Cell modules, their forms, and radicals

In the quotient $A^{\ge \lambda }\mathrm{/}{A}^{>\lambda }$, the classes of $C_{s,t}^{\lambda }$ form a basis by [F1]. Define $c_{t}a$ by the coefficients in the cell axiom. For any fixed $s$, this is exactly the unique coefficient vector of the class of $C_{s,t}^{\lambda }a$. Associativity in the quotient gives $((c_{t}a)b)=c_t(ab)$ by comparing the coefficients of that basis. The class of $C_{s,t}^{\lambda }1$ is $C_{s,t}^{\lambda }$, so $c_{t}1=c_t$. Thus the action is well-defined and unital.

If $T(\lambda )$ is empty, then $\mathrm{\Delta }(\lambda )=0$ and the unique bilinear form is the zero form, so the assertions for this label are vacuous. Otherwise fix $s,t,u,v$. The cell axiom gives$C_{s,t}^{\lambda }{C}_{u,v}^{\lambda }\equiv {\sum }_w{r}_{t,w}^{\lambda }(C_{u,v}^{\lambda })C_{s,w}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{>\lambda }),$ whose terms have fixed first index $s$. Apply the involution and the cell axiom to $C_{v,u}^{\lambda }{C}_{t,s}^{\lambda }$ to obtain a second expression for the same product, with terms $C_{w,v}^{\lambda }$ and fixed second index $v$. Since the cell-layer classes are a basis, the intersection of the span of $\{C_{s,w}^{\lambda }{\}}_w$ and the span of $\{C_{w,v}^{\lambda }{\}}_w$ is the rank-one span of $C_{s,v}^{\lambda }$. The product is therefore a scalar multiple of this basis vector. The first expansion makes that scalar independent of $s$; the involuted expansion makes it independent of $v$. It depends only on $t,u$, defining the unique coefficient $⟨c_t,c_u{⟩}_{\lambda }$.

Applying $\ast$ to the defining product congruence gives$C_{v,u}^{\lambda }{C}_{t,s}^{\lambda }\equiv ⟨c_t,c_u{⟩}_{\lambda }{C}_{v,s}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{>\lambda })\mathrm{.}$ Applying step 2.1 directly to its left-hand side gives coefficient $⟨c_u,c_t{⟩}_{\lambda }$ on the same basis vector. Independence forces $⟨c_t,c_u{⟩}_{\lambda }=⟨c_u,c_t{⟩}_{\lambda }$. Bilinearity follows from the bilinearity of multiplication and uniqueness of the coefficient.

For $x,y\in \mathrm{\Delta }(\lambda )$ and $a\in A$, choose any fixed $s,v\in T(\lambda )$. Associativity gives $(C_{s,x}^{\lambda }a)C_{y,v}^{\lambda }=C_{s,x}^{\lambda }(a{C}_{y,v}^{\lambda })$ modulo $A^{>\lambda }$, where $C_{s,x}^{\lambda }$ and $C_{y,v}^{\lambda }$ denote the corresponding linear combinations when $x,y$ are not basis vectors. The left coefficient of $C_{s,v}^{\lambda }$ is $⟨xa,y{⟩}_{\lambda }$. Applying $\ast$ to the right cell law shows that left multiplication by $a$ on the second factor has coefficients equal to the right action of $a^{\ast }$ on $\mathrm{\Delta }(\lambda )$; hence the right coefficient is $⟨x,y{a}^{\ast }{⟩}_{\lambda }$. Coefficient uniqueness proves the asserted invariance.

Define $\mathrm{rad}\mathrm{\Delta }(\lambda )=\{x:⟨x,y{⟩}_{\lambda }=0\text{ for every }y\}$. If $x$ lies in this radical, then for all $a\in A$ and all $y\in \mathrm{\Delta }(\lambda )$, step 3.2 gives $⟨xa,y{⟩}_{\lambda }=⟨x,y{a}^{\ast }{⟩}_{\lambda }=0$. Thus $xa$ is again in the radical. The quotient is a right module. The coefficient-defined module action and the form use only the cell-layer basis and therefore commute with scalar extension to any commutative $R$-algebra. The radical over the new ring is defined afresh as the kernel of its specialized Gram pairing; radicals need not commute with arbitrary scalar extension, and no such assertion is used.

For $A=H_{r,n}$, [F2] supplies the Murphy cellular basis, so the preceding steps construct its cell modules, forms, and radicals over the universal ring and after every commutative specialization.