Chapter 21 Norton Algebras

Wednesday, March 17, 1993

Proof (Proof of Lemma 20.4).

$(i)$ Suppose

\[v = \sum_{x\in X}\alpha_x \hat{x}.\] Pick a vertex $z\in X$ and compare $z$-coordinate of each side in $(i)$. \[\begin{align} (A^*_j(x)v)_z & = (A^*_j(x))_{zz}v_z = |X|(E_j)_{xz}\alpha_z.\\ |X|(E_{\hat{j}}\hat{x}\circ v)_z & = |X|(E_{\hat{j}}\hat{x})_z\cdot \alpha_z = |X|(E_j)_{xz}\alpha_z. \end{align}\] Note that $E_{\hat{j}}\hat{x}$ is the column $x$ of $E_{\hat{j}}$, which is the row $x$ of $E_j$.

$(ii)$ Fix $i, j, h$ such that $q^h_{ij} = 0$.

Claim. $E_h(E_iV \circ E_jV) = 0$.

\[\begin{align} E_h(E_iV \circ E_jV) & = E_h(\mathrm{Span}(v\circ w\mid v\in E_iV, w\in E_jV))\\ & = E_h(\mathrm{Span}(E_i\hat{y}\circ E_j\hat{z}\mid y,z\in X))\\ & = \mathrm{Span}(E_h(E_j\hat{z}\circ E_i\hat{y}\mid y,z\in X)\\ & = \mathrm{Span}((E_hA^*_{\hat{j}}(z)E_i)\hat{y}\mid y,z\in X) && \text{by $(i)$.} \end{align}\] But $q^h_{ij} = 0$ implies $q^{\hat{h}}_{\hat{j}\hat{i}} = 0$.

So, by Lemma 20.3 $(ii)$, \[ 0 = (E_{\hat{i}}A^*_{\hat{j}}E_{\hat{h}})^\top = E_h A^*_{\hat{j}}E_i.\] Hence, $E_h(E_iV\circ E_jV) = 0$.

$(iii)$ Fix $i, j, h$ such that $q^h_{ij}\neq 0$. Then,

\[E_h(E_iV \circ E_jV)\subseteq E_hV\] is clear. We show the other inclusion. Since \[\begin{align} E_i\hat{y} \circ E_j\hat{y} &= (\text{column $y$ of $E_i$}\circ \text{column $y$ of $E_j$}) \\ & = \text{column $y$ of $E_i\circ E_j$}\\ & = (E_i\circ E_j)\hat{y}\\ & = \left(\frac{1}{|X|}\sum_{h=0}^D q^h_{ij}E_h\right)\hat{y}, \end{align}\] we have, \[\begin{align} E_h(E_iV\circ E_jV) & = E_h\mathrm{Span}(E_i\hat{y}\circ E_j\hat{z}\mid y,z\in X)\\ & \supseteq E_h \mathrm{Span}(E_i\hat{y}\circ E_j\hat{y}\mid y\in X)\\ & = \mathrm{Span}(q^h_{ij}E_h\hat{y}\mid y\in X)\\ & = \mathrm{Span}(E_h\hat{y}\mid y\in X) && \text{since $q^h_{ij}\neq 0$}\\ & = E_hV. \end{align}\] This proves the assertion.

Lemma 21.1 Given a commutative scheme $Y = (X, \{R_i\}_{0\leq i\leq D})$, fix $j$ $(0\leq j\leq D)$. Define a binary multiplication: \[E_jV \times E_jV \longrightarrow E_jV \quad ((v,w) \mapsto v\ast w = E_j(v\circ w)).\] Then,

$(i)$ $v\ast w = w\ast v$, for all $v,w\in E_jV$,

$(ii)$ $v\ast (w + w') = v\ast w + v\ast w'$ for all $v,w,w'\in E_jV$, and

$(iii)$ $(\alpha v)\ast w = \alpha(v\ast w)$ for all $\alpha \in \mathbb{C}$.

In particular, the vector space $E_jV$ together with $\ast$ is a commutative $\mathbb{C}$-algebra, (not associative in general).

($N_j: (E_jV, \ast)$ is called the Norton algebra on $E_jV$.)

$(iv)$ $v\ast w = 0$ for all $v, w\in E_jV$ if and only if $q^j_{jj} = 0$.

Proof.

$(i)-(iii)$ Immediate.

$(iv)$ Immediate from Lemma 20.4 $(ii)$, $(iii)$.

Let $Y$, $j$, $N_j$ be as in Lemma 21.1, and $M$ Bose-Mesner algebra of $Y$. Let \[\begin{align} \mathrm{Aut}Y & = \{\sigma\in \mathrm{Mat}_X(\mathbb{C}) \mid \sigma: \text{ permutation matrix }, \sigma \cdot m = m\cdot \sigma \;\text{ for all }m\in M\}\\ & = \{\sigma\in \mathrm{Mat}_X(\mathbb{C}) \mid \sigma: \text{ permutation matrix },\\ & \qquad (x,y)\in R_i \to (\sigma x, \sigma y)\in R_i, \text{ for all } i, \text{ and for all } x,y\in X\}\\ \mathrm{Aut}(N_j) & = \{\sigma: E_jV \to E_jV \mid \sigma \text{ is a $\mathbb{C}$-algebra isomorphim},i.e.,\\ & \qquad \sigma(v\ast w) = \sigma(v)\ast\sigma(w) \text{ for all }v, w\in E_jV\}. \end{align}\]

Lemma 21.2 Let $Y, j, \ast$ be as in Lemma 21.1.

$(i)$ $E_jV$ is a module for $\mathrm{Aut}(Y)$.

$(ii)$ $\sigma|_{E_jV}\in \mathrm{Aut}(N_j)$ for all $\sigma \in \mathrm{Aut}(Y)$.

$(iii)$ $\mathrm{Aut}Y \longrightarrow \mathrm{Aut}(N_j), \; (\sigma \mapsto \sigma|_{E_j})$ is a homomorphism of groups,

(i.e., a representation of $\mathrm{Aut}(Y)$).

$(iv)$ Suppose $R_0, \ldots, R_D$ are orbits of $\mathrm{Aut}(Y)$ acting on $X\times X$, (so, we are in Example 17.2) then above representation is irreducible.

Proof.

$(i)$ Pick $\sigma\in \mathrm{Aut}Y$ and $v\in V$. Then,

\[\sigma E_j v = E_j\sigma v,\] since $\sigma$ commutes with each element of $M$.

$(ii)$ $\sigma|_{E_jV}: E_jV \to E_jV$ is an isomorphism of a vector space. Since $\sigma$ is invertible,for all $v,w\in E_jV$,

\[\sigma(v\ast w) = \sigma(E_j(E_jv\circ E_jw)) = E_j\sigma(E_jv\circ E_jw) = E_j(E_j\sigma v\circ E_j\sigma w) = \sigma(v)\ast \sigma(w).\]

$(iii)$ Immediate from $(i)$ and $(ii)$.

$(iv)$ Here, Bose-Mesner algebra $M$ is the full commuting algebra, i.e.,

\[M = \{m\in \mathrm{Mat}_X(\mathbb{C})\mid \sigma\cdot m = m\cdot \sigma, \text{ for all }\sigma\in \mathrm{Aut}(Y)\}.\] Suppose there sia a nonzero proper subspace $0\neq W\subsetneq E_jV$ that is $\mathrm{Aut}(Y)$-invariant.

Set \[W^\bot = \{v\in E_jV\mid \langle w, v\rangle = 0, \text{ for all }w\in W\}.\] Then, $W^\bot$ is a module for $\mathrm{Aut}(Y)$, since $\mathrm{Aut}(Y)$ is closed under transpose conjugate.

Let $e: V\to W$ and $f: V\to W^\bot$ be the orthogonal projection such that $e + f = E_j$, \[e^2 = e, f^2 = f, ef = fe = 0, eE_h = 0, \text{ if } h\neq j.\]

Since $e$ commutes with all $\sigma\in \mathrm{Aut}(Y)$, $e\in M$ and \[e = \sum_{i=0}^D \alpha_i E_i.\] If $h\neq j$, then $0 = eE_h$ and $\alpha_h = 0$. Thus, $e = \alpha_jE_j$, i.e., $e=0$ or $f=0$.

A contradiction.

Norton algebras were used in original construction of Monster, a finite simple group $G$.

Compute character table of $G$,

$\quad\to$ $p^h_{ij}$, $q^h_{ij}$ of group scheme on $G$,

$\quad\to$ find $j$ where $m_j = \dim E_jV$ is small and $q^j_{jj}\neq 0$,

$\quad\to$ guess abstract structure of $N_j$ using the knowlege of $p^h_{ij}$’s and $q^h_{ij}$’s,

$\quad\to$ compute $\mathrm{Aut}(N_j)$,

$\quad\to$ $G$.