Nonzero cell heads classify the simple modules of a cellular algebra

For $\lambda \in \mathrm{\Lambda }$ the spans $A^{⊵\lambda }$ and $A^{⊳\lambda }$ are two-sided ideals: by [F2] both $c_{st}^{\lambda }a$ and $a{c}_{st}^{\lambda }$ are congruent modulo $A^{⊳\lambda }$ to combinations of the $c^{\lambda }$-elements, so $A^{⊵\lambda }\cdot A\subseteq A^{⊵\lambda }$ and $A\cdot A^{⊵\lambda }\subseteq A^{⊵\lambda }$, and $A^{⊳\lambda }$, the sum of the ideals $A^{⊵\mu }$ over $\mu ⊳\lambda$, is a two-sided ideal as well. Consequently $A^{⊵\nu }{A}^{⊵\lambda }\subseteq A^{⊵\nu }\cap A^{⊵\lambda }$ and $A^{⊵\lambda }{A}^{⊵\nu }\subseteq A^{⊵\nu }\cap A^{⊵\lambda }$, because a product of a $c^{\nu }$-element and a $c^{\lambda }$-element lies in $A^{⊵\nu }$ by the ideal property of $A^{⊵\nu }$ and in $A^{⊵\lambda }$ by that of $A^{⊵\lambda }$. The intersection $A^{⊵\nu }\cap A^{⊵\lambda }$ is spanned by the $c^{\gamma }$ with $\gamma ⊵\nu$ and $\gamma ⊵\lambda$; if $\lambda ⋭\nu$ then no such $\gamma$ equals $\lambda$, since $\gamma =\lambda$ would force $\lambda ⊵\nu$, so all of them satisfy $\gamma ⊳\lambda$ and $A^{⊵\nu }\cap A^{⊵\lambda }\subseteq A^{⊳\lambda }$. Hence, when $\lambda ⋭\nu$, every product $c_{tt}^{\lambda }b$ with $b\in A^{⊵\nu }$ lies in $A^{⊳\lambda }$; the right action of $b$ on $C^{\lambda }$ is read off from $c_{tt}^{\lambda }b\equiv {\sum }_v{r}_v{c}_{tv}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳\lambda })$ with unique coefficients $r_v$, so $b$ acts as the zero operator on $C^{\lambda }$, that is, $A^{⊵\nu }\cdot C^{\lambda }=0$.

Generation by vectors outside the radical: let $x\in C^{\lambda }$ with $x\mathrm{\notin }\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$. Since $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }=\{z:⟨z,-{⟩}_{\lambda }=0\}$ there is $y\in C^{\lambda }$ with $⟨x,y{⟩}_{\lambda }\mathrm{\ne }0$, and after rescaling $y$ we may assume $⟨x,y{⟩}_{\lambda }=1$. Write $y={\sum }_u{a}_u{c}_u^{\lambda }$ and put $y_v:={\sum }_u{a}_u{c}_{uv}^{\lambda }\in A$ for $v\in T(\lambda )$. Then [F1] gives $x{y}_v={\sum }_u{a}_u(x{c}_{uv}^{\lambda })={\sum }_u{a}_u⟨x,c_u^{\lambda }{⟩}_{\lambda }{c}_v^{\lambda }=⟨x,y{⟩}_{\lambda }{c}_v^{\lambda }=c_v^{\lambda }$, so every basis vector of $C^{\lambda }$ lies in the submodule $xA$, hence $xA=C^{\lambda }$.

Nontrivial action of the upward ideal: suppose $D^{\mu }\mathrm{\ne }0$. Since $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }\mathrm{\ne }{C}^{\mu }$ there is $t\in T(\mu )$ with $c_t^{\mu }\mathrm{\notin }\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$; by definition of the radical there is $u\in T(\mu )$ with $⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }\mathrm{\ne }0$, and then $c_u^{\mu }\mathrm{\notin }\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$ as well. By [F1] the element $c_{ut}^{\mu }\in A^{⊵\mu }$ acts on $C^{\mu }$ by $c_t^{\mu }\cdot c_{ut}^{\mu }=⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }{c}_t^{\mu }$, so on the quotient $D^{\mu }$ it sends the nonzero class of $c_t^{\mu }$ to its nonzero multiple $⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }[c_t^{\mu }]\mathrm{\ne }0$. Thus $A^{⊵\mu }$ does not annihilate $D^{\mu }$.

Consequences for the radical: every proper submodule $M⊊C^{\lambda }$ is contained in $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$, since $x\in M$ with $x\mathrm{\notin }\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ would give $C^{\lambda }=xA\subseteq M$ by step 1.2, a contradiction. Hence, if $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }\mathrm{\ne }{C}^{\lambda }$, then $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ contains every proper submodule, is the unique maximal proper submodule, and equals the Jacobson radical of $C^{\lambda }$; then $D^{\lambda }=C^{\lambda }\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ is simple, because a nonzero submodule of the quotient has the form $N\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ with $N⊋\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ a submodule, and $N=C^{\lambda }$ by maximality. Moreover $D^{\lambda }\mathrm{\ne }0$ if and only if $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }\mathrm{\ne }{C}^{\lambda }$ if and only if the form $⟨-,-{⟩}_{\lambda }$ is not identically zero: the radical is the kernel of $x\mapsto ⟨x,-{⟩}_{\lambda }$, so it is all of $C^{\lambda }$ exactly when $⟨x,y{⟩}_{\lambda }=0$ for all $x,y$, and otherwise this kernel is a proper subspace containing $0$.

Pairwise distinctness: suppose $D^{\lambda }\cong D^{\mu }\mathrm{\ne }0$. If $\lambda$ does not dominate $\mu$, then step 1.1 gives $A^{⊵\mu }\cdot C^{\lambda }=0$, hence $A^{⊵\mu }\cdot D^{\lambda }=0$, and transporting this along the isomorphism gives $A^{⊵\mu }\cdot D^{\mu }=0$, contradicting step 1.3. Hence $\lambda ⊵\mu$; the same argument with the roles of $\lambda$ and $\mu$ exchanged gives $\mu ⊵\lambda$, so $\lambda =\mu$ by antisymmetry of the poset. Thus the nonzero cell heads are pairwise non-isomorphic.

Absolute irreducibility: let $F^{\mathrm{\prime }}\mathrm{/}F$ be a field extension and let $C^{\mathrm{\prime }\lambda }=C^{\lambda }{\otimes }_F{F}^{\mathrm{\prime }}$ be the base-changed cell module of the base-changed cellular algebra $A^{\mathrm{\prime }}=A{\otimes }_F{F}^{\mathrm{\prime }}$. The cellular axioms and the cell form base change, so the Gram matrix of the form on $C^{\mathrm{\prime }\lambda }$ is the scalar extension of the Gram matrix on $C^{\lambda }$; the rank of a matrix is the largest size of a minor with nonzero determinant, a field-independent computation with the polynomial determinants of the entries, so ${\dim }_{F^{\mathrm{\prime }}}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mathrm{\prime }\lambda }={\dim }_F\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$. If $D^{\lambda }\mathrm{\ne }0$ then $D^{\mathrm{\prime }\lambda }:=C^{\mathrm{\prime }\lambda }\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mathrm{\prime }\lambda }\cong D^{\lambda }{\otimes }_F{F}^{\mathrm{\prime }}$ is nonzero, and steps 1.2 and 2.1 applied over $F^{\mathrm{\prime }}$ show that every element of $C^{\mathrm{\prime }\lambda }$ outside $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mathrm{\prime }\lambda }$ generates $C^{\mathrm{\prime }\lambda }$, so $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mathrm{\prime }\lambda }$ is the unique maximal submodule of $C^{\mathrm{\prime }\lambda }$ and $D^{\mathrm{\prime }\lambda }$ is simple over $F^{\mathrm{\prime }}$. Hence $D^{\lambda }$ remains simple after every extension of scalars, i.e. it is absolutely irreducible.

Completeness: every simple right $A$-module $S$ is isomorphic to some $D^{\lambda }$ with $D^{\lambda }\mathrm{\ne }0$. Order $\mathrm{\Lambda }$ as ${\lambda }_1,\dots ,{\lambda }_N$ so that ${\lambda }_i⊳{\lambda }_j$ implies $i<j$, so larger labels come first, and put $A_j:=⟨c_{uv}^{{\lambda }_i}:i\le j{⟩}_F$; then each $A_j$ is a two-sided ideal (if $\mu$ occurs among ${\lambda }_1,\dots ,{\lambda }_j$ and $\gamma ⊵\mu$, then $\gamma$ occurs there too, so $A^{⊵\mu }\subseteq A_j$ and the claim follows from step 1.1), $A_0=0$, $A_N=A$. Moreover $A_j\mathrm{/}{A}_{j-1}\cong A^{⊵{\lambda }_j}\mathrm{/}{A}^{⊳{\lambda }_j}$, which as a right $A$-module is the direct sum over $s\in T({\lambda }_j)$ of the column spans $⟨c_{sv}^{{\lambda }_j}+A^{⊳{\lambda }_j}:v\in T({\lambda }_j)⟩\cong C^{{\lambda }_j}$ (the right action preserves the first index by [F2], and $A^{⊳{\lambda }_j}\subseteq A_{j-1}$ while $A^{⊵{\lambda }_j}\cap A_{j-1}=A^{⊳{\lambda }_j}$). Let $S\mathrm{\ne }0$ be simple and let $j$ be least with $S{A}_j\mathrm{\ne }0$ (such $j$ exists since $S{A}_N=SA=S$ by unitality). Then $S{A}_j$ is a nonzero submodule of $S$, so $S{A}_j=S$, while $S{A}_{j-1}=0$ by minimality of $j$; choosing $s\in S$ with $s{A}_j\mathrm{\ne }0$, the map $A_j\mathrm{/}{A}_{j-1}\to S$, $a+A_{j-1}\mapsto sa$, is a well-defined nonzero $A$-module surjection. As $A_j\mathrm{/}{A}_{j-1}$ is a direct sum of copies of $C^{{\lambda }_j}$, the images of these copies sum to $S\mathrm{\ne }0$, so some copy surjects onto the simple module $S$; thus $S$ is a simple quotient of $C^{{\lambda }_j}$. If the form on $C^{{\lambda }_j}$ were identically zero, then $c_{st}^{{\lambda }_j}{c}_{uv}^{{\lambda }_j}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳{\lambda }_j})$ for all $s,t,u,v$ by [F1]; since $A^{⊵{\lambda }_j}{A}^{⊵{\lambda }_i}\subseteq A^{⊵{\lambda }_i}\subseteq A_{j-1}$ for $i<j$ and $A^{⊵{\lambda }_j}{A}^{⊳{\lambda }_j}+A^{⊳{\lambda }_j}{A}^{⊵{\lambda }_j}\subseteq A^{⊳{\lambda }_j}\subseteq A_{j-1}$, this gives $A_j^2\subseteq A_{j-1}$, whence $S=S{A}_j=S{A}_j^2\subseteq S{A}_{j-1}=0$, a contradiction. Hence the form on $C^{{\lambda }_j}$ is not identically zero, so by step 2.1 $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{{\lambda }_j}$ is the unique maximal submodule and the simple quotient $S$ of $C^{{\lambda }_j}$ satisfies $S\cong D^{{\lambda }_j}$ with $D^{{\lambda }_j}\mathrm{\ne }0$.

Assembly: part 1 is step 3.1; part 2 is step 3.2 (exhaustion) together with step 2.2 (pairwise non-isomorphism) and step 2.1 (each nonzero $D^{\lambda }$ is simple); part 3 is step 2.1. The proof classifies right modules, and since the anti-automorphism $i$ identifies right and left $A$-modules, the statement holds on either side. The classification applies to $A=H_{F,q}(S_n)$ because the Murphy basis of The Murphy tableau elements form a cellular basis is a cellular basis of the type-$A$ Hecke algebra over every field. No semisimplicity was assumed, all posets and tableau sets are finite, and no choice principle is used.