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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] with identity and let $A$ be an [[associative algebra|3CTW1P4HBP]] over $R$.\nA *cell datum* for $A$ consists of four pieces of data.\nThey are a finite partially ordered set $\\Lambda$, a finite set $M(\\lambda)$ for each $\\lambda \\in \\Lambda$, and an $R$-linear map $* \\colon A \\to A$.\nThe fourth is an injective map $(\\lambda, S, T) \\mapsto C^{\\lambda}_{S,T} \\in A$, defined for $\\lambda \\in \\Lambda$ and $S, T \\in M(\\lambda)$.\nThe data must satisfy:\n\n1. The elements $C^{\\lambda}_{S,T}$, for $\\lambda \\in \\Lambda$ and $S, T \\in M(\\lambda)$, form an $R$-[[basis|0KCQ9Y3QKA]] of $A$.\n2. The map $*$ is an anti-involution, that is $(ab)^* = b^* a^*$ and $(a^*)^* = a$ for all $a, b \\in A$, and $(C^{\\lambda}_{S,T})^* = C^{\\lambda}_{T,S}$.\n3. For all $a \\in A$, $\\lambda \\in \\Lambda$ and $S, T \\in M(\\lambda)$,\n$$a\\, C^{\\lambda}_{S,T} \\equiv \\sum_{S' \\in M(\\lambda)} r_a(S', S)\\, C^{\\lambda}_{S',T} \\pmod{A(\u003c \\lambda)},$$\nwhere $r_a(S', S) \\in R$ does not depend on $T$, and $A(\u003c \\lambda)$ is the $R$-span of the $C^{\\mu}_{U,V}$ with $\\mu \u003c \\lambda$.\n\nThe algebra $A$ is *cellular* if it has a cell datum, and the basis in clause 1 is then a *cellular basis*.\n\n## Remarks\n\nExample: $A = M_n(R)$, with $\\Lambda = \\set{\\lambda}$, $M(\\lambda) = \\set{1, \\dots, n}$, $C^{\\lambda}_{i,j} = E_{ij}$ and $*$ the transpose.\nThe check of clause 3: for $a = (a_{kl})$, $a E_{ij} = \\sum_k a_{ki} E_{kj}$, whose coefficients $a_{ki}$ do not depend on $j$, and $A(\u003c \\lambda) = 0$.\nNon-example of a datum: the same basis of $M_2(R)$ with $*$ the identity map fails clause 2, since $(E_{12} E_{21})^* = E_{11}$ but $E_{21}^* E_{12}^* = E_{22}$.\nNon-example of an algebra: the upper triangular $2 \\times 2$ matrices over a field are not cellular.\nGraham and Lehrer showed that the [[Cartan matrix|ZBQ4KRC98K]] of a cellular algebra over a field is symmetric.\nThis algebra has the Cartan matrix $\\begin{pmatrix} 1 \u0026 0 \\\\ 1 \u0026 1 \\end{pmatrix}$, which is not symmetric.\nApplying $*$ to clause 3 gives the mirror condition for right multiplication.\n",
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