Nonzero-form cellular heads, cell purity and shared-factor linkage

The cell-form scalar and its symmetry. For basis elements $c_{us}^{\lambda },c_{tv}^{\lambda }$ the product $c_{us}^{\lambda }{c}_{tv}^{\lambda }$ equals a scalar $G_{\lambda }(s,t)$ times $c_{uv}^{\lambda }$ modulo $A^{>\lambda }$ with $G_{\lambda }(s,t)$ independent of $u,v$: the right cell axiom fixes the first index and the left cell axiom fixes the last one, and comparing the two evaluations leaves the same scalar. The cellular involution gives $G_{\lambda }(s,t)=G_{\lambda }(t,s)$. Associativity of the threefold product yields $⟨xa,y⟩=⟨x,y{a}^{\ast }⟩$, since both sides compute the coefficient in the cell layer of the same product.

Cell purity and the zero cell. If $T(\lambda )$ is empty, then $C_{\lambda }=0$, its Gram matrix and form quotient are zero, and every central block projection acts with zero image; thus its block support is empty and it has no simple factors. The corresponding cell layer $J_{\lambda }$ is zero as well. Now assume $C_{\lambda }\mathrm{\ne }0$ and let $z$ be a central idempotent of $A$. On the cell layer $J_{\lambda }\cong C_{\lambda }^{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}{\otimes }_F{C}_{\lambda }$ the equality of left and right multiplication by $z$ gives $B\otimes \mathrm{Id}=\mathrm{Id}\otimes A_{\lambda }$ for the first- and second-factor coefficient matrices; comparing matrix coefficients shows every off-diagonal entry of $B$ and $A_{\lambda }$ vanishes and all diagonal entries of $B$ and $A_{\lambda }$ are equal, so both are a single scalar ${\theta }_{\lambda }$ times the identity. Since $z$ is idempotent, ${\theta }_{\lambda }\in \{0,1\}$. Applying this to all primitive central idempotents, whose sum is $1$, exactly one acts as $1$ on the nonzero cell module $C_{\lambda }$ and the others as $0$; hence $C_{\lambda }$, and therefore every composition factor of it, lies in one block. This covers labels with zero-form quotient because the tensor argument never divides by the Gram matrix.

Nonzero form gives a simple head. Let $R_{\lambda }:=\{x\in C_{\lambda }:⟨x,C_{\lambda }⟩=0\}$ be the radical of the form. By step 1.1 the rank-one action is $x{c}_{uv}^{\lambda }=⟨x,c_u^{\lambda }⟩c_v^{\lambda }$, which shows that $R_{\lambda }$ is a submodule and that every vector $x\mathrm{\notin }{R}_{\lambda }$ generates $C_{\lambda }$: if $⟨x,c_u^{\lambda }⟩\mathrm{\ne }0$, then scaling $x{c}_{uv}^{\lambda }$ by the inverse scalar produces $c_v^{\lambda }$ for every $v$. Hence every proper submodule is contained in $R_{\lambda }$ and $D_{\lambda }=C_{\lambda }\mathrm{/}{R}_{\lambda }$ is simple when the form is nonzero; the criterion is nonzero form, and a degenerate but nonzero form is allowed. Extending scalars preserves the Gram rank and the rank-one action, so $D_{\lambda }$ is absolutely simple.

Every simple occurs. Let $D$ be a simple $A$-module, and choose a cell ideal $I_{\lambda }$ acting nontrivially on $D$, maximal among such ideals in the finite cell order; then $A^{>\lambda }$ annihilates $D$, while $J_{\lambda }=I_{\lambda }\mathrm{/}{A}^{>\lambda }$ does not, so $D{J}_{\lambda }=D$. If the Gram matrix of $C_{\lambda }$ were zero, the cell-layer multiplication would satisfy $J_{\lambda }^2=0$ and $D{J}_{\lambda }^2=0$, a contradiction; hence the form is nonzero. Choose $s,u$ and $d\in D$ with $d{c}_{su}^{\lambda }\mathrm{\ne }0$; the map $c_v^{\lambda }\mapsto d{c}_{sv}^{\lambda }$ has nonzero image, which is all of $D$ by simplicity and factors through the nonzero head $D_{\lambda }$ of step 2.1, so $D\cong D_{\lambda }$. This exhausts the simple labels without classifying which multipartitions have nonzero forms.

Order and multiplicity. If $\lambda$ does not dominate $\mu$, then $C_{\lambda }$ annihilates $I_{\mu }$: products in $I_{\lambda }\cap I_{\mu }$ involve only shapes strictly above $\lambda$. Since $D_{\mu }$ does not annihilate $I_{\mu }$ by the rank-one action, $[C_{\lambda }:D_{\mu }]\mathrm{\ne }0$ forces $\lambda \ge \mu$. The radical $R_{\lambda }$ annihilates $I_{\lambda }$, while $D_{\lambda }$ does not, so $[C_{\lambda }:D_{\lambda }]=1$ when $D_{\lambda }\mathrm{\ne }0$. Two distinct nonzero heads are not isomorphic: their ideal supports would force $\lambda \ge \mu$ and $\mu \ge \lambda$ simultaneously.

The corner identity. Choose one primitive idempotent $e_{\mu }$ for each simple $D_{\mu }$ with $\dim D_{\nu }{e}_{\mu }={\delta }_{\mu \nu }$. The exact functor $V\mapsto V{e}_{\mu }$ (lift a vector along a surjection and multiply by $e_{\mu }$; restrict along an injection) gives $\dim C_{\lambda }{e}_{\mu }=[C_{\lambda }:D_{\mu }]$. The nonzero heads are self-dual under the involution by the symmetric associative nondegenerate quotient form of steps 1.1 and 2.1, so the corresponding left factor has the same multiplicities. Filtering $A$ by a linear extension of the cellular ideals and applying the two exact projections $e_{\mu }(-)e_{\nu }$ to the cell layers $J_{\lambda }\cong C_{\lambda }^{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}\otimes C_{\lambda }$, dimension additivity gives ${\dim }_F{e}_{\mu }A{e}_{\nu }={\sum }_{\lambda }[C_{\lambda }:D_{\mu }][C_{\lambda }:D_{\nu }]\mathrm{.}$ An empty cell layer contributes zero to both the filtration dimension and every product of its composition multiplicities.

Blocks and linkage. A finite split algebra has blocks equal to the connected components of the graph on its primitive idempotents with an edge $e$–$f$ whenever $eAf\mathrm{\ne }0$ or $fAe\mathrm{\ne }0$: a sum of full primitive idempotents over a union of components is central, and conversely a central idempotent cannot separate a nonzero corner, while repeated primitive idempotents of the same simple are linked through the matrix units of $A\mathrm{/}\mathrm{rad}A$. By step 4.1 two simple labels have a nonzero corner exactly when they occur together as factors of some cell module, so two simples share a block exactly when a chain of cell modules links their factors. Since by step 1.2 every nonzero cell module is pure and by step 3.1 every simple occurs in a nonzero-form head, two nonzero cell modules lie in the same block exactly when there is a finite chain of nonzero cell modules each neighbouring pair of which shares a simple composition factor. Each nonzero cell has a simple composition factor because it has positive finite dimension. A zero cell has no such factor and has empty block support by step 1.2, so it is absent from this support graph and cannot create a chain edge or a common supported block.

Conclusion. Steps 1.1, 2.1, 3.1 and 3.2 prove the head, exhaustion, order and multiplicity statement (1); step 1.2 proves cell purity (2); step 4.1 proves the corner identity (4); and step 5.1 proves the exact shared-factor linkage criterion (3). All arguments work over every field and allow degenerate nonzero cell forms, nonzero cells with zero-form quotient, and empty index sets with zero cell modules and empty block support.