Chapter 21 Norton Algebras
Wednesday, March 17, 1993
Proof (Proof of Lemma 20.4).
v=∑x∈Xαxˆx. Pick a vertex z∈X and compare z-coordinate of each side in (i). (A∗j(x)v)z=(A∗j(x))zzvz=|X|(Ej)xzαz.|X|(Eˆjˆx∘v)z=|X|(Eˆjˆx)z⋅αz=|X|(Ej)xzαz. Note that Eˆjˆx is the column x of Eˆj, which is the row x of Ej.
Claim. Eh(EiV∘EjV)=0.
Eh(EiV∘EjV)=Eh(Span(v∘w∣v∈EiV,w∈EjV))=Eh(Span(Eiˆy∘Ejˆz∣y,z∈X))=Span(Eh(Ejˆz∘Eiˆy∣y,z∈X)=Span((EhA∗ˆj(z)Ei)ˆy∣y,z∈X)by (i). But qhij=0 implies qˆhˆjˆi=0.
So, by Lemma 20.3 (ii), 0=(EˆiA∗ˆjEˆh)⊤=EhA∗ˆjEi. Hence, Eh(EiV∘EjV)=0.
Eh(EiV∘EjV)⊆EhV is clear. We show the other inclusion. Since Eiˆy∘Ejˆy=(column y of Ei∘column y of Ej)=column y of Ei∘Ej=(Ei∘Ej)ˆy=(1|X|D∑h=0qhijEh)ˆy, we have, Eh(EiV∘EjV)=EhSpan(Eiˆy∘Ejˆz∣y,z∈X)⊇EhSpan(Eiˆy∘Ejˆy∣y∈X)=Span(qhijEhˆy∣y∈X)=Span(Ehˆy∣y∈X)since qhij≠0=EhV. This proves the assertion.
Lemma 21.1 Given a commutative scheme Y=(X,{Ri}0≤i≤D), fix j (0≤j≤D). Define a binary multiplication: EjV×EjV⟶EjV((v,w)↦v∗w=Ej(v∘w)). Then,
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.)
Let Y, j, Nj be as in Lemma 21.1, and M Bose-Mesner algebra of Y. Let AutY={σ∈MatX(C)∣σ: permutation matrix ,σ⋅m=m⋅σ for all m∈M}={σ∈MatX(C)∣σ: permutation matrix ,(x,y)∈Ri→(σx,σy)∈Ri, for all i, and for all x,y∈X}Aut(Nj)={σ:EjV→EjV∣σ is a C-algebra isomorphim,i.e.,σ(v∗w)=σ(v)∗σ(w) for all v,w∈EjV}.
Lemma 21.2 Let Y,j,∗ be as in Lemma 21.1.
(i.e., a representation of Aut(Y)).
Proof.
σEjv=Ejσv, since σ commutes with each element of M.
σ(v∗w)=σ(Ej(Ejv∘Ejw))=Ejσ(Ejv∘Ejw)=Ej(Ejσv∘Ejσw)=σ(v)∗σ(w).
M={m∈MatX(C)∣σ⋅m=m⋅σ, for all σ∈Aut(Y)}. Suppose there sia a nonzero proper subspace 0≠W⊊ 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.