The Hecke Jucys-Murphy action is dominance triangular

Braid identities. Put $T_i=T_{(i,i+1)}$. The reduced word for $(j,k+1)$ gives $L_{k+1}=q^{-1}(T_k{L}_k{T}_k+T_k)$. We prove $T_i{L}_k=L_k{T}_i$ whenever $i\mathrm{\ne }k-1,k$. For $i\ge k+1$, each generator in the displayed words defining $L_k$ commutes with $T_i$. For $i\le k-3$ the recurrence and induction apply. For the remaining case $k=i+2$, put $a=T_i$, $b=T_{i+1}$ and $C=L_i$; then $bC=Cb$ by the already established disjoint case and $L_{i+2}=q^{-1}b+q^{-2}bab+q^{-2}baCab$. The braid and quadratic relations give $[a,bab]=-q[a,b]$ and $a(baCab)=(baCab)a$ (replace $aba$ by $bab$ and move $C$ past $b$), so its commutator with $a$ vanishes. Thus $L_k$ centralizes $H(S_{k-1})$; since $L_j\in H(S_{k-1})$ for $j<k$, all $L_j$ commute. Multiplying the recurrence by $T_i$ and using $T_i^2=(q-1)T_i+q$ gives $T_i{L}_{i+1}=L_i{T}_i+1+(q-1)L_{i+1}$; starring gives $T_i{L}_i=L_{i+1}{T}_i-1-(q-1)L_{i+1}$. Hence $T_i$ commutes with $L_i+L_{i+1}$ and $L_i{L}_{i+1}$: for the product substitute the first identity into $(L_{i+1}{T}_i-1-(q-1)L_{i+1})L_{i+1}$. Every symmetric polynomial is a polynomial in the elementary symmetric polynomials over the coefficient ring: in each homogeneous degree, its lexicographically leading exponent vector satisfies $a_1\ge \cdots \ge a_n$ by symmetry. Subtract its coefficient times $e_1^{a_1-a_2}\cdots e_{n-1}^{a_{n-1}-a_n}{e}_n^{a_n}$, whose leading monomial has that vector and coefficient $1$, and repeat on the finite set of degree-$d$ monomials. No coefficient is divided. Each elementary symmetric polynomial splits, at the adjacent pair, as $e_d({\widehat{L}}_i,{\widehat{L}}_{i+1})+(L_i+L_{i+1})e_{d-1}({\widehat{L}}_i,{\widehat{L}}_{i+1})+L_i{L}_{i+1}{e}_{d-2}({\widehat{L}}_i,{\widehat{L}}_{i+1})$. It therefore commutes with every $T_i$ and is central.

Lifting smaller-rank congruences. Here is the precise ideal check used in the induction below. Let $\tau$ be either a whole-row prefix $({\lambda }_1,\dots ,{\lambda }_r)$ of size $m<n$, or the prefix $({\lambda }_1,\dots ,{\lambda }_r,1)$ ending at the first box of the last row, of size $m<n$. In both cases $x_{\lambda }=h{x}_{\tau }=x_{\tau }h$, where $h$ is the row symmetrizer of the remaining blocks, embedded in $H_n$; in the second case its last block starts at the singleton label $m$, and no generator of $x_{\tau }$ meets that block. If $x_{\tau }{L}_k-c{x}_{\tau }\in I^{⊳\tau }$ in $H_m$, expand this difference in its smaller Murphy basis. Every shape $\eta ⊳\tau$ lies in $H_m{x}_{\eta }{H}_m$. Embedded in $H_n$, $x_{\eta }$ is $x_{{\kappa }_{\eta }}$ for the partition ${\kappa }_{\eta }=(\eta ,1^{n-m})$, so [F2] puts its two-sided ideal in $I^{⊵{\kappa }_{\eta }}$. Some prefix $j$ before the final row of $\tau$ has ${\sum }_{a\le j}{\eta }_a>{\sum }_{a\le j}{\tau }_a={\sum }_{a\le j}{\lambda }_a$: strict dominance cannot first differ at or after the final row, where $\tau$ has total $m$. In the whole-row case the same argument permits $j<r$. Hence $\lambda$ does not dominate ${\kappa }_{\eta }$, and none of these ideals has a shape-$\lambda$ component. Multiplying the smaller congruence on the appropriate side by $h$ gives $x_{\lambda }{L}_k-c{x}_{\lambda }\in {\sum }_{\eta }{I}^{⊵{\kappa }_{\eta }}$. This difference also belongs to $I^{⊵\lambda }$ because both terms are in $x_{\lambda }{H}_n$. Their intersection, read in the Murphy basis, has only shapes strictly dominating $\lambda$, and the lifted congruence therefore holds modulo $I^{⊳\lambda }$.

An integral scalar for the central sum. For this step work over $A=\mathbb{Z}[v^{\pm 1}]$, with fraction field $K=\mathbb{Q}(v)$. The matrix of the regular trace pairing $B(a,b)={\mathrm{Tr}}_K({\mathrm{\ell }}_{ab})$ in the standard basis has entries in $A$. At $v=1$ it is the matrix $n!{\delta }_{u,w^{-1}}$, since a nonidentity permutation has no fixed point in the regular permutation action. Its determinant is nonzero, so $B$ is nondegenerate over $K$. If $a$ belongs to the Jacobson radical $J$ then $ab\in J$ is nilpotent for every $b$, by [F3]; thus ${\mathrm{\ell }}_{ab}$ is nilpotent and has trace zero. Nondegeneracy forces $J=0$, so $H_K$ is semisimple. In its cell layer $I_K^{⊵\lambda }\mathrm{/}{I}_K^{⊳\lambda }$, [F2] identifies multiplication with $X\cdot Y=X{G}_{\lambda }Y$ on $d\times d$ matrices, $d=\mathrm{\mid }\mathrm{Std}(\lambda )\mathrm{\mid }$. If $G_{\lambda }$ were singular, the nonzero subspace $N=\{X:G_{\lambda }X=0\}$ would be a two-sided ideal of this layer with $N^2=0$, contradicting [F3]. Hence $G_{\lambda }$ is invertible and $X\mapsto G_{\lambda }X$ identifies the layer with ${\mathrm{Mat}}_d(K)$. A central element of $H_K$ acts on this layer by a bimodule endomorphism of that matrix algebra, necessarily a scalar: its value at the identity commutes with every matrix and therefore is a scalar matrix. Applied to $Z_n={\sum }_k{L}_k$, central by step 1.1, this shows that its matrix on the cell module is ${\alpha }_{\lambda }{I}_d$ over $K$. The cell action matrix already has entries in $A$, so ${\alpha }_{\lambda }\in A$, and freeness over $A$ makes the identity hold integrally and after every base change to $R$. Thus $m_t{Z}_n={\alpha }_{\lambda }{m}_t$ for every $t$, without any assumption on the specialized cell form or its head.

Transport along standard adjacent swaps. Suppose $s^{\mathrm{\prime }}=s{s}_i$ is standard with $s⊳s^{\mathrm{\prime }}$ and length increasing by one, and suppose $m_s{L}_k=r_s(k)m_s+{\sum }_{u⊳s}{a}_u{m}_u$ for all $k$. By [F1], $m_{s^{\mathrm{\prime }}}=m_s{T}_i$. For $k\mathrm{\ne }i,i+1$ step 1.1 gives $m_{s^{\mathrm{\prime }}}{L}_k=m_s{L}_k{T}_i$, and [F2] puts every error strictly above $s^{\mathrm{\prime }}$. For $k=i+1$, the same step gives $m_{s^{\mathrm{\prime }}}{L}_{i+1}=m_s(L_i{T}_i+1+(q-1)L_{i+1})$; its coefficient at $m_{s^{\mathrm{\prime }}}$ is $r_s(i)$, since all remaining terms are strictly above $s^{\mathrm{\prime }}$. The second adjacent identity similarly gives coefficient $r_s(i+1)$ at $m_{s^{\mathrm{\prime }}}$ for $L_i$. Thus the diagonal data are transposed in positions $i,i+1$ and unchanged elsewhere. Every standard tableau is reachable from $t^{\lambda }$ by such standard length-increasing swaps: reverse a right-descent reduction of $d(t)$; the two consecutive entries at a descent lie in incomparable cells, so their interchange remains standard. Consequently diagonal data at the row-filled tableau determine the asserted diagonals at every standard tableau.

Row-filled induction away from the final singleton. Induct on $n$, the cases $n=0,1$ being immediate, and on $k$ within the row-filled tableau. If $k$ is in a row which is not the last row, choose the whole-row prefix $\tau$ ending with that row. It has size $m<n$, its row-filled tableau has no more dominant standard tableaux, and the induction on rank gives $x_{\tau }{L}_k=[{\mathrm{content}}_{t^{\tau }}(k){]}_q{x}_{\tau }$ modulo $I^{⊳\tau }$. Lift by step 1.2. If $k$ is in the last row but not its first column, $T_{k-1}$ belongs to its row subgroup, and the recurrence gives $m_{t^{\lambda }}{L}_k=(1+q[{\mathrm{content}}_{t^{\lambda }}(k-1){]}_q)m_{t^{\lambda }}=[{\mathrm{content}}_{t^{\lambda }}(k){]}_q{m}_{t^{\lambda }}$, by induction on $k$ and $[c+1{]}_q=1+q[c{]}_q$. If $k<n$ is in the first column of the last row, use the singleton-ended prefix $\tau$ of step 1.2 and induction on its smaller rank. These arguments leave only $k=n$ for a shape $\lambda =({\lambda }_1,\dots ,{\lambda }_r,1)$. All $L_k$ with $k<n$ have already been shown to act on $m_{t^{\lambda }}$ by their claimed scalars. Step 2.1 then supplies $m_{t^{\lambda }}{L}_n=r_n{m}_{t^{\lambda }}$ for some $r_n\in R$. Step 2.2 consequently applies with these diagonal data to every standard tableau; at this point all $L_k$ are triangular, though $r_n$ remains to be identified.

The final singleton is fixed by one Garnir calculation. In the remaining shape let $i$ be the first entry in row $r$ of $t^{\lambda }$ and put $t=t^{\lambda }{s}_{n-1}\cdots s_{i+1}$. Then $t$ is standard, its last row contains $i+1$, and its penultimate row contains $i,i+2,\dots ,n$. Step 2.2 gives $m_t{L}_i=[1-r{]}_q{m}_t+{\sum }_{v⊳t}{a}_v{m}_v$ and $m_t{L}_{i+1}=r_n{m}_t+{\sum }_{v⊳t}{b}_v{m}_v$. The tableau $g:=t{s}_i$ is exactly the Garnir tableau for the first-column belt $i,\dots ,n$ between the last two rows. For $j=i,\dots ,n$, let $u_j$ have last-row entry $j$ and the other belt labels increasingly in row $r$. The finite belt identity of [F2] gives ${\sum }_{j=i}^n{x}_{\lambda }{T}_{d(u_j)}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{⊳\lambda })$: its other side is a sum of multiples of the symmetrizer obtained by merging these two rows, whose sorted shape strictly dominates $\lambda$. Here $u_i=g$, $u_{i+1}=t$, and all $u_j$ with $j>i+1$ are standard and strictly dominate $t$. The length-increasing row rule gives $m_t{T}_i=x_{\lambda }{T}_{d(g)}+I^{⊳\lambda }$, so the belt identity explicitly yields $m_t{T}_i=-m_t-{\sum }_{j>i+1}{m}_{u_j}$. Put $U=\mathrm{span}\{m_v:v⊳t\}$. Step 3.1 and transport show $U{L}_i\subseteq U$. Also $U{T}_i\subseteq U$: dominance forces each $v⊵t$ to agree with $t$ on the filled first $r-1$ rows and to place $i$ in node $(r,1)$. If $i+1$ occupies the singleton last row, all remaining labels must occupy row $r$ increasingly, forcing $v=t$. Thus every $v⊳t$ places $i+1$ in $(r,2)$, and the row rule gives $m_v{T}_i=q{m}_v$. (If ${\lambda }_r=1$, the higher span is zero.) Working modulo $U$, the recurrence now gives $m_t{L}_{i+1}=q^{-1}{m}_t(T_i{L}_i{T}_i+T_i)=q^{-1}([1-r{]}_q-1)m_t=[-r{]}_q{m}_t$. Comparing its coefficient at $m_t$ proves $r_n=[-r{]}_q$, exactly the content of the final singleton. This completes the rank induction and identifies all row-filled diagonals.

The required flag and the initial left generator. Step 2.2 now gives $m_t{L}_k=[{\mathrm{content}}_t(k){]}_q{m}_t+{\sum }_{v⊳t}{a}_v{m}_v$ for every standard $t$, over the given ring. Order the standard tableaux by any linear extension with the most dominant first. The spans of initial segments are stable under all $L_k$, and their successive rank-one quotients have exactly the q-content diagonals; this proves the claimed common triangular flag. Commutativity and symmetric-polynomial centrality were established independently in step 1.1. In particular the maximal tableau $t^{\lambda }$ has the exact eigenrelation $m_{t^{\lambda }}{L}_k=[{\mathrm{content}}_{t^{\lambda }}(k){]}_q{m}_{t^{\lambda }}$. Since $L_k^{\ast }=L_k$, starring its defining cell congruence gives the corresponding left-cell generator relation $L_k{z}_{\lambda }=[{\mathrm{content}}_{t^{\lambda }}(k){]}_q{z}_{\lambda }$. For the multiplicative normalization $X_k=1+(q-1)L_k$ this reads $X_k{z}_{\lambda }=q^{{\mathrm{content}}_{t^{\lambda }}(k)}{z}_{\lambda }$, using $1+(q-1)[c{]}_q=q^c$ for all integral $c$. These identities concern the cell generator, hence hold after taking the indicated cell quotient; they do not assert the same equality for the raw symmetrizer in $H$. All formulas specialize from $A$ to $R$, and no quantum integer or factorial is inverted in the specialized algebra.