Ariki Categorification and Graded Decomposition Numbers

The permutation criterion. Let $t=(ab)$ with $a<b$ act on values and write $p=x^{-1}(a)$, $q=x^{-1}(b)$. If $p<q$, direct inversion counting gives $\mathrm{\ell }(tx)-\mathrm{\ell }(x)=1+2\mathrm{\#}\{j:p<j<q,a<x(j)<b\}>0$; if $p>q$ the difference is its negative. A positive difference need not equal one. The rank inequalities in [F5] show $x<tx$ precisely in the first case: the rank count decreases by one for $p\le h<q$ and $a\le k<b$, and is unchanged elsewhere. Thus every displayed earlier-node transposition lowers strong Bruhat order. Conversely a saturated strong-Bruhat chain from $w_S$ to $w_T$ consists of transposition covers by [F6], each of whose inverse lowering move has the earlier larger value by the formula just proved. Reading this chain backwards yields the displayed transposition criterion.

Agreement outside the Garnir belt. In the row-reading initial labelling the two pieces of the belt form one contiguous interval of positions $[L,U]$: the upper-row tail is immediately followed by the lower-row prefix. A Garnir tableau changes labels only inside this interval, so $w_T$ is in the parabolic subgroup generated by $s_L,\dots ,s_{U-1}$. Sorting its interval permutation gives a reduced expression using only these generators. The Bruhat subword criterion applied to this expression says that every $w_S\le w_T$ likewise uses only these generators. Hence $w_S$ fixes all positions outside $[L,U]$, proving agreement with $t^{\mu }$ outside the belt. This applies even to nonstandard $S$.

The prefix-shape criterion. For a row-strict $W$ and a value $c$, row-strictness makes the entries $\le c$ in each row a prefix, so $\mathrm{sh}(W^{\le c})$ is a row sequence of size $c$, and for $p=P_k+t$ with $P_k$ the number of nodes in the first $k$ rows and $0\le t<{\mu }_{k+1}$ we get $N_W(p,c):=\mathrm{\#}\{i\le p:\text{entry }i\text{ of }W\le c\}=A_k(W,c)+\min (t,s_{k+1}(W,c)),$ where $A_k(W,c)$ is the number of entries $\le c$ in the first $k$ rows and $s_{k+1}(W,c)$ the prefix length in row $k+1$. By [F5], $w_S\le w_T$ iff $N_S(p,c)\ge N_T(p,c)$ for all $p,c$. At $p=P_k$ this says $A_k(S,c)\ge A_k(T,c)$ for all $k,c$, which is exactly $\mathrm{sh}(S^{\le c})⊵\mathrm{sh}(T^{\le c})$ for all $c$; hence the prefix criterion is necessary. Conversely assume all these prefix inequalities, fix $p=P_k+t$ and put $a=A_k(S,c)-A_k(T,c)\ge 0$, $s=s_{k+1}(S,c)$, $s^{\mathrm{\prime }}=s_{k+1}(T,c)$; the inequality at $k+1$ gives $a+s\ge s^{\mathrm{\prime }}$. If $t\le s$ then $a+\min (t,s)=a+t\ge \min (t,s^{\mathrm{\prime }})$, because $t\le s^{\mathrm{\prime }}$ gives $a+t\ge t$ and $t>s^{\mathrm{\prime }}$ gives $a+t\ge t\ge s^{\mathrm{\prime }}$; if $t>s$ then $a+\min (t,s)=a+s\ge s^{\mathrm{\prime }}\ge \min (t,s^{\mathrm{\prime }})$. Thus $N_S(p,c)\ge N_T(p,c)$ for all $p,c$ and $S⊴T$. Since the first $k$ rows of $t^{\mu }$ contain $\min (c,P_k)$ entries $\le c$, the maximum possible, $\mathrm{sh}(t^{\mu \le c})$ dominates every $\mathrm{sh}(W^{\le c})$, so $t^{\mu }$ is the least element of $⊴$ on row-strict tableaux.

Consecutive swaps and the edges of the weak graph. Let $T$ be standard. If $r,r+1$ share a row, row-strictness puts $r$ immediately left of $r+1$ and the swap reverses an adjacent pair, so $s_{r}T$ is not row-strict; if they share a column, column-strictness puts $r$ immediately above $r+1$ and the swap reverses an adjacent column pair, so $s_{r}T$ is not column-strict. If in one component $r+1$ lies strictly above and strictly left of $r$, let $x$ be the entry of $T$ in the node in the row of $r$ and the column of $r+1$: row-strictness gives $x<r$ while column-strictness gives $x>r+1$, impossible, and symmetrically $r$ cannot lie strictly above and strictly left of $r+1$. The only remaining same-component configurations are $r{\nearrow }_{T}r+1$ and $r+1{\nearrow }_{T}r$; in either, the nodes sharing a row or column with the two positions are unchanged except for the interchange of the consecutive values $r,r+1$, so all strict neighbour inequalities survive and $s_{r}T$ is standard. If $r,r+1$ lie in different components the swap is standard as well. This proves the stated equivalence. For the edge criterion note that $S=s_{r}T$ with both standard and $r$ earlier than $r+1$ in $T$: in one component the alternatives of the equivalence leave only $r+1{\nearrow }_{T}r$ (in the other alternative $r+1$ is in a higher row and hence earlier, and same-row or same-column swaps are excluded), while in different components $r$ is earlier exactly when its component is earlier.

Row sorting. Let $S,T$ be $\mu$-tableaux with $S⊴T$; by steps 1.1 and 2.1 it suffices to treat a single transposition $S=(ab)T$ with $a<b$ and $b$ earlier than $a$ in $T$, the general case following by composing the finitely many single-swap comparisons. Write $S^{+},T^{+}$ for the tableaux obtained by sorting every row increasingly. If $a,b$ share a row of $T$, then $S$ and $T$ have the same row sets and $S^{+}=T^{+}$. Otherwise the node of $b$ lies in a strictly earlier row than the node of $a$: they are in different rows by the previous sentence, and $b$ earlier than $a$ forces $b$'s row to be the earlier one. Entries other than $a,b$ occupy the same nodes in $S$ and $T$. For a value $c$, the prefix row lengths of $S^{+}$ and $T^{+}$ at $c$ are the rows' counts of entries $\le c$; if $c<a$ or $c\ge b$ these counts agree, while if $a\le c<b$ the count vector of $S^{+}$ is obtained from that of $T^{+}$ by moving one unit from the row of $a$ to the earlier row of $b$. Moving a unit to an earlier row raises every partial sum of the row sequence, so $\mathrm{sh}(S^{+\le c})⊵\mathrm{sh}(T^{+\le c})$ for every $c$; both tableaux are row-strict, so step 2.1 gives $S^{+}⊴T^{+}$. Hence $S⊴T$ implies $S^{+}⊴T^{+}$ in general.

The vertical consecutive swap and its exceptional prefix. Assume $\mu$ is a multipartition, $T$ standard, $r{\downarrow }_{T}r+1$, so $r$ and $r+1$ occupy adjacent rows $a,a+1$ of one component in the same column $b$; the neighbour check of step 3.1 shows that $U:=s_{r}T$ is row-strict. Let $S$ be standard with $S⊲U$; we prove $S⊴T$. By step 2.1, $\mathrm{sh}(S^{\le c})⊵\mathrm{sh}(U^{\le c})$ for all $c$, and for $c\mathrm{\ne }r$ the tableaux $U^{\le c}$ and $T^{\le c}$ occupy the same nodes, so their shapes agree. For $c=r$ let $A_k$ be the number of entries $\le r-1$ in the first $k$ rows of $T$ (the same for $U$) and note that $\mathrm{sh}(U^{\le r})$ has row lengths $A_k-A_{k-1}+[k=a+1]$ while $\mathrm{sh}(T^{\le r})$ has $A_k-A_{k-1}+[k=a]$; thus $\mathrm{sh}(T^{\le r})$ is obtained from $\mathrm{sh}(U^{\le r})$ by moving one node from row $a+1$ up to row $a$. Because $T$ is standard, the node of $r$ in row $a$ has column $b$, so exactly $b-1$ entries of row $a$ lie in columns $<b$, all of them $\le r-1$, and the same count holds in row $a+1$ whose node in column $b$ is occupied by $r+1$; hence $A_a-A_{a-1}=b-1=A_{a+1}-A_a$ and the switch hypothesis $\mathrm{sh}(U^{\le r}{)}_a-\mathrm{sh}(U^{\le r}{)}_{a+1}=(b-1)-((b-1)+1)=-1$ holds. Lemma 3.1 of the source, proved here: if $\nu$ is a row sequence and ${\nu }_a-{\nu }_{a+1}=-1$, and $\widehat{\nu }$ interchanges the two consecutive parts ${\nu }_a,{\nu }_{a+1}$, then $\widehat{\nu }$ has the same partial sums as $\nu$ except at index $a$, where they are larger by one; consequently any multipartition $\lambda$ with $\lambda ⊵\nu$ also satisfies $\lambda ⊵\widehat{\nu }$: if its partial sum at $a$ equalled that of $\nu$, the inequalities at $a-1$ and $a+1$ would force ${\lambda }_a\le {\nu }_a<{\nu }_{a+1}\le {\lambda }_{a+1}$, contradicting the weakly decreasing parts in that component. Thus its partial sum at $a$ is at least one larger; other partial sums are unchanged. Apply this with $\lambda =\mathrm{sh}(S^{\le r})$, $\nu =\mathrm{sh}(U^{\le r})$, $\widehat{\nu }=\mathrm{sh}(T^{\le r})$: the row sequence of a standard tableau's prefix is a multipartition, and the computed switch identity verifies the hypothesis, so $\mathrm{sh}(S^{\le r})⊵\mathrm{sh}(T^{\le r})$. With the equalities for $c\mathrm{\ne }r$, step 2.1 gives $S⊴T$. In the same-row case $r{\to }_{T}r+1$ one has $T=U^{+}$ and $S=S^{+}$, so step 3.2 gives $S⊴T$ directly; this proves both clauses of the vertical exceptional-prefix fact.

The four terminal alternatives (horizontal). Let $\mu$ be a multipartition and $T$ be standard with $r{\to }_{T}r+1$. Consider the three properties: (a) there is $t$ with $\mathrm{\mid }t-r\mathrm{\mid }>1$ such that $t{\nearrow }_{T}t+1$ or $t$ lies in a later component than $t+1$; (b) $r+1{\nearrow }_{T}r+2$ or $r+1$ lies in a later component than $r+2$; (c) $r-1{\nearrow }_{T}r$ or $r-1$ lies in a later component than $r$. By step 3.1 an incoming edge at $t$ (that is, $s_{t}T$ standard with $t$ earlier than $t+1$ in $s_{t}T$) is equivalent to $t{\nearrow }_{T}t+1$ or $t$ in a later component than $t+1$. If (a) holds, then $X:=s_{t}T$ is standard and $X\stackrel{t}{\to }T$ is an edge, so $T$ is $t$-terminal, and since $t\mathrm{\notin }\{r,r+1\}$ the relation $r{\to }_{T}r+1$ persists in $X$; this is alternative (1). If (b) holds with $r+1{\nearrow }_{T}r+2$, then in $X:=s_{r+1}T$ the entry $r+1$ occupies the higher node of $r+2$, so $X\stackrel{r+1}{\to }T$ is an edge; since $r,r+1$ are adjacent in one row and $r+2$ is above and right, $r+1$ is above and right of $r$ in $X$, so $Y:=s_{r}X$ is standard and $Y\stackrel{r}{\to }X$ is an edge with $r$ moving to the higher node, and in $Y$ the entries $r+1,r+2$ are adjacent in one row with $r+1$ to the left; this is alternative (2). If (b) holds with $r+1$ in a later component than $r+2$, the same construction with component arguments gives (2). If (c) holds, the mirror construction gives (3). It remains to observe that if none of (a),(b),(c) holds, then $T$ has no incoming edge: indeed (a) is exactly the incoming-edge alternative at any $t\mathrm{\notin }\{r-1,r,r+1\}$, (c) at $t=r-1$, and (b) at $t=r+1$, while at $t=r$ both alternatives of step 3.1 fail because $r$ and $r+1$ lie in one row with $r$ to the left. A standard tableau with no incoming edge is $t^{\mu }$: if $T\mathrm{\ne }{t}^{\mu }$, choose the least $t$ with the node of $t+1$ earlier than the node of $t$ in $T$ (such $t$ exists because otherwise the entries occur in the order $1,2,\dots ,d$); then standardness forces $t{\nearrow }_{T}t+1$ or an earlier component for $t+1$, so the incoming-edge alternative at $t$ holds and gives the existence of an edge into $T$, a contradiction. Hence in this final case $T=t^{\mu }$, which is alternative (4).

The vertical terminal alternatives. Let $\mu$ be a multipartition and $r{\downarrow }_{T}r+1$ at $(a,b,n)$, so the two nodes are vertically adjacent. Use the following precise tests: (a) an incoming edge $t$ with $\mathrm{\mid }t-r\mathrm{\mid }>1$; (b) $r{\nearrow }_{T}r+2$ or $r+1$ is in a later component than $r+2$; (c) $r-1{\nearrow }_{T}r+1$ or $r-1$ is in a later component than $r$. Test (a) gives $s_{t}T\stackrel{t}{\to }T$ and preserves the vertical pair. In test (b), standardness and the consecutiveness of the three entries force $s_{r+1}T$ and $s_r{s}_{r+1}T$ to be standard: the node of $r+2$ is above and right of $r$, hence above and right of the node of $r+1$, and after the first swap the nodes of $r,r+1$ are incomparable. Thus $s_r{s}_{r+1}T\stackrel{r}{\to }{s}_{r+1}T\stackrel{r+1}{\to }T$, while $r+1,r+2$ occupy the original vertical pair in $s_r{s}_{r+1}T$. The different-component case uses the same two swaps without any within-component inequality. Test (c) gives the analogous $s_r{s}_{r-1}T\stackrel{r}{\to }{s}_{r-1}T\stackrel{r-1}{\to }T$, with $r-1,r$ in the original vertical pair. If none of the tests holds, labels other than the triple adjacent to $r$ cannot have an inversion between consecutive row-reading positions: any such inverse-order consecutive pair is in incomparable nodes or distinct components and is exactly an incoming edge. Reading $r-1,r-2,\dots ,1$ backwards and $r+2,r+3,\dots ,d$ forwards consequently places them in row-reading order except for the lower-row prefix before the vertical pair and the upper-row tail after it; tests (c) and (b) exclude any further excursion. More explicitly, if $L$ is the initial label at $(a,b,n)$ and $U$ the initial label at $(a+1,b,n)$, the forced belt filling is $L,\dots ,L+b-2$ in the first $b-1$ lower-row positions, $r=L+b-1$ in $(a,b,n)$, $r+1=L+b$ in $(a+1,b,n)$, and $L+b+1,\dots ,U$ in the remaining upper-row tail; all outside entries are unchanged. Among standard belt fillings this is Bruhat-maximal: the $b$ lower-row entries are the lexicographically smallest possible increasing subset, namely $L,\dots ,L+b-2,L+b$, because choosing the $b$ smallest would violate column $b$. Every other valid subset has no smaller prefix counts in the upper row, hence is below it by step 2.1. This is exactly the Garnir tableau, including $b=1={\mu }_a$. A horizontal brick is a consecutive same-row string. All selections, chains and comparisons are finite, so no choice principle is used.

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.