Generic nondegenerate local inverse tables on integer colors

All coefficients are legal. For $a\mathrm{\ne }b$, $x-z$ has constant $q^a-q^b\mathrm{\ne }0$. In the last case of $Q$, $x-qz$ has nonzero constant because $a\mathrm{\ne }b+1$. In the backward case $-q^a\mathrm{/}(x-z)$ is a unit. In the equal case $1-q\mathrm{\ne }0$. Thus every denominator is a unit and each $Q$ is a unit. The formulas depend only on the adjacent pair, so they make sense in every rank and in every finite color window.

Use an explicit polynomial representation. On the $i$-component let dots multiply, and let a permutation simultaneously permute coordinates and colors. With ${\mathrm{\partial }}_r=(s_r-1)\mathrm{/}(y_r-y_{r+1})$, define ${\psi }_r$ on an equal pair to be ${\mathrm{\partial }}_r$; on an unequal source pair $(a,b)$ let its output in the swapped component be $h_{ab}(u,v)s_r$, where $h_{ab}(u,v)=v-u$ if $a=b+1$ and $h_{ab}=1$ otherwise. The equal-color quotient is a power series because the numerator vanishes on $u=v$. Its product rule is $\mathrm{\partial }(fg)=\mathrm{\partial }(f)g+(sf)\mathrm{\partial }(g)$. Consequently $\mathrm{\partial }v=u\mathrm{\partial }+1$, $v\mathrm{\partial }=\mathrm{\partial }u+1$, and ${\mathrm{\partial }}^2=0$. On unequal pairs the square is multiplication by $h_{ba}(u,v)h_{ab}(v,u)=K_{ab}(u,v)$. These computations verify all mixed and quadratic relations, including the native sign $\tau =-\psi$.

Three distinct colors. In both three-crossing words the coefficient of the longest permutation is the product of the three pair factors, transported to the same coordinates. Each pair is crossed once. The factors commute and $s_1{s}_2{s}_1=s_2{s}_1{s}_2$, so the two words agree. This argument does not require nonadjacency; some of the factors can be linear.

First two colors equal. For $(a,a,b)$ write $h=h_{ab}$. The left word is $h(u,v)h(u,w)s_1{s}_2{\mathrm{\partial }}_1$, and the right word is ${\mathrm{\partial }}_{2}h(u,v)h(u,w)s_1{s}_2$. The product of factors is invariant under swapping $v,w$, so its ${\mathrm{\partial }}_2$ is zero. Also ${\mathrm{\partial }}_2{s}_1{s}_2=s_1{s}_2{\mathrm{\partial }}_1$ by conjugating the displayed quotient. The words agree. For $(a,b,b)$ the left word is ${\mathrm{\partial }}_{1}h(v,w)h(u,w)s_2{s}_1$, and the right word is $h(v,w)h(u,w)s_2{s}_1{\mathrm{\partial }}_2$. Here the product is invariant under swapping $u,v$, and ${\mathrm{\partial }}_1{s}_2{s}_1=s_2{s}_1{\mathrm{\partial }}_2$. This verifies the third color pattern separately.

Outer colors equal. On $(a,b,a)$ with $a\mathrm{\ne }b$, put $h=h_{ab}$ and $g=h_{ba}$. Expanding the one equal-color crossing gives left and right words $\frac{g(u,v)}{u-w}(h(v,w)s_1{s}_2{s}_1-h(v,u)),\frac{h(v,w)}{u-w}(g(u,v)s_2{s}_1{s}_2-g(w,v))\mathrm{.}$ Their longest-permutation terms cancel. Their difference is $[K_{ab}(w,v)-K_{ab}(u,v)]\mathrm{/}(u-w)$, the stated braid correction. This equals $1$ when $b=a+1$, $-1$ when $a=b+1$, and zero otherwise. Thus the apparent denominator cancels identically.

All three colors equal. Put $D=(u-v)(u-w)(v-w)$. Expanding the quotients and collecting each of the six permutations gives ${\mathrm{\partial }}_1{\mathrm{\partial }}_2{\mathrm{\partial }}_1={\mathrm{\partial }}_2{\mathrm{\partial }}_1{\mathrm{\partial }}_2=-D^{-1}{\sum }_{t\in S_3}\mathrm{sgn}(t)t\mathrm{.}$ The numerator is alternating and hence divisible by each of the three pair differences, so this is an operator on power series. This explicitly supplies the all-equal case omitted as an exercise in the source. Together with steps 2.1, 2.2, 2.3 it proves all five braid patterns.

Compute the Hecke generators in the same representation. At an output component set $U=X_r$, $V=X_{r+1}$. For unequal colors the swapped input series $Q$ times $h$ is $(qU-V)\mathrm{/}(U-V)$: in the backward case $qU-V=q^a(v-u)$, and the swapped input series is $-q^a\mathrm{/}(V-U)=q^a\mathrm{/}(U-V)$, so multiplying by $h=v-u$ gives exactly the stated coefficient. In every other unequal case this is the last row of $Q$ directly. Also $-P=(1-q)V\mathrm{/}(U-V)$. For equal colors, expand $\mathrm{\partial }Q-1$; its permutation coefficient is $(1-q+qu-v)\mathrm{/}(u-v)$ and its identity coefficient is $-(1-q)(1-v)\mathrm{/}(u-v)$. Thus in every case $T_r=A(U,V)s_r+B(U,V),A(U,V)=\frac{qU-V}{U-V},B(U,V)=\frac{(1-q)V}{U-V}\mathrm{.}$ The equal-color poles cancel on power series, as follows already from $T=\mathrm{\partial }Q-1$.

Mixed and quadratic Hecke identities. Write $A_{ij}=A(X_i,X_j)$ and $B_{ij}=B(X_i,X_j)$. Direct common-denominator expansion gives $B_{ij}+B_{ji}=q-1$, $A_{ij}=q-B_{ij}$, and $A_{ij}{A}_{ji}=q+(q-1)B_{ij}-B_{ij}^2$. Squaring $A_{ij}s+B_{ij}$ therefore gives $(q-1)T+q$. The identity $T{X}_i-X_{i+1}T=(1-q)X_{i+1}$ follows by cancellation of its permutation term. Combining this identity with the quadratic identity gives $T{X}_{i}T=q{X}_{i+1}$. Distant commutations follow from disjoint variables.

The Hecke braid holds in every color pattern, including all equal. Set $c=q-1$. Clearing the three pair denominators gives the rational identity $B_{13}(c-B_{12}-B_{23})+B_{12}{B}_{23}=0$. The coefficients of the longest permutation, $s_1{s}_2$, and $s_2{s}_1$ in $T_1{T}_2{T}_1$ are $A_{12}{A}_{13}{A}_{23}$, $A_{12}{A}_{13}{B}_{23}$, and $B_{12}{A}_{23}{A}_{13}$; the right word has the same coefficients. The $s_1$ coefficients are $A_{12}(B_{13}{B}_{21}+B_{12}{B}_{23})$ and $A_{12}{B}_{23}{B}_{13}$; the displayed identity makes them equal. The $s_2$ comparison is its transposed version. The identity coefficients differ by $(B_{12}-B_{23})(B_{13}(c-B_{12}-B_{23})+B_{12}{B}_{23})=0.$ This uses the quadratic product identity of step 3.1. These are all six permutations, so the braid is proved. It is a rational identity before any color specialization; the equal-color operators preserve power series by step 2.5.

Faithful actual affine Hecke presentation before completion or quotient. Work first over $C=k[X_1^{\pm 1},\dots ,X_d^{\pm 1}]$, not over a cyclotomic quotient or an assumed local algebra. On its rational function field use $T_r=A(X_r,X_{r+1})s_r+B(X_r,X_{r+1})$, whose mixed, quadratic and full six-permutation braid identities were proved in steps 3.1–4.1. The action preserves C: $T_{r}f=(s_{r}f)T_r+B(f-s_{r}f)$, and the difference $f-s_{r}f$ is divisible by $X_r-X_{r+1}$ for Laurent monomials, then all Laurent polynomials. This formula also follows in the actual affine presentation from its generator mixed relation by its product rule. It moves coefficients past crossings; Coxeter word reductions using the quadratic and braid relations span by $T_{w}f(X)$ with chosen reduced words w. In the rational action, $T_w$ has the product of the transported nonzero A factors times its own permutation w, and other terms have strictly shorter permutation length. Distinct variable permutations are independent over the rational field: a minimal nonzero relation evaluated on fg minus a selected permutation's value on g times its evaluation on f gives a shorter nonzero relation, choosing a coordinate g on which two permutations differ. Hence these spanning words are independent and the affine presented algebra is finite free of rank d! as a right C-module. This proves actual affine faithfulness BEFORE local completion and before any cyclotomic quotient; no specialization or generic-rank dimension argument is used.

Faithful actual affine KLR presentation. On the finite word set the polynomial operators in step 1.2 satisfy the native dot/quadratic relations and all FIVE equality patterns of three colors by steps 2.1, 2.2, 2.3, 2.4. Move dots to the right, then reduce crossing words to fixed reduced permutation words. A repeated crossing is a polynomial by its square relation; a braid replacement has a polynomial correction with fewer crossings. The adjacent-transposition sorting proof of the symmetric-group Coxeter reductions thus gives finite spanning words ${\psi }_{w}f(y)e(\nu )$. In the direct sum of the rational-function fields of the word components, each such crossing word has a nonzero polynomial-or-divided-difference leading coefficient times its permutation w, and all other terms have shorter permutation length. Isolate an input and output component; distinct permutations stabilizing those components remain distinct automorphisms of their independent dot variables. The minimal-relation argument of step 5.1 makes these leading permutations independent. Descend through crossing length to prove the displayed spanning words independent. Hence this is a faithful representation of the actual affine KLR presentation and gives its finite coefficientwise PBW module. Fractions are only used to prove independence of regular polynomial operators, not to invert dots in the presented algebra.

Genuine central completions and spectral projectors. Fix one spectral multiset $\beta$ and let J in the symmetric Laurent polynomial center be generated by $f(X)-f(q^{\beta })$. Symmetric coefficients are central by step 5.1's mixed formula. Each coordinate in C/JC is a root of ${\prod }_{i\in \beta }(t-q^i)$, so this coefficient quotient is finite-dimensional and its geometric points are exactly the permutations of that multiset. All points lie over k. Its radical is the intersection of the corresponding finitely many maximal ideals; the finite-dimensional quotient has nilpotent radical, so its powers are cofinal with the powers of JC. Chinese remainders identify the coefficient completion with ${⨁}_{\nu \in \mathrm{Orb}(\beta )}k[\text{\hspace{0.17em}⁣}[y_1,\dots ,y_d]\text{\hspace{0.17em}⁣}]e(\nu )$, with $X_r=q^{{\nu }_r}(1-y_r)$. Step 5.1's affine PBW module then identifies the ACTUAL central completion with the same finite sum of PBW words and formal coefficients, and its spectral projectors are the displayed completed coefficient idempotents. On the KLR side the positive symmetric dot polynomials are central by the equal divided-difference rule and unequal permutation rule. Their ideal has radical $(y_1,\dots ,y_d)$: each $y_r^d$ belongs to that symmetric ideal by the characteristic-polynomial identity, giving cofinal powers explicitly. Step 6.1 therefore gives its actual central completion as the same finite formal PBW module. A finite permutation-stable set with several contents is the finite direct sum of these completions; distinct spectral multisets give comaximal central ideals. Formal multiplication is legitimate: a permutation preserves dot order and each divided difference lowers it by at most one, while the finite PBW crossing lengths give uniform finite order loss. The faithful rational triangular argument continues over the fraction fields of the formal rings. Thus both completions have the coefficient ring assumed in the local formulas, with actual completed spectral projectors, represented by finite polynomial projectors in each nilpotent quotient, and continuous multiplication, independently of any cyclotomic faithfulness assertion.

Both directions and both composites. In the faithful KLR action, steps 2.5–4.1 prove the Hecke relations of the displayed inverse generators. In the faithful local Hecke action, solving step 2.5 gives precisely the polynomial crossings of step 1.2, whose relations were proved in steps 1.2 and 2.1–2.4. Since $Q$ is a unit, the two operator subalgebras generated by coefficients and their respective crossings coincide. Thus the presentations are isomorphic. On idempotents and dots the two composites are the identity by the spectral decomposition and $X_r=q^{i_r}(1-y_r)$. On crossings the composite is $({\psi }_{r}Q-P+P)Q^{-1}={\psi }_r$; on Hecke crossings it is $(T_r+P)Q^{-1}Q-P=T_r$. These are ordered products; no coefficient has been commuted past a crossing. This completes the local inverse comparison in both directions.