The weight-one block decomposition pattern

Enumeration and ordering. Choose a conjugate core beta set long enough that every runner contains a bead, and list its bottom beads $x_1<\cdots <x_e$. Take this beta length divisible by $e$. Every partition with this core has, at the same beta length, the same packed set: its packed set represents $\gamma$ and the length uniquely determines a beta-set for $\gamma$. Weight one means its total runner displacement from packing is one, so exactly one bottom bead moves down one row. Conversely each such move adds exactly $e$ boxes and preserves the core. Thus replacing $x_k$ by $x_k+e$ gives precisely the $e$ weight-one partitions, denoted ${\lambda }_k$. For $j>k$, moving $x_j$ instead of $x_k$ gives the more dominant conjugate partition: the two bead lists differ by replacing $x_k+e,x_j$ with $x_k,x_j+e$. The new maximum is larger than both old values and the new minimum smaller than both, with the same sum. For each number of selected entries, the largest possible sum therefore does not decrease, and at least one sorted prefix increases. The sorted conjugate parts differ from these prefixes by the same beta offsets. Conjugation reverses dominance, hence ${\lambda }_1⊳\cdots ⊳{\lambda }_e$. For the empty core put ${\gamma }_i=0$. The largest original partition is $({\gamma }_1+e,{\gamma }_2,\dots )$, since its added horizontal first-row strip dominates every other placement.

Choose a lift with valuation one. If $q\mathrm{\ne }1$, its finite quantum characteristic $e$ is its multiplicative order, and is prime to the field characteristic; use $R=F[X{]}_{(X-q)}$ and $\widehat{q}=X$. The derivative of $X^e-1$ at $q$ is $e{q}^{e-1}\mathrm{\ne }0$, so $\nu ([e{]}_X)=1$; also $\nu ([h{]}_X)=0$ when $e\nmid h$. If $q=1$, then $\mathrm{char}F=p=e$. First work over ${\mathbb{F}}_p$ with $R={\mathbb{Z}}_{(p)}$, $\widehat{q}=1$, giving $\nu ([e{]}_1)=\nu (p)=1$ and $\nu ([h{]}_1)=0$ for $p\nmid h$, and then extend scalars to $F$. This extension preserves the decomposition multiplicities because the q-Schur simple modules are absolutely irreducible and its integral Weyl modules commute with scalar extension. In both cases the generic factorial is nonzero and [F2] applies.

Restriction. If $\lambda$ is not restricted, some ${\lambda }_i-{\lambda }_{i+1}\ge e$, so removing the last $e$ boxes of row $i$ leaves a partition. Its size is $\mathrm{\mid }\gamma \mathrm{\mid }$ and its core is $\gamma$, so it equals $\gamma$. Thus $\lambda$ is obtained by adding $e$ boxes to row $i$ of $\gamma$. An $e$-core has ${\gamma }_r-{\gamma }_{r+1}<e$ for every $r$, since otherwise it has a removable horizontal $e$-strip. Consequently $i>1$ is impossible: the partition inequality ${\gamma }_{i-1}\ge {\gamma }_i+e$ would contradict this core property. Therefore the only nonrestricted member is ${\lambda }_1=({\gamma }_1+e,{\gamma }_2,\dots )$; it is indeed nonrestricted because its first-row difference is at least $e$.

Evaluate the balanced formula. A nonzero coefficient involves a hook of length divisible by $e$. A weight-one abacus has exactly one bead-gap pair of equal residue with the gap below the bead: $x_k+e$ and $x_k$; thus its only $e$-divisible hook has length $e$. Any surviving term in [F2] replaces two bead values, and its new runner counts equal the old ones: either $e$ divides the transfer $h_{ac}$, leaving residues fixed, or $e$ divides $h_{ab}=h_{ac}+{\beta }_b-{\beta }_c$, exchanging their residues. Hence its sorted symbol is one of the ${\lambda }_j$. For $j>k$, the distinct bead sets of ${\lambda }_k$ and ${\lambda }_j$ differ exactly by replacing $x_k+e,x_j$ with $x_k,x_j+e$. There is exactly one balanced transfer yielding this change: if $x_j<x_k+e$, its two hook lengths are $h_{ab}=e$, $h_{ac}=x_j-x_k$; if $x_j>x_k+e$, they are $h_{ab}=x_j-x_k$, $h_{ac}=e$. The difference $x_j-x_k$ is not divisible by $e$ since bottom beads are on distinct runners. Thus the respective unsigned coefficients are $+1$ and $-1$ by step 1.2. Sorting gives in either case the signed coefficient $(-1{)}^{j-k+1}$: in the packed core set, exactly one bead from each runner with bottom bead $>x_r$ lies between $x_r$ and $x_r+e$, so moving $x_r$ crosses $e-r$ beads; the relative sorting parity for the two moves, together with their opposite orientation, is $(-1{)}^{(e-j)-(e-k)+1}$. There are no other terms and no repeated contributions: the two lost and two gained bead values determine the pair and transfer uniquely. Therefore ${\sum }_{i>0}[W^{{\lambda }_k}(i)]={\sum }_{j>k}(-1{)}^{j-k+1}[W^{{\lambda }_j}]$.

Descending induction and the matrix. For $k=e$, step 2.2 gives zero on the left; all summands are actual submodule classes with nonnegative simple multiplicities, so $W^{{\lambda }_e}(1)=0$ and $W^{{\lambda }_e}=L^{{\lambda }_e}$. Suppose the stated rows are known for $j>k$. Substitution into step 2.2 telescopes to ${\sum }_{i>0}[W^{{\lambda }_k}(i)]=[L^{{\lambda }_{k+1}}]$. Since $W(1)=\mathrm{rad}W$ and each later term is a submodule of $W(1)$, positivity shows $W(1)\cong L^{{\lambda }_{k+1}}$ and $W(i)=0$ for $i\ge 2$. The head quotient is $L^{{\lambda }_k}$, giving $[W^{{\lambda }_k}]=[L^{{\lambda }_k}]+[L^{{\lambda }_{k+1}}]$. Notice that the identity $[W]=[L]+{\sum }_{i>0}[W(i)]$ is used only after proving all later filtration submodules vanish; it is false for a general Jantzen filtration. Reading simple multiplicities yields the complete $e\times e$ bidiagonal matrix.

Specht rows. Apply the exact functor of [F4] to step 3.1 and use step 2.1: $D_{\mathrm{M}}^{{\lambda }_1}=0$, while the other $D_{\mathrm{M}}^{{\lambda }_k}$ are the nonzero simples. This gives exactly the three displayed Murphy Specht class identities, hence the full $e\times (e-1)$ Murphy matrix. Conjugation sends the core to ${\gamma }^{\mathrm{\prime }}$ and the restricted simple labels to regular ones, and the same $\theta$ pullback applied to every module preserves all composition multiplicities. Thus the ordinary Specht matrix has the same entries under the accepted conjugate/sign label dictionary.