Cyclotomic Schur weight Gram valuations equal signed rim-hook Kostka sums

Characters as finite symmetric polynomials. Fix $d\ge n$ row labels per type component with ordered variables $x_{(a,r)}$, put $Y_a=\{x_{(b,r)}:b\ge a\}$ with the inherited order, and for a shape $\lambda$ let ${\mathrm{ch}}_{\lambda }(x)={\sum }_{T\in \mathrm{SStd}(\lambda )}{x}^{\mathrm{wt}(T)}$. The flag condition says that the physical component $a$ of a semistandard tableau is an ordinary semistandard tableau over the alphabet $Y_a$, and the components are chosen independently, so ${\mathrm{ch}}_{\lambda }(x)={\prod }_a{s}_{{\lambda }^{(a)}}(Y_a)$ with $s_{\rho }(Y)=\det (h_{{\rho }_i-i+j}(Y))$. The determinant equals the ordinary semistandard generating polynomial: expand it as signed families of east/north lattice paths from $(-j,1)$ to $({\rho }_i-i,m)$ with east steps at height $r$ weighted $y_r$, choose the first intersection in a fixed order and exchange the two tails, an involution that preserves the monomial weight and reverses the permutation sign, so all intersecting families cancel; a nonintersecting family forces the identity matching, and the surviving families are exactly the semistandard tableaux, sign $+1$. Hence the coefficient of $x^{\nu }$ in ${\mathrm{ch}}_{\lambda }$ is the Kostka number $K_{\lambda \nu }$; empty and short-row cases are included.

Beta determinants and rim-hook signs. Choose $N\ge n$, put ${\beta }_{a,i}={\lambda }_i^{(a)}+N-i$ with zero parts appended, and for a nonnegative integer array $v$ define $\chi (v;x)={\prod }_a\det (h_{v_{a,i}-N+j}(Y_a){)}_{i,j}$. If two entries of one component of $v$ agree the determinant has equal rows and is zero; otherwise sorting that component into strictly decreasing order and writing ${\rho }_i=v_i^{\mathrm{\prime }}-N+i$ gives $\chi (v;x)=\epsilon (v){\prod }_a{s}_{{\rho }^{(a)}}(Y_a)$ with $\epsilon$ the product of the row-sorting signs, so every coefficient is a signed Kostka multiplicity. A beta move lowering a bead from $v+L$ to an unoccupied $v$ is exactly the removal of a rim hook of length $L$: tracing the southeast boundary shows the strip between the two boundary steps has $L$ boxes, its leg is the number of occupied positions strictly between $v$ and $v+L$, its foot content is $v-N+1$, and the associated sorting sign is $(-1{)}^{\mathrm{leg}}$; the correspondence is a bijection by the reverse move.

The hook side as a complete finite alternating polynomial. Define $B_{\lambda }(x)$ as the finite alternating sum of step 2.1: for crosscomponent terms, over $a<b$, over beads $i,j$ and lengths $h\le {\beta }_{a,i}$, of $\chi (\beta -h{e}_{a,i}+h{e}_{b,j};x)V(U_a{z}^{{\beta }_{a,i}-h}-U_b{z}^{{\beta }_{b,j}})$; and for same-component terms, over $a$, $i<j$ and $1\le h\le {\beta }_{a,i}-{\beta }_{a,j}-1$, of $\chi (\beta -h{e}_{a,i}+h{e}_{a,j};x)V(U_a{z}^{{\beta }_{a,i}-h}-U_a{z}^{{\beta }_{a,j}})$; terms whose $\chi$ is the zero polynomial are omitted before any valuation is evaluated. Then $B_{\lambda }(x)={\sum }_{\mu :\lambda ▹\mu }{J}_{\lambda \mu }{\mathrm{ch}}_{\mu }(x)$: a crosscomponent move removing from the earlier component $a$ and adding to a later component $b>a$ produces a strictly dominated $\mu$ and occurs exactly once by the bead description; occupied targets give repeated rows and zero; sorting contributes $(-1{)}^{\text{removed leg}+\text{added leg}}$; and the prefactor $z^{-N+1}$ converts the beta difference to the difference of the deformed foot residues. In one component the two moved beads $i<j$ give a strict dominance decrease precisely for $1\le h<{\beta }_i-{\beta }_j$, the exchange case $h={\beta }_i-{\beta }_j$ giving $\mu =\lambda$ and the outward case giving the reverse dominance; occupied intermediate positions again give zero rows, and the range exceeding a component's size gives a repeated or negative bead, hence a zero determinant. Thus no hook pair is lost or added, and the surviving summands are exactly the $J_{\lambda \mu }$ terms.

Nonvanishing of every retained foot difference. For a removable rim hook of length $L$ from a partition of size $M+L$ with foot $(h,j)$, one has $-(M+L-1)\le j-h\le M$: every box left of the foot in its row survives, so $j-1\le M$, and $h\le M+L$ gives the lower bound. If $\lambda$ and $\mu$ remove equal-length hooks in distinct components $a\mathrm{\ne }b$ to the same remainder with remainder sizes $M_a,M_b$, their foot contents satisfy $\mathrm{\mid }{\gamma }_x-{\gamma }_y\mathrm{\mid }\le M_a+M_b+L-1\le n-1$, so the exponent $n(a-b)+{\gamma }_x-{\gamma }_y$ is a nonzero integer; in the same component equal foot content and equal length reconstruct the same inverse hook addition, hence would force $\lambda =\mu$, which the strict-dominance terms exclude. Therefore every retained Laurent difference $U_a{z}^u-U_b{z}^v$ is nonzero in $F(t)$: it is a unit times a nonzero Laurent monomial times $(z^N-1)$-type factors and, by [F2], its valuation is finite, being $0$ when the constant residues differ, $1$ in characteristic zero when they agree, and $p^{v_p(N)}$ in characteristic $p$ when they agree.

The same-component operator. In the same-component beta sum put $D={\beta }_i-{\beta }_j$ and $k=D-h$, so the valuation is $f(k)=V(z^k-1)$ up to unit factors; swapping the two changed determinant rows shows $\chi (\beta -(D-k)e_i+(D-k)e_j;x)=-\chi (\beta -k{e}_i+k{e}_j;x)$. Hence the complete coefficient of $f(k)$ is $-{\sum }_{i<j:{\beta }_i-{\beta }_j>k}\chi (\beta -k{e}_i+k{e}_j;x)$, which is the strictly dominance-lowering part of $-p_k{p}_k^{\perp }$ on the abstract shape basis, where $p_k^{\perp }$ lowers an occupied bead by $k$ into an unoccupied nonnegative position with sign $(-1{)}^{\mathrm{\#}\text{crossed beads}}$ and $p_k$ raises a bead by $k$ with the same sign rule; the bead-raising formula follows from the alternant identity $\det (y_r^{{\beta }_i}){\sum }_r{y}_r^k={\sum }_i\det (y_r^{{\beta }_i+k{\delta }_{ij}})$ divided by the Vandermonde, and the removal map is its transpose. After one removal and one addition a non-diagonal lower-dominance result comes only from distinct beads $i<j$ with $0<k<{\beta }_i-{\beta }_j$, occupied intermediate positions giving zero repeated rows and the diagonal being removed by the strict projection; this proves the operator identification integrally, in every characteristic.

The crosscomponent operator as exterior currents. On the $N$-fold exterior power of a space with basis $e_0,e_1,\dots$, represent a partition by the ordered wedge of its beta beads and let $E(p,q)$ insert $e_p$ after removing $e_q$, with exterior signs; it vanishes if $q$ is absent or $p$ is occupied, and its sign is the bead-sorting sign. Expanding $E=\text{creation}\cdot \text{contraction}$ gives $[E(p,q),E(r,s)]={\delta }_{q,r}E(p,s)-{\delta }_{p,s}E(r,q)$. The crosscomponent matrix acting on column shape vectors is $J^{\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}}={\sum }_{a<b}{\sum }_{h>0}{\sum }_{q\ge h,w\ge 0}V(U_a{z}^{q-h}-U_b{z}^w)E_a(q,q-h)E_b(w,w+h)$, a finite operator on each size-$n$ sector; the reverse hook pair has forward current $E_a(q-h,q)E_b(w+h,w)$, and transposing gives exactly this operator with the same signs and valuation.

Horizontal strips and their conjugation. With $S{e}_q=e_{q+1}$ and $H(x)=(I-xS{)}^{-1}={\sum }_{r\ge 0}{x}^r{S}^r$, whose entry $H_{p,q}$ is $x^{p-q}$ for $p\ge q$ and $0$ otherwise, the matrix of ${\wedge }^{N}H(x)$ between ${\beta }_{\mu }$ and ${\beta }_{\rho }$ is $x^{\mathrm{\mid }\mu \mathrm{\mid }-\mathrm{\mid }\rho \mathrm{\mid }}\det (1_{{\beta }_{\mu ,i}\ge {\beta }_{\rho ,j}})$. Subtracting consecutive columns of the $0\mathrm{/}1$ matrix makes its determinant $1$ exactly when ${\beta }_{\mu ,i}\ge {\beta }_{\rho ,i}>{\beta }_{\mu ,i+1}$ for all $i$, i.e. ${\mu }_i\ge {\rho }_i\ge {\mu }_{i+1}$, the horizontal-strip criterion, and $0$ otherwise; so $\mathrm{\Gamma }(x)={\wedge }^{N}H(x)$ adds every horizontal strip once with weight $x^{\text{strip size}}$. For a type component $a$, ${\mathrm{\Gamma }}_a(x)={\prod }_{b\le a}{\mathrm{\Gamma }}_b(x)$ adds a horizontal strip in every permitted physical component $b\le a$ and leaves the rest unchanged, so the successive ${\mathrm{\Gamma }}_a$ from the empty shape enumerate exactly the flagged semistandard tableaux and give the branching matrix. Conjugation is explicit: since $H^{-1}=I-xS$ has column vector $e_p-x{e}_{p+1}$ and row covector ${\sum }_{r\le q}{x}^{q-r}{e}_r^{\ast }$, $\mathrm{\Gamma }(x{)}^{-1}E(p,q)\mathrm{\Gamma }(x)={\sum }_{r=0}^q{x}^{q-r}(E(p,r)-xE(p+1,r)),$ and the two-current conjugation is the ordered product of two such sums.

Strip factors and exact minor expressions. For a permitted strip transition $\rho \to \lambda$ for the next type component $c$, let $\eta =[\lambda ]\setminus [\rho ]$, $r_x=U_{\mathrm{comp}(x)}{z}^{\mathrm{content}(x)}$, and put $F_{\lambda \rho }^{(c)}={\prod }_{x\in \eta }{\prod }_{\mathrm{comp}(x)<b\le c}(r_x-U_b){\prod }_{x\in [\rho ],y\in \eta \text{ in one physical row}}{C}_{+}(r_x,r_y){\prod }_{x<y,x,y\in \eta \text{ in different physical rows}}{C}_{+}(r_x,r_y){\prod }_{x\in \eta ,y\in [\rho ],x<y}R(r_x,r_y)$. Splitting the pair lists of [F1] into old/old, old/new and new/new gives this factor, and their product along a semistandard chain is ${\gamma }_{t_T}\mathrm{/}(C_{t_T}{u}_T)$. Choose $N>n$, write $q_i={\beta }_{\lambda ,i}$, $p_i={\beta }_{\rho ,i}$ in each component, and $H(q,p{)}_{ij}=1_{q_i\ge p_j}$, so $q_N=p_N=0$. Write $\varphi (k)=V(z^k-1)$ for $k>0$ and ${\psi }_{ab}(d)=V(U_a{z}^d-U_b)$. First regard these as independent formal symbols; canceled symbols are never evaluated. By steps 3.1 and 5.1, the same-component row contribution to $J\mathrm{\Gamma }$ is ${\sum }_{i<j}{\sum }_{0<h<q_i-q_j}\varphi (q_i-q_j-h)\det H(q-h{e}_i+h{e}_j,p)$; the column contribution to $\mathrm{\Gamma }J$ is ${\sum }_{i<j}{\sum }_{h>0}(\varphi (p_i-p_j+h)-\varphi (h))\det H(q,p+h{e}_i-h{e}_j)$. Negative or repeated shifted beads give zero determinants. For the second formula, the outward source has beads $p_i+h,p_j-h$; its two hook pairs returning to $p$ have lengths $h$ and $p_i-p_j+h$, foot gaps $p_i-p_j+h$ and $h$, and opposite signs because the second pair exchanges the two target beads. This proves the difference of potentials, including crossed inverse sources. A crosscomponent row term replaces $q_{a,i},q_{b,j}$ by $q_{a,i}-h,q_{b,j}+h$ with potential ${\psi }_{ab}(q_{a,i}-h-q_{b,j})$; its column term replaces $p_{a,i},p_{b,j}$ by $p_{a,i}+h,p_{b,j}-h$ with potential ${\psi }_{ab}(p_{a,i}-p_{b,j}+h)$. The determinants in the original indexed positions retain all sorting signs.

Norm endpoints. On support $q_i\ge p_i>q_{i+1}$. Put $M_i=q_i-p_i$, $K_j=p_j-q_{j+1}-1$ for $j<N$, $K_N=0$, and $S(m)={\sum }_{k=1}^m\varphi (k)$. The same-component norm valuation is ${\sum }_i(S(q_i)-S(p_i)-S(M_i))+{\sum }_{i<j}{\sum }_{u=p_i}^{q_i-1}(\varphi (u-q_j)-\varphi (u-p_j+1))$. To verify this directly, put $b_i=N-i$: same-row old/new pairs sum to $S(q_i-b_i)-S(M_i)-S(p_i-b_i)$. For $i<j$, new/new and new-upper/old-lower pairs sum to ${\sum }_{u=p_i}^{q_i-1}(\varphi (u-q_j)-\varphi (u-p_j+1)+\varphi (u-b_j+1)-\varphi (u-b_j))$. The last two terms telescope in $u$ to $\varphi (q_i-b_j)-\varphi (p_i-b_j)$; summing over $j>i$ cancels the baseline $b_i$ terms and gives the formula. Every displayed $\varphi$ argument is positive. For $a<b\le c$, the crosscomponent endpoint expression is ${\sum }_i{\sum }_{u=p_{a,i}}^{q_{a,i}-1}({\psi }_{ab}(u+1)+{\sum }_{j=1}^N({\psi }_{ab}(u-q_{b,j})-{\psi }_{ab}(u-p_{b,j}+1)))$. Indeed the new/new $C_{+}$ factors telescope in the lower-component node index $v$ to $\psi (u-q_{b,j})-\psi (u-p_{b,j})$, and the new/old $R$ factors add $\psi (u-p_{b,j})-\psi (u-b_j)$ and the rectangle difference ${\sum }_{v=b_j}^{p_{b,j}-1}(\psi (u+1-v)-\psi (u-v))$. Telescoping this rectangle in the other direction and adding the boundary factor $\psi (u-N+1)$ gives the displayed formula; the vacuum differences telescope and cancel that boundary. This padded formula is an equality of formal symbols: its $j=N$ term cancels $\psi (u+1)$ before evaluation.

All on-support minors. Here $H$ is the unit upper triangular step matrix. Expanding a replacement row or column in its basis gives $d_i^{\mathrm{r}\mathrm{o}\mathrm{w}}(v)=1_{p_i\le v<p_{i-1}}$ and $d_i^{\mathrm{c}\mathrm{o}\mathrm{l}}(v)=1_{q_{i+1}<v\le q_i}$, with $p_0=\mathrm{\infty }$ and no positive lowering from $p_N=0$. A two-row minor is its direct product minus its crossed product. For the same-component row expression, the direct term has $1\le h\le \min (M_i,K_{j-1})$ and the crossed term, with $k=q_i-q_j-h$, has $1\le k\le \min (M_i,K_{j-1})$. The outward two-column minor has no crossed term and its direct bound is $1\le h\le \min (M_i,K_j)$. Pair direct row index $j+1$ with first-potential column index $j$. For arbitrary symbols $\psi$ the interval identity ${\sum }_{s=0}^{M-1}(\psi (A+s+K+1)-\psi (A+s+1))={\sum }_{h=1}^{\min (M,K)}(\psi (A+M+K+1-h)-\psi (A+h))$ follows by canceling the overlap of the two integer intervals (when $M\le K$ they are disjoint, and when $M>K$ only $K$ endpoints remain). Applying it with $A=p_i-p_j$, the paired contribution is ${\sum }_{j=i+1}^{N-1}{\sum }_{u=p_i}^{q_i-1}(\varphi (u-q_{j+1})-\varphi (u-p_j+1))$. Crossed rows cancel the second column potentials except $-S(\min (M_i,K_i))$; the unpaired direct row $j=i+1$, combined with that residual by the same identity at $A=0$, gives ${\sum }_{u=p_i}^{q_i-1}\varphi (u-q_{i+1})-S(M_i)$. Summing yields exactly the same-component endpoints of step 7.1. For a crosscomponent pair, row $j+1$ and column $j$ share the bound $\min (M_i,K_{b,j})$ and $A=p_{a,i}-p_{b,j}$. The unpaired first row contributes ${\sum }_{u=p_{a,i}}^{q_{a,i}-1}\psi (u-q_{b,1})$; the final column vanishes. The same interval identity converts the paired terms to the remaining endpoints of step 7.1. Thus $(J\mathrm{\Gamma }-\mathrm{\Gamma }J{)}_{\lambda \rho }=x^{\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\rho \mathrm{\mid }}V(F_{\lambda \rho }^{(c)})$ on support.

Crosscomponent off support. The row of $H$ at bead $v$ is the tail vector starting at threshold $\kappa (v)$, whose interval is $[p_r,p_{r-1})$. Different nonzero tail vectors are independent; hence rank equals the number of distinct thresholds. At rank $N-1$ there is one missing threshold $r$ and two adjacent duplicated rows $i,i+1$. Put $Q=q_i>Q^{\mathrm{\prime }}=q_{i+1}$, $[P,P^{\mathrm{\prime }})=[p_r,p_{r-1})$, $u=e_i-e_{i+1}$ and $v=e_r-e_{r-1}$, omitting $e_0$. The adjugate is $\epsilon v{u}^{\mathsf{T}}$: its nullspaces have those generators, and removing a duplicate row and the appropriate column gives a minor $\pm 1$. Thus replacement determinants are $\epsilon u_l{1}_{P\le x<P^{\mathrm{\prime }}}$ and $\epsilon v_l{1}_{Q^{\mathrm{\prime }}<x\le Q}$ with the same sign. Rank $\le N-2$ kills every one-row/column replacement. A defective earlier component can be repaired only when $Q^{\mathrm{\prime }}\ge P^{\mathrm{\prime }}$ (missing below), and a defective later component only when $Q<P$ (missing above); opposite directions give zero. If only the earlier component is defective, pair normal later row $j+1$ and column $j$ and write $L=q_{b,j+1}$, $M=p_{b,j}$. Their coefficients both give $-\epsilon {\sum }_{d\in [P-L,P^{\mathrm{\prime }}-L)\cap (Q^{\mathrm{\prime }}-M,Q-M]}\psi (d)$; the first unpaired row vanishes because its two source signs sum to zero. If only the later component is defective, write its endpoints $R>R^{\mathrm{\prime }}$, $[S,S^{\mathrm{\prime }})$ and let $M=q_{a,i}-p_{a,i}$. The row coefficient at restored position $w$ is $\epsilon 1_{R^{\mathrm{\prime }}+M<w\le R+M}$ on $[S,S^{\mathrm{\prime }})$, and the column coefficient at $t$ is $\epsilon 1_{S\le t+M<S^{\mathrm{\prime }}}$ on $(R^{\mathrm{\prime }},R]$. The translation $w=t+M$ matches their potentials because $q_{a,i}=p_{a,i}+M$. With both components defective, the row coefficient of $\psi (d)$, divided by ${\epsilon }_a{\epsilon }_b$, is $(1_{S\le Q-d<S^{\mathrm{\prime }}}-1_{S\le Q^{\mathrm{\prime }}-d<S^{\mathrm{\prime }}})(1_{P\le R+d<P^{\mathrm{\prime }}}-1_{P\le R^{\mathrm{\prime }}+d<P^{\mathrm{\prime }}})$; the column coefficient is $(1_{R^{\mathrm{\prime }}<P-d\le R}-1_{R^{\mathrm{\prime }}<P^{\mathrm{\prime }}-d\le R})(1_{Q^{\mathrm{\prime }}<S+d\le Q}-1_{Q^{\mathrm{\prime }}<S^{\mathrm{\prime }}+d\le Q})$. They agree by the endpoint identity $1_{B_0\le A_1-d<B_1}-1_{B_0\le A_0-d<B_1}=1_{A_0<B_0+d\le A_1}-1_{A_0<B_1+d\le A_1}$, obtained by regrouping four half-line indicators. Infinite top endpoints simply omit the virtual source. An untouched defective component kills the product. Components beyond $c$ are empty in the reachable prefix: the forward transfer would create a nonempty forbidden intermediate component, and the inverse transfer would remove a hook from an empty component, so both vanish. This exhausts crosscomponent off-support cases.

Same-component deficiency one, missing below. Keep the rank-one data of step 9.1, with duplicated rows $s,s+1$ and missing threshold $r>s$, so $Q^{\mathrm{\prime }}\ge P^{\mathrm{\prime }}$. A nonzero inward two-row replacement must use one duplicate source and a unique normal row $j\ge r+1$: one new row fills the missing interval and the other restores the unique source threshold. The direct and exchanged assignments have signs $\epsilon u_l$ and $-\epsilon u_l$; any other assignment leaves an unchanged duplicate or missing threshold, or violates the strict inward range. Nonzero outward two-column repairs use one endpoint $P,P^{\mathrm{\prime }}$ and a unique column $j\ge r+1$, or the endpoint pair itself. Put $K_j=p_j-q_{j+1}-1$ for $j\ge r$, $K_N=0$. The direct row contribution is $R_j=-\epsilon {\sum }_{d\in [P-q_j,P^{\mathrm{\prime }}-q_j)\cap (Q^{\mathrm{\prime }}-p_{j-1},Q-p_{j-1}]}\varphi (d)$. Split each unique column contribution into its first and second potentials $C_j^{\mathrm{f}\mathrm{i}\mathrm{r}\mathrm{s}\mathrm{t}}-C_j^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}$ from step 6.1. Then $R_{j+1}=C_j^{\mathrm{f}\mathrm{i}\mathrm{r}\mathrm{s}\mathrm{t}}$, by the same interval intersection, while $C_j^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\epsilon {\sum }_{h=1}^{K_j}\varphi (h)(1_{Q^{\mathrm{\prime }}<P+h\le Q}-1_{Q^{\mathrm{\prime }}<P^{\mathrm{\prime }}+h\le Q})$. The exchanged row is $X_j=-\epsilon {\sum }_{k=1}^{K_{j-1}}\varphi (k)(1_{P\le Q-k<P^{\mathrm{\prime }}}-1_{P\le Q^{\mathrm{\prime }}-k<P^{\mathrm{\prime }}})=-C_{j-1}^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}$ by endpoint reciprocity. All terms cancel except $R_{r+1}-C_r^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}-C^{\mathrm{e}\mathrm{n}\mathrm{d}}$. Put $A=Q^{\mathrm{\prime }}-P^{\mathrm{\prime }}\ge 0$, $B=P^{\mathrm{\prime }}-P>0$, $L=Q-Q^{\mathrm{\prime }}>0$, $I_0=(A,A+L]$, $I_1=I_0+B$, $K=K_r$. These three terms are respectively $-\epsilon {\sum }_{d\in I_1\cap (K,K+B]}\varphi (d)$, $\epsilon {\sum }_{h=1}^K\varphi (h)(1_{I_1}(h)-1_{I_0}(h))$, and $-\epsilon {\sum }_{h\in I_0,1\le h\le K}(\varphi (B+h)-\varphi (h))$; the endpoint-pair sign is negative because the raised endpoint is $P^{\mathrm{\prime }}$. Their residual is $-\epsilon {\sum }_{d\in I_1\cap [1,K+B]}\varphi (d)+\epsilon {\sum }_{h\in I_0,1\le h\le K}\varphi (B+h)=0$ by $d=h+B$. This proves every coefficient cancels, including $K=0$.

Same-component deficiency one, missing above. Here $r\le s$, $P>Q$, and $\epsilon =(-1{)}^{s-r}$, obtained by moving the inserted missing threshold from row $s$ to position $r$; the column cofactor has the same sign. If $r=1$, all inward row positions are at most $q_1<p_1$ and cannot fill the missing top threshold; the first column is zero and any outward repair raises it further, so both sums vanish. For $r\ge 2$ the only unique row sources are $i\le r-1$, with $M_i=q_i-p_i\ge 0$. The direct rows have $R_i=\epsilon {\sum }_{d\in (q_i-P^{\mathrm{\prime }},q_i-P]\cap [p_i-Q,p_i-Q^{\mathrm{\prime }})}\varphi (d)$. The first column potentials equal $R_i$ for $i\le r-2$, while $C_i^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\epsilon {\sum }_{h=1}^{M_i}\varphi (h)(1_{Q^{\mathrm{\prime }}<P-h\le Q}-1_{Q^{\mathrm{\prime }}<P^{\mathrm{\prime }}-h\le Q})$. Exchanged rows are $X_i=-\epsilon {\sum }_{k=1}^{M_i}\varphi (k)(1_{P\le Q+k<P^{\mathrm{\prime }}}-1_{P\le Q^{\mathrm{\prime }}+k<P^{\mathrm{\prime }}})=-C_i^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}$, again by endpoint reciprocity. This exhausts repairs: the duplicate source must be paired with a head row above the missing interval, and outward column repairs must lower an endpoint paired with a unique head column, or use the endpoint pair. The sole residual is at $i=r-1$. Put $M=q_{r-1}-P^{\mathrm{\prime }}\ge 0$, $B=P^{\mathrm{\prime }}-P>0$, $A=P-Q>0$, $L=Q-Q^{\mathrm{\prime }}>0$, $I_0=[A,A+L)$, $I_1=I_0+B$. Then $R_{r-1}=\epsilon {\sum }_{d\in I_1\cap (M,M+B]}\varphi (d)$ and $C_{r-1}^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}=\epsilon {\sum }_{h=1}^M\varphi (h)(1_{I_0}(h)-1_{I_1}(h))$; the endpoint-pair column term is $C^{\mathrm{e}\mathrm{n}\mathrm{d}}=\epsilon {\sum }_{h\in I_0,1\le h\le M}(\varphi (B+h)-\varphi (h))$, with positive cofactor because the lowered endpoint is $P$. Thus $R_{r-1}-C_{r-1}^{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}-C^{\mathrm{e}\mathrm{n}\mathrm{d}}=\epsilon {\sum }_{d\in I_1\cap [1,M+B]}\varphi (d)-\epsilon {\sum }_{h\in I_0,1\le h\le M}\varphi (B+h)=0$. The translation $d=h+B$ is valid since $A>0$.

Same-component deficiency two and higher. At rank $N-2$, the excess threshold copies are one triple or two pairs. Inward row repairs of a triple cannot reach missing thresholds between its source beads. For two pairs with thresholds $r<s$, both missing thresholds must lie strictly between them, say $r<m_1<m_2<s$; otherwise both row and column sums are zero. For columns this follows from $\mathrm{\ell }(p_j)=\mathrm{\#}\{i:\kappa (q_i)\le j\}$: duplicated row thresholds are jumps, missing thresholds are flats, and outward column repairs require two separate size-two jumps with both flats between. A zero first column also cannot be repaired by raising it. In the active pattern put $Q=q_r>Q^{\mathrm{\prime }}=q_{r+1}$, $R=q_{s-1}>R^{\mathrm{\prime }}=q_s$, $I_1=[P,P^{\mathrm{\prime }})=[p_{m_1},p_{m_1-1})$, $I_2=[S,S^{\mathrm{\prime }})=[p_{m_2},p_{m_2-1})$, $J_Q=(Q^{\mathrm{\prime }},Q]$, $J_R=(R^{\mathrm{\prime }},R]$. The common row/column sign is $\epsilon =(-1{)}^{m_1-r+s-1-m_2}$, counted by moving the inserted thresholds to their sorted positions. Switching a source copy or switching the two target thresholds reverses it. Define $A_I(k)=1_{R+k\in I}-1_{R^{\mathrm{\prime }}+k\in I}$ and $B_I(k)=1_{Q-k\in I}-1_{Q^{\mathrm{\prime }}-k\in I}$; define $D_P^{\pm }(J;k)=1_{P\pm k\in J}-1_{P^{\mathrm{\prime }}\pm k\in J}$ and $D_S^{\pm }(J;k)=1_{S\pm k\in J}-1_{S^{\mathrm{\prime }}\pm k\in J}$. The complete row coefficient of $\varphi (k)$ is $\epsilon (A_{I_1}(k)B_{I_2}(k)-A_{I_2}(k)B_{I_1}(k))$; the column coefficient is $\epsilon (D_S^{+}(J_Q;k)D_P^{-}(J_R;k)-D_P^{+}(J_Q;k)D_S^{-}(J_R;k))$. Indeed putting $k=q_i-q_j-h$ gives row targets $q_j+k,q_i-k$, while the positive and negative column potentials give targets $p_j+k,p_i-k$ and $p_i+k,p_j-k$ respectively. Interval membership implies all required positive-length bounds. Endpoint reciprocity gives $A_{I_1}=D_P^{-}(J_R)$, $A_{I_2}=D_S^{-}(J_R)$, $B_{I_1}=D_P^{+}(J_Q)$, $B_{I_2}=D_S^{+}(J_Q)$, proving equality. If $m_2=m_1+1$, the repeated column groups share an endpoint; its formal double-source term is zero because its potential is $\varphi (h)-\varphi (h)$, so the formula still includes adjacent missing thresholds. At rank $\le N-3$, two replacements cannot restore full rank. Together with steps 9.1–10.2 this proves off-support vanishing for every rank and orientation. All identities are formal coefficient equalities; retained terms are actual valid hook currents, so step 3.2 permits their finite arbitrary-characteristic valuation specialization.

Telescoping and the valuation formula. Multiply the $\mathrm{\Gamma }$ matrices in increasing type-row-label order and telescope $J\mathrm{\Gamma }-\mathrm{\Gamma }J$ from $J_0=0$; by steps 8.1–11.1 each commutator inserts exactly $V(F)$ at its strip position, so the $\lambda$-coordinate of the resulting identity is ${\sum }_{T}V({\gamma }_{t_T}\mathrm{/}(C_{t_T}{u}_T))x^{\mathrm{wt}(T)}=L_{\lambda }(x)$ on one side and $B_{\lambda }(x)$ on the other. All padded artificial symbols cancel before the surviving coefficients are evaluated, and every surviving coefficient corresponds to a valid current, hence by step 3.2 to a nonzero foot difference with finite valuation. Comparing with the exact monic product [F1] and the hook side of step 3.1, and taking the coefficient of $x^{\nu }$ at an actual multipartition weight $\nu$ (auxiliary compositions only index the same polynomials and never enter the Schur conclusion), gives $V(\det G_{\lambda \nu })={\sum }_{\mu }{J}_{\lambda \mu }{K}_{\mu \nu };$ the empty case $n=0$ gives $0$, all ranks keep the fixed parameters $z=q_{H}t$, $U_a=Q_a{t}^{na}$, and arbitrary-composition coefficients are combinatorial only.