Integral type-A Garnir straightening and the Murphy cell law

Put $R_A^{\mathrm{\prime }}=\mathbb{Z}[v^{\pm 1}]$ and map $R_A\to R_A^{\mathrm{\prime }}$ by $Q\mapsto v^2$. Every Laurent polynomial in $v$ has a unique expression $f(Q)+vg(Q)$, so $R_A^{\mathrm{\prime }}$ is free over $R_A$ on $1,v$ and $R_A\hookrightarrow R_A^{\mathrm{\prime }}$. In the symmetric normalization of [F1], setting $T_i=v{\tilde{T}}_i$ gives $(T_i-v^2)(T_i+1)=0$ and preserves braid relations; the generator rescaling multiplies $T_w$ by the unit $v^{\mathrm{\ell }(w)}$. Thus the base-changed algebra over $R_A^{\mathrm{\prime }}$ has basis $\{T_w\}$ by [F1]. The quadratic relation spans products over $R_A$; any relation among the $T_w$ remains a relation over $R_A^{\mathrm{\prime }}$, where they are independent. Injectivity of $R_A\to R_A^{\mathrm{\prime }}$ proves that $\{T_w\}$ is an $R_A$-basis.

For a composition $\alpha$, let $x_{\alpha }={\sum }_{w\in S_{\alpha }}{T}_w$. The elements $x_{\alpha }{T}_d$ for distinguished right coset representatives $d$ span $x_{\alpha }{H}_n$ because each group element factors uniquely as $yd$ with $y\in S_{\alpha }$ and $\mathrm{\ell }(yd)=\mathrm{\ell }(y)+\mathrm{\ell }(d)$, while $x_{\alpha }{T}_y=Q^{\mathrm{\ell }(y)}{x}_{\alpha }$. Their standard-basis supports are disjoint cosets and each has coefficient $1$ at its distinguished representative, so they are independent. The one-generator rules of [F1] give $x_{\alpha }{T}_i=Q{x}_{\alpha }$ by pairing $w$ and $w{s}_i$ in the row sum. Consequently, on the row-module basis $e_t=x_{\alpha }{T}_{d_{\alpha }(t)}$, multiplication by $T_i$ has the three cases: $Q{e}_t$ if $t{s}_i$ is not row standard, $e_{t{s}_i}$ if it is row standard and length increases, and $Q{e}_{t{s}_i}+(Q-1)e_t$ if it is row standard and length decreases. The left-action formulas follow by applying $\ast$.

Define $\tau (T_w)={\delta }_{w,1}$. The two Hecke multiplication cases give $\tau (T_w{T}_i)=\tau (T_i{T}_w)=Q{\delta }_{w,s_i}$; moving the generator letters of a word around the trace one at a time therefore proves $\tau (ab)=\tau (ba)$ for arbitrary $a,b$. Define $B(T_u,T_v)={\delta }_{u,v}{Q}^{\mathrm{\ell }(u)}$. On each right-$s_i$ pair with $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$, right multiplication has matrix $A=\left(\begin{array}{cc}0& Q\\ 1& Q-1\end{array}\right)$ and $B$ has matrix $D=\mathrm{diag}(Q^{\mathrm{\ell }(w)},Q^{\mathrm{\ell }(w)+1})$; direct multiplication gives $A^{\mathsf{T}}D=DA$. Distinct pairs are orthogonal, hence $B(a{T}_i,b)=B(a,b{T}_i)$. Moving the letters of a reduced word for $u$ from the first argument to the second now gives $B(T_u,T_v)=\tau (T_v{T}_u^{\ast })=\tau (T_u^{\ast }{T}_v)$. Thus $\tau (a^{\ast }b)$ is a unimodular pairing: its standard-basis Gram matrix is diagonal with the units $Q^{\mathrm{\ell }(w)}$. The pairing $\tau (ab)$ is also unimodular, since $\ast$ permutes the standard basis.

For $\alpha$ a composition and $\bar{\alpha }$ the partition obtained by sorting its parts, reorder the consecutive row blocks by a permutation $d$ so that $S_{\alpha }d=d{S}_{\bar{\alpha }}$. The blocks are intervals, so the intersections of the two block decompositions are intervals; parabolic length additivity gives $x_{\alpha }{T}_d=T_d{x}_{\bar{\alpha }}$ and hence $x_{\alpha }=T_d{x}_{\bar{\alpha }}{T}_d^{-1}$. The row-module bases from step 2.1 show that $H{x}_{\bar{\alpha }}H$ is spanned by row-standard Murphy elements of partition shape $\bar{\alpha }$. Thus every row-standard Murphy element of composition shape $\alpha$ lies in that span.

Fix $\lambda ⊢n$ and a node $(i,j)$ with ${\lambda }_{i+1}\ge j$. Set $a=t^{\lambda }(i,j)$, $b=t^{\lambda }(i+1,j)$, and let $g$ be the Garnir tableau that fills the belt (row $i$ from column $j$ to its end and row $i+1$ from its start through column $j$) increasingly, first along the lower row segment and then along the upper segment. Put $\nu =({\lambda }_1,\dots ,{\lambda }_{i-1},j-1,{\lambda }_i+1,{\lambda }_{i+1}-j,{\lambda }_{i+2},\dots )$, a composition. The two parabolic decompositions of the interval-block product $S_{\lambda }{S}_{\nu }$ give ${\sum }_{w\in S_{\nu }\cap D_{\lambda }}{x}_{\lambda }{T}_w={\sum }_{x\in S_{\lambda }{S}_{\nu }}{T}_x={\sum }_{v\in S_{\lambda }\cap D_{\nu }}{T}_v^{\ast }{x}_{\nu }\mathrm{.}$ Indeed each side lists every element of $S_{\lambda }{S}_{\nu }$ exactly once, with length-additive factors, by [F2]. The tableaux $t^{\lambda }w$ on the left consist of $g$ and the tableaux obtained by assigning any other $j$-element subset of belt labels to the lower segment; every other assignment gives a standard tableau: rows are filled increasingly, columns before $j$ compare unchanged upper entries below all belt labels, columns after $j$ compare belt entries above unchanged lower suffix labels, and in column $j$ some smaller belt label remains in the complement than a selected lower label, so the top entry (the smallest complementary label) is less than the bottom entry (the largest selected lower label). Each is strictly more dominant than $g$, since at the first label where the subset differs a smaller label moves to the upper row. Isolating the $g$ term expresses $x_{\lambda }{T}_{d_{\lambda }(g)}$ as a linear combination of row-standard Murphy elements of shape $\nu$ and row-standard right-tableau terms strictly more dominant than $g$. By step 3.1, the shape-$\nu$ terms reduce to partition shapes lexicographically greater than $\lambda$: the sorted $i$-tuple of ${\lambda }_1,\dots ,{\lambda }_{i-1},{\lambda }_i+1$ dominates $({\lambda }_1,\dots ,{\lambda }_i)$, and the first $i$ parts of $\bar{\nu }$ dominate that tuple, so $\bar{\nu }$ is strictly lexicographically greater than $\lambda$.

Let $t$ be a nonstandard row-standard $\lambda$-tableau and choose a column violation $t(i,j)>t(i+1,j)$. Order the boxes first by row order and add the relation $(i+1,j)<(i,j)$. Both $t$ and the Garnir tableau $g$ of step 4.1 are linear extensions. The inversions of $g$ relative to row reading are exactly the pairs consisting of a lower-belt box through column $j$ followed by an upper-belt box from column $j$ onward; the added relation and the two row orders force every such pair in every extension. Thus the inversion set of $g$ is contained in that of $t$. For row-standard tableaux the Coxeter length of $d_{\lambda }(t)$ counts these inversions, and $t=g\cdot w$ for the relative permutation $w$ has $\mathrm{\ell }(w)$ equal to the number of pairs whose order differs between $g$ and $t$. Therefore $d_{\lambda }(t)=d_{\lambda }(g)w$ and $\mathrm{\ell }(d_{\lambda }(t))=\mathrm{\ell }(d_{\lambda }(g))+\mathrm{\ell }(w)$.

For fixed shape, define tableau dominance by comparing, for every label cutoff $m$ and row cutoff $k$, the number of entries at most $m$ in the first $k$ rows. This relation agrees with Bruhat order on the distinguished permutations: at row boundaries the Bruhat rank numbers are exactly these counts by [F7]; inside a row, row standardness makes the count the minimum of the preceding-row count plus the partial row length and the count at the next row boundary, so boundary inequalities give all rank inequalities. Hence $u{▹}_{\mathrm{t}\mathrm{a}\mathrm{b}}t$ implies $d_{\lambda }(u)<d_{\lambda }(t)$. In step 4.1, each other row-standard tableau $u{▹}_{\mathrm{t}\mathrm{a}\mathrm{b}}g$ has $d_{\lambda }(u)$ a proper reduced subword of $d_{\lambda }(g)$. Iterating the one-generator rules of [F1] along the reduced word $T_w$ of step 5.1 shows that every standard-basis term $T_z$ in $T_{d_{\lambda }(u)}{T}_w$ is represented by a subword of the concatenated word $d_{\lambda }(u)w$, hence by a proper subword of the reduced expression $d_{\lambda }(t)=d_{\lambda }(g)w$. If that subword is nonreduced, the deletion property in [F2] reduces it to a reduced subword representing $z$, so $z<d_{\lambda }(t)$ by [F7]. Write $z=y{d}_{\lambda }(v)$ using the distinguished row-coset factorization; length additivity makes $d_{\lambda }(v)$ a reduced suffix of a reduced expression for $z$, hence $d_{\lambda }(v)\le z<d_{\lambda }(t)$. The fixed-shape Bruhat comparison therefore makes every same-shape row-standard output tableau strictly dominate $t$. Induction first on the finite lexicographic shape order and then on the finite tableau order, and applying $\ast$ for the first index, straightens every row-standard Murphy element to standard Murphy elements; higher-shape terms are handled before lower shapes.

When $\lambda =(1^n)$, $x_{\lambda }=1$ and the row-standard elements include every standard Hecke basis element $T_w$. Step 6.1 therefore shows that the standard Murphy elements span $H_n$. Their number is ${\sum }_{\lambda ⊢n}\mathrm{\mid }\mathrm{Std}(\lambda ){\mathrm{\mid }}^2=n!$ by [F3], equal to the rank of $H_n$ by step 1.1. The square coordinate matrix of this spanning family has a right inverse, hence unit determinant by [F4], so the Murphy elements form an $R_A$-basis. Reversal fixes $x_{\lambda }$ and sends $T_{d(s)}^{\ast }{x}_{\lambda }{T}_{d(t)}$ to $m_{ts}$.

Let $\lambda$ be maximal in lexicographic order. For $x_{\lambda }h\in x_{\lambda }{H}_n$, the row-module basis and step 6.1 express it modulo higher lexicographic shapes as ${\sum }_v{r}_v{x}_{\lambda }{T}_{d(v)}$, with coefficients independent of a later first tableau. Assume by descending induction that the higher lexicographic Murphy span is a two-sided ideal. Multiplying this congruence on the left by $T_{d(s)}^{\ast }$ gives the fixed-first-index law for shape $\lambda$; the higher terms remain in the higher ideal, and the same-shape terms become $m_{sv}$. The induction proves the lexicographic cell law for every shape. Together with the basis and involution, this gives a lexicographic cell datum, so the higher lexicographic spans are two-sided ideals by [F5].

Index a square matrix by standard Murphy pairs, grouping its rows by shape $\eta$ and its columns by a partition $\nu$, and set $P_{(\eta ,s,t),(\nu ,a,b)}=\tau (m_{st}^{\eta }\theta (m_{ab}^{{\nu }^{\mathrm{\prime }}}))$, where $a,b$ range over standard ${\nu }^{\mathrm{\prime }}$-tableaux. Transposition of tableaux gives $\mathrm{\mid }\mathrm{Std}({\nu }^{\mathrm{\prime }})\mathrm{\mid }=\mathrm{\mid }\mathrm{Std}(\nu )\mathrm{\mid }$, so each diagonal shape block is square. Both the Murphy family and its $\theta$-image with conjugate shape labels are bases, by step 7.1 and [F8]; step 2.2 therefore makes $P$ invertible over $R_A$. Cyclically moving the leading factors inside the trace displays the factor $x_{\eta }{H}_n\theta (x_{{\nu }^{\mathrm{\prime }}})$, which is zero unless $\eta ⊴\nu$ by [F8]–[F9]. Thus $P_{\eta ,\nu }=0$ unless $\eta ⊴\nu$. In any increasing lexicographic ordering of the partitions, $P$ is block upper triangular. Its determinant is the product of the determinants of its square diagonal blocks; since that product is a unit in a commutative ring, each factor is a unit. Every diagonal block $P_{\nu ,\nu }$ is consequently invertible over $R_A$.

Fix a partition $\kappa$ and $h\in H_n{x}_{\kappa }{H}_n$. Whenever $\nu$ does not dominate $\kappa$, cyclicity and [F8]–[F9] give $\tau (h\theta (m_{ab}^{{\nu }^{\mathrm{\prime }}}))=0$ for every pair $a,b$: expand $h$ as a finite sum of $c{x}_{\kappa }d$ and move $c$ to the end, leaving $x_{\kappa }{H}_n\theta (x_{{\nu }^{\mathrm{\prime }}})$ as a factor. Expand $h$ in the Murphy basis. Process the partitions $\nu$ not dominating $\kappa$ in increasing lexicographic order. In its pairing against the $\nu$ column, the only potentially nonzero row blocks have $\eta ⊴\nu$ by step 8.2. If $\eta \mathrm{\ne }\nu$, then $\eta$ is lexicographically smaller; moreover $\eta$ cannot dominate $\kappa$, since $\kappa ⊴\eta ⊴\nu$ would make $\nu$ dominate $\kappa$. Their coefficient blocks have therefore already vanished. The remaining equation is the coefficient row of shape $\nu$ times the invertible matrix $P_{\nu ,\nu }$ equal to zero, forcing every shape-$\nu$ coefficient to vanish. Finite induction proves $H_n{x}_{\kappa }{H}_n\subseteq H_n^{⊵\kappa }$. This proves, in particular, the requested dominance support of every $x_{\kappa }h$.

Each Murphy element of shape $\eta$ belongs to $H_n{x}_{\eta }{H}_n$ by definition, while step 9.1 places the latter ideal in the span of shapes dominating $\eta$. Summing these inclusions and the reverse generator inclusions gives $H_n^{⊵\lambda }={\sum }_{\eta ⊵\lambda }{H}_n{x}_{\eta }{H}_n$ and $H_n^{▹\lambda }={\sum }_{\eta ▹\lambda }{H}_n{x}_{\eta }{H}_n$. Thus both dominance spans are two-sided ideals. In the lexicographic cell law of step 8.1, put $j=m_{st}h-{\sum }_v{r}_{t,v}(h)m_{sv}$. It lies in the strictly higher lexicographic span by step 8.1 and in $H_n^{⊵\lambda }$ by step 9.1, since every displayed term belongs to $H_n{x}_{\lambda }{H}_n$. These are spans of subsets of the same basis, so their intersection is the span of the common subset: shapes strictly lexicographically larger than $\lambda$ and dominating $\lambda$, which are precisely the shapes strictly dominating $\lambda$. Hence $j\in H_n^{▹\lambda }$, proving the dominance cell law with the same coefficients, independent of $s$.

Steps 7.1 and 10.1 prove every assertion of the Statement. All identities were proved over $R_A$ using only unit powers of $Q$ and finite free matrices with unit determinants. The extension $Q=v^2$ in step 1.1 is free on $1,v$ and hence faithfully flat; it is used only to identify the standard basis, and no inference from generic semisimplicity is used. Tensoring the established basis and its structure constants along any homomorphism $R_A\to R$ proves the same result for every commutative ring with unit image of $Q$, including $Q=-1$. No division or cancellation of a specialized quantum integer occurs.