q-Schur Weyl modules are in the same block iff their labels have the same e-core

Lift and valuation. Use $R=F[X{]}_{(X-q)}$ with $\widehat{q}=X$ and uniformizer $X-q$. The element $X$ is a unit since $q\mathrm{\ne }0$, and every polynomial $[m{]}_X=1+X+\cdots +X^{m-1}$ is nonzero, so the generic factorial is nonzero. For $n=0$, $S_F(d,0)=F$, with its one empty-label module, so all assertions hold directly; hence assume $n>0$. Its valuation $v(m)$ is positive exactly when $[m{]}_q=0$, equivalently $e\mid m$ for finite $e$; for $e=\mathrm{\infty }$ all valuations vanish. In characteristic $p$ the positive valuation may exceed1, which causes no difficulty below.

Block support and Jantzen edges. A finite-dimensional module with simple head is indecomposable: every nonzero summand has nonzero head, so two summands would make the head nonsimple. Thus $W^{\lambda }$ has just one nonzero block summand, and every Jantzen submodule is supported in that block. By [F4], any nonzero accumulated coefficient $J_{\lambda \mu }$ of $[W^{\mu }]$ in [F2] puts $W^{\lambda },W^{\mu }$ in the same block. This uses the sum of submodule classes, without identifying it with the radical class.

Balanced terms preserve the core and descend in dominance. Put $s=h_{ab}$ and $t=h_{ac}$, so $s-t={\beta }_b-{\beta }_c$. If a hook coefficient is nonzero, at least one of $s,t$ is divisible by $e$. If $e\mid t$, the shifted bead residues are unchanged. Otherwise $e\mid s$, so $t\equiv {\beta }_c-{\beta }_b(\mathrm{m}\mathrm{o}\mathrm{d}e)$ and the two shifted residues exchange. Thus every surviving symbol has the same runner counts and core as $\lambda$. Moreover replacing the two conjugate beta entries $u>v$ by $u+t,v-t$ strictly increases conjugate majorization; the sorted target therefore strictly decreases original dominance. Accumulating equal target labels preserves both conclusions.

No outgoing edge implies a binary row condition. Suppose every accumulated outgoing coefficient from $\mu$ is zero. For row $i$, put $g_i=N+i-1-{\mu }_i$, a gap in its conjugate beta set; its row hooks are exactly ${\beta }_j-g_i>0$. If that row contained a divisible hook $s$ and a nondivisible hook $t$, repeatedly replace $t$ by $s+t$ while their sum is another row hook. This finite climb ends with $s+t$ absent from the hook set. The balanced term for the nodes of hook lengths $s,t$ then has lower shifted bead $g_i$, which is a gap, and upper shifted bead $g_i+s+t$, which is absent from the beta set. Its symbol survives and its coefficient is $\pm v(s)\mathrm{\ne }0$. There is no cancellation: any surviving balanced term replaces two lost values by a new value larger than both and a new value smaller than both, so those four values determine the pair and its row gap uniquely; the gaps $g_i$ strictly increase with $i$. This contradicts the zero accumulated coefficient. Therefore each row has either all hook lengths divisible by $e$ or none; call this the binary row condition.

Composition-factor cores. An off-head simple occurs in $W^{\lambda }$ exactly when it occurs in its radical, equivalently in the positive Jantzen sum: the first submodule is the radical and every later one is contained in it. Its nonzero multiplicity in that sum implies it occurs in some $W^{\mu }$ with $J_{\lambda \mu }\mathrm{\ne }0$, since otherwise its coefficient in the sum would be zero. Step 2.1 gives the same core and strict dominance descent. Induction on the finite dominance poset proves that every factor $L^{\nu }$ of $W^{\lambda }$ has $C_e(\nu )=C_e(\lambda )$. No identity equating the radical class with the positive sum is asserted.

Vertical core removal under the binary row condition. A removable $e$-hook has hook length $e$ at its defining node. Every hook in that row is then divisible, so its last-column hook, a positive multiple of $e$ no larger than $e$, equals $e$. Hook lengths strictly decrease across the row, forcing the defining node to be at the row end: the strip is vertical on $e$ equal-length rows $i,\dots ,i+e-1$ in column $j$. Removing it decreases that column height by $e$. In unaffected rows hook divisibility is unchanged. For an affected row $i+t$, hooks in the retained columns were all congruent to $-t$ modulo $e$, because row $i$ had all hooks divisible and those rows had equal length. Their new hook lengths are all congruent to $-t-1$, so the binary row condition persists. Induction shows every core-removal strip is vertical.

The Cartan graph locally. Choose a complete orthogonal primitive idempotent system of $S$, with one representative $f_{\mu }$ for each simple label and repeated representatives as necessary. By [F5] and the absolute simplicity in [F1], the semisimple quotient is a product of full matrix rings over $F$. Lift its matrix diagonal idempotents: the radical is nilpotent and the iteration $u\mapsto u+(1-2u)(u^2-u)$ changes an error $a=u^2-u$ to $-3a^2+4a^3$, so finite iteration lifts an idempotent; successively working in complementary corners lifts an orthogonal system. Absolute simplicity gives $\dim L^{\lambda }{f}_{\mu }={\delta }_{\lambda \mu }$, hence exact multiplication by $f_{\mu }$ gives $\dim W^{\lambda }{f}_{\mu }=d_{\lambda \mu }$. The successive cellular ideal bimodules are the contragredient cell module tensor the right cell module. Apply the exact projection $f_{\mu }(-)f_{\nu }$ and add their dimensions; [F1]'s duality fixes the simple labels, giving $\dim f_{\mu }S{f}_{\nu }={\sum }_{\lambda }{d}_{\lambda \mu }{d}_{\lambda \nu }$. This is the Cartan entry of the primitive projective $f_{\mu }S$. A disconnected graph on the full primitive idempotent system, with edges whenever either mixed corner is nonzero, yields central idempotents by summing each component: all mixed corners vanish in both directions. Conversely a central block idempotent forbids mixed corners. Thus simple labels in one block are joined by pairs sharing a Weyl factor, and step 3.1 proves that labels in the same block have the same core.

The minimum label and the bridge shapes. Assume $e<\mathrm{\infty }$. Fix core $\gamma$ and weight $w$, with $l=\mathrm{\ell }(\gamma )$, and set $\rho =({\gamma }_1,\dots ,{\gamma }_l,1^{ew})$. It has core $\gamma$ by removing its bottom vertical $e$-strips, and is dominance-minimal in the class: every same-core partition contains $\gamma$ as a diagram, and appending all additional boxes as rows of size1 minimizes each dominance partial sum. Let $\mu \mathrm{\ne }\rho$ satisfy the binary row condition. Choose the first row $i$ with divisible hooks and put $j={\mu }_i$, $h=h_{ij}^{\mu }={\mu }_j^{\mathrm{\prime }}-i+1$, a positive multiple of $e$. If $j=1$, every divisible hook is in column1, so step 3.2 would give $\mu =\rho$; thus $j>1$. Remove the vertical $h$-strip, obtaining $\sigma$, then add $h$ boxes horizontally to its first row to get $\nu$, and vertically to its first column to get $\eta$. All have the same core, since $h$ is a multiple of $e$; $\nu$ dominates $\mu ,\eta$, and $\eta$ is strictly below $\mu$ because the removed nodes are moved into new bottom rows.

The bridge coefficients are nonzero. Write $l=\mathrm{\ell }(\mu )=\mathrm{\ell }(\sigma )$. If $i>1$, then ${\sigma }_1={\mu }_1$ and the row1 hook pair in $\nu$ at columns $j,{\mu }_1+1$ has lengths $A=h_{1j}^{\mu }$ and $h$: the first is nondivisible because row1 precedes the first divisible row, and the second is divisible. Its balanced symbol sorts to $\mu$, so its coefficient is nonzero. If $i=1$, then ${\sigma }_1=j-1$, and the row1 pair at columns $j,j+1$ has lengths $h,h-1$; its balanced transfer $h-1$ sorts to $\mu$, again giving a nonzero coefficient. For $\eta$, the pair at columns $1,{\sigma }_1+1$ has lengths $A^{\mathrm{\prime }}={\sigma }_1+h+l-1$ and $h$. When $i>1$, $A^{\mathrm{\prime }}-h_{i1}^{\mu }=A$, so $A^{\mathrm{\prime }}$ is nondivisible; when $i=1$, $A^{\mathrm{\prime }}=h_{11}^{\mu }+h-1$, also nondivisible. Its balanced symbol sorts to $\eta$. Each claimed target has precisely two lost and two gained beta values, so the uniqueness argument in step 2.2 rules out cancellation. Thus both $J_{\nu \mu }$ and $J_{\nu \eta }$ are nonzero.

Connectivity and sufficiency. Induct upwards from $\rho$ in the finite dominance poset. For $\mu \mathrm{\ne }\rho$, if an outgoing coefficient $J_{\mu \tau }$ is nonzero, then $\tau$ is strictly below $\mu$ in the same core class by step 2.1, so induction connects $\tau$, and hence $\mu$, to $\rho$. If no outgoing coefficient exists, steps 2.2, 3.2, 4.2 and 5.1 connect $\mu$ to $\nu$ and then $\eta$, with $\eta$ strictly below $\mu$, and induction again connects it to $\rho$. Every such edge is a same-block implication by step 1.2. Thus every pair of labels of the same core lies in the same block. Together with step 4.1 this proves the equivalence.

Parameterization and infinite characteristic. Every simple is the head of a Weyl module, so every block has a partition label. The core sizes are $n-ew$, and each admissible core occurs via the partition $\rho$ of step 4.2; step 6.1 therefore gives the stated indexing. If $e=\mathrm{\infty }$, all coefficients vanish by step 1.1, so positivity forces every first Jantzen submodule, hence every radical, to vanish. All Weyl modules are simple; their decomposition matrix is the identity. Step 4.1 then gives identity Cartan matrix, so every primitive projective is simple and $S$ is semisimple, with each partition its own block and core. This proves all assertions.