The q-Schur Weyl weight-space Gram determinant is a hook product

Preserve small alphabets. Enlarge to $D=\max (d,n)$ temporarily. The sum of permutation summands with types supported on the first $d$ letters is a literal summand of the $D$-alphabet permutation module, with projection $p$ and $pS(D,n)p=S(d,n)$. The normalized maps whose two types have that support, their cell multiplication and their Weyl form coefficients are unchanged. Every semistandard shape with these types has at most $d$ rows. Thus proving the weight norm in the enlarged algebra, where the regular corner exists, proves the original norm as well. No hypothesis $d\ge n$ is added.

Compute the standard norms. For the canonical standard tableau, ${\gamma }_{t^{\lambda }}$ telescopes rowwise to $[\lambda ]{!}_Q$, also its norm by $x_{\lambda }^2=[\lambda ]{!}_Q{x}_{\lambda }$. If $s$ dominates $t=s(i,i+1)$, the adjacent action gives $f_t=f_s(T_i+\alpha )$ with $\alpha =1\mathrm{/}[\rho {]}_Q$. Contravariance, orthogonality and the quadratic relation imply $⟨f_t,f_t⟩=(Q+\alpha )(1-\alpha )⟨f_s,f_s⟩$. The two swapped predecessor ratios give ${\gamma }_t=Q^{-1}(Q+\alpha )(1-\alpha ){\gamma }_s$: use $[\rho +1]=1+Q[\rho ]$ and $[\rho -1]=Q^{-1}([\rho ]-1)$. Increasing-length adjacent swaps of incomparable boxes reach every standard tableau, so $⟨f_t,f_t⟩=Q^{\mathrm{\ell }(d(t))}{\gamma }_t$.

Define the cross-weight maps. In the enlarged algebra of step 1.1 let $\omega =(1^n,0^{D-n})$, so $M^{\omega }=H$. Let $a_{\mu }:M^{\mu }\to H$ be inclusion, $a_{\mu }(m_{\mu }h)=m_{\mu }h$, and let $b_{\mu }:H\to M^{\mu }$ be $b_{\mu }(h)=m_{\mu }h$; both are zero on the other permutation summands. These are actual $H$-linear maps, not maps depending on an unspecified $T$. Under the Hom/intersection coordinate identification of the q-Schur construction, both maps have image coordinate $m_{\mu }$. Since $m_{\mu }^{\ast }=m_{\mu }$, the index-swapping star in [F1] gives $a_{\mu }^{\ast }=b_{\mu }$; this works for compositions as well as partitions. For a standard $s$, composition of its Weyl map with $a_{\mu }$ evaluates at $m_{\mu }$ as $m_s{m}_{\mu }$, whereas composition of a type-$\mu$ Weyl map with $b_{\mu }$ evaluates at $1$ as $m_T$. We use right-action composition throughout.

Establish the support and the initial scalar. Write $m_s=m_{\lambda }{T}_{d(s)}$ in the Murphy cell. For $s=\mathrm{first}(T)$, $d(s)$ is the least-length double-coset representative for the row blocks of $\lambda$ and $\mu$. The intersection subgroup has one symmetric-group factor of size $c_{iv}=\mathrm{\mid }\{(i,j):T(i,j)=v\}\mathrm{\mid }$ for every nonempty row/label intersection. Decomposing both row subgroups into their distinguished intersection cosets, every reduced coset word occurs once; the remaining intersection sum acts on $m_{\lambda }$ by its row character $Q^{\mathrm{\ell }(w)}$. Hence the exact identity is $m_s{m}_{\mu }=({\prod }_{i,v}[c_{iv}]{!}_Q)m_T\mathrm{.}$ No factorial is divided out. The same-entry predecessor ratios for this first refinement are $[2]\mathrm{/}[1],[3]\mathrm{/}[2],\dots ,[c_{iv}]\mathrm{/}[c_{iv}-1]$ within each intersection row, so their product ${\pi }_s$ is exactly that factorial. We also need support after multiplication by $m_{\mu }$. Filter standard Murphy vectors by the shapes of their prefixes at the ends of the $\mu$ label blocks. Every generator $T_i$ inside such a block preserves this filtration: its row case is multiplication by $Q$; its standard-swap terms have the same block-end prefixes; in its column case the Garnir correction terms have larger tableau-prefix dominance. The last assertion follows by isolating the distinguished term in the length-additive row-coset Garnir identity; all other same-shape terms increase prefix dominance, as proved in Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations. Iterating over each word in $m_{\mu }$ shows that $m_v{m}_{\mu }$ has only coarsened types dominating $\mu (v)$. Its expansion has constant coefficients on semistandard fibers and zero coefficients elsewhere, by the right-hand (starred) permutation-module basis in that supplier. Thus only semistandard $U⊵\mu (v)$ occur, strictly if $\mu (v)$ is not semistandard. In particular, if $v▹\mathrm{first}(T)$, every surviving $U$ is strictly above $T$: equality would put $v$ in the fiber of $T$, whose most dominant member is $\mathrm{first}(T)$. This proves the needed support assertion including non-semistandard types.

Determine the scalar on every standardization. Sums of JM elements over a label block commute with that block's row symmetrizer: the adjacent JM relation makes $L_i+L_{i+1}$ commute with $T_i$, and every other $L_j$ commutes with it. The dominance-triangular standard JM action and the support filtration of step 3.1 consequently give triangular block-sum operators on the semistandard maps, with diagonal entries ${\sum }_{T(i,j)=v}[j-i{]}_Q$. Their joint eigenvalues distinguish $T$: the Laurent coefficients and the known prefix sizes recover every initial-label content multiset, whose diagonal counts recover its Young diagram. They are self-adjoint, so finite triangular diagonalization gives orthogonal ${\psi }_T={\phi }_T+$ strictly more dominant terms. The same block-sum identities show that $f_t{a}_{\mu }$ lies in the $T$ joint eigenspace whenever $\mu (t)=T$, hence equals $r_t{\psi }_T$. At $s=\mathrm{first}(T)$, expand $f_s=m_s+$ more dominant standard vectors. Step 3.1 gives coefficient ${\pi }_s$ at ${\phi }_T$ and only higher types in the remaining terms. Thus $r_s={\pi }_s$. Now connect standardizations of $T$ by increasing-length adjacent swaps inside each label block; these are the linear extensions of the repeated-label node posets, and bubble sorting gives such a chain from their most dominant extension. If $t=s(i,i+1)$ is one step, [F2] gives $f_t=f_s(T_i+\alpha )$, $\alpha =1\mathrm{/}[\rho {]}_Q$, while $T_i{a}_{\mu }=Q{a}_{\mu }$ because $T_i{m}_{\mu }=Q{m}_{\mu }$. Therefore $r_t=(Q+\alpha )r_s$. The only newly included same-entry predecessor pair has content difference $\rho$, so its ratio is $[\rho +1{]}_Q\mathrm{/}[\rho {]}_Q=Q+\alpha$, proving ${\pi }_t=(Q+\alpha ){\pi }_s$. Induction proves the exact identity $f_t{a}_{\mu }={\pi }_t{\psi }_T$ for every standardization.

Determine the last-standardization coefficient and the norm. For $t=\mathrm{last}(T)$, evaluation at $1$ gives ${\phi }_T{b}_{\mu }=m_T={\sum }_{\mu (v)=T}{m}_v$. Every other member of this fiber is more dominant than $t$. Also every refinement of $U▹T$ dominates $t$: compare at label-block ends, then use that $t$ is the least-dominant refinement within its own blocks. Consequently the two unitriangular basis changes give ${\psi }_T{b}_{\mu }=f_t+$ more dominant standard eigenvectors, with coefficient exactly $1$ at $f_t$. At this last refinement the standard predecessor set partitions into the strict semistandard boundary ${\mathrm{\Gamma }}_T$ and same-entry predecessors, so ${\gamma }_t={\pi }_t{\gamma }_T$. By adjointness of the explicit maps in step 2.1 and orthogonality, ${\pi }_t⟨{\psi }_T,{\psi }_T⟩=⟨f_t{a}_{\mu },{\psi }_T⟩=⟨f_t,{\psi }_T{b}_{\mu }⟩=Q^{\mathrm{\ell }(d(t))}{\gamma }_t\mathrm{.}$ All factors of ${\pi }_t$ are nonzero over $K$, so $⟨{\psi }_T,{\psi }_T⟩=Q^{\mathrm{\ell }(d(t))}{\gamma }_T$. The transition determinant is $1$, yielding $G_{\mu }^{\lambda }=Q^N{g}_{\mu }^{\lambda }$ with $g_{\mu }^{\lambda }={\prod }_T{\gamma }_T$. This calculates the cross-weight scalar rather than inferring it from the standard norm alone.

Prove the normalized beta product at the level of formal exponents. For $h\ge 1$ put $A_h(\beta ,\mu )=-{\sum }_{b\mathrm{\ne }c,{\beta }_b-h<{\beta }_c}{d}_{\mu }(\beta -h{\mathbf{e}}_b+h{\mathbf{e}}_c)\mathrm{.}$ A nonzero term requires $0\le {\beta }_b-h$, so the product is finite. Repeated beta entries cancel by exchanging the equal indices; a negative entry cannot become a valid term because, if it is the receiver, the lowered donor is still smaller. Thus these invalid sequences contribute zero exponents. Delete the largest semistandard label, occurring $z$ times in column feet $C$, to get $\beta -{\mathbf{1}}_C,\bar{\mu }$. The strict predecessor boundary excludes every column also in $C$. Consequently the literal exponent $E_h$ in $g$ obeys $E_h(\beta ,\mu )={\sum }_{\mathrm{\mid }C\mathrm{\mid }=z}{E}_h(\beta -{\mathbf{1}}_C,\bar{\mu })+{\sum }_{\mathrm{\mid }C\mathrm{\mid }=z}{\sum }_{b\mathrm{\notin }C,c\in C}({\mathbf{1}}_{{\beta }_b-{\beta }_c+1=h}-{\mathbf{1}}_{{\beta }_b-{\beta }_c=h})d_{\bar{\mu }}(\beta -{\mathbf{1}}_C)\mathrm{.}$ The final sum is exactly the axial ratio contributed by each deleted largest-label box.

Verify the same recurrence for $A_h$. Apply [F3] to its summands and compare with ${\sum }_C{A}_h(\beta -{\mathbf{1}}_C,\bar{\mu })$. Set $\delta ={\beta }_b-{\beta }_c-h$. The original condition is $\delta <0$ and the deleted one $\delta -{\mathbf{1}}_C(b)+{\mathbf{1}}_C(c)<0$. They differ only when $b\in C,c\mathrm{\notin }C,\delta =0$ or $b\mathrm{\notin }C,c\in C,\delta =-1$. In the first case, exchanging the two shifted beta entries turns the contribution into $-d_{\bar{\mu }}(\beta -{\mathbf{1}}_{C^{\mathrm{\prime }}})$ with $C^{\mathrm{\prime }}=C-b+c$, giving the negative term of step 6.1. In the second case, the exchanged entries give $+d_{\bar{\mu }}(\beta -{\mathbf{1}}_C)$, giving its positive term. Every other case cancels. At degree $0$ both exponent families vanish: a nonzero shifted empty beta sequence would have to exchange two entries with $h={\beta }_b-{\beta }_c$, excluded by the strict condition. Induction gives $E_h=A_h$ and therefore $g_{\mu }^{\lambda }={\prod }_{h\ge 1}{\prod }_{b\mathrm{\ne }c,{\beta }_b-h<{\beta }_c}[h{]}_Q^{-d_{\mu }(\beta -h{\mathbf{e}}_b+h{\mathbf{e}}_c)}\mathrm{.}$ This proves the regrouping without copying the inconsistent signs in the printed display.

Convert to hooks. A surviving lowered bead ${\beta }_b-h$ is a nonnegative gap, so the bead-gap correspondence makes $h=h_{ab}$ for a node $(a,b)$. The remaining condition ${\beta }_c>{\beta }_b-h_{ab}=n-1-{\lambda }_a+a$ is equivalent to $c\le {\lambda }_a$; the paired nodes therefore lie in the same row. For $b<c$ write $h=h_{ab}$ and $k=h_{ac}$. Since $h-k={\beta }_b-{\beta }_c$, the two ordered displaced sequences $\beta -h{\mathbf{e}}_b+h{\mathbf{e}}_c$ and $\beta -k{\mathbf{e}}_c+k{\mathbf{e}}_b$ exchange their two changed entries. Their signed numbers are negatives of each other, and their factors are exactly $([h{]}_Q\mathrm{/}[k{]}_Q{)}^{d_{\mu }(\beta +k{\mathbf{e}}_b-k{\mathbf{e}}_c)}$. This proves the stated balanced hook product. For the strip interpretation, first lower ${\beta }_c$ to the gap $g={\beta }_c-k=n-1-{\lambda }_a+a$, whose sorted position is $r\ge c$. Recovering conjugate parts from the sorted sequence replaces ${\lambda }_j^{\mathrm{\prime }}$ by ${\lambda }_{j+1}^{\mathrm{\prime }}-1$ for $c\le j<r$ and by $g-n+r$ for $j=r$, leaving other columns unchanged. The removed nodes form the rim strip $R_{ac}$ in columns $c,\dots ,r$, with $k$ boxes. Since $b<c$, its original bead ${\beta }_b$ is unaffected and is still in position $b$ in this intermediate diagram $\kappa$. Raising it to ${\beta }_b+k$ and sorting puts it in position $s\le b$; applying the same column calculation in reverse shows that $\eta \mathrm{/}\kappa$ is a $k$-strip in columns $s,\dots ,b$, with lowest node in column $s$. The two sorting permutations make $r-c$ and $b-s$ crossings. The strip-height counting identity gives heights $k-1-(r-c)$ and $k-1-(b-s)$, so their sum has the same parity as the total crossings and gives exactly the stated sign. Colliding or negative sequences vanish by definition.

Specialize integrally. The semistandard form matrix has entries in $\mathbb{Z}[Q,Q^{-1}]$, so its determinant and $Q^{-N}{G}_{\mu }^{\lambda }$ are Laurent polynomials. Steps 5.1 and 7.1 identify the rational products with these same polynomials in $\mathbb{Q}(Q)$. Base change of the free semistandard basis and its form now proves the identity over the given domain, including singular specializations and weights of dimension zero. No individual vanishing ratio is evaluated. All constructions are finite; no choice principle is used.