Chapter 13 Modules of a DRG

Wednesday, February 17, 1993

Lemma 13.1 Let $\Gamma = (X, E)$ be any graph. Pick an edge $xy\in E$.

Assume the trivial $T(x)$-module $T(x)\delta$ is thin with measure $m_x$,

and the trivial $T(y)$-module $T(y)\delta$ is thin with measure $m_y$.

Then,

$(ia)$ ${\displaystyle \frac{m_x(\theta)}{k_x} = \frac{m_y(\theta)}{k_y} \text{ for all } \theta\in \mathbb{R \smallsetminus \{0\}}}$.

$(ib)$ ${\displaystyle \frac{m_x(0) -1}{k_x} = \frac{m_y(0)-1}{k_y} \text{ for all } \theta\in \mathbb{R \smallsetminus \{0\}}}$.

\[(\delta = \sum_{y\in X}\delta_y \hat{y} \quad \text{eigenvector corresponding to the maximal eigenvalue})\]

Proof. Apply Theorem 12.1, \[\begin{align} W & = T(x)\delta \quad r = 0, \quad d = d(x)\\ W' & = T(y)\delta \quad r' = 0, \quad d' = d(y). \end{align}\] Take $w = \hat{x}$, $w' = \hat{y}$.

Claim. $\mathrm{proj}_{T(y)\delta}\hat{x} = k^{-1}_yA\hat{y}$.

Pf. Since \[\hat{y}\in T(y)\delta, \quad A\hat{y}\in T(y)\delta.\] Show \[(\hat{x} - {k_y}^{-1} A\hat{y}) \bot (T(y)\delta).\] Recall \[A\hat{y} = \sum_{z\in X, yz\in E}\hat{z}.\] \[\hat{x} - {k_y}^{-1}Ay \in E^*_1(y)V.\] So, \[\hat{x} - \frac{1}{k_y}A\hat{y} \; \bot \; E^*_j(y)T(y)\delta \quad \text{if $j\neq 1$}\; (0\leq j\leq k(y)).\] And we have, \[\begin{align} \left\langle \hat{x} - \frac{1}{k_y}A\hat{y}, A\hat{y}\right\rangle & = \left\langle \hat{x}, \sum_{z\in X, yz\in E}\hat{z}\right\rangle - \frac{1}{k_y}\left\|\sum_{z\in X, yz\in E}\hat{z}\right\|^2\\ & = 1 - 1\\ & = 0 \end{align}\] This proves Claim.

Similarly, \[\mathrm{proj}_{T(x)\delta}\hat{y} = {k_x}^{-1}A\hat{x}.\] Hence, the polynomials $p, p'\in \mathbb{C}[\lambda]$ from Theorem 12.1 equal \[\frac{\lambda}{k_y} \quad \text{ and }\quad \frac{\lambda}{k_x}\] respectively.

By Theorem 12.1, \[\frac{m_x(\theta)\theta}{k_x} = m_x(\theta)\overline{p'(\theta)} = m_y(\theta)\overline{p(\theta)} = \frac{m_y(\theta)\theta}{k_y}.\] If $\theta\neq 0$, we have $(ia)$.

Also, \[\begin{align} \frac{1-m_x(0)}{k_x} & = \left(\sum_{\theta\in \mathbb{R}\setminus \{0\}}m_x(0)\right)\frac{1}{k_x} && \text{by $(ia)$}\\ & = \left(\sum_{\theta\in \mathbb{R}\setminus \{0\}}m_y(0)\right)\frac{1}{k_y}\\ & = \frac{1 - m_y(0)}{k_y} \end{align}\] Hence, we have $(ib)$.

Theorem 13.1 Suppose any graph $\Gamma = (X, E)$ is distance-regular with respect to every vertex $x\in X$. (So $\Gamma$ is regular or biregular by Lemma 12.1.)

Then,

Case $\Gamma$ is regular: the diameter $d(x)$ and the intersection numbers $a_i(x)$, $b_i(x)$, $c_i(x)$ $(0\leq i\leq d(x))$ are independent of $x\in X$.

(And $\Gamma$ is called distance-regular.)

Case $\Gamma$ is biregular: $(X = X^+\cup X^-)$

$d(x)$ and $a_i(x)$, $b_i(x)$, $c_i(x)$ $(0\leq i\leq d(x))$ are constant over $X^+$ and $X^-$. (And $\Gamma$ is called distance-biregular.)

Proof. We apply Lemma 13.1.

Case $\Gamma$: regular.

Then $m_x = m_y$ for all $xy\in E$. Hence, the measure of the trivial $T(x)$-module is independent of $x\in X$.

Case $\Gamma$ is biregular.

Then $m_x = m_{x'}$ for all $x, x'\in X$ with $\partial(x,x') = 2$.

Hence, the measure of the trivial $T(x)$-module is constant over $x\in X^+$, $X^-$.

Fix $x\in X$. Write $T\equiv T(x)$, $E^*_i \equiv E^*_i(x)$, $W = T\delta$ with measure $m$, diameter $d = d(x)$.

We know by Corollary 10.1 that $m$ determines \[d, \quad a_i(W) \; (0\leq i\leq d), \quad x_i(W) \; (1\leq i\leq d)\] (as $d = D(x) = d(W)$ by Lemma 11.1.)

We shall show that $m$ determines \[a_i(x), \; c_i(x), \; b_i(x) \quad (0\leq i\leq d).\] Observe: \[\begin{align} a_i(W) & = a_i(x) \quad (0\leq i\leq d)\\ x_i(W) & = b_{i-1}(x)c_i(x) \quad (1\leq i\leq d) \end{align}\]

HS MEMO

$a_i = a_i(W)$ is an eigenvalue of \[E^*_iAE^*_i \text{ on } E^*_iW = \langle \sum_{y\in \Gamma_i(x)}\hat{y}\rangle. \] (See Lemma 12.1.)

$x_i = x_i(W)$ is an eigenvalue of \[E^*_{i-1}AE^*_iAE^*_{i-1} \text{ on } E^*_{i-1}W,\] and \[\begin{align} A\sum_{y\in X,\partial(x,y)}\hat{y} & = b_{i-1}(x)\sum_{y\in X, \partial(x,y)=i-1}\hat{y} \\ & \quad + a_i(x)\sum_{y\in X, \partial(x,y)=i}\hat{y} \\ & \quad + c_{i+1}(x)\sum_{y\in X, \partial(x,y)=i+1}\hat{y} \end{align}\] So $x_i = b_{i-1}(x)c_i(x)$.

Set $k^+ = k_x$. Define \[k^- = \frac{{\theta_0}^2}{k^+},\] where $\theta_0$ is the maximal eigenvalue. (See Lemma 11.1.)

(So, $k^+ = k^-$ is the valency, if $\Gamma$ is regular.)

For every $i \; (0\leq i\leq d)$ and for every $z\in X$ with $\partial(x,z) = i$, \[\begin{align} k_z & = c_i(x) + a_i(x) + b_i(x)\\ & = \begin{cases} k^+ & \text{if $i$ is even,}\\ k^- & \text{ if $i$ is odd.} \end{cases} \end{align}\]

Now $m$ determines \[c_0(x) = a_0(x) = 0,\quad c_1(x) = 1,\] \[b_0(x) = b_0(x)c_1(x) = x_1(W).\] \[\begin{align} k^+ & = b_0(x)\\ k^- & = {\theta_0}^2/k^+\\ c_i(x) & = x_i(W)/b_{i-1}(x) \quad (1\leq i\leq d)\\ b_i(x) & = \begin{cases} k^+ - a_i(x) - c_i(x) \quad \text{$i$; even,}\\ k^--a_i(x)-c_i(x) \quad \text{$i$: odd}. \end{cases} \end{align}\] This proves the assertions.

Proposition 13.1 Under the assumption of Theorem 13.1, the following hold.

Case $\Gamma$: regular.

$(i)$ $\dim E_iV = |X|m(\theta_i)$.
$(ii)$ $\Gamma$ has exactly $d+1$ distinct eigenvalues

($d = \mathrm{diam}\Gamma = d(x), \; \text{ for all }\; x\in X$).

Case $\Gamma$: biregular.

$(i)$ $\dim E_iV = |X^+| m^+(\theta_i) + |X^-|m^-(\theta_i)$.
$(ii)$ $\Gamma$ has exactly $d^++1$ distinct eigenvalues $(d^+\geq d^-)$.
$(iii)$ If $d^+$ is odd, the $\Gamma$ is regular.
$(iv)$ $d^+ = d^-$, or $d^+ = d^-+1$ is even.
$(v)$ $a_i(x) = 0$ for all $i$ and for all $x$.

Proof. $(i)$ Suppose $\Gamma$ is regular.

Let $m_x$ be the measure of the trivial $T(x)$-module, \[m_x(\theta_i) = \|E_i\hat{x}\|^2, \quad \text{as}\; \|\hat{x}\| = 1.\] Now, \[\begin{align} |X|m_x(\theta_i) & = \sum_{x\in X}m_x(\theta_i)\\ & = \sum_{x\in X}\|E_i\hat{x}\|^2\\ & = \sum_{y,z\in X}|(E_i)_{yz}|^2\\ & = \mathrm{trace} E_i\overline{E_i}^\top. \end{align}\] Since $A$ is real symmetric and \[E_i\overline{E_i}^\top = E_i^2 = E_i\] with $E_i$ symmetric \[E_i \sim \begin{pmatrix} I & O \\ O & O\end{pmatrix}.\] \[\mathrm{trace} E_i = \mathrm{rank} E_i = \dim E_iV.\] Thus, we have the assertion in this case.

Suppose $\Gamma$ is biregular.

Then, same except, \[\sum_{x\in X} m_x(\theta_i) = |X^+|m^+(\theta_i) + |X^-|m^-(\theta_i).\]

$(ii)$ $\Gamma$: regular. Immediately, if $\theta$ is an eigenvalue of $\Gamma$, then $m(\theta) \neq 0$.

$\Gamma$: biregular. For each $\theta = \theta_i \in \mathbb{R}\setminus\{0\}$, \[\begin{align} m^-(\theta) \neq 0 &\Leftrightarrow m^+(\theta) \neq 0\\ & \Leftrightarrow \theta \; \text{ is an eigenvalue of $\Gamma$}\\ & \quad\quad \left(\frac{m^+(\theta)}{k^+} = \frac{m^-(\theta)}{k^-} \right) \end{align}\]

$(iv)$ and $(v)$ are clear.

HS MEMO

$(iii)$ If $d^+$ is odd, $d^+ = d^-$ and $\Gamma$ has even number of eigenvalues, i.e., $0$ is not an eigenvalue. So $A$ is nonsingular, and $\Gamma$ is regular.