Polytabloid specialization and the exact signed multiplicity dictionary

The group algebra over every coefficient ring. Let $R$ be any commutative ring and specialize at $q=1$. The Hecke relations become $T_i^2=1$, braid and distant commutation. The Coxeter presentation in [F5] therefore gives the homomorphism $H_{R,1}(S_n)\to R[S_n]$ carrying $T_i$ to $s_i$. A reduced word shows that it carries every $T_w$ to $w$. These are bases of the source and target over $R$, so the map is an isomorphism over $R$, including in positive characteristic and with zero divisors. Equivalently the assignment $s_i\mapsto T_i$ gives its inverse directly from the presentation. No complex-field specialization, dimension comparison over a field or semisimplicity is used.

Multiplicities and zero quotients. For a field and unit $q$, the unit form identity in [F1] identifies $D_{\mathrm{D}\mathrm{J}}^{\mu }$ with $(D_{\mathrm{M}}^{{\mu }^{\mathrm{\prime }}}{)}^{\theta }$, including zero quotients. A nonzero cell-form quotient is simple: the cell rank-one action shows that every vector outside its radical generates the cell module, so the radical is its unique maximal submodule. Thus when $D_{\mathrm{M}}^{{\mu }^{\mathrm{\prime }}}\mathrm{\ne }0$, the module $D_{\mathrm{D}\mathrm{J}}^{\mu }$ is a legitimate simple composition-factor label. Applying the exact twist equivalence in [F2] to a composition series and using [F1] gives $[P(\lambda ):D_{\mathrm{D}\mathrm{J}}^{\mu }]=[C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }}{)}^{\theta }:D_{\mathrm{D}\mathrm{J}}^{\mu }]=[C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }}):(D_{\mathrm{D}\mathrm{J}}^{\mu }{)}^{\theta }]=[C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }}):D_{\mathrm{M}}^{{\mu }^{\mathrm{\prime }}}]\mathrm{.}$ This proves clause 2. If a form is identically zero, only its zero form quotient is asserted; no claim is made that the actual module head vanishes.

The actual polytabloids and the forms. Give the classical tabloid module the right action $\{t\}\cdot w=\{w^{-1}\cdot t\}=\{t\cdot w\}$. The map $b_d=x_{\lambda }{T}_d\mapsto \{t^{\lambda }\cdot d\}$ sends the distinguished basis to the tabloid basis. It respects the right action: at $q=1$ the row-module multiplication rule is simply the permutation of these cosets, with a same-row transposition acting as $1$. Every diagonal value of $\beta$ becomes $1$, so this is an isometry to the orthonormal tabloid form. The column group of the column-filled tableau $t_{\lambda }$ is exactly $S_{{\lambda }^{\mathrm{\prime }}}$, because its columns carry the consecutive row blocks of ${\lambda }^{\mathrm{\prime }}$. Hence $z_{\lambda }=x_{\lambda }{w}_{\lambda }{\sum }_{u\in S_{{\lambda }^{\mathrm{\prime }}}}\mathrm{sgn}(u)u$ maps to ${\sum }_u\mathrm{sgn}(u)\{t_{\lambda }\cdot u\}$, the classical column polytabloid of $t_{\lambda }$ (replace $u$ by $u^{-1}$ to compare with the left-action formula). Covariance shows $e_t=z_{\lambda }{d}_{{\lambda }^{\mathrm{\prime }}}(t)$ maps to the polytabloid of $t_{\lambda }\cdot d_{{\lambda }^{\mathrm{\prime }}}(t)=t^{\mathrm{\prime }}$, the transpose of the standard ${\lambda }^{\mathrm{\prime }}$-tableau $t$. Thus the actual standard bases correspond bijectively. By the split integral classical basis in [F3] and the split Hecke basis in [F5], this identifies $P(\lambda )$ with $R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ over every $R$, and the restricted DJ form with the restricted tabloid form.

The exact position of the twist. At $q=1$, [F2] gives $\theta (T_i)=-T_i$, hence $\theta (w)=\mathrm{sgn}(w)w$. Therefore the right module $P(\lambda {)}^{\theta }$ is the sign tensor twist. Applying the involution $\theta$ to [F1] yields $P(\lambda {)}^{\theta }\cong C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }}),$ without a remaining twist on the Murphy cell. Combining this with the independent same-shape pairing in [F6] gives $P(\lambda {)}^{\theta }\cong P({\lambda }^{\mathrm{\prime }}{)}^{\circ }$. Under step 2.1, star dual is the ordinary group contragredient dual, so this is exactly the module identity $S_F^{\lambda }\otimes \mathrm{sgn}\cong (S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\ast }$ of [F4] after converting the published left modules to right modules. This asserts the same compatible module dictionary, not an unproved equality of separately normalized explicit maps. In characteristic $2$ the sign becomes trivial at $q=1$; conjugation remains part of the identity.

Conditional Schur dictionary and conclusion. Under the stated separate hypothesis on a Schur corner, its images at ${\lambda }^{\mathrm{\prime }}$ are $C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }})$ and $D_{\mathrm{M}}^{{\lambda }^{\mathrm{\prime }}}$. The untwisted Murphy identities from step 3.1 and [F1], valid for arbitrary unit $q$ by involutivity of $\theta$, identify these with $P(\lambda {)}^{\theta }$ and $(D_{\mathrm{D}\mathrm{J}}^{\lambda }{)}^{\theta }$, including the zero quotient. This proves clause 3 without a Schur theorem as a prerequisite. The actual nonzero Schur head and the Hecke cell-form quotient are distinct notions when the corner kills the former. For the empty shape, the generators, forms, twists and all isomorphisms have value $1$. All bases, maps and scalar identities are explicit and use no choice principle.