Chapter 21 Norton Algebras

Wednesday, March 17, 1993

Proof (Proof of Lemma 20.4).

(i) Suppose

v=xXαxˆx. Pick a vertex zX and compare z-coordinate of each side in (i). (Aj(x)v)z=(Aj(x))zzvz=|X|(Ej)xzαz.|X|(Eˆjˆxv)z=|X|(Eˆjˆx)zαz=|X|(Ej)xzαz. Note that Eˆjˆx is the column x of Eˆj, which is the row x of Ej.

(ii) Fix i,j,h such that qhij=0.

Claim. Eh(EiVEjV)=0.

Eh(EiVEjV)=Eh(Span(vwvEiV,wEjV))=Eh(Span(EiˆyEjˆzy,zX))=Span(Eh(EjˆzEiˆyy,zX)=Span((EhAˆj(z)Ei)ˆyy,zX)by (i). But qhij=0 implies qˆhˆjˆi=0.

So, by Lemma 20.3 (ii), 0=(EˆiAˆjEˆh)=EhAˆjEi. Hence, Eh(EiVEjV)=0.

(iii) Fix i,j,h such that qhij0. Then,

Eh(EiVEjV)EhV is clear. We show the other inclusion. Since EiˆyEjˆy=(column y of Eicolumn y of Ej)=column y of EiEj=(EiEj)ˆy=(1|X|Dh=0qhijEh)ˆy, we have, Eh(EiVEjV)=EhSpan(EiˆyEjˆzy,zX)EhSpan(EiˆyEjˆyyX)=Span(qhijEhˆyyX)=Span(EhˆyyX)since qhij0=EhV. This proves the assertion.

Lemma 21.1 Given a commutative scheme Y=(X,{Ri}0iD), fix j (0jD). Define a binary multiplication: EjV×EjVEjV((v,w)vw=Ej(vw)). Then,

(i) vw=wv, for all v,wEjV,
(ii) v(w+w)=vw+vw for all v,w,wEjV, and
(iii) (αv)w=α(vw) for all αC.

In particular, the vector space EjV together with is a commutative C-algebra, (not associative in general).

(Nj:(EjV,) is called the Norton algebra on EjV.)

(iv) vw=0 for all v,wEjV if and only if qjjj=0.

Proof.

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

Let Y, j, Nj be as in Lemma 21.1, and M Bose-Mesner algebra of Y. Let AutY={σMatX(C)σ: permutation matrix ,σm=mσ for all mM}={σMatX(C)σ: permutation matrix ,(x,y)Ri(σx,σy)Ri, for all i, and for all x,yX}Aut(Nj)={σ:EjVEjVσ is a C-algebra isomorphim,i.e.,σ(vw)=σ(v)σ(w) for all v,wEjV}.

Lemma 21.2 Let Y,j, be as in Lemma 21.1.

(i) EjV is a module for Aut(Y).
(ii) σ|EjVAut(Nj) for all σAut(Y).
(iii) AutYAut(Nj),(σσ|Ej) is a homomorphism of groups,

(i.e., a representation of Aut(Y)).

(iv) Suppose R0,,RD are orbits of Aut(Y) acting on X×X, (so, we are in Example 17.2) then above representation is irreducible.

Proof.

(i) Pick σAutY and vV. Then,

σEjv=Ejσv, since σ commutes with each element of M.

(ii) σ|EjV:EjVEjV is an isomorphism of a vector space. Since σ is invertible,for all v,wEjV,

σ(vw)=σ(Ej(EjvEjw))=Ejσ(EjvEjw)=Ej(EjσvEjσw)=σ(v)σ(w).

(iii) Immediate from (i) and (ii).
(iv) Here, Bose-Mesner algebra M is the full commuting algebra, i.e.,

M={mMatX(C)σm=mσ, for all σAut(Y)}. Suppose there sia a nonzero proper subspace 0W 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.