Chapter 23 Representation of a Scheme

Monday, March 22, 1993

Theorem 23.1 Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ be a symmetric scheme. (View the standard module $V$ as an algebra of functions from $X$ to $\mathbb{C}$.) Then the following are equivalent.

$(i)$ $Y$ is $Q$-polynomial with respect to ordering $E_0, E_1, \ldots, E_D$ of primitive idempotents.

$(ii)$ For all $i$ $\:(0\leq i\leq D)$,

\[E_0V + E_1V + (E_1V)^2 + \cdots + (E_1V)^i = E_0V + E_1V + \cdots + E_iV.\]

Proof.

By Lemma 20.4 $(ii)$, $(iii)$.

\[E_h(E_iV\circ E_jV) = 0 \text{ if and only if } q^h_{ij} = 0 \quad (0\leq i,j,h\leq D).\]

$(i)\to(ii)$ By our assumption,

\[q^h_{1j} = 0 \text{ if } |h-j|>1, \text{ and } q^{j+1}_{1j}\neq 0.\] So, \[\begin{equation} E_1V\circ E_jV \subseteq E_{j-1}V + E_jV + E_{j+1}V \quad (0\leq j\leq D), \tag{23.1} \end{equation}\] \[\begin{equation} E_{j+1}(E_1V\circ E_jV) = E_{j+1}V \quad (0\leq j\leq D-1), \tag{23.2} \end{equation}\] by Lemma 20.4.

Also $E_0V \subseteq \mathrm{Span}(\delta)$, where $\delta$ is all 1’s vector, i.e., $1$ as a function $X\to \mathbb{C}$. So, \[\begin{equation} E_0V\circ E_jV = E_jV \quad (0\leq j\leq D). \tag{23.3} \end{equation}\] Show $(ii)$ by induction on $i$.

The cases $i=0, 1$ are trivial.

$i>1$: $\subseteq$. \[\begin{align} & E_0V + E_1V + (E_1V)^2 + \cdots + (E_1V)^i\\ & \quad = E_0V + E_1V\circ (E_0V + E_1V + \cdots + (E_1V)^{i-1})\\ & \quad = E_0V + E_1V\circ (E_0V + E_1V + \cdots + E_{i-1}V)\\ & \quad \subseteq E_0V + E_1V + \cdots + E_{i}V \end{align}\] by (23.1).

$\supseteq$.

Claim. $E_iV\subseteq E_1V\circ E_{i-1}V + E_{i-1}V + E_{i-2}V \quad (2\leq i\leq D)$.

Proof of Claim. By (23.2), \[E_i(E_1V \circ E_{i-1}V) = E_iV.\] For all $v\in E_i V$, there exists $u\in E_1V\circ E_{i-1}V$ such that $E_iu = v$.

On the other hand, by (23.1), \[E_1V\circ E_{i-1}V \subseteq E_{i-2}V + E_{i-1}V + E_{i-2}V.\] So, $u = w+v$, where $w\in E_{i-2}V + E_{i-1}V$. We have, \[w = u-v \in E_1V \circ E_{i-1}V + E_{i-1}V + E_{i-2}V\] as desired.

HS MEMO

\[E_iV \circ E_jV = \mathrm{Span}(u\circ v \mid u\in E_iV, v\in E_jV).\]

By claim, \[\begin{align} & E_0V + E_1V + \cdots + E_iV\\ & \quad \subseteq E_0V + E_1V + \cdots + E_{i-1}V + E_1V\circ E_{i-1}V\\ & \quad \subseteq E_0V + E_1V + \cdots + (E_{1}V)^{i-1} + E_1V(E_0V + E_1V + \cdots + (E_{1}V)^{i-1})\\ & \quad \subseteq E_0V + E_1V + \cdots + (E_{1}V)^{i-1} + (E_1V)^{i}. \end{align}\]

$(ii)\to(i)$

Claim 1. Pick $i, j$ $(0\leq i,j\leq D)$ with $j>i+1$. Then $q^j_{1i} = 0$.

Proof of Claim 1. \[\begin{align} E_j(E_1\circ E_jV) & \subseteq E_j(E_1V\circ(E_0V + E_1V + (E_1V)^2 + \cdots + (E_1V)^i))\\ & \subseteq E_j(E_0V + E_1V + (E_1V)^2 + \cdots + (E_1V)^{i+1})\\ & = E_j(E_0V + E_1V + \cdots + E_{i+1}V)\\ & = 0. \end{align}\] So $q^j_{1i}=0$ by Lemma 20.4.

Claim 2. $q^{i+1}_{1i} \neq 0$ $(0\leq i < D)$.

Proof of Claim 2. \[\begin{align} & E_0V + E_1V + \cdots + E_{i+1}V\\ & \quad = E_0V + E_1V + \cdots + (E_1V)^{i+1}\\ & \quad = E_0V + E_1V\circ(E_0V + E_1V + \cdots + (E_{1}V)^{i})\\ & \quad = E_0V + E_1V\circ(E_0V + E_1V + \cdots + E_iV)\\ & \quad = E_0V + E_1V\circ(E_0V + \cdots + E_iV). \end{align}\] So, \[\begin{align} E_{i+1}V & = E_{i+1}(E_1V\circ (E_0V + \cdots + E_iV))\\ & = E_{i+1}(E_1V \circ E_iV) \end{align}\] by Claim 1 and Lemma 20.4.

Hence, $q^{i+1}_{1i}\neq 0$ by Lemma 20.4.

Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ be a commutative scheme with standard module $V$.

Definition 23.1 A representation of $Y$ is a pair $(\rho, H)$, where $H$ is a non-zero Hermitean space (with inner product $\langle \;, \;\rangle$) and $\rho: X\to H$ is a map satisfying the following.

$\mathrm{R1}$. $H = \mathrm{Span}(\rho(x)\mid x\in X)$.

$\mathrm{R2}$. $\langle \rho(x), \rho(y)\rangle$ depends only on $i$ for which $(x,y)\in R_i$ $(x,y\in X)$.

$\mathrm{R3}$. For every $x\in X$ and for all $i$ $(0\leq i\leq D)$,

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

Above representation is nondegenerate if $\{\rho(x)\mid x\in X\}$ are distinct.

Example 23.1 $Y = H(D,2)$, $X = \{a_1\cdots a_D\mid a_i\in \{1,-1\}, 1\leq i\leq D\}$. Let $H = \mathbb{C}^D$ and $\langle \;, \;\rangle$ usual Hermitean dot product.

For a vertex $x = a_1\cdots a_D\in X$, define \[\rho(x) = a_1\cdots a_D\in H.\] Then, $\mathrm{R1}-\mathrm{R3}$ hold.

HS MEMO

$\mathrm{R1}, \mathrm{R2}$ are obvious. For $\mathrm{R3}$, we may assume that $x = 1\cdots 1$. Restrict \[\sum_{y\in X, (y,x)\in R_i}\rho(y)\] on the first coordinate. Then, \[\begin{align} -1 & \quad \text{appears }\; \binom{D-1}{i-1} \;\text{ times}\\ 1 & \quad \text{appears } \;\binom{D-1}{i} \;\text{ times}. \end{align}\] Hence, \[\sum_{y\in X, (y,x)\in R_i}\rho(y) = \left(\binom{D-1}{i} - \binom{D-1}{i-1}\right)\rho(x).\]

Let $(\rho, H)$ be a representation of arbitrary commutative scheme $Y$. Set \[E = (\langle \rho(x),\rho(y)\rangle)_{x,y\in X}\] Gram matrix of the representation.

Definition 23.2 Representations $(\rho, H)$, $(\rho', H')$ of $Y$ are equivalent, whenever, Gram matrices are related by \[E'\in \mathrm{Span} E.\] We do not distinguish between equivalent representations.

Note. Suppose $(\rho, H)$ is a representation of a symmetric scheme $Y$. Pick $x,y\in X$ with $(x,y)\in R_j$.

Then $(y,x)\in R_j$. So, by $\mathrm{R2}$, \[\langle \rho(x), \rho(y)\rangle = \langle \rho(y),\rho(x)\rangle = \overline{\langle \rho(x), \rho(y)\rangle},\] since $\langle \;, \;\rangle$ is Hermitean.

Hence, the Gram matrix $E$ of $\rho$ is real symmetirc. Without loss of generality, we can view $H$ as a real Euclidean space in this case.

Lemma 23.1 Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ be a commutative scheme and $V$ a standard module.

Let $E_j$ be any primitive idempotent of $Y$.

$(i)$ $(\rho, H)$ is a representation of $Y$, where $H = E_jV$ (with inner product inherited from $Y$).

\[\rho: X \to H \quad (x\mapsto E_j\hat{x})\] (i.e., $\rho(x)$ is the $x$-th column of $E_j$.)

$(ii)$ $\langle \rho(x),\rho(y)\rangle = |X|^{-1}q_j(i)$, if $(x,y)\in R_i$, $(x,y\in X)$.

$(iii)$ For $0\leq i\leq D$ and $x,y\in X$,

\[\sum_{y\in X, (y,x)\in R_i}\rho(y) = p_i(j)\rho(x).\]

$(iv)$ $(\rho,H)$ is nondegenerate if and only if $q_j(i) \neq q_j(0)$ for all $i$, $(0\leq i\leq D)$.

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

Proof.

$(i)-(iii)$.

$\mathrm{R1}$: $\mathrm{Span}(\rho X)$ is the column space of $E_j$ which is equal to $H$.

$\mathrm{R2}$: \[\begin{align} \langle \rho(x),\rho(y)\rangle & = \langle E_j\hat{x}, E_j\hat{y}\rangle \\ & = (\overline{E_j\hat{x}})^\top E_j\hat{y}\\ & = \hat{x}^\top \overline{E_j}^\top E_j\hat{y}\\ & = \hat{x}^\top E_j \hat{y}\\ & = (E_j)_{xy}. \end{align}\] Note that $\overline{E_j}^\top = E_j$ by Lemma 19.1.

Recall \[E_j = |X|^{-1}(q_j(0)A_0 + \cdots + q_j(D)A_D).\] So, \[(E_j)_{xy} = |X|^{-1}q_j(i), \quad \text{ where } \; (x,y)\in R_i.\]

$\mathrm{R3}$: Recall \[A_i = p_i(0)E_0 + \cdots + p_i(D)E_D.\] So, $E_jA_i = p_i(j)E_j$, and \[p_i(j)\rho(x) = p_i(j)E_j\hat{x} = E_jA_i\hat{x} = E_j\sum_{y\in X, (y,x)\in R_i}\hat{y} = \sum_{y\in X, (y,x)\in R_i}\rho(y).\]

Note. \[A_i\hat{x} = \sum_{y\in X, (x,y)\in R_{i'}}\hat{y}.\]

Pf. \[\begin{align} \text{$z$ entry of LHS} & = (A_i\hat{x})_z \\ & = \sum_{w\in X}(A_i)_{zw}\hat{x}_w\\ & = (A_i)_{zx}\\ & = \begin{cases} 1 & \text{if $(x,z)\in R_{i'}$}\\ 0 & \text{else}. \end{cases} \end{align}\] \[\begin{align} \text{$z$ entry of RHS} & = \sum_{y\in X, (x,y)\in R_{i'}, z = y}1\\ & = \begin{cases} 1 & \text{if $(x,z)\in R_{i'}$}\\ 0 & \text{else}. \end{cases} \end{align}\]

$(iv)$ By $(ii)$,

\[\begin{align} \|\rho(x)\|^2 & = \langle \rho(x), \rho(y)\rangle\\ & = |X|^{-1}q_j(0)\\ & = |X|^{-1}m_j, \end{align}\] as $m_j = \dim E_jV$, and is independent of $x\in X$.

Pick distinct $x,y\in X$ such that $(x,y)\in R_i$ with $i\neq 0$.

Then, \[\begin{align} \rho(x) = \rho(y) & \Leftrightarrow \langle \rho(x),\rho(y)\rangle = \|\rho(x)\|^2 = |X|^{-1}q_j(0)\\ & \Leftrightarrow |X|^{-1}q_j(i) = |X|^{-1}q_j(0)\\ & \Leftrightarrow q_j(i) = q_j(0). \end{align}\]

Hence, we have $(iv)$. To be continued.