Finite type-A tensor Hecke bars and complementary coefficient charts

The actual reciprocal action. The ordinary bar preserves total coordinate weight, so its triangular corrections and both coefficient matrices preserve the entry multiset. Work henceforth in one such weight space. Set $q=s^{-1}$. On adjacent letters define $M_a{H}_i=M_{a{s}_i}$ if $a_i<a_{i+1}$, $M_a{H}_i=s{M}_a$ if they are equal, and $M_a{H}_i=M_{a{s}_i}+(s-s^{-1})M_a$ if $a_i>a_{i+1}$. On an unequal two-letter span its matrix has square $1+(s-s^{-1})H_i$; on an equal span this is $s^2=1+(s-s^{-1})s$. For three adjacent letters, applying the two triple products to their six possible strict orders, or their three possible orders with one equality, gives the same swapped tuple and the same coefficients; with all equal both are $s^3$. Distant actions commute. Thus these are the generators of $H_q$. Substitution of the natural $E_j,F_j$ replacement actions and the source coproduct shows that they commute with the quantum action: on the two active letters the toral coefficients are $s^{{\mathbf{1}}_{a=j}-{\mathbf{1}}_{a=j+1}}$, and the unequal two-term expression cancels the extra $s-s^{-1}$ term; equal active letters give the sum $s+s^{-1}$ on both sides. All other letters are unaffected. The same two-letter calculation gives $\psi M_{ba}=M_{ba}+(s^{-1}-s)M_{ab}$ for $a<b$ and fixes increasing and equal tuples. Therefore $\psi (x{H}_i)=\psi (x){\bar{H}}_i$; adjoining spectator factors follows from quasi-$R$ naturality and factorization in [F1].

Cyclic induced bar and signed transport. If $b$ is the sorted tuple, $M_b$ is bar fixed by weight transfer. Every tuple is $b{d}_a$, where $d_a$ is shortest left modulo $K$, and along a reduced expression all successive swaps increase inversions, so $M_b{H}_{d_a}=M_a$. Equal-position generators act as $s=q^{-1}$, giving a surjection ${\mathbf{1}}_K\otimes H_q\to \mathrm{span}\{M_a\}$; the same shortest representatives index both bases, so it is an isomorphism. Step 1.1 and the fixed cyclic vector identify the induced and tensor bars. Change generators to $G_i=-H_i$: they satisfy the normalization with $v=s$, their stabilizer character is $-s$, and their standard vectors are $n_{d_a}=(-1{)}^{\mathrm{\ell }(d_a)}{M}_a$. The positive-$s$ sign canonical vector at $d_b$, multiplied by $(-1{)}^{\mathrm{\ell }(d_b)}$, is exactly $L_b$ by triangular uniqueness.

Primitive exterior embedding. For one decreasing column $a_1>\cdots >a_m$ put $Y_a={\sum }_{b\sim a}(-s{)}^{\mathrm{\ell }(a,b)}{M}_b$. Pairing the two orders of any adjacent unequal letters using step 1.1 gives $Y_a{H}_i=-s^{-1}{Y}_a$; repeated entries have no sign eigenvector since their scalar is $s$. Replacing adjacent alphabet entries with $E_j,F_j$ sends $Y_a$ to the identically normalized $Y_{a^{\mathrm{\prime }}}$ or to zero on collision: all unaffected permutations match term by term, and the terms causing a collision cancel in adjacent pairs. Thus $w_{\{a_1,\dots ,a_m\}}\mapsto Y_a$ realizes the elementary wedge, with coefficient one on the decreasing tuple, integrally and injectively. The sign eigen-equations and the unique column top vector make its factor bar the ordinary restriction of the tensor bar; naturality and parenthesization in [F1] give the product embedding for arbitrary ordered column heights. Its displayed expansion is a product of these sums, hence exactly the sum over independent column permutations in the Statement.

The positive tensor coefficient. The sign reduction in [F3] gives $U_{a,b}=(-1{)}^{\mathrm{\ell }(d_a)+\mathrm{\ell }(d_b)}{\sum }_z(-s{)}^{\mathrm{\ell }(z)}{s}^{\mathrm{\ell }(d_b)-\mathrm{\ell }(z{d}_a)}{P}_{z{d}_a,d_b}(s^{-2})$. Shortest-left length additivity cancels the powers of $s$ inside the sum. Inversion of permutations preserves ordinary KL coefficients: the algebra anti-involution $H_x\mapsto H_{x^{-1}}$ preserves bar and positive triangularity, so uniqueness in [F2] proves $P_{x,y}=P_{x^{-1},y^{-1}}$. Replace $z$ by $z^{-1}$ to obtain the stated formula with $w_{a}z,w_b$. This is a change of generators and standard-vector signs, not a reciprocal substitution in a positive chart.

Reverse-position indices and pairing. For $x\in S_N$ put $R_x(m,h)=\mathrm{\mid }\{i\le m:x(i)\le h\}\mathrm{\mid }$. Reversing values gives $R_{w_{0}x}(m,h)=m-R_x(m,N-h)$, so the rank inequalities in [F4] reverse under $x\mapsto w_{0}x$. Inversion symmetry then gives reversal under $x\mapsto x{w}_0$ as well. Put $M_a^{\ast }=M_{a{w}_0}$ as actual vectors in the same tensor space. The shortest-left representative of $a{w}_0$ is $c_a=w_K{d}_a{w}_0$: reversal interchanges increasing and decreasing places, and reversal within each equal-value block restores shortest-left status. Its length is $\mathrm{\ell }(w_0)-\mathrm{\ell }(w_K)-\mathrm{\ell }(d_a)$. Define the bilinear form $(M_a,\psi M_b^{\ast })={\delta }_{ab}$; its Gram matrix is the inverse of the bar matrix after reversing the second label, so it is not the raw permutation matrix. The three generator cases of step 1.1 and $\psi (x{H}_i)=\psi (x){\bar{H}}_i$ give $(x{H}_i,y)=(x,y{H}_{N-i})$, where $N$ is the number of tensor positions. The coproduct replacement calculation likewise gives $(E_{j}x,y)=(x,F_{j}y)$ and $(D_{j}x,y)=(x,D_{j}y)$. The two-letter bar identity iterated through a reduced word for $w_0$ gives $\psi M_a=s^{{\sum }_u\left(\genfrac{}{}{0px}{}{{\mathrm{mult}}_u(a)}{2}\right)}{M}_a^{\ast }{H}_{w_0}^{-1}\mathrm{.}$ The exponent depends only on the fixed multiset. This formula follows by successively moving the last letter past its predecessors: unequal pairs contribute the two-letter bar above, while each equal pair contributes $s^{-1}$ from $H_i^{-1}$, cancelled by the displayed power. It also proves symmetry: writing $m(a)={\sum }_u\left(\genfrac{}{}{0px}{}{{\mathrm{mult}}_u(a)}{2}\right)$, generator adjunction and the invariance of $H_{w_0}$ under $H_i\mapsto H_{N-i}$ give $(\psi M_b^{\ast },M_a)=s^{m(b)}(M_b{H}_{w_0}^{-1},M_a)=s^{m(b)-m(a)}(M_b,\psi M_a^{\ast })={\delta }_{ab}$. Since the two displayed families are bases, the form is symmetric.

Complementary chart and inverse, with all indices. Let $L_b^{\ast }=M_b^{\ast }+{\sum }_{a}q\mathbb{Z}[q]M_a^{\ast }$ be the ordinary-bar fixed complementary chart. In the induced module of step 2.1 this is the spherical positive-$q$ basis at $c_b$, so $L_b^{\ast }={\sum }_a{V}_{a,b}(q)M_a^{\ast }$ with $V_{a,b}(q)=h_{w_K{c}_a,w_K{c}_b}(q)$ by [F3]. Write $U_{a,b}=U_{a,b}(s)$ and $V_{a,b}=V_{a,b}(s^{-1})$. Fixedness gives $L_b^{\ast }={\sum }_a{\bar{V}}_{a,b}\psi M_a^{\ast }$, while $L_b={\sum }_a{\bar{U}}_{a,b}\psi M_a={\sum }_a{\bar{U}}_{a,b}\psi M_{a{w}_0}^{\ast }$. Also $L_b^{\ast }={\sum }_a{V}_{a,b}{M}_{a{w}_0}$. Accordingly $(L_a,L_b^{\ast })={\sum }_c{U}_{c,a}{\bar{V}}_{c,b}$ lies in ${\delta }_{ab}+s\mathbb{Z}[s]$, while the symmetric evaluation $(L_b^{\ast },L_a)={\sum }_c{V}_{c,b}{\bar{U}}_{c,a}$ lies in ${\delta }_{ab}+s^{-1}\mathbb{Z}[s^{-1}]$. Their equality forces $(L_a,L_b^{\ast })={\delta }_{ab}$. Thus $U^{\mathsf{T}}\bar{V}=I$ and $D=U^{-1}={\bar{V}}^{\mathsf{T}}$. Since $w_K{c}_b=d_b{w}_0$, the spherical formula at parameter $s$ yields $D_{a,b}=h_{d_b{w}_0,d_a{w}_0}(s)=s^{\mathrm{\ell }(d_b)-\mathrm{\ell }(d_a)}{P}_{d_b{w}_0,d_a{w}_0}(s^{-2})\mathrm{.}$ This supplies the complementary longest-element identification, rather than only an abstract inverse-matrix identity.

Wedge charts and both orders. Each nontrivial column permutation replaces a decreasing tuple by a smaller tuple and has coefficient in $s\mathbb{Z}[s]$. Expanding a wedge $L_A$ under step 2.2 therefore gives a bar-fixed ambient vector with leading term $M_{a(A)}$ and all corrections in $s\mathbb{Z}[s]$ on smaller tuple labels. Triangular uniqueness identifies it with the ambient $L_{a(A)}$. Expanding $\iota (M_B)$ in the ambient $L$ basis using step 4.1, and then expressing the same vector in the embedded wedge $L$ basis, gives precisely the claimed signed inverse coefficient; all ambient labels outside the decreasing-column subset have zero coefficients because the embedded $L$ vectors are a basis of the image. None of the arguments compares the heights of different columns, so the construction works unchanged in reversed factor order. At $s=1$ the displayed primitive exterior sum becomes the ordinary alternating column sum, and $D_{a,b}(1)=P_{d_b{w}_0,d_a{w}_0}(1)$ in both orders.

Conclusion. The generator and signed-chart steps prove the explicit generator/bar and signed antispherical transport; the reverse-position pairing and complementary-chart steps prove the spherical coefficient and its exact longest indices; the primitive exterior and wedge-chart steps prove the ordinary-bar embedding and signed inverse matrix in either ordered tensor. All constructions use finite tensor and Hecke data, independently of any categorical identification.