Cyclotomic Hecke and Quiver Hecke Algebras — Examples

The common degree of the two words. The three crossings of ${\psi }_r{\psi }_{r+1}{\psi }_{r}e(i)$ cross the colour pairs $(i_r,i_{r+1})$, $(i_r,i_{r+2})$, $(i_{r+1},i_{r+2})$ of the successive permuted sequences, so its degree is $d(i)$; the other braid word has the same multiset of crossed pairs, hence the same degree $d(i)$.

The four cases of the correction. Substituting the cyclic Cartan entries: (a) if $i_{r+2}=i_r\to i_{r+1}$ then $d(i)=-(-1)-2-(-1)=0$ and the correction is $+1$; (b) if $i_{r+2}=i_r\leftarrow i_{r+1}$ then $d(i)=0$ and the correction is $-1$; (c) if $e=2$ and $i_{r+2}=i_r\rightleftarrows i_{r+1}$, then $a_{i_r{i}_{r+1}}=a_{i_{r+1}{i}_{r+2}}=-2$ and $a_{i_r{i}_{r+2}}=a_{00}=2$, so $d(i)=2+2-2=2$, matching the degree $2$ of $y_r+y_{r+2}-2y_{r+1}$; (d) in all other cases the correction is $0$. This is the source's case list, and the nonzero correction requires equal outer colours, while zero corrections remain homogeneous without asserting that their degree is zero.

Conclusion. In each nonzero case the correction term has degree $d(i)$, equal to the common degree of the two braid words computed in step 1.1; in the remaining cases the correction vanishes, so the identity is homogeneous at degree $d(i)$. Hence the single-edge corrected braid relations have degree $0$ and the $e=2$ double-edge corrected braid relation has degree $2$, both matching the two sides. The displayed source word order is ${\psi }_r{\psi }_{r+1}{\psi }_r-{\psi }_{r+1}{\psi }_r{\psi }_{r+1}$, which equals the native opposite-order $\tau$-difference under $\psi =-\tau$. No choice is used.

The two residue projectors and the $y$-values. Put $C=c+a+s$. Group multiplication in ${\mathbb{F}}_3{S}_3$ gives $C^2=0$, because the three squares contribute $3=0$ and the six cross products contribute three copies of each 3-cycle; and $Cc=Cs=1+(123)+(132)$. Put $e_{+}=(1+c)\mathrm{/}2$ and $e_{-}=(1-c)\mathrm{/}2$; since $c^2=1$ and $2$ is invertible in ${\mathbb{F}}_3$, these are idempotents, and they are the actual joint CRT projectors $e(0,1,2)$ and $e(0,2,1)$ of the two standard residue words. With $x_3=C-c$ one computes $y_1=y_2=0$ and $y_3=C$: indeed $(c-1)e_{+}=(c-1)(1+c)\mathrm{/}2=0$, $(c-2)e_{-}=(c+1)(1-c)\mathrm{/}2=0$, and $x_3-i_3$ equals $C$ on both projectors. All other residue words vanish, and both projectors are nonzero by the group basis.

There is exactly one block. The full center has basis $1$, $C$, $1+(123)+(132)$ and its augmentation ideal is spanned by the last two elements, with $(1+(123)+(132){)}^2=0$ and $C(1+(123)+(132))=0$; hence the augmentation ideal is square-zero and contains no nonzero idempotent. Therefore the only central idempotents are $0$ and $1$, so ${\mathbb{F}}_3{S}_3$ is a single block and the two projectors combine as $e_{\alpha }=e_{+}+e_{-}=1$.

The $\psi$-generators and the explicit identities. For the explicit source choice of the power series, ${\psi }_1=0$. On $e_{+}$ one has $p_2=-1+C$ and $q_2=-1$; on $e_{-}$ one has $p_2=1+C$ and $q_2=1$. Thus ${\psi }_2{e}_{+}=(1-s-C)e_{+}$ and ${\psi }_2{e}_{-}=(s+1+C)e_{-}$. Direct group multiplication using $c^2=s^2=1$ and $csc=scs$ gives ${\psi }_2{e}_{+}=e_{-}{\psi }_2$, ${\psi }_2{e}_{-}=e_{+}{\psi }_2$, ${\psi }_2^2{e}_{+}=C{e}_{+}$, ${\psi }_2^2{e}_{-}=-C{e}_{-}$, and $C{\psi }_2={\psi }_{2}C=0$. These are exactly the claimed mixed and quadratic identities with $y_3=C$ and $y_2=0$; the first-crossing mixed identities vanish on every surviving unequal-colour word, equal-first-colour words are absent, so the universal equal-colour formula also holds. The six group elements $1,c,s,cs,sc,csc$ give dimension $6$, with no primitive-block citation or Hecke-basis dimension loop hidden.

The cyclotomic relation on $I^2$. Enumerate $I^2=\{0,1,2{\}}^2$. The cyclotomic exponent of the sequence $i$ is the pairing $({\mathrm{\Lambda }}_0,{\alpha }_{i_1})={\delta }_{i_1,0}$, so the defining cyclotomic relation reads $y_1^{{\delta }_{i_1,0}}e(i)=0$. For the three sequences with $i_1=0$ this is $y_{1}e(i)=0$; for the six sequences with $i_1\mathrm{\ne }0$ the exponent is zero and the relation is $y_1^{0}e(i)=e(i)=0$, so those idempotents vanish rather than merely a degree-one dot relation.

The quadratic relation. The KLR quadratic relation distinguishes the colour cases: $i_1=i_2$, $i_2=i_1+1$, $i_2=i_1-1$. For $e=3$ and $I=\{0,1,2\}$ every ordered pair of distinct colours is adjacent in the $3$-cycle, so these are all cases, and substituting gives ${\psi }_1^{2}e(i)=0$, $(y_2-y_1)e(i)$, and $(y_1-y_2)e(i)$ respectively.

The two nonzero blocks. By Cyclotomic Hecke algebras of level one are finite Hecke algebras the level-one algebra $H_2^{\mathrm{\Lambda }}$ is the finite Hecke algebra $H_{F,q}(S_2)$ with standard basis $\{1,T_1\}$, so it is two-dimensional, and $T_1^2-(q-1)T_1-q=0$ gives $H_2^{\mathrm{\Lambda }}\cong F[T]\mathrm{/}((T-q)(T+1))\cong F\times F$, the two factors being distinct because $q\mathrm{\ne }1$ and $q\mathrm{\ne }-1$ for a primitive cube root in characteristic $\mathrm{\ne }3$. Hence there are exactly two nonzero one-dimensional blocks. Their labels come from the polynomial generator: on the trivial module $X_1=q^0=1$ and $X_2=q^{-1}{T}_1{X}_1{T}_1=q^{-1}{q}^2=q$, giving residue sequence $(0,1)$; on the sign module $X_1=1$ and $X_2=q^{-1}(-1{)}^2=q^{-1}=q^2$, giving $(0,2)$; equivalently these are the $q$-contents of the standard tableaux of $(2)$ and $(1,1)$ (The Hecke Jucys-Murphy action is dominance triangular). Every nonzero block contains at least one of these surviving idempotents, so the remaining four blocks of height $2$ are zero.

The positive-rank affine algebra is infinite dimensional. At rank one the affine presentation is $F[X_1^{\pm 1}]$ in NG and $F[x_1]$ in DG by [F1], and their distinct powers are independent. More generally for any $d\ge 1$, the NG assignments $T_r\mapsto q_H$, $X_r\mapsto q_H^{r-1}t$ give a surjection onto $F[t,t^{-1}]$: the quadratic relation reads $q_H^2=(q_H-1)q_H+q_H$, the Bernstein relation reads $q_H^2{q}_H^{r-1}t=q_H{q}_H^{r}t$, and the commuting and braid relations are scalar identities. In DG the assignments $s_r\mapsto 1$, $x_r\mapsto t+(r-1)1_F$ similarly give a surjection onto $F[t]$, since the adjacent mixed relation reads $t+(r-1)=(t+r)-1$ and all other relations hold. The images contain $t$, so the affine algebras are infinite-dimensional at every positive rank. Every cyclotomic block is finite-dimensional by [F2], and its KLR partner has the same dimension by [F4]; thus neither is isomorphic to the whole positive-rank affine algebra. At rank zero both presentations give $F$, and the obstruction does not apply.

The whole cyclotomic quotient is not one block. Take $\mathrm{\Lambda }={\mathrm{\Lambda }}_0$ (level one), $d=2$, $e=3$ and $q_H$ a primitive cube root of unity. By the level-one identification [F3] the algebra is $H_{F,q}(S_2)\cong F[T]\mathrm{/}((T-q_H)(T+1))\cong F\times F$, since $q_H\mathrm{\ne }-1$ in characteristic $\mathrm{\ne }3$. Here $X_1=1$ and $X_2=q_H^{-1}{T}_1^2$, so the two factors $T_1=q_H$ and $T_1=-1$ have polynomial eigenwords $(1,q_H)$ and $(1,q_H^2)$, hence residue words $(0,1)$ and $(0,2)$. Thus there are two nonzero blocks $H_{{\alpha }_0+{\alpha }_1}^{\mathrm{\Lambda }}$ and $H_{{\alpha }_0+{\alpha }_2}^{\mathrm{\Lambda }}$, each one dimensional with residue sequences $(0,1)$ and $(0,2)$. Therefore $H_2^{\mathrm{\Lambda }}$ has dimension $2$ and is not isomorphic to either one-dimensional block; on the KLR side [F4] gives $R_2^{\mathrm{\Lambda }}=R_{{\alpha }_0+{\alpha }_1}^{\mathrm{\Lambda }}\oplus R_{{\alpha }_0+{\alpha }_2}^{\mathrm{\Lambda }}$, which is not isomorphic to a single block.

Conclusion. Both branches of the attempted replacement fail: the affine algebra is infinite dimensional, and the whole cyclotomic quotient decomposes into more than one block. Summing the central-sector isomorphisms on both sides remains valid.