Componentwise initial-row straightening

Let $n_c=\mathrm{\mid }{\alpha }^{(c)}\mathrm{\mid }$ and $a_c={\sum }_{d<c}{n}_d$. Write $S_{\eta }=S_{n_1}\times \cdots \times S_{n_r}$ for the subgroup permuting labels inside the consecutive component intervals. Every row-standard $s$ has a unique factorization $s=s_0\cdot w_s$ with $s_0$ of initial kind and $w_s$ the distinguished representative for the component coset; similarly $t=t_0\cdot w_t$. By [F4], $d_{\alpha }(s)=d_{\alpha }(s_0)w_s$, $d_{\alpha }(t)=d_{\alpha }(t_0)w_t$, and both lengths add. For any finite generator $T_i$ inside component interval $k$, the factors of $u_{\alpha }$ with index $c\le k$ use only $L_j$ with $j<i$ and commute distantly; factors with $c>k$ contain both $L_i,L_{i+1}$ symmetrically and commute by the Murphy divided-difference relation. Thus $u_{\alpha }$ commutes with all of $H(S_{\eta })$.

For initial-kind $s_0,t_0$, the row group and distinguished tableau permutations split into their component intervals. Since $u_{\alpha }$ commutes with $S_{\eta }$, the finite part of $m_{s_0{t}_0}^{\alpha }$ is a product of ordinary type-A row Murphy elements, one for each component. Apply the ordinary type-A composition-to-partition straightening [F3] separately in each component. Every output component partition dominates the sorted input component partition. The row-tableau prefix inequalities are exactly the componentwise inequalities from [F3]. The component factors commute because their simple-generator intervals are disjoint, so multiplying their finite expansions gives the asserted initial-kind expansion and componentwise dominance.

Restore arbitrary $s,t$ by multiplying each initial-kind term on the left by $T_{w_s}^{\ast }$ and on the right by $T_{w_t}$. Length additivity makes these products the Murphy elements indexed by the relabeled row-standard tableaux, without lower-length corrections. The component assignments are preserved because the same component coset representatives act on every term. To check tableau-prefix dominance after relabeling, fix a global label cutoff $m$. Since each $w_s$ preserves order within an initial component interval, the labels of that interval sent to values at most $m$ form an initial segment. Every cumulative row count of a relabeled tableau is therefore the corresponding componentwise row-prefix count at that segment length, where the inequalities have already been proved in step 2.1. The same argument applies to $w_t$.

For each multipartition $\lambda$, the exact finite row-module action expresses every $m_{st}{T}_i$ as a combination of row-standard elements of shape $\lambda$ with first index $s$. Apply the componentwise expansion just proved: all output shapes dominate $\lambda$ component by component, so $N^{⊵\lambda }$ is a right ideal for the finite Hecke subalgebra. Applying $\ast$ gives left closure. The strict span is the sum of these ideals over shapes strictly dominating $\lambda$, so it is also a two-sided ideal for that subalgebra. This argument proves ideal closure before using a fixed-first-index congruence.

Now take $s=t^{\lambda }$. In every same-shape standard output term from that componentwise expansion the first tableau $u$ dominates $t^{\lambda }$. The row-filled tableau is maximal for every row-prefix count among standard tableaux of this fixed shape, so equality of all the counts forces $u=t^{\lambda }$. Thus $m_{t^{\lambda },t}{T}_i$ is a linear combination of the $m_{t^{\lambda },v}$ modulo $N^{▹\lambda }$. Its coefficients depend only on $t,T_i$ and $\lambda$. Multiply this congruence on the left by $T_{d(s)}^{\ast }$ for arbitrary standard $s$. The error remains in $N^{▹\lambda }$ by step 4.1, and the displayed terms become $m_{sv}$. This proves the asserted finite-generator law with coefficients independent of $s$.