Ordinary Murphy straightening preserves both tableau prefixes

Work first over $K$. The cell forms are nonsingular by the separated supplier’s cellwise norm argument, before its faithfulness argument. If $\mu$ does not dominate $\lambda$, the product of a shape-$\mu$ Murphy element and a shape-$\lambda$ element lies in $H^{⊵\mu }\cap H^{⊵\lambda }\subseteq H^{▹\mu }$ by [F1], so every shape-$\lambda$ element acts as zero on $W_{\mu }$. Order shapes by a linear extension of dominance. The action map $H_{n,K}\to {⨁}_{\mu }{\mathrm{End}}_K(W_{\mu })$ is therefore block triangular in the Murphy basis, and its diagonal blocks are invertible by [F4]. It is a linear isomorphism, proving faithfulness without the supplier’s semisimple-quotient inference. For standard $u,v$ of shape $\mu$, define the endomorphism $R_{uv}$ by $w{R}_{uv}=⟨w,f_u⟩f_v$ on $W_{\mu }$, and by zero on other $W_{\nu }$. Nonsingularity makes these a basis of ${⨁}_{\mu }{\mathrm{End}}_K(W_{\mu })$. The action isomorphism just proved identifies these operators with elements of $H_{n,K}$. Thus the $R_{uv}$ are coordinates for $H_{n,K}$. Invariance gives $w(T_i{R}_{uv})=⟨w,f_u{T}_i⟩f_v$ and $w(R_{uv}{T}_i)=⟨w,f_u⟩(f_v{T}_i)$. Left multiplication therefore changes only $u$, right multiplication only $v$, and each changes its index only to itself or its standard adjacent swap; both directions of a standard swap have nonzero coefficients.

We need a comparison valid across shapes. Let $b$ be row standard with $i$ in an upper row and $i+1$ in a lower row, and put $b^{\mathrm{\prime }}=b(i,i+1)$. If standard $a{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}b$, then $a{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}{b}^{\mathrm{\prime }}$ and every standard swap $a^{\mathrm{\prime }}=a(i,i+1)$ also dominates $b^{\mathrm{\prime }}$. Only the cutoff $m=i$ changes. A swap raising $a$'s counts causes no problem. For a lowering swap, a failure at row cutoff $j$ would require $A_a(j,i)=A_b(j,i)$ while $b$'s swap does not lower its count at $j$. If both labels lie above $j$ in $b$, then $A_a(j,i+1)=A_b(j,i)<A_b(j,i+1)$, contradicting $a{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}b$. If both lie below $j$, then $A_a(j,i-1)=A_b(j,i)-1<A_b(j,i-1)$, again a contradiction. These exhaust the possible failures.

Let $B_r$ be the interval of labels occupying row $r$ of $t^{\alpha }$, and write $x_{\alpha }=\sum a_{uv}{R}_{uv}$. Fix a right index $v$ and an adjacent pair $i,i+1$ inside $B_r$. Comparing coefficients in $T_i{x}_{\alpha }=Q{x}_{\alpha }$, if $i,i+1$ share a column of $u$, gives $-a_{uv}=Q{a}_{uv}$; no other first index can contribute because $u(i,i+1)$ is not standard. Hence $a_{uv}=0$. If $u^{\mathrm{\prime }}=u(i,i+1)$ is standard, its coefficient equation has the form $Q{a}_{uv}=c_u{a}_{uv}+a_{u^{\mathrm{\prime }}}{a}_{u^{\mathrm{\prime }}v}$, where $a_{u^{\mathrm{\prime }}}\mathrm{\ne }0$ is the seminormal off-diagonal scalar. Thus $a_{uv}=0$ forces $a_{u^{\mathrm{\prime }}v}=0$, and the reverse implication follows from the equation at $u^{\mathrm{\prime }}$. Zeros therefore propagate across every standard swap inside a block $B_r$.

Replace each label of $u$ in $B_r$ by $r$, obtaining a weakly increasing row and column tableau $S$. The standard refinements of a fixed $S$ are the linear extensions of the node-order posets in its equal-label classes. Any two linear extensions of a finite poset are connected by adjacent incomparable swaps: move the first desired element left across its predecessors in the other extension, all of which must be incomparable, and continue inductively. Here these are standard swaps of consecutive integers inside a block $B_r$. If $S$ repeats a label in a column, weak column order supplies vertically adjacent equal-label nodes, a cover pair $x<y$ in that class. A cover pair occurs consecutively in some linear extension: first extend the order ideal $\{z:z<y,z\mathrm{\ne }x\}$, then place $x,y$, then complete; the displayed set is an ideal because $x<z<y$ would contradict the cover condition. This refinement has consecutive labels in one column, so step 2.1 and connectivity force every coefficient in its fibre to vanish. Therefore every supported $u$ refines a column-strict, hence semistandard, $S$.

A positive-integer column-strict tableau has label at least its row index. Put $b_j={\alpha }_1+\cdots +{\alpha }_j$. Every entry $1,\dots ,b_j$ in a surviving refinement $u$ therefore lies in its first $j$ rows, giving $A_u(j,m)\ge \min (m,b_j)=A_{t^{\alpha }}(j,m)$ for all $j,m$. The right parabolic equation and the same argument give $v{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}{t}^{\alpha }$. Also choose the $k$ row labels whose multiplicities are the $k$ largest parts of $\alpha$. Column-strictness permits at most $k$ chosen labels per column, so their total multiplicity is at most ${\sum }_c\min (k,\mathrm{height}(c))={\mu }_1+\cdots +{\mu }_k$. Thus each surviving spectral shape satisfies $\mu ⊵\bar{\alpha }$.

Choose a reduced expression $d_{\alpha }(s)=s_{i_1}\cdots s_{i_k}$. Every reduced prefix is a distinguished row representative: write $d_{\alpha }(s)=ab$ with $a$ that prefix and $\mathrm{\ell }(d_{\alpha }(s))=\mathrm{\ell }(a)+\mathrm{\ell }(b)$. If $a$ were not minimal in its left row coset, some $y\in S_{\alpha }$ would have $\mathrm{\ell }(ya)<\mathrm{\ell }(a)$, giving $\mathrm{\ell }(y{d}_{\alpha }(s))\le \mathrm{\ell }(ya)+\mathrm{\ell }(b)<\mathrm{\ell }(d_{\alpha }(s))$, contrary to the minimality of $d_{\alpha }(s)$. Hence $b_r=t^{\alpha }{s}_{i_1}\cdots s_{i_r}$ is row standard, and each step moves $i_r$ from an upper row to a lower one. Apply $T_{i_1},\dots ,T_{i_k}$ successively on the left of $x_{\alpha }$. Their final product is $T_{d_{\alpha }(s)}^{\ast }{x}_{\alpha }$. By steps 1.1 and 1.2, spectral indices initially dominating $t^{\alpha }$ continue to dominate $b_r$ at each stage; an unchanged index also works because $b_{r-1}{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}{b}_r$. The right index and shape do not change. Applying the right-handed argument along a reduced word of $d_{\alpha }(t)$ now proves that the spectral expansion of $m_{st}^{\alpha }$ has support only at $R_{uv}$ with $u{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}s$, $v{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}t$, and shape dominating $\bar{\alpha }$.

Let $P$ be the finite set of pairs of standard tableaux of a common shape, ordered by $(u,v)\ge (s,t)$ when both tableau-prefix comparisons hold. This is a partial order across shapes: equality of all counts determines the row of each label, hence the row-standard tableau. Apply step 5.1 to every standard Murphy element. Its transition matrix $C$ to the spectral basis has entry zero unless its output pair dominates its input pair. Both families are bases by [F1] and step 1.1, so $C$ is invertible. In a linear extension of $P$ it is triangular, hence its diagonal part $D$ is invertible. Write $C=D(I+N)$; $N$ has only strict-order entries, so $N^{\mathrm{\mid }P\mathrm{\mid }}=0$, since a nonzero product would require a strict chain with $\mathrm{\mid }P\mathrm{\mid }+1$ elements. Therefore $C^{-1}=(I-N+\cdots +(-N{)}^{\mathrm{\mid }P\mathrm{\mid }-1})D^{-1}$. Every term has the same ordered-pair support, because the partial order is transitive. Each $R_{uv}$ thus expands only in $m_{wz}$ with $w{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}u$, $z{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}v$.

Substitute these inverse transitions into step 5.1. Every surviving standard Murphy term has $w{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}u{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}s$ and $z{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}v{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}t$. At the final label cutoff $n$, $w{⊵}_{\mathrm{t}\mathrm{a}\mathrm{b}}u$ implies that its shape dominates the shape of $u$, so the shape also dominates $\bar{\alpha }$. This proves all assertions over $K$. By [F1] the unique standard-Murphy coordinates of $m_{st}^{\alpha }$ already lie in $A$; injectivity of $A\hookrightarrow K$ makes every coordinate proved zero over $K$ equal to zero in $A$. The resulting integral identities specialize under any map $A\to R$ sending $Q$ to a unit. No semisimplicity or quantum-integer division is required in $R$.