Faithful polynomial representation of a KLR algebra

(Shifts.) Write $p_{ij}:={\mathrm{deg}}_{[i,j]}{P}_{ij}$ and ${\mathrm{\Delta }}_{ij}:=-b_{ij}-p_{ij}$ for $i\mathrm{\ne }j$. By [F2], $p_{ij}+p_{ji}=-2b_{ij}$, so ${\mathrm{\Delta }}_{ji}=-{\mathrm{\Delta }}_{ij}$. Fix the given finite order on $I$ and set $w(\nu ):={\sum }_{a<b,{\nu }_a>{\nu }_b}{\mathrm{\Delta }}_{{\nu }_b{\nu }_a}$. An adjacent unequal swap only changes the contribution of that pair, hence $w(s_a\nu )-w(\nu )={\mathrm{\Delta }}_{{\nu }_a{\nu }_{a+1}}$. This explicit formula supplies the shifts on every sequence; any two shift functions differ by a constant on each fixed-content orbit, since adjacent unequal swaps connect it.

(Spanning.) Move dots to the right using relation (6), inducting on the number of dot/crossing inversions and the crossing length of correction terms. By [F4], any nonreduced transposition word admits finitely many braid moves exposing an adjacent repeat, and any reduced word connects to the chosen one by braid moves. Each braid correction has three fewer crossings and each quadratic substitution removes two; induction on crossing length therefore reduces all correction terms, while the finite chosen sequence of braid moves handles the leading term. Hence the monomials ${\tau }_w{x}^{a}e(\nu )$ span.

(Definition and homogeneity.) Define $e(\nu )$ and $x_a$ as stated and ${\tau }_a$ by the two displayed formulas. Each formula is a finite sum of polynomials times a permutation of the variables, hence a well-defined $k$-linear endomorphism; the equal-colour quotient lies in $k[x_1,\dots ,x_n]$ because $f-s_{a}f$ vanishes on $x_a=x_{a+1}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, Polynomial rings in finitely many commuting indeterminates by iteration). With the shifts of step 1.1, on each word component $x_a$ is homogeneous of degree $2d_{{\nu }_a}$ (the shifts are constant on a summand and do not change dot degrees), and ${\tau }_a$ is homogeneous of degree $-b_{{\nu }_a{\nu }_{a+1}}$: on unequal colours in the target summand $s_a\nu$, the variable $x_{a+1}$ has color $p$ and $x_a$ has color $q$, so the factor $P_{pq}(x_{a+1},x_a)$ has degree ${\mathrm{deg}}_{[p,q]}{P}_{pq}$, so multiplying by it and passing to the summand $s_a\nu$ with shift $w(s_a\nu )=w(\nu )+{\mathrm{\Delta }}_{pq}$ changes the total degree by ${\mathrm{deg}}_{[p,q]}{P}_{pq}+{\mathrm{\Delta }}_{pq}=-b_{pq}$; on equal colours the divided difference has degree $-2d_p=-b_{pp}$ because its numerator has degree $\mathrm{deg}f$ and the denominator has degree $2d_p$.

(Relations (1), (2), (3), (5), (6).) Relations (1)-(3) are immediate from the definitions: the $e(\nu )$ are orthogonal idempotents summing to $1$, $x$'s commute and commute with the $e(\nu )$, and both displayed ${\tau }_a$'s map the $\nu$-summand to the $s_a\nu$-summand. Relation (5) holds because ${\tau }_a$ and ${\tau }_b$ with $\mathrm{\mid }a-b\mathrm{\mid }>1$ act on disjoint sets of variables and their polynomial factors commute. For (6): if $b\mathrm{\notin }\{a,a+1\}$ the multiplication by $x_b$ commutes with $s_a$ and with the factor $P_{{\nu }_a{\nu }_{a+1}}(x_{a+1},x_a)$, giving ${\tau }_a{x}_b=x_b{\tau }_a=x_{s_a(b)}{\tau }_a$; if $b=a$ and ${\nu }_a\mathrm{\ne }{\nu }_{a+1}$ then ${\tau }_a{x}_{a}f=P(x_{a+1},x_a)x_{a+1}{s}_{a}f=x_{a+1}{\tau }_{a}f$, and $b=a+1$ is the same with $x_a$; if ${\nu }_a={\nu }_{a+1}$ the two identities ${\mathrm{\partial }}_a{x}_a-x_{a+1}{\mathrm{\partial }}_a=1$ and ${\mathrm{\partial }}_a{x}_{a+1}-x_a{\mathrm{\partial }}_a=-1$ of [F3] are exactly the two displayed dot moves.

(Relation (4).) On equal colours ${\tau }_a^2={\mathrm{\partial }}_a^2=0=Q_{pp}(x_a,x_{a+1})e(\nu )$ by [F3] and $Q_{pp}=0$. On unequal colours $(p,q)$ a double application gives ${\tau }_a^2(fe(\nu ))=P_{qp}(x_{a+1},x_a)s_a(P_{pq}(x_{a+1},x_a)s_{a}f)=P_{pq}(x_a,x_{a+1})P_{qp}(x_{a+1},x_a)f=Q_{pq}(x_a,x_{a+1})f$, which is relation (4) because $Q_{pq}(u,v)=P_{pq}(u,v)P_{qp}(v,u)$ evaluated at $(u,v)=(x_a,x_{a+1})$ equals $P_{pq}(x_a,x_{a+1})P_{qp}(x_{a+1},x_a)$.

(Relation (7), equal outer colours $(i,j,i)$.) Work on $e(i,j,i)$ in variables $x_1,x_2,x_3=x_a,x_{a+1},x_{a+2}$, write $P=P_{ij}$, $P^{\mathrm{\prime }}=P_{ji}$, $Q=P{P}^{\mathrm{\prime }}(v,u)$, let $s=(12)$, $t=(23)$ and $r=tst=sts=(13)$; write $A={\tau }_{a+1}{\tau }_a{\tau }_{a+1}$ and $B={\tau }_a{\tau }_{a+1}{\tau }_a$. Expanding the definitions on $fe(i,j,i)$ and collecting the outer factor, $Af=P(x_3,x_2)\cdot t\left[{\mathrm{\partial }}_a\right(P^{\mathrm{\prime }}(x_3,x_2)tf\left)\right]$ and $Bf=P^{\mathrm{\prime }}(x_2,x_1)\cdot s\left[{\mathrm{\partial }}_{a+1}\right(P(x_2,x_1)sf\left)\right]$. Using $t{\mathrm{\partial }}_{a}t=(1-r)\mathrm{/}(x_1-x_3)$ and ${\mathrm{\partial }}_{a+1}(hg)={\mathrm{\partial }}_{a+1}(h)g+t(h){\mathrm{\partial }}_{a+1}(g)$ with $h=P^{\mathrm{\prime }}(x_3,x_2)$ first, then the mirrored expansion of $B$ with $h=P(x_2,x_1)$, gives $Af=P(x_3,x_2)[P^{\mathrm{\prime }}(x_2,x_3)f-P^{\mathrm{\prime }}(x_2,x_1)rf]\mathrm{/}(x_1-x_3)$ and $Bf=P^{\mathrm{\prime }}(x_2,x_1)[P(x_1,x_2)f-P(x_3,x_2)rf]\mathrm{/}(x_1-x_3)$. Subtracting, the two $rf$-terms cancel and the remaining coefficient is $P(x_3,x_2)P^{\mathrm{\prime }}(x_2,x_3)-P(x_1,x_2)P^{\mathrm{\prime }}(x_2,x_1)=Q(x_3,x_2)-Q(x_1,x_2)$, so $(A-B)f={\displaystyle \frac{Q(x_1,x_2)-Q(x_3,x_2)}{x_3-x_1}}f$, which is relation (7) on $e(i,j,i)$.

(Relation (7), remaining colours.) If ${\nu }_a={\nu }_{a+1}={\nu }_{a+2}=i$, then all three operators are divided differences and both braid words equal the same antisymmetrizer expression ${\sum }_{w\in S_3}\mathrm{sgn}(w)w\mathrm{/}{\prod }_{i<j}(x_i-x_j)$, so $A-B=0$, consistent with $Q_{ii}=0$; if exactly the outer two colours agree, step 4.1 applies with the two adjacent colours distinct; if only the first two colours agree, say $(i,i,j)$, then with $h=P_{ij}(x_2,x_1)P_{ij}(x_3,x_1)$, which is symmetric in $x_2,x_3$, the two expansions give $Af={\mathrm{\partial }}_{a+1}(h\cdot stf)$ and $Bf=h\cdot st{\mathrm{\partial }}_a(f)$, and the product rule plus the Demazure identity ${\mathrm{\partial }}_{a+1}st=st{\mathrm{\partial }}_a$ valid for the $3$-cycle $w=st$ yield $A-B=0$; the case $(i,j,j)$ is the mirror image, obtained by reading the same identities from the other end. If the three colours are distinct, both $A$ and $B$ act as the same product $P_{ij}{P}_{ik}{P}_{jk}$ (with arguments placed by the two expansions, all three factors read off the three crossed pairs) times the longest permutation $r$, and $A=B$; the expansion is by the same substitution $t{\mathrm{\partial }}_{a}t=(1-r)\mathrm{/}(x_1-x_3)$ as in step 4.1 applied to three distinct colours, where the two middle expressions coincide because the two factorizations of the long element have the same length. Hence relation (7) holds in every case.

(The homomorphism.) Steps 3.1, 3.2, 4.1 and 5.1 verify relations (1)-(7) for the operators, and step 2.1 shows each word-component generator acts homogeneously of its assigned degree, so the universal property of the presented algebra gives a homomorphism $\rho :R_n(Q)\to {\mathrm{E}\mathrm{n}\mathrm{d}}_k^{\mathrm{g}\mathrm{r}}({\mathrm{P}\mathrm{o}\mathrm{l}}_n\{w\})$ of graded algebras with $\rho (R_n(Q{)}_m){\mathrm{P}\mathrm{o}\mathrm{l}}_{n,d}\subseteq {\mathrm{P}\mathrm{o}\mathrm{l}}_{n,d+m}$.

(Faithfulness.) Localize for this step only at the nonzero elements $\{P_{ij}(x_a,x_b):{\nu }_a\mathrm{\ne }{\nu }_b\}$ and $\{x_a-x_b:{\nu }_a={\nu }_b\}$; the localized operators of a reduced word $w=s_{i_1}\cdots s_{i_r}$ on a fixed source sequence $\nu$ have the form $c_w\cdot w+{\sum }_{u<w}{c}_{u}u$ where $c_w\mathrm{\ne }0$ is the product of the nonzero $P$-factors (unequal crossings) and the nonzero coefficient $-1\mathrm{/}(x_a-x_{a+1})$ (equal crossings), and the terms $u<w$ are strictly shorter permutations. Suppose ${\sum }_{\nu }{\sum }_{w,a}{c}_{w,a}{\tau }_w{x}^{a}e(\nu )$ acted as zero; for the maximal $w$ occurring in a fixed source/destination block the leading terms give a nontrivial linear relation ${\sum }_w{d}_{w}w=0$ among distinct automorphisms $w$ of the field $k(x_1,\dots ,x_n)$. Distinct automorphisms are linearly independent: take a relation with the least positive number of nonzero coefficients; one term cannot vanish on $1$. With at least two terms, choose $w\mathrm{\ne }{w}_0$ and $g$ with $w(g)\mathrm{\ne }{w}_0(g)$, evaluate on $fg$, and subtract $w_0(g)$ times the evaluation on $f$. The $w_0$ term disappears and the $w$ coefficient stays nonzero, contradicting minimality; hence no nonzero combination of the spanning monomials of step 1.2 acts as zero and $\rho$ is injective.

(Localized isomorphism and finiteness.) Inverting the factors and differences makes every unequal-colour crossing ${\tau }_a$ equal to $s_a$ up to an invertible scalar, and every equal-colour crossing satisfy $s_a=1-(x_a-x_{a+1}){\mathrm{\partial }}_a$, so the localized image contains all variable permutations $s_a$ and all multiplications $x_a$, i.e. the localization of the skew group algebra $A_n(I)=({⨁}_{\nu \in I^n}k[x_1,\dots ,x_n]e(\nu ))⋊S_n$; with injectivity from step 7.1 and the evident containment of the image in the localized algebra, $\rho$ becomes an isomorphism after localization. For finiteness of homogeneous components: by step 1.2 the monomials with $w\in S_n$ and dot exponents $a_j\ge 0$ span $R_n(Q)$; for fixed weight $\beta$ and degree $m$ the finitely many $w$ and $\nu$ occur, each crossing degree is one of finitely many integers bounded below, and each dot has positive degree $2d_{{\nu }_a}$, so the total dot degree of a spanning monomial of degree $m$ is bounded, leaving only finitely many exponent tuples; hence ${\dim }_{k}R(\beta {)}_m<\mathrm{\infty }$.