\right\rangle = \bar{\alpha}\langle \rho(x), \rho(x)\rangle = \bar{\alpha}\sigma_0.\] Hence, $\alpha$ is independent of $x$. In maatrix form (24.1) becomes \[EA_i\hat{x} = \alpha E\hat{x}.\]

HS MEMO

\[Eu = Ev \Leftrightarrow \langle z, Eu\rangle = \langle z, Ev\rangle \text{ for all }z\in X \Leftrightarrow (Eu)_z = (Ev)_z \text{ for all }z\in X.\] \[\begin{align} (EA_i\hat{x})_z & = \left\langle \rho(z), \sum_{y\in X, (y,x)\in R_i}\rho(y)\right\rangle\\ & = \alpha \langle \rho(z), \rho(x)\rangle\\ & = (\alpha E\hat{x})_z. \end{align}\] Hence, \[EA_i\hat{x} = \alpha E\hat{x}.\]

Since $x$ is arbitrary, \[EA_i = \alpha E.\] So, \[EA_i \in \mathrm{Span} E\; \text{ and }\; EM = \mathrm{Span} E.\] We have $E\in \mathrm{E_j}$ for unique $j$ $(0\leq j\leq D)$.

HS MEMO

\[E = \tau_0 E_0 + \cdots + \tau_D E_D, \; \tau_j\in \mathbb{C}\quad (0\leq j\leq D).\] And, at least one of $\tau_j$ is nonzero, and \[\tau_jE_j = EE_j \in \mathrm{Span}E.\] So, \[\tau_jE_j = E\] as $E_0, \ldots, E_D$ are linearly independent.

Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ be a symmetric scheme, and let $E$ be a primitive idempotent.

Definition 24.1 $Y$ is $Q$-polynomial with respect to $E$, if and only if $Y$ is $Q$-polynomial with respect to some ordering $E_0, E_1, \ldots, E_D$ of primitive idempotents, where $E_0 = |X|^{-1}J$, and $E_1 = E$.

Theorem 24.1 Assume $Y = (X, \{R_i\}_{0\leq i\leq D})$ is $P$-polynomial (i.e., $(X, R_1)$ is distance-regular). Let $E$ be any primitive idempotent of $Y$. Let $(\rho, H)$ be the corresponding representation.

$(i)$ The following are equivalent.

$\quad (ia)$ $Y$ is $Q$-polyonimial with respect to $E$.

$\quad (ib)$ $(\rho, H)$ is nondegenerate and for all $x,y\in X$, and for all $i,j$ $(0\leq i,j\leq D)$,

\[\sum_{z\in X, (x,z)\in R_i, (y,z)\in R_j}\rho(z) - \sum_{z'\in X, (x,z')\in R_j, (y,z')\in R_i}\rho(z')\in \mathrm{Span}(\rho(x)-\rho(y)).\]

$\quad (ic)$ $(\rho, H)$ is nondegenerate and for all $x,y\in X$,

\[\sum_{z\in X, (x,z)\in R_1, (y,z)\in R_2}\rho(z) - \sum_{z'\in X, (x,z')\in R_2, (y,z')\in R_1}\rho(z')\in \mathrm{Span}(\rho(x)-\rho(y)).\]

$(ii)$ Write

\[E = |X|^{-1}\sum_{j=0}^D \theta^*_j A_j,\] and suppose $(ia)-(ic)$ hold. Then the coefficient in $(ib)$ is \[p^h_{ij}\frac{\theta_i^*-\theta_j^*}{\theta^*_0-\theta^*_h} \quad (1\leq h\leq D, 0\leq i,j\leq D).\]

Proof.

$(ia)\to(ib)$ Without loss of generality, assume $E \equiv E_1$, and $Y$ is $Q$-polynomial with respect to $E$.

Then by Lemma 22.2, $\theta_0^*, \ldots, \theta^*_D$ are distinct. So $\theta^*_h\neq \theta^*_0$ for all $h\in \{1,2\ldots, D\}$, and $(\rho, H)$ is nondegenerate.

Fix $x\in X$, write $E^*_i\equiv E^*_i(x)$, $A^*_i \equiv A^*_i(x)$, $A^* \equiv A_1^*$.

Let $M$ be the Bose-Mesner algebra. Set \[L = \{mA^*n - nA^*m\mid m, n\in M\}.\]

Claim 1. $\dim L \leq D$.

Proof of Claim 1. \[\begin{align} L & = \mathrm{Span}(E_iA^*E_j - E_jA^*E_i \mid 0\leq i<j\leq D)\\ & = \mathrm{Span}(E_iA^*E_{i+1} - E_{i+1}A^*E_i \mid 0\leq i\leq D-1). \end{align}\] Since $E_iA^*E_j = 0$ if $q^1_{ij} = 0$ by Lemma 20.2 and Lemma 20.3, and this occurs if $|i-j|>1$ by $Q$-polynomial property.

Hence, $\dim L \leq D$.

Claim 2. $(i)$ $\{A^*A_h - A_hA^*\mid 1\leq h\leq D\}$ is a basis for $L$. In particular,

$(ii)$ there exist $r^h_{ij}\in \mathbb{C}$ $(1\leq h\leq D, 0\leq i,j\leq D)$ such that

\[A_iA^*A_j - A_jA^*A_i = \sum_{h=1}^D r^h_{ij}(A^*A_h - A_hA^*).\]

Proof of Claim 2.

$(i)$ The column $x$ of $A^*A_h - A_hA^*$ is a nonzero scalar $\theta^*_h - \theta^*_0$ times the column $x$ of $A_h$.

HS MEMO

\[ ((A^*A_h - A_hA^*)\hat{x})_y = E_{xy}(A_h)_{yx}- (A_h)_{yx}E_{xx} = (\theta^*_h-\theta^*_0)(A_h)_{yz}.\]

Also the column $x$ of $A_0, A_1, \ldots, A_D$ are linearly independent.

Hence, the matrices given are linearly independent.

They are in $L$ by construction, so they form a basis for $L$ by Claim 1.

$(ii)$ This is immediate since

\[A_iA^*A_j - A_jA^*A_i\in L, \quad \text{for all $i,j$}.\]

Claim 3. \[r^\ell_{ij} = p^\ell_{ij}\left(\frac{\theta^*_i-\theta^*_j}{\theta^*_0 - \theta^*_\ell}\right)\quad (1\leq \ell\leq D, 0\leq i,j\leq D).\]

Proof of Claim 3. Fix $i,j$, \[A_iA^*A_j - A_jA^*A_i - \sum_{h=1}^D r^h_{ij}(A^*A_h - A_hA^*) = 0.\] Pick $\ell$ $(1\leq \ell \leq D)$. Pick $y\in X$ such that $(x,y)\in R_\ell$. \[\begin{align} (A_iA^*A_j)_{xy} & = \sum_{z\in X}(A_i)_{xz}(A^*)_{zz}(A_j)_{zy}\\ & = \sum_{z\in X, (x,z)\in R_i, (y,z)\in R_j}(A^*)_{zz}\\ & = |X|^{-1}p^\ell_{ij}\theta^*_i. \end{align}\] Similarly, \[(A_jA^*A_i)_{xy} = |X|^{-1}p^\ell_{ij}\theta^*_j.\] \[\begin{align} (A^*A_h-A_hA^*)_{xy} & = (A_0A^*A_h - A_hA^*A_0)_{xy}\\ & = |X|^{-1}p^\ell_{0h}(\theta^*_0 - \theta^*_h)\\ & = \begin{cases} 0 & \text{ if }\; \ell \neq h\\ |X|^{-1}(\theta^*_0-\theta^*_h) & \text{ if } \ell = h. \end{cases} \end{align}\] Hence, \[\sum_{h=1}^D r^h_{ij}(A^*A_h - A_hA^*)_{xy} = |X|^{-1}r^\ell_{ij}(\theta^*_0-\theta^*_\ell).\] Comparing terms, we have \[p^\ell_{ij}(\theta^*_i-\theta^*_j) - r^\ell_{ij}(\theta^*_0-\theta^*_\ell) = 0.\]

Claim 4. For all $h$ $(1\leq h\leq D)$, for all $i,j$ $(0\leq i,j\leq D)$, for all $w,y\in X$, $(w,y)\in R_h$, \[\begin{equation} \sum_{z\in X,(w,z)\in R_i, (y,z)\in R_j}\rho(z)-\sum_{z'\in X, (w,z')\in R_j, (y,z)\in R_i}\rho(z') - r^h_{ij}(\rho(w)-\rho(y))=0. \tag{24.2} \end{equation}\]

Proof of Claim 4. Set $L = \langle \mathrm{LHS}$ of (24.2), $\rho(x)\rangle$ It suffices to show that $L = 0$.

Note that since $x$ is arbitrary, if $\mathrm{LHS}$ of (24.2) is zero. \[\begin{align} L & = \sum_{z\in X,(w,z)\in R_i, (y,z)\in R_j}\langle \rho(z), \rho(x)\rangle -\sum_{z'\in X, (w,z')\in R_j, (y,z)\in R_i}\langle\rho(z'),\rho(x)\rangle \\ & \qquad - r^h_{ij}\langle \rho(w)-\rho(y), \rho(x)\rangle\\ & = |X|^{-1}(A_iA^*A_j)_{wy} - |X|^{-1}(A_jA^*A_i)_{wy}-|X|^{-1}\sum_{\ell=1}^Dr^\ell_{ij}(A^*A_\ell - A_\ell A^*)_{wy}\\ & = |X|^{-1} \text{times $wy$ entry of a matrix known to be zero by Claim 2}\\ & = 0. \end{align}\] Thus we have the claim.

HS MEMO

\[\begin{align} |X|^{-1}\sum_{\ell=1}^D r^\ell_{ij}(A^*A_\ell - A_\ell A^*)_{wy} & = |X|^{-1}r^h_{ij}(A^*A_h- A_hA^*)_{wy}\\ & = r^h_{ij}(\langle \rho(x),\rho(w)\rangle - \langle \rho(x),\rho(y)\rangle) \end{align}\]