The q-Schaper calculation for (3,2) at e = 2

Check the lift. Localization makes $\widehat{q}$ a unit since its residue is $-1$, so the exact PID hypotheses of [F1] hold; the generic factorial is nonzero. The identities $[2]=1+\widehat{q}$, $[4]=(1+\widehat{q})(1+{\widehat{q}}^2)$ and evaluation of the odd quantum integers at $-1$ prove [F2]. These values refer to this lift, not an arbitrary field or lift with quantum characteristic $2$.

Compute the balanced symbols explicitly. For $(3,2)$, $\beta =(6,5,3,1,0)$ and its hook array is $\begin{array}{ccc}4& 3& 1\\ 2& 1\end{array}$. The pair $(1,1),(1,2)$ gives $(9,2,3,1,0)$, which sorts to the beta sequence of $(1^5)$ with sign $-1$ and coefficient $v(4)-v(3)=1$. The pair $(1,1),(1,3)$ gives $(7,5,2,1,0)$, shape $(2,2,1)$ with sign $+1$ and coefficient $v(4)-v(1)=1$. The pair $(2,1),(2,2)$ gives $(7,4,3,1,0)$, shape $(3,1^2)$ with sign $+1$ and coefficient $v(2)-v(1)=1$. The remaining pair has repeated beta entries and vanishes. Thus the first displayed Weyl identity follows by direct substitution.

Verify the two zero forms directly at $q=-1$. For each $\eta \in \{(3,2),(3,1,1)\}$, write $v_j$ for the Murphy vector whose row-reading word is, respectively, $12345,12435,12534,13425,13524$ for $j=0,\dots ,4$; the shape $(3,1,1)$ has the additional word $14523$, giving $v_5$. The row action gives $v_0{T}_1=v_0{T}_2=-v_0$, $v_1{T}_2=v_3$ and $v_2{T}_2=v_4$. The first-column Garnir belt between the first two rows gives $v_3{T}_1=-v_0-v_1-v_3$: its row-standard terms have lower first entry $1,2,3,4$, and its other side has shape $(4,1)$, strictly above either $\eta$. Also $v_3{T}_4=v_4$, $v_1{T}_4=v_2$ and $v_0{T}_4=-v_0$. The last identity is the row action for $(3,2)$; for $(3,1,1)$ it follows from $x_{(3,1,1)}(1+T_4)=x_{(3,2)}$, a higher-shape element. Since $T_1{T}_4=T_4{T}_1$, these identities give $v_4{T}_1=v_0-v_2-v_4$. These are finite row-action and Garnir calculations from the supplied integral Murphy construction.

Put $r_j=⟨v_0,v_j⟩$. Self-adjointness of $T_1,T_2$ and $v_0{T}_i=-v_0$ imply $⟨v_0,v_j{T}_i⟩=-r_j$. The four identities of step 1.3 consequently give $r_1=-r_0$, $r_2=r_0$, $r_3=-r_1$ and $r_4=-r_2$. But $r_0=[\eta ]{!}_{-1}=0$, since $x_{\eta }^2=[\eta ]{!}_q{x}_{\eta }$ and $[3]{!}_{-1}$ contains $[2{]}_{-1}=0$. Thus $r_0,\dots ,r_4$ are zero for both shapes.

Compute the smaller rows rather than assuming them. The same calculation for $(3,1^2)$ gives ${\sum }_{i>0}[W^{(3,1^2)}(i)]=[W^{(2,2,1)}]+[W^{(1^5)}]$: the pairs with hook lengths $(5,2)$ and $(2,1)$ give those two terms, while the $(5,1)$ coefficient is zero. For $(2,2,1)$, the only nonzero symbols arise from pairs $(4,2)$ and $(3,1)$, both of valuation difference zero. For $(2,1^3)$ the sole pair is $(5,1)$, also zero; $(1^5)$ has no pair. Hence these three Weyl modules have zero positive sums and zero radicals, so they equal their simple heads. For $(3,1^2)$ each displayed simple occurs once in the positive sum; because its first submodule is the radical and later ones lie inside it, both radical multiplicities are exactly $1$. Its class is consequently $[L^{(3,1^2)}]+[L^{(2,2,1)}]+[L^{(1^5)}]$. Substitution into step 1.2 now gives $[L^{(3,1^2)}]+2[L^{(2,2,1)}]$, proving the stated preliminary multiplicity conclusions without reading a signed individual Weyl coefficient as a simple multiplicity.

For the remaining vector of $(3,1,1)$, put $x=x_{(3,1,1)}$. Its distinguished inverse word is $T_3{T}_4{T}_2{T}_3$. Expanding the finite row symmetrizers with $T_i^2=-2T_i-1$ and the braid relations gives the exact identity $x{T}_3{T}_4{T}_2{T}_{3}x=x_{(5)}-(1+T_4)x_{(4,1)}(1+T_4)-x_{(3,2)}\mathrm{.}$ Every term on the right belongs to the higher dominance ideal for $(3,1,1)$, so the cell-product coefficient is $r_5=0$. Together with step 2.1, this puts the canonical vector $v_0$ in the form radical of each cell. That radical is a submodule, and $v_{0}H$ is the entire Murphy cell, since its standard basis is $x_{\eta }{T}_{d(t)}$ modulo the higher ideal. Hence both forms are identically zero. Their Schur-simple images are zero by the exact radical-corner identity in [F3]. This uses no general Gram-entry divisor or finite-$e$ classification.

Supply the actual dimension argument. The regular-weight Gram product for the Murphy cell $(2,2,1)$ is ${\widehat{q}}^7\frac{[4{]}_{\widehat{q}}}{[2{]}_{\widehat{q}}}[3{]}_{\widehat{q}}^4={\widehat{q}}^7(1+{\widehat{q}}^2)[3{]}_{\widehat{q}}^4,$ since its two signed symbols have shapes $(1^5)$ and $(2,1^3)$ with standard-tableau counts $1$ and $4$. The five standardizations have lengths $0,1,1,2,3$, giving exponent $7$. At $\widehat{q}=-1$ this determinant is $-2\mathrm{\ne }0$. Thus $S_{\mathrm{M}}^{(2,2,1)}=D_{\mathrm{M}}^{(2,2,1)}$ is simple of dimension $5$, counted by its five standard tableaux. Both Schur simples $L^{(3,1^2)}$ and $L^{(3,2)}$ are killed by the local zero-form calculation of step 3.1 and the exact corner/form identity [F3]. Applying the functor to the positive sum of step 2.2 therefore gives ${\sum }_{i>0}[S_{\mathrm{M}}^{(3,2)}(i)]=2[D_{\mathrm{M}}^{(2,2,1)}]$. The form of $(3,2)$ is identically zero by step 3.1, so its first submodule is the whole five-dimensional cell. Every composition factor of that cell must appear in the positive sum, hence must be this five-dimensional simple. Dimension forces exactly one factor, proving $S_{\mathrm{M}}^{(3,2)}\cong D_{\mathrm{M}}^{(2,2,1)}$. Exact projection of a composition series, with distinct nonzero simple images as in [F3], gives $d_{(3,2),(2,2,1)}=1$.

A further consistency check is the regular-weight Gram product ${\widehat{q}}^8[3{]}_{\widehat{q}}[4{]}_{\widehat{q}}^4[2{]}_{\widehat{q}}^6$ for $(3,2)$, whose valuation is $10$, agreeing with two five-dimensional positive Specht submodules. This is a check of the complete balanced formula; it is not the unsupported full-hook product used in the former classification proof.

Translate simultaneously by $\theta$ and conjugation. The actual ordinary/Murphy module-and-form comparison sends the isomorphism of step 4.1 to $S_{\mathrm{D}\mathrm{J}}^{(2,2,1)}\cong D_{\mathrm{D}\mathrm{J}}^{(3,2)}$. The partition $(2,2,1)$ is not regular, so its own cell-form quotient is zero; this does not say its actual head is zero, since the module itself is simple. The dominance inequality $(3,2)⊵(2,2,1)$ agrees with the ordinary-convention unitriangular bound. This proves every claimed numeric conclusion.