The finite q-wedge space and the normal-ordering relations

Symmetric Laurent multiplication preserves $\ker (S+1)$. If $g=s(g)$, [F2] gives $S(gx)=gS(x)$, since the divided difference of $g$ is zero; this applies also to negative common loop shifts $(Y_1{Y}_2{)}^r$. On a diagonal $d_j=v_j\otimes v_j$ one has $S(d_j)=-d_j$. For $j<k$, set $w=v_j\otimes v_k$ and $w^{\mathrm{\prime }}=v_k\otimes v_j$; substitution gives $S(w+q^{-1}{w}^{\mathrm{\prime }})=-(w+q^{-1}{w}^{\mathrm{\prime }})$. For $j>k$ the divided differences of $Y_1,Y_2$ give $S(Y_{1}w)=-q{Y}_2{w}^{\mathrm{\prime }}$ and $S(Y_2{w}^{\mathrm{\prime }})=-q{Y}_{1}w+(Q-1)Y_2{w}^{\mathrm{\prime }}$, so $Y_{1}w+q^{-1}{Y}_2{w}^{\mathrm{\prime }}$ is a $-1$ eigenvector. These calculations take place in the original Laurent tensor space. Their images vanish in every adjacent pair of slots by [F1].

Diagonal and equal-residue relations. A repeated-index tensor is $(Y_1{Y}_2{)}^p{d}_j$ and hence vanishes. If $l<m$ have the same residue, write $l=j-(r+a)e$, $m=j-re$ with $a>0$. The symmetric multiple $(Y_1{Y}_2{)}^r(Y_1^a+Y_2^a)d_j$ is in $\ker (S+1)$ by step 1.1; its image is $u_l\wedge u_m+u_m\wedge u_l$, giving the equal-residue relation. No individual nonsymmetric loop multiplier of a kernel vector has been assumed to preserve the kernel.

Distinct residues without wrapping. Write $l=j-pe$, $m=k-re$ with $j<k$. Then $m-l=(p-r)e+(k-j)=ae+b$, so $p=r+a$ and $b=k-j$. If $a=0$, the symmetric common shift $(Y_1{Y}_2{)}^r(w+q^{-1}{w}^{\mathrm{\prime }})$ gives $u_l\wedge u_m=-q^{-1}{u}_m\wedge u_l$. If $a>0$, multiply the same kernel vector by the symmetric Laurent polynomial $(Y_1{Y}_2{)}^r(Y_1^a+Y_2^a)$. The four resulting images are, with coefficients $1,q^{-1},1,q^{-1}$, respectively $(l,m)$, $(m-ae,l+ae)$, $(l+ae,m-ae)$, $(m,l)$. Their sum is zero, which is precisely the fourth displayed relation.

Distinct residues across the representative boundary. If $j>k$, the same expressions have $p-r=a+1$ and $b=e+k-j$. The wrapped base kernel vector of step 1.1 is $h=Y_{1}w+q^{-1}{Y}_2{w}^{\mathrm{\prime }}$. For $a=0$, its common shift $(Y_1{Y}_2{)}^{r}h$ gives the third displayed relation. For $a>0$, its symmetric multiple $(Y_1{Y}_2{)}^r(Y_1^a+Y_2^a)h$ has images $(l,m)$, $(m,l)$, $(l+ae,m-ae)$ and $(m-ae,l+ae)$ with coefficients $1,q^{-1},1,q^{-1}$. Thus it gives the same fourth relation in this case also. Together with step 2.1 this exhausts all residue and Laurent-exponent configurations.

Pair recursion and whole-word termination. In the mixed case the correction indices are $m-b$ and $l+b$, strictly between $l$ and $m$, with gap $ae-b<ae+b$; the first correction pair is already decreasing, and the reverse pair can be rewritten recursively. For a word of length $N$, fix its original interval $[L,U]$ and total sum $T$. Every branch stays among the finitely many integer tuples in $[L,U{]}^N$ of sum $T$. Sort a tuple decreasingly and list its first $N-1$ prefix sums. Balancing a pair of extremes strictly decreases this list in lexicographic order: the sums of the largest $r$ coordinates cannot increase, and at least one decreases since the sorted multisets differ. A leading adjacent swap preserves this list and decreases by one the number of pairs of positions whose indices are increasing. Hence the lexicographic pair consisting of prefix-sum list and inversion count strictly decreases along each nonzero branch. Its range is finite, so rewriting terminates. Equal adjacent indices vanish; a terminal nonzero word is strictly decreasing.

Charts and energy. All coefficients in the displayed relations belong to $A=\mathbb{Z}[q,q^{-1}]$. Step 4.1 therefore expresses every Laurent tensor-lattice image in the $A$-span of the normal wedges; conversely each normal wedge is such an image. These two lattices equal $W_A$. For independence, take finitely many distinct normal tuples $I_1,\dots ,I_t$ and their raw tensors $t_{I_k}$. By [F2], the Laurent Hecke action is regular at $q=1$, where its residue formula gives $S_i=-P_i$ with $P_i$ the factor flip and its divided-difference term vanishes. Thus $A_N$ specializes to the signed antisymmetrizer. The coefficient matrix of $A_N{t}_{I_k}$ on the decreasing raw rows $t_{I_j}$ specializes to the identity: a different normal tuple has a different multiset, and only the identity permutation contributes to its own decreasing row. This minor is nonzero over $\mathbb{Q}(q)$. A relation ${\sum }_k{c}_k{u}_{I_k}=0$ in the quotient implies ${\sum }_k{c}_k{A}_N{t}_{I_k}=0$ by its definition, so the minor forces every $c_k=0$. Consequently the positive chart is the internal direct sum $W_0={⨁}_I\mathbb{Z}[q]u_I$ as asserted. Since $q$ is a unit on $W_A$, $W_A\mathrm{/}q{W}_A=0$. The energy criterion is exactly [F3] and concerns the opposite channel. These computations establish every displayed relation and chart assertion in the Definition, using the kernel contract in [F1].