Chapter 8 Thin Graphs
Friday, February 5, 1993
Proof (Proof of Theorem 7.1 continued).
Fix \(i\) and pick \(a, b\in E_i^*TE_i^*\).
Since \(a\), \(b\) and \(ab\) are symmetric, \[ab = (ab)^\top = b^\top a^\top = ba.\] Hence \(E_i^*TE_i^*\) is commutative.
Fix \(i\) and pick \(a \in E_i^*TE_i^*\). Pick vertices \(y, z\in X\).
We want to show that \[a_{yz} = a_{zy}.\] We may assume that \[\partial(x, y) = \partial(x,z) = i,\] otherwise \[a_{yz} = a_{zy} = 0.\] By our assumption, there exists \(g\in G\) such that \[g(y) = z, \quad g(z) = y, \quad g(x) = x.\] Let \(\hat{g}\) denote the permutation matrix representing \(g\), i.e., \[\hat{g}\hat{y} =\widehat{g(y)} \quad \text{for all }\ y\in X, \quad \hat{y} = \begin{pmatrix}0\\\vdots \\ 1 \\\vdots \\0\end{pmatrix}\begin{matrix} \\ \\ \leftarrow y \\ \\ \text{ } \end{matrix}.\] If \(g\in \mathrm{Aut}(\Gamma)\), then \[\hat{g}A = A\hat{g} \quad \text{(Exercise)}.\] Also, we have \[\hat{g}E_j^* = E^*_j\hat{g} \quad (0\leq j\leq D),\] since \[\partial(x,y) = \partial(g(x), g(y)) = \partial(x, g(y)).\] Hence, \(\hat{g}\) commutes with each element of \(T\). We have \[\begin{align} a_{yz} & = (\hat{g}^{-1}a\hat{g})_{yz}, \quad (\hat{g})_{yz} = \begin{cases} 1 & g(z) = y\\ 0 & \text{else.}\end{cases}\\ & = \sum_{y', z'}(\hat{g}^{-1})_{yy'}a_{y'z'}\hat{g}_{z'z}\\ & \quad (\text{zero except for $g^{-1}(y') = y, \; g(z) = z'$}.)\\ & = a_{g(y)g(z)}\\ & = a_{zy}. \end{align}\] This proves Theorem 7.1.
Open Problem: Find all the graphs that satisfy the condition (G) for every vertex \(x\).
\(H(N, 2)\) is one example, because \[\mathrm{Aut}\Gamma_{1\cdots 1} \simeq S_\Omega, \quad x = (1\cdots 1), \quad \Gamma_i(x) = \{\hat{S} \mid |S| = i\}.\]
Property (G) is clearly related to the distance-transitive property.
Definition 8.1 Let \(\Gamma = (X, E)\) be any graph. \(\Gamma\) with \(G\subseteq \mathrm{Aut}(\Gamma)\) is said to be distance-transitive (or two-point homogeneous), whenever \[\text{for all } x, x', y, y'\in X \; \text{ with } \partial(x,y) = \partial(x',y'),\] there exists \(g\in G\) such that \[g(x) = x',\quad g(y) = y'.\] (This means \(G\) is as close to being doubly transitive as possible.)
Lemma 8.1 Suppose a graph \(\Gamma = (X, E)\) satisfies the property for every \(x\in X\). Then,
\(\quad (ia)\) \(\Gamma\) is vertex transitive; or
\(\quad (iia)\) \(\Gamma\) is bipartite \((X = X^+ \cup X^-)\) with \(X^+\), \(X^-\) each an orbit of \(\mathrm{Aut}(\Gamma)\).
Proof. \((i)\) Claim. Suppose \(y, z\in X\) are conneced by a path of even length. Then \(y, z\) are in the same orbit of \(\mathrm{Aut}(\Gamma)\).
Pf of Claim. It suffices to assume that the path has lenght \(2\), \(y \sim w\sim z\).
Now \(\partial(y,w) = \partial(w,z) = 1\). So there exits \(g\in \mathrm{Aut}(\Gamma)\) such that \[gw = w, \quad gy = z, \quad gz = y.\] This proves Claim.
Fix \(x\in X\). Now suppose that \(\Gamma\) is not vertex transitive, and we shall show \((ib)\).
Observe that \(X = X^+ \cup X^-\), where \[\begin{align} X^+ & = \{y\in X\mid \text{there exists a path of even length connecting $x$ and $y$}\},\\ X^- & = \{y\in X\mid \text{there exists a path of odd length connecting $x$ and $y$}\}. \end{align}\] Also, \(X^+\) is contained in an orbit \(O^+\) of \(\mathrm{Aut}(\Gamma)\), and \(X^-\) is contained in an orbit \(O^-\) of \(\mathrm{Aut}(\Gamma)\).
Now \(O^+\cap O^- = \emptyset\) (else \(O^+ = O^- = X\) and vertex transitive). So, \(X = O^+\), and \(X^- = O^-\).
Also \(X^+ \cup X^- = X\) is a bipartition by construction.
\((ii)\) Fix \(x, y, x', y'\) with \(\partial(x,y) = \partial(x',y')\).
By vertex transitivity, there exists an element \[g_1\in G \text{ such that } g_1x = x'.\] Observe that \[\partial(x', y') = \partial(x,y) = \partial(g_1x, g_1y) = \partial(x', g_1y).\] Hence, there exisits an element \[g_2\in G \text{ such that } g_1x' = x', g_2y' = g_1y, g_2g_1y = y'\] by (G(\(x'\))) property.
Set \(g = g_2g_1\). Then \[gx = x', gy = y'\] by construction.
The following graphs \(\Gamma = (X, E)\) are vertex transitive, and satisfy the property (G(\(x\))) for all \(x\in X\). \[J(D, N), \quad H(D, r), \quad J_q(D,N),\] where
\[\begin{align} X & = \{a_1\cdots a_D\mid a_i\in F, 1\leq i\leq D\}\\ & \quad F: \text{ any set of cardinality $r$}\\ E & = \{xy\mid y, x\in X, \; \text{$x$ and $y$ differ in exactly one coordiate}\}. \end{align}\]
\[\begin{align} X & = \text{the set of all $D$-dimensional subspaces of $N$-dimensional vector space over $GF(q)$.}\\ & \quad F: \text{ any set of cardinality $r$}\\ E & = \{xy\mid y, x\in X, \; \dim (x\cap y) = D-1\}. \end{align}\]
The following graph is distance-transitive but does not satisify (G(\(x\))) for any \(x\in G\).
\[\begin{align} X & = \text{the set of all $D\times N$ matrices with entries in $GF(q)$}.\\ E & = \{xy\mid y, x\in X, \; \mathrm{rank}(x-y) = 1\}. \end{align}\]
HS MEMO
For \(x, y\in X\) with \(\partial(x, y) = \partial(x,z) = i\), \[\begin{align} Y = \{j\in \Omega \mid x_j\neq y_j\} & \leftrightarrow Z = \{j\in \Omega \mid x_j\neq z_j\}\\ (y_{j_1}, \ldots, y_{j_i}) & \leftrightarrow (z_{\ell_1}, \ldots, z_{\ell_i}) \end{align}\]
\[\begin{align} X\cap Y & \leftrightarrow X \cap Z\\ (\Omega \setminus X)\cap Y & \leftrightarrow (\Omega \setminus X)\cap Z. \end{align}\]
\[X\cap Y \leftrightarrow X \cap Z.\]
The theory of a single thin irreducible \(T\)-module.
Let \(\Gamma = (X, E)\) be any graph. \[\begin{align} M & = \text{Bose-Mesner algebra over $K/\mathbb{C}$ generated by the adjacency matrix $A$.}\\ & = \mathrm{Span}(E_0, \ldots, E_R). \end{align}\]
\(M\) acts on the standard module \(V = \mathbb{C}^{|X|}\).
Fix \(x\in X\), let \(D \equiv D(x)\) be the \(x\)-diameter, and \(k = k(x)\) be the valency of \(x\).