Chapter 5 \(T\)-Modules of \(H(D,2)\), I

Friday, January 29, 1993

Let \(\Gamma = (X, E)\) be a graph, \(A\) the adjacency matrix, and \(V\) the standard module over \(K = \mathbb{C}\).

Fix a base \(x\in X\) and write \(E_i^* \equiv E_i^*(x)\), and \(T \equiv T(x)\).

Let \(W\) be an irreducible \(T\)-module with endpoint \(r:= \min\{i\mid E_i^*W \neq 0\}\) and diameter \(d:=|\{i\mid E_i^*W\neq 0\}|-1\).

We have \[\begin{align} {E^*_i}W & \neq 0 & r\leq i \leq r+d\\ & = 0 & 0 \leq i < r \;\text{ or }\; r+d < i \leq d(x). \end{align}\]

Claim: \(E_i^*AE_j^*W \neq 0\) if \(|i-j| = 1\) for \(r\leq i,j\leq r+d\). (See Lemma 4.1.)

Suppose \(E_{j+1}^*AE_j^*W = 0\) for some \(j\) with \(r \leq j < r+d\). Observe that \[\tilde{W} = E^*_rW + \cdots + E^*_jW\] is \(T\)-invariant with \[0 \subsetneq \tilde{W} \subsetneq W.\] Becase \(A\tilde{W} \subseteq \tilde{W}\) since \(AE_j^*W \subseteq E^*_{j-1}W + E^*_jW\), \[E_k^*\tilde{W} \subseteq \tilde{W} \quad\text{for all }\; k,\] we have \(T\tilde{W} \subseteq{W}\).

Suppose \(E_{i-1}^*AE_i^*W = 0\) for some \(i\) with \(r \leq i < r+d\).

Similarly, \[\tilde{W} = E^*_iW + \cdots + E^*_{r+d}W\] is a \(T\)-module with \(0\subsetneq \tilde{W} \subsetneq W\).

Definition 5.1 Let \(\Gamma\), \(E^*_i\), and \(T\) be as above. Irreducible \(T\)-modules \(W\) and \(W'\) are isomorphic whenever there is an isomorphism \(\sigma: W \to W'\) of vector spaces such that \(a\sigma = \sigma a\) for all \(a\in T\).

Recall that the standard module \(V\) is an orthogonal direct sum of irreducible \(T\)-modules \[W_1 \oplus W_2 \oplus \cdots \oplus W_{\ell}, \; \text{for some }\ell.\] Given \(W\) in this list, the multiplicity of \(W\) in \(V\) is \[|\{j \mid W_j \simeq W\}|.\]

HS MEMO

It is known that the multiplicity does not depend on the decomposition.

Now assume that \(\Gamma\) is the \(D\)-cube, \(H(D,2)\) with \(D\geq 1\). View \[\begin{align} X & = \{a_1\cdots a_D\mid a_i\in \{1, -1\}, 1\leq i\leq D\},\\ E & = \{xy\mid x, y\in X, \; x, y \;\text{ differ in exactly 1 coordinate}\}. \end{align}\] Find \(T\)-modules.

Claim: \(H(D,2)\) is bipartite with a partition \(X = X^+ \cup X^-\), where \[\begin{align} X^+ & = \{a_1\cdots a_D\in X\mid \prod a_i > 0\}\\ X^- & = \{a_1\cdots a_D \in X \mid \prod a_i < 0\} \end{align}\]

Observe: for all \(y, z\in X\), \[\partial(y,z) = i \Leftrightarrow y, z \; \text{ differ in exactly in }\; i\; \text{ coorinates with }\; 0\leq i\leq D.\] Here, the diameter of \(H(D, 2) = D = d\) for all \(x\in X\).

Theorem 5.1 Let \(\Gamma = H(D,2)\) be as above. Fix \(x\in X\), and write \(E_i^* = E^*_i(x)\), and \(T = T(x)\).

Let \(W\) be an irreducible \(T\)-module with endpoint \(r\), and diameter \(d\) with \(0\leq r \leq r+d\leq D\).

\((i)\) \(W\) has a basis \(w_0, w_1, \ldots, w_d\) with \(w_i\in E^*_{i+r}W\) for \(0\leq i\leq d\). With respect to which the matrix representing \(A\) is

\[ \begin{pmatrix} 0 & d & 0 & \cdots & 0 & 0 & 0\\ 1 & 0 & d-1 & \cdots & 0 & 0 & 0\\ 0 & 2 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & 3 & \cdots & 0 & 0 & 0\\ \cdots & \cdots & \cdots & \ddots & \ddots & \cdots & \cdots \\ 0 & 0 & 0 & \ddots & 0 & 2 & 0\\ 0 & 0 & 0 & \cdots & d-1 & 0 & 1\\ 0 & 0 & 0 & \cdots & 0 & d & 0 \end{pmatrix} \]

\((ii)\) \(d= D-2r\). In particular, \(0\leq r\leq D/2\).

\((iii)\) Let \(W'\) denote an irreducible \(T\)-module with endpoint \(r'\). Then \(W\) and \(W'\) are isormorphic as \(T\)-modules if and only if \(r = r'\).

\((iv)\) The multiplicity of the irreducible \(T\)-module with endpoint \(r\) is

\[\binom{D}{r} - \binom{D}{r-1} \quad \text{if } 1\leq r \leq R/2,\] and \(1\) if \(r = 0\).

Proof. Recall that \(\Gamma\) is vertex transitive. It is a Cayley graph.

Hence without loss of generality, we may assume that \(x = \overbrace{11\cdots 1}^{D}\).

Notation: Set \(\Omega = \{1, 2, \ldots, D\}\). For every subset \(S \subseteq \Omega\), let \[\hat{S} = a_1\cdots a_d \in X \quad a_i = \begin{cases} -1 & \text{if }\; i\in S\\ 1 & \text{if } i\not\in S.\end{cases}\] In particular, \(\hat{\emptyset} = x\) and \[|S| = i \Leftrightarrow \partial(x, \hat{S}) = i \Leftrightarrow \hat{S}\in E^*_iV.\] For all \(S, T\subseteq \Omega\), we say \(S\) covers \(T\) if and only if \(S\supseteq T\) and \(|S| = |T| +1\).

Observe that \(\hat{S}, \hat{T}\) are adjacent in \(\Gamma\) if and only if either \(T\) coverse \(S\) or \(S\) coverr \(T\).

Define the ‘raising matrix’ \[R = \sum_{i=0}^D E^*_{i+1}AE^*_i.\]

Observe that \[RE_i^*V \subseteq E^*_{i+1} V \; \text{ for }\; 0\leq i \leq D, \; \text{ and }E^*_{D+1}V = 0.\] Indeed for any \(S\subseteq \Omega\) with \(|S| = i\), \[\begin{align} R\hat{S} & = RE^*_i\hat{S} \\ & = E^*_{i+1}A\hat{S} \\ & = \sum_{T_1 \subseteq \Omega, S \text{ covers }T_1} E^*_{i+1}\widehat{T_1} + \sum_{T \subseteq \Omega, T \text{ covers }S} E^*_{i+1}\hat{T}\\ & = \sum_{T \subseteq \Omega, T \text{ covers }S} E^*_{i+1}\hat{T}. \end{align}\]

Define the ‘lowering matrix’ \[L = \sum_{i=0}^D E^*_{i-1}AE^*_i.\]

Observe that \[LE_i^*V \subseteq E^*_{i-1}V \; \text{ for }\; 0\leq i \leq D, \; \text{ and }E^*_{-1}V = 0.\] Indeed for any \(S\subseteq \Omega\), \[L\hat{S} = \sum_{T\subseteq \Omega, S \text{ covers }T} \hat{T}.\]

Observe that \(A = L + R\).

For convenience, set \[A^* = \sum_{i=0}^D (D-2i)E_i^*.\]

Claim: The following hold.

\((a)\) \(LR - RL = A^*\).
\((b)\) \(A^*L - LA^* = 2L\).
\((c)\) \(A^*R - RA^* = -2R\).

In particular \(\mathrm{Span}(R,L, A^*)\) is a ’representation of Lie algebra \(\mathrm{sl}_2(\mathbb{C})\).

HS MEMO

\[\mathrm{sl}_2(\mathbb{C}) = \{X\mid \mathrm{Mat}(\mathbb{C} \mid \mathrm{tr}(X) = 0\}.\] For \(X, Y\in \mathrm{sl}_2(\mathbb{C})\), define a binary operation \([X, Y] = XY - YX\). \[A^*\sim \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}, \quad L\sim \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, \quad R\sim \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}.\] Then these satisfy the relations \((a)\) - \((c)\) above.

Proof of Claim. Apply both sides to \(\hat{S}\) \((S\subseteq \Omega)\). Say \(|S| = i\).

Proof of \((a)\): \[\begin{align} (LR - RL)\hat{S} & = L\left(\sum_{\substack{T \subseteq \Omega, T \text{ covers }S\\(D-i \text{ of them})}}\hat{T}\right) - R \left(\sum_{\substack{U \subseteq \Omega, S \text{ covers }U\\(i \text{ of them})}}\hat{U}\right)\\ & = (D-i)\hat{S} + \sum_{V \subseteq \Omega, |V| = i, |S\cap V| = i-1}\hat{V}\\ & \qquad - \left(i\hat{S} + \sum_{V \subseteq \Omega, |V| = i, |S\cap V| = i-1}\hat{V}\right)\\ & = (D-2i)\hat{S}\\ & = A^*\hat{S}. \end{align}\]

Proof of \((b)\): \[\begin{align} (A^*L - LA^*)\hat{S} & = (D-2(i-1))L\hat{S} - (D-2i)L\hat{S} \quad (\text{since} \; L\hat{S}\in E^*_{i-1}V)\\ & = 2L\hat{S}. \end{align}\]

Proof of \((c)\): \[\begin{align} (A^*R - RA^*)\hat{S} & = (D-2(i+1))R\hat{S} - (D-2i)R\hat{S} \quad (\text{since} \; R\hat{S}\in E^*_{i+1}V)\\ & = -2R\hat{S}. \end{align}\]

Let \(W\) be an irreducible \(T\)-module with endpoint \(r\) and diameter \(d\) \((0\leq r \leq r+d \leq D)\).

Proof of \((i)\) and \((ii)\):

Pick \(0\neq w \in E^*_rW\).

Claim: \(LRw = (D-2r)w\).

Pf. \[\begin{align} LRw & = (A^*+RL)w \quad (\text{by Claim }(a))\\ & = A^*w \quad (Lw \in E^*_{r-1}W = 0)\\ & = (D-2r)w. \end{align}\] Define \[w_i = \frac{1}{i!}R^iw \in E^*_{r+i}W \quad (0\leq i \leq d).\] Then, \[\begin{align} Rw_i & = (i+1)w_{i+1}\quad (0\leq i \leq d)\\ Rw_d & = 0 \quad (\text{by definition of }d) \end{align}\]

Claim: \(Lw_0 = 0\) and \[Lw_i = (D-2r-i+1)w_{i-1} \quad (1\leq i\leq d).\]

Pf. We prove by induction on \(i\). The case \(i=0\) is trivial, and the case \(i=1\) follows from above claim. Let \(i\geq 2\), \[\begin{align} Lw_i & = \frac{1}{i}LRw_{i-1} = \frac{1}{i}(A^*+RL)w_{i-1} \quad (\text{by Claim (a)})\\ & \quad \text{(by induction hypothesis)}\\ & = \frac{1}{i}((D-2(r+i-1))w_{i-1} + (D-2r-(i-1)+1)Rw_{i-2} \quad (Rw_{i-2} = (i-1)w_{i-1})\\ & = \frac{1}{i}i(D-2r-i+1)w_{i-1}\\ & = (D-2r-i+1)w_{i-1}. \end{align}\]

Claim: \(w_0, \ldots, w_d\) is a basis for \(W\).

Pf. Let \(W' = \mathrm{Span}\{w_0, \ldots, w_d\}\). Then \(W'\) is \(R\) and \(L\) invariant. So it is \(A = R+L\) invariant.

Also it is \(E^*_i\)-invariant for every \(i\).

Hence \(W'\) is a \(T\)-module.

Since \(W\) is irreducible, \(W' = W\).

As \(w_i\)’s are orthogonal, they are linearly independent. Note that \(w_i\neq 0\) by the definition of \(d\) and Lemma 4.1 \((iv)\).

Claim: \(d = D-2r\).

Pf. By \((a)\), \[\begin{align} 0 & = (LR - RL - A^*)w_d \\ & = 0 - (D-2r-d+1)Rw_{d-1} - (D-2(r+d))w_d\\ & = -d(D-2r-d+1)w_d - (D-2(r+d))w_d\\ & = (-dD + 2rd + d^2 - d - D + 2r + 2d)w_d\\ & = (d^2 + (2r-D+1)d + 2r - D)w_d\\ & = (d+2r-D)(d+1)w_d. \end{align}\] Hence \(d = D-2r\).

Therefore, with respect to a bais \(w_0, w_1, \ldots, w_d\), \(A = L+R\), \(w_{-1} = w_{d+1} = 0\), \[Lw_i = (d-i+1)w_{i-1}, \quad Rw_i = (i+1)w_{i+1}.\] \[L = \begin{pmatrix} 0 & d & 0 & \cdots & 0 & 0\\ 0 & 0 & d-1 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ \vdots & \vdots & \cdots & \cdots & 0 & 1\\ 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}, \qquad R = \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 0\\ 1 & 0 & 0 & \cdots & 0 & 0\\ 0 & 2 & 0 & \cdots & \vdots & \vdots\\ \vdots & \vdots & \ddots & \ddots & 0 & 1\\ 0 & 0 & 0 & \cdots & d & 0 \end{pmatrix}.\] This completes the proof of \((i)\) and \((ii)\).