The Murphy cellular basis, the cell law and the Murphy form

Compositions. A composition of $n$ is a finite sequence $\alpha =({\alpha }_1,\dots ,{\alpha }_k)$ of nonnegative integers with $\sum {\alpha }_i=n$; its row blocks $B_r$, row group $S_{\alpha }$, row-filled tableau $t^{\alpha }$, row-standard tableaux, set $D_{\alpha }$, distinguished elements $d_{\alpha }(u)$, and elements $x_{\alpha }={\sum }_{w\in S_{\alpha }}{T}_w$, $b_u:=x_{\alpha }{T}_{d_{\alpha }(u)}$ are defined exactly as for partitions in Hecke row permutation modules and ordinary Dipper-James polytabloids, and for row-standard $\alpha$-tableaux $s,t$ put $m_{st}^{\alpha }:=T_{d_{\alpha }(s)}^{\ast }{x}_{\alpha }{T}_{d_{\alpha }(t)}$, so that $b_t=m_{t^{\alpha }t}^{\alpha }$. Every statement of [F2] was proved for partitions but uses only that the row group is a product of symmetric groups on consecutive blocks, so it applies verbatim to compositions.

Dominance of tableaux by counting. For $\alpha$-tableaux $u,v$ and integers $m\ge 0$, $k\ge 1$ put $c_{m,k}(u):=\mathrm{\#}\{x\in B_1\cup \cdots \cup B_k:u(x)\le m\}$; write $u⊵v$ when $c_{m,k}(u)\ge c_{m,k}(v)$ for all $m,k$, and $u\equiv v$ when all these numbers agree. Passing from $u$ to $u\cdot s_i$ exchanges the labels $i$ and $i+1$, so it changes only the counts at $m=i$, and $c_{i,k}(u\cdot s_i)=c_{i-1,k}(u)+[i+1\text{ lies in the first }k\text{ blocks of }u]$ because the label $i$ is replaced by $i+1$ and all other labels $\le i$ stay put. In particular $u\equiv v$ implies $u\cdot s_i\equiv v\cdot s_i$, and if $u,v$ are both row standard then $u\equiv v$ if and only if $u=v$, because the numbers $c_{m,k}$ determine the multiset of entries of each block and a row-standard tableau is its block multisets sorted increasingly. Suppose now that $s$ is row standard, that $s^{\mathrm{\prime }}:=s\cdot s_i$ is row standard, and that the labels $i,i+1$ lie in rows $a<b$ of $s$, so that $s^{\mathrm{\prime }}$ is obtained from $s$ by moving $i$ down into row $b$ and $i+1$ up into row $a$. Then for every row-standard $r$ with $r⊵s$ one has $r⊵s^{\mathrm{\prime }}$ and $r\cdot s_i⊵s^{\mathrm{\prime }}$. Indeed for $m\mathrm{\ne }i$ the counts are unchanged by both swaps, so $c_{m,k}(r)\ge c_{m,k}(s)=c_{m,k}(s^{\mathrm{\prime }})$ and the same for $r\cdot s_i$; for $m=i$ one has $c_{i,k}(s^{\mathrm{\prime }})=c_{i-1,k}(s)+[k\ge b]$, while $c_{i,k}(r)\ge c_{i,k}(s)\ge c_{i-1,k}(s)+[k\ge b]=c_{i,k}(s^{\mathrm{\prime }})$ for every $k$, and $c_{i,k}(r\cdot s_i)=c_{i-1,k}(r)+[i+1\text{ lies in the first }k\text{ blocks of }r]\ge c_{i-1,k}(r)\ge c_{i-1,k}(s)=c_{i,k}(s^{\mathrm{\prime }})$ for $k<b$, while for $k\ge b$ the sharper bound $c_{i,k}(r\cdot s_i)\ge c_{i+1,k}(r)-1\ge c_{i+1,k}(s)-1=c_{i-1,k}(s)+1=c_{i,k}(s^{\mathrm{\prime }})$ holds, the first inequality by $[i\text{ lies in the first }k\text{ blocks of }r]\le 1$ and the last because both labels then lie in the first $k$ blocks of $s$.

The trace pairing. Define $\tau :H\to R$ by $\tau (T_w)={\delta }_{w,1}$. The multiplication rules give $\tau (T_w{T}_i)=\tau (T_i{T}_w)=q{\delta }_{w,s_i}$ for every $w$ and every generator $T_i$: in the two-case formulas the identity $T_1$ occurs only in the descent case $w=s_i$, with coefficient $q$. Moving generator letters across one another successively gives $\tau (ab)=\tau (ba)$ for all $a,b\in H$, and $\tau (a^{\ast })=\tau (a)$ since $\tau$ is $R$-linear and $\ast$ permutes the standard basis. Let $B_0$ be the bilinear form defined on the standard basis by $B_0(T_u,T_v):={\delta }_{u,v}{q}^{\mathrm{\ell }(u)}$. Every right $s_i$-orbit $\{T_w,T_{w{s}_i}\}$ has two elements; writing $w$ for the element with $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$, right multiplication by $T_i$ on the block spanned by the orbit has matrix $A=\left(\begin{array}{cc}0& q\\ 1& q-1\end{array}\right)$, and the restriction of $B_0$ to the block has matrix $D=\mathrm{diag}(q^{\mathrm{\ell }(w)},q^{\mathrm{\ell }(w)+1})$ with $A^{\mathsf{T}}D=DA$; distinct blocks are orthogonal, so $B_0(x{T}_i,y)=B_0(x,y{T}_i)$ for all $x,y\in H$. Iterating along a reduced word $T_u=T_{i_1}\cdots T_{i_r}$ gives $B_0(T_u,T_v)=B_0(T_1,T_v{T}_{i_r}\cdots T_{i_1})=B_0(T_1,T_v{T}_u^{\ast })$, and $B_0(T_1,y)=\tau (y)$ for every $y$ by the definitions; hence $B_0(T_u,T_v)=\tau (T_v{T}_u^{\ast })=\tau (T_{u^{-1}}{T}_v)={\delta }_{u,v}{q}^{\mathrm{\ell }(u)}$. In particular the bilinear form $(a,b)\mapsto \tau (a^{\ast }b)$ equals $B_0$ and has diagonal Gram matrix with unit entries $q^{\mathrm{\ell }(u)}$ in the standard basis, so it is unimodular.

The support lemma. Let $g$ be a row-standard $\alpha$-tableau, let $w=s_{i_1}\cdots s_{i_r}$ be a reduced word, put $t:=g\cdot w$ and $g_k:=g\cdot (s_{i_1}\cdots s_{i_k})$, and assume $\mathrm{\ell }(d_{\alpha }(t))=\mathrm{\ell }(d_{\alpha }(g))+\mathrm{\ell }(w)$. If $v$ is a row-standard $\alpha$-tableau with $v⊳g$, then every row-standard tableau $u$ whose basis element $b_u$ is involved in $b_v{T}_w$ satisfies $u⊳t$. The assumed length equality makes the concatenation of a reduced word for $d_{\alpha }(g)$ and the displayed word for $w$ reduced, since $d_{\alpha }(t)=d_{\alpha }(g)w$. Starting at the row-standard $t$ and removing its final letters backwards, each removed letter is a right descent. A right descent exchanges consecutive labels in reversed row-reading order; they cannot lie in one increasing row, so the swap preserves row standardness by [F2]. Thus every $g_k$ is row standard, $d_{\alpha }(g_k)=d_{\alpha }(g)s_{i_1}\cdots s_{i_k}$, and $\mathrm{\ell }(d_{\alpha }(g_k))=\mathrm{\ell }(d_{\alpha }(g))+k$. Hence each step replaces $g_k$ by the row-standard tableau $g_k\cdot s_{i_{k+1}}$ with $\mathrm{\ell }(d_{\alpha }(g_k\cdot s_{i_{k+1}}))=\mathrm{\ell }(d_{\alpha }(g_k))+1$, so the three-case rule of [F2] gives $b_{g_k}{T}_{i_{k+1}}=b_{g_{k+1}}$: the swapped tableau is row standard and the length increases. The labels $i,i+1$ therefore lie in different rows of $g_k$ (otherwise the swap would stay in one row and the rule would return a multiple of $b_{g_k}$), and since $g_{k+1}$ is row standard with $\mathrm{\ell }(d_{\alpha }(g_k)s_{i_{k+1}})=\mathrm{\ell }(d_{\alpha }(g_k))+1$, the row-reading position of $i$ precedes that of $i+1$, so $i$ lies in a strictly upper row of $g_k$; this is exactly the configuration of step 1.2. We now show by induction on $k$ that every row-standard $u$ with $b_u$ involved in $b_v{T}_{i_1}\cdots T_{i_k}$ satisfies $u⊵g_k$ and $u\equiv ̸g_k$. For $k=0$ we have $v⊵g$ and $v\mathrm{\ne }g$, and $v\equiv g$ would give $g⊵v$, so $v\equiv ̸g$. For the step, the action rule of [F2] shows that $b_u{T}_{i_{k+1}}$ is a combination of $b_u$ and, when $u\cdot s_{i_{k+1}}$ is row standard, $b_{u\cdot s_{i_{k+1}}}$; step 1.2 applied to $s:=g_k$, $s^{\mathrm{\prime }}=g_{k+1}$ gives $u⊵g_{k+1}$ and $u\cdot s_{i_{k+1}}⊵g_{k+1}$ whenever the swapped tableau is row standard. Neither output is $\equiv g_{k+1}$: if $u\equiv g_{k+1}$ then $u⊵g_k⊵g_{k+1}\equiv u$ forces $u\equiv g_k$, and if $u\cdot s_{i_{k+1}}\equiv g_{k+1}$ then applying the swap to the two equivalent tableaux gives $u\equiv g_k$, both contradicting the induction hypothesis. At the end $t$ is row standard, so $u⊵t$ together with $u\equiv ̸t$ says exactly that $u⊳t$.

Compositions to partitions. Let $\alpha$ be a composition and $\beta$ the partition obtained by rearranging its parts. Choose $d$ mapping each row block of $\beta$ increasingly onto a row block of $\alpha$ of the same size. Then $d^{-1}$ is increasing on each $\alpha$-block, $d$ is increasing on each $\beta$-block, $d\in D_{\alpha }$, $d^{-1}\in D_{\beta }$, and $d^{-1}{S}_{\alpha }d=S_{\beta }$. Hence $x_{\alpha }{T}_d={\sum }_{w\in S_{\alpha }}{T}_{wd}={\sum }_{u\in S_{\alpha }d}{T}_u$ and $T_d{x}_{\beta }={\sum }_{v\in S_{\beta }}{T}_{dv}={\sum }_{u\in d{S}_{\beta }}{T}_u$, and these agree because $S_{\alpha }d=d(d^{-1}{S}_{\alpha }d)=d{S}_{\beta }$. Thus $x_{\alpha }=T_d{x}_{\beta }{T}_d^{-1}$ and $T_d$ is invertible. Since $H{x}_{\beta }$ is free with basis $\{T_{d_{\beta }(s)}^{\ast }{x}_{\beta }\}$ (the star of the basis of $x_{\beta }H$) and $x_{\beta }H$ is free with basis $\{x_{\beta }{T}_{d_{\beta }(u)}\}$ by [F2], expanding $T_{d_{\alpha }(s)}^{\ast }{T}_d$ and $T_d^{-1}{T}_{d_{\alpha }(t)}$ expresses $m_{st}^{\alpha }$ as an $R$-linear combination of $m_{uv}^{\beta }$ with $u,v$ row standard; all coefficients are Laurent polynomials in $q$, with no quantum factorial inverted.

The Garnir belt identity. Let $\rho$ be a partition and $(i,j)$ a node with $j\le {\rho }_{i+1}$. Let $\alpha$ replace the two consecutive parts ${\rho }_i,{\rho }_{i+1}$ by $j-1,{\rho }_i+1,{\rho }_{i+1}-j$, deleting zeros; its blocks are those of $\rho$ except that the belt $a,\dots ,b$, where $a=t^{\rho }(i,j)$ and $b=t^{\rho }(i+1,j)$, is cut into three consecutive blocks of sizes $j-1$, ${\rho }_i+1$ and ${\rho }_{i+1}-j$; the middle block is the interval $a,\dots ,b$ itself, of size ${\rho }_i+1=b-a+1$. Put $D:=S_{\alpha }\cap D_{\rho }$ and $E:=S_{\rho }\cap D_{\alpha }$. Then $x_{\rho }{\sum }_{d\in D}{T}_d={\sum }_{v\in E}{T}_v^{\ast }{x}_{\alpha }\mathrm{.}$ Put $K=S_{\rho }\cap S_{\alpha }$. Sorting a permutation of $S_{\alpha }$ within each $\rho$-block preserves the $\alpha$-blocks, because every intersection of the two interval blocks is an interval. The unique sorted result is in $D$, so $D$ is a transversal for the right cosets $K\mathrm{\backslash }{S}_{\alpha }$; likewise $E$ is a transversal for $K\mathrm{\backslash }{S}_{\rho }$. Two products $ud$ and $u^{\mathrm{\prime }}{d}^{\mathrm{\prime }}$ coincide only if $d,d^{\mathrm{\prime }}$ lie in the same $K$-coset, which forces $d=d^{\mathrm{\prime }}$ and then $u=u^{\mathrm{\prime }}$. Inverting the representatives in $E$ gives the corresponding transversal for the cosets of $K$ on the other side. Thus the products $S_{\rho }\times D\to S_{\rho }{S}_{\alpha }$, $(u,d)\mapsto ud$, and $E\times S_{\alpha }\to S_{\rho }{S}_{\alpha }$, $(v,u)\mapsto v^{-1}u$, are bijections with length-additive products, because $D\subseteq D_{\rho }$, $E\subseteq D_{\alpha }$ and $\mathrm{\ell }(u^{-1}v)=\mathrm{\ell }(v^{-1}u)$ for all $u\in S_{\alpha }$, $v\in D_{\alpha }$; expanding both sides in the standard basis therefore gives ${\sum }_{w\in S_{\rho }{S}_{\alpha }}{T}_w$ for the left side and the same sum for the right side, since $T_v^{\ast }=T_{v^{-1}}$.

The Garnir tableau. With $\rho ,\alpha ,(i,j)$ as in step 2.3, let $g$ be the row-standard tableau agreeing with $t^{\rho }$ outside the belt $a,\dots ,b$, whose row $i+1$ carries the $j$ smallest belt entries increasingly in its first $j$ positions and whose row $i$ carries the remaining belt entries increasingly in its last ${\rho }_i-j+1$ positions. Among the tableaux $t^{\rho }\cdot d$, $d\in D$, exactly one fails to be standard, namely $g$, and every other one strictly dominates $g$: each of them agrees with $t^{\rho }$ outside the belt and is determined by the $j$-element subset $S$ of belt entries that it places increasingly in the first $j$ positions of row $i+1$, the complementary belt entries occupying increasingly the last ${\rho }_i-j+1$ positions of row $i$; column $j$ fails exactly when $\min (\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{t}\setminus S)>\max (S)$, that is, when $S$ consists of the $j$ smallest belt entries, which is $g$. For the dominance comparison, the cumulative counts $c_{m,k}$ of step 1.2 for $k\le i-1$ and $k\ge i+1$ do not depend on $S$; for $k=i$ row $i$ carries the belt complement of $S$, and $g$ carries the largest ${\rho }_i-j+1$ belt entries, whose count function is pointwise minimal, so every other belt tableau strictly dominates $g$.

The straightening identity. Let $\nu$ be the partition obtained by sorting the parts of $\alpha$. Rewriting the left-hand side of the belt identity of step 2.3 as row-module basis elements through step 3.1, and applying step 2.2 to its right-hand side, the belt identity becomes $x_{\rho }{T}_{d(g)}=h-{\sum }_{d\in D,d\mathrm{\ne }d(g)}{x}_{\rho }{T}_d,$ where $h$ is an $R$-linear combination of elements $T_{d_{\nu }(s^{\mathrm{\prime }})}^{\ast }{x}_{\nu }{T}_{d_{\nu }(t^{\mathrm{\prime }})}$ with $s^{\mathrm{\prime }},t^{\mathrm{\prime }}$ row standard, and the tableaux $t^{\rho }\cdot d$ with $d\in D$, $d\mathrm{\ne }d(g)$, are all standard and strictly more dominant than $g$; here $\nu$ is strictly greater than $\rho$ in lexicographic order because the part ${\rho }_i+1$ exceeds ${\rho }_i$ while all earlier parts are retained and sorting can only move such an increase earlier.

Length-additive reduction to a Garnir tableau. Suppose the row-standard $\rho$-tableau $t$ violates column order at $(i,j)$, so $t(i,j)>t(i+1,j)$. Order the cells partially by their left-to-right order within each row and the extra relation $(i+1,j)<(i,j)$. The cells in row $i+1$ through column $j$ must then precede every cell in row $i$ from column $j$ onward. The label order of $t$ is a linear extension of this partial order. The label order of the Garnir tableau $g$ in step 3.1 is the row-reading order with precisely these two consecutive groups interchanged. Its inverted cell pairs relative to row reading are exactly the forced pairs between these groups. Sort the label-ordered cell list of $t$ into the label order of $g$ by swapping adjacent entries that occur in the opposite order in $g$. The swapped entries are incomparable in the partial order, because both lists are linear extensions; thus every intermediate list remains a linear extension and its tableau remains row standard with the chosen column violation. Each swapped pair is reversed relative to row reading, since the only reversals of $g$ are forced pairs which cannot be reversed in $t$. Consequently every swap exchanges consecutive labels at a right descent of the distinguished permutation and decreases its inversion length by one. The finite sorting ends at $g$. Reversing the $r$ swaps gives $t=g\cdot w$ and $d(t)=d(g)w$, with $\mathrm{\ell }(d(t))=\mathrm{\ell }(d(g))+r$. The triangle inequality gives $r\le \mathrm{\ell }(w)$, while its $r$-letter expression gives $\mathrm{\ell }(w)\le r$; hence $\mathrm{\ell }(d(t))=\mathrm{\ell }(d(g))+\mathrm{\ell }(w)$.

Two-index straightening. Every $m_{st}^{\rho }$ with $s,t$ row standard and $t$ non-standard is an $R$-linear combination of elements $m_{uv}^{{\rho }_0}$ with either ${\rho }_0{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho$, or ${\rho }_0=\rho$ and $u⊵s$, $v⊳t$. Indeed, writing $t=g\cdot w$ with $\mathrm{\ell }(d(t))=\mathrm{\ell }(d(g))+\mathrm{\ell }(w)$ at an offending Garnir node by step 4.2, multiply the identity of step 4.1 on the left by $T_{d(s)}^{\ast }$ and on the right by $T_w$; step 2.1 pushes every same-shape term strictly above $t$ while the higher-shape terms are handled by descending lexicographic induction, in all same-shape terms this right-index reduction retains the first index $s$, because only the row-module identity $x_{\rho }{T}_{d(t)}$ is straightened. After the second index is standard, the left index is treated symmetrically using $m_{st}^{\rho }=m_{ts}^{\rho }{}^{\ast }$, and the finite dominance order on row-standard tableaux makes the inner induction terminate. Composing with step 2.2, every $m_{st}^{\alpha }$ with $\alpha$ a composition is a combination of partition-indexed $m_{uv}^{\nu }$ with $u,v$ standard.

The Murphy basis. For $\alpha =(1^n)$ one has $x_{(1^n)}=1$, so every standard basis element is a Murphy element and the standard Murphy family spans $H$ by step 5.1. Its cardinality is ${\sum }_{\rho }(f^{\rho }{)}^2=n!$ by [F3], equal to the rank of the free module $H$, so writing the family in the standard basis gives a square matrix that, having a right inverse, has unit determinant; hence the family including the empty shape is an $R$-basis of $H$, and $m_{st}^{\ast }=m_{ts}$ follows from $x_{\rho }^{\ast }=x_{\rho }$ and [F1]. This is clause (1).

Lexicographic ideals and the first-index law. Let $J^{{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho }$ be the span of all $m_{uv}$ whose shape is strictly lexicographically larger than $\rho$. Expand $x_{\rho }{T}_{d(t)}h$ in the row-module basis of [F2] and multiply on the left by $T_{d(s)}^{\ast }$. Step 5.1 straightens the resulting row-standard Murphy elements using only shapes ${⊵}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho$; applying $\ast$ gives the analogous assertion for left multiplication. Since the family is a basis by step 6.1, the subspaces $J^{{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho }$ are two-sided ideals, and for standard $s,t$ and $h\in H$ one has $m_{st}h\equiv {\sum }_{v\in \mathrm{Std}(\rho )}{r}_{tv}(h)m_{sv}(\mathrm{m}\mathrm{o}\mathrm{d}{J}^{{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho })$ with coefficients independent of the first index $s$, because right straightening retains the first index and the same-shape coefficients of step 5.1 depend only on the column index.

The mixed pairing and its diagonal. For partitions $\eta ,\nu$ and standard tableaux index the matrix $Q_{(\eta ,s,t),(\nu ,a,b)}:=\tau (m_{st}^{\eta }\theta (m_{ab}^{{\nu }^{\mathrm{\prime }}}))$. By [F4] and step 1.3, $Q$ vanishes unless $\eta ⊴\nu$: cyclically permuting the trace moves the factors until $x_{\eta }$ meets $H\theta (x_{{\nu }^{\mathrm{\prime }}})$, and $\theta (x_{{\nu }^{\mathrm{\prime }}})=q^N{y}_{{\nu }^{\mathrm{\prime }}}$ while $x_{\eta }H{y}_{{\nu }^{\mathrm{\prime }}}=0$ unless $\eta$ is dominated by $({\nu }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\nu$. Ordering shapes by lexicographic order, $Q$ is block upper triangular with square diagonal blocks, since transposition bijects $\mathrm{Std}(\nu )$ with $\mathrm{Std}({\nu }^{\mathrm{\prime }})$ and $f^{\nu }=f^{{\nu }^{\mathrm{\prime }}}$; and $Q$ is invertible because by step 1.3 it is the matrix of the unimodular pairing $\tau (a^{\ast }b)$ between two bases of $H$, the standard Murphy basis and the $\theta$-image of the standard Murphy basis of conjugate shapes. Hence every diagonal block is perfect over $R$, and this remains true after every base change.

Dominance ideals. Let $\kappa$ be a partition and $h\in H{x}_{\kappa }H$. If $\nu$ does not dominate $\kappa$, then $\tau (h\theta (m_{ab}^{{\nu }^{\mathrm{\prime }}}))=0$ for all $a,b$: cyclic permutation of the trace moves the factors of $h$ until $x_{\kappa }$ is followed by $H\theta (x_{{\nu }^{\mathrm{\prime }}})$, that is, by a multiple of $H{y}_{{\nu }^{\mathrm{\prime }}}$, and $x_{\kappa }H{y}_{{\nu }^{\mathrm{\prime }}}=0$ by [F4] because $\kappa$ is not dominated by $({\nu }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\nu$. Expanding $h$ in the lex Murphy basis and inspecting shapes in increasing lexicographic order, only shapes $\eta ⊴\nu$ in dominance pair nontrivially against the column labelled $\nu$ by step 7.2; for $\eta \mathrm{\ne }\nu$ these shapes are lexicographically smaller and do not dominate $\kappa$, since $\kappa ⊴\eta ⊴\nu$ would contradict $\kappa ⋬\nu$, so their coefficients have already vanished; the perfect diagonal block at $\nu$ then forces all coefficients of shape $\nu$ to vanish. Therefore $H{x}_{\kappa }H$ lies in the span of the Murphy elements of shapes $⊵\kappa$; since conversely $m_{ab}^{\eta }\in H{x}_{\eta }H$ by definition, $I^{⊵\kappa }={\sum }_{\eta ⊵\kappa }H{x}_{\eta }H$ and $I^{⊳\kappa }={\sum }_{\eta ⊳\kappa }H{x}_{\eta }H$. Both are therefore two-sided ideals of $H$.

The cell law and the cell module. Let $s,t$ be standard $\rho$-tableaux and $h\in H$. Since $x_{\rho }{T}_{d(t)}h\in M(\rho )={⨁}_{v}R{x}_{\rho }{T}_{d(v)}$ with $v$ running over the row-standard $\rho$-tableaux, there are unique $c_{tv}(h)\in R$ with $x_{\rho }{T}_{d(t)}h={\sum }_v{c}_{tv}(h)x_{\rho }{T}_{d(v)}$; multiplying on the left by $T_{d(s)}^{\ast }$ gives the exact identity $m_{st}h={\sum }_v{c}_{tv}(h)m_{sv}$, whose coefficients do not involve $s$. For $v$ non-standard, step 5.1 writes $m_{sv}={\sum }_{w\in \mathrm{Std}(\rho )}{b}_{vw}{m}_{sw}+j_v$ with $b_{vw}\in R$ independent of the first index $s$ and $j_v\in J^{{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho }$; step 8.1 puts $m_{sv}$ and the $m_{sw}$ in $I^{⊵\rho }$, so $j_v\in J^{{⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho }\cap I^{⊵\rho }$. Both spans are spanned by Murphy basis elements, of shapes ${⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho$ and $⊵\rho$ respectively, and the Murphy basis is independent, so their intersection is spanned by the basis elements whose shapes satisfy both conditions; a shape $\mu \mathrm{\ne }\rho$ satisfying $\mu ⊵\rho$ and $\mu {⊳}_{\mathrm{l}\mathrm{e}\mathrm{x}}\rho$ is exactly a shape $\mu ⊳\rho$. Hence $j_v\in I^{⊳\rho }$ and $m_{st}h\equiv {\sum }_{v\in \mathrm{Std}(\rho )}{r}_{tv}(h)m_{sv}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{⊳\rho })$ with coefficients $r_{tv}(h)$ independent of the first index. By step 6.1 the quotient $I^{⊵\rho }\mathrm{/}{I}^{⊳\rho }$ is free with basis the classes of $\{m_{uv}:u,v\in \mathrm{Std}(\rho )\}$, and $x_{\rho }{T}_{d(v)}=m_{t^{\rho }v}^{\rho }$; the congruence with $s=t^{\rho }$ shows that the image of $M(\rho )$ in this quotient is spanned by the $f^{\rho }$ independent classes $m_t=x_{\rho }{T}_{d(t)}+I^{⊳\rho }$, with kernel $M(\rho )\cap I^{⊳\rho }$. Therefore $C_{\mathrm{M}}(\rho )=M(\rho )\mathrm{/}(M(\rho )\cap I^{⊳\rho })$ is free with basis $\{m_t:t\in \mathrm{Std}(\rho )\}$ and satisfies the displayed cell law. This is clauses (2) and (3).

The cell form. Let $u,v,s,t\in \mathrm{Std}(\rho )$. Applying the cell law of step 9.1 to $m_{us}{m}_{tv}$ with $h=m_{tv}$ gives $m_{us}{m}_{tv}\equiv {\sum }_w{r}_{sw}(m_{tv})m_{uw}$, and applying it to $m_{vt}{m}_{su}=m_{tv}^{\ast }{m}_{us}^{\ast }$ with $h=m_{su}$ and starring gives $m_{us}{m}_{tv}\equiv {\sum }_w{r}_{tw}(m_{su})m_{wv}$. Comparing coefficients in the basis $\{m_{ab}:a,b\in \mathrm{Std}(\rho )\}$ of $I^{⊵\rho }\mathrm{/}{I}^{⊳\rho }$ yields $r_{sw}(m_{tv})=0$ for $w\mathrm{\ne }v$ and $r_{sv}(m_{tv})=r_{tu}(m_{su})$, a scalar $G_{\rho }(m_s,m_t)\in R$ independent of $u$ and $v$; hence $m_{us}{m}_{tv}\equiv G_{\rho }(m_s,m_t)m_{uv}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{⊳\rho })$, and $G_{\rho }$ is the unique bilinear form with this property. Starring the defining congruence and exchanging $(s,t)$ with $(u,v)$ gives $G_{\rho }(m_t,m_s)=G_{\rho }(m_s,m_t)$. Comparing the two bracketings of $m_{us}h{m}_{tv}$ through step 9.1 gives $G_{\rho }(m_{s}h,m_t)=G_{\rho }(m_s,m_t{h}^{\ast })$, which shows that the radical of $G_{\rho }$ is a submodule of $C_{\mathrm{M}}(\rho )$. Every coefficient is a Laurent polynomial in $q$ obtained from the finite multiplication rules, so the form and all clauses survive every base change; the form may be degenerate and no determinant or quantum factorial is inverted, and for $n=0,1$ the unique basis vector has form $1$. This is clauses (4) and (5), and since only the presentation of $H$, inversion counting, the local double-coset vanishing, the trace computation and finite free duality were used, no choice principle is used.