Graded Specht heads classify the graded simple modules

The intrinsic radical is homogeneous. For restricted $\lambda$, let $R_{\lambda }$ be the ungraded cell-form radical in $S(\lambda )$. By [F1] it is the unique maximal proper submodule, and the same is true after scalar extension to the infinite field $K=F(u)$: the Gram kernel extends to $K{\otimes }_F{R}_{\lambda }$ and the quotient remains simple. The same argument also works over an infinite algebraic closure; using $F(u)$ requires no algebraic-closure choice. For $t\in K^{\times }$ define $s_t(a)=t^{j}a$ on $B_j$ and $g_t(v)=t^{k}v$ on $S(\lambda {)}_k$. The grading gives $g_t(av)=s_t(a)g_t(v)$, so $g_t$ takes submodules to submodules and preserves the unique maximal one. If $v={\sum }_{k\in E}{v}_k$ belongs to this radical, all ${\sum }_k{t}^k{v}_k$ do. The distinct Laurent monomials $t^k$ are linearly independent as functions on the infinite field: multiply a vanishing relation by a large power of $t$ and use the root bound for a polynomial. Thus one can select $\mathrm{\mid }E\mathrm{\mid }$ values of $t$ with invertible evaluation matrix, and solve for each $v_k$ in the radical. It is homogeneous over $K$, hence over $F$, since $v_k\otimes 1\in K\otimes R_{\lambda }$ implies $v_k\in R_{\lambda }$. Therefore the quotient $\dot{D}(\lambda )=S(\lambda )\mathrm{/}{R}_{\lambda }$ has its actual quotient grading and is irreducible even after forgetting that grading. No homogeneity or degree of the cellular form has been assumed.

Categorical duality transports to the adjoint upper bar in both branches. The projective $Be(\mathbf{i})$ is $\mathrm{\#}$-self-dual: evaluation sends ${\mathrm{HOM}}_B(Be(\mathbf{i}),B)$ to $e(\mathbf{i})B$, and twisting by the generator-fixing $\psi$ gives $Be(\mathbf{i})$ with no shift. Under [F7] its image is a word $F_{i_d}\cdots F_{i_1}{v}_{{\mathrm{\Lambda }}_0}$, which lower bar fixes since it fixes the vacuum and every $F_i$. These words span each rational weight space. Thus the semilinear projective duality transported by $\varphi$ equals lower bar over $\mathbb{Q}(q)$, and hence on the integral lattice as well; ordinary words need not generate that lattice integrally. Evaluation of a tensor functional gives $D(P^{\psi }{\otimes }_{B}M)={\mathrm{HOM}}_B(P,D_{\psi }M)$, so $\overline{T(P,M)}=h(P,D_{\psi }M)=T(P^{\mathrm{\#}},D_{\psi }M)$. Consequently $\mathrm{\Psi }(D_{\psi }M)(x)=\overline{\mathrm{\Psi }(M)(\stackrel{‾}{x})}$, the adjoint upper bar on the restricted dual. This uses only [F6, F7], so holds in (CP) as well as (C0), without identifying positive-characteristic simples with a global basis.

Completeness of the graded list and uniqueness of shifts. Let $M$ be any finite-dimensional graded simple $B$-module. Its underlying module has a simple submodule, which is isomorphic to the underlying $\dot{D}(\lambda )$ for some restricted $\lambda$ by [F1]. An ungraded embedding $f:\dot{D}(\lambda )\to M$ decomposes into finitely many homogeneous maps $f_r$: for homogeneous $a,v$, comparing degree components in $f(av)=af(v)$ gives $f_r(av)=a{f}_r(v)$. A nonzero $f_r$ has zero kernel since its source is ungraded simple, and its image is a nonzero graded submodule, hence all of $M$. It is therefore a graded isomorphism from the suitable shift of $\dot{D}(\lambda )$. The same decomposition applied to an ungraded isomorphism between two graded lifts proves that they differ by a shift. Distinct labels cannot become isomorphic by [F1]; a nonzero finite-dimensional graded vector space cannot equal a nonzero shift of itself because its finite degree support would have to be invariant under a nonzero translation. Thus the list in (1) is complete and pairwise distinct.

The standard-class bar difference is strictly lower. By [F3], for restricted $\lambda$ the upper expansion is ${\mathrm{Std}}_{\lambda }=G^{\ast }(\lambda )+{\sum }_{\nu ⊲\lambda }{d}_{(\nu ,\lambda )}{G}^{\ast }(\nu )$. Upper bar fixes every $G^{\ast }(\nu )$ and reverses coefficients, so its difference from ${\mathrm{Std}}_{\lambda }$ is a combination of strictly smaller upper vectors. The finite dominance-unitriangular restricted matrix has a triangular inverse over $A$; hence each smaller upper vector is itself a combination of restricted ${\mathrm{Std}}_{\eta }$ with $\eta ⊴\nu ⊲\lambda$. Applying [F3] and step 1.2 gives $[S(\lambda {)}^{⊛}]-[S(\lambda )]\in {\sum }_{\eta ⊲\lambda }A[S(\eta )]$. This is a statement in the upper integral dual lattice, not in an asserted self-dual lower lattice.

The actual graded triangular expansion. A finite graded composition series exists by repeatedly taking a proper graded submodule of maximal dimension. By step 2.1 its factors, after forgetting grading, are the absolutely simple heads of [F1]. Their ungraded multiplicities and dominance support therefore give $[S(\mu )]={\sum }_{\lambda ⊴\mu }{c}_{\lambda \mu }(q)[\dot{D}(\lambda )]$. If $\mu$ is unrestricted, every such restricted $\lambda$ is strictly smaller. If $\mu$ is restricted, the actual quotient map in step 1.1 has degree zero and supplies $\dot{D}(\mu )$ once; ungraded diagonal multiplicity $1$ in [F1] excludes all further shifts of that label, so $c_{\mu \mu }=1$. This proves every clause of (2), without extracting labels or grading shifts from a Grothendieck-group rank count.

The intrinsic head has zero duality shift. By [F6] the orbit of $\dot{D}(\lambda )$ contains a $D_{\psi }$-self-dual representative $L$. Write $\dot{D}(\lambda )\cong L⟨r⟩$, so $\dot{D}(\lambda {)}^{⊛}\cong \dot{D}(\lambda )⟨-2r⟩$; denote this integer shift by $m$. This uses the separately proved normalization supplier, not an inference about a homogeneous cell form or its half-degree. Applying exact duality to the expansion in step 3.1 preserves every smaller label's orbit by the same [F6]. Thus the coefficient of $[\dot{D}(\lambda )]$ in $[S(\lambda {)}^{⊛}]-[S(\lambda )]$ is $q^m-1$, while step 2.2 and the triangular expansions in step 3.1 make it zero. The simple shift-orbit classes are a free $A$-basis (graded Jordan--Hölder, step 2.1), so $q^m=1$ in $\mathbb{Z}[q,q^{-1}]$, forcing $m=0$. This proves the asserted self-duality in both branches and finishes (1).

The (C0) crystal label is the intrinsic head label. In (C0), [F4] gives $[S(\mu )]={\sum }_{\lambda ⊴\mu }{d}_{(\lambda ,\mu )}[D(\lambda )]$ with diagonal $1$. Both the intrinsic heads of step 4.1 and the $D(\lambda )$ are normalized self-dual representatives of the same simple orbits, so their bases differ by a permutation, with no shifts. Induct through a finite linear extension of dominance: after the strictly smaller labels have been identified, subtract their terms from the two expansions of $S(\lambda )$. The remaining leading class is both $[\dot{D}(\lambda )]$ and $[D(\lambda )]$, proving equality and hence graded isomorphism. This establishes only the (C0) crystal-label clause of (3).

The ordinary dictionary and Choice. Independently, the intrinsic nondegenerate form on the ungraded cell head gives its raw-Murphy-star duality by $v\mapsto ⟨v,-⟩$. The Laurent-unit form identity of [F5] identifies the cell radicals and quotients, including the zero case, after every stated field specialization. Through the right-to-left star conversion and involutive sign pullback this gives the ungraded identification $\dot{D}(\lambda )\cong (D_{\mathrm{D}\mathrm{J}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\theta }$ in both branches, with the reflected ordinary residue content described in the Statement. This raw cellular-star comparison is separate from the transported $\psi$ duality used in the categorical duality argument; no equality of those anti-involutions has been inferred. The (CP) crystal-label assertion remains unasserted. All new grading and interpolation arguments are finite; Choice is precisely inherited through [C1].