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The degenerate level-one Hecke–KLR map for S_3

Example

Work over F=F3 with q=1 (so e=3), Λ=Λ0 at level one and d=3: then H3Λ≅F3S3 (Cyclotomic Hecke algebras of level one are finite Hecke algebras, Affine and cyclotomic type-A Hecke algebras) and exactly one block is nonzero, namely α=α0+α1+α2 with eα=1 (Residue-sequence idempotents and cyclotomic Hecke blocks). The residue sequences realised by standard tableaux all have residue multiset {0,1,2}. The degenerate KLR generators are the elements yr=∑i(xr−ir)e(i) with xr the Jucys–Murphy elements of F3S3 (so x1=0, x2=(12), x3=(13)+(23)) and ψr=∑iφrqr(i)−1e(i) with φr=sr+pr(i) and pr(i)=(ir−ir+1+yr−yr+1)−1 on unequal components (Cyclotomic Hecke intertwiners and local power series, Hecke intertwiners produce the local KLR generators). In this block the mixed relations ψ1y2e(i)=y1ψ1e(i)+e(i) for i1=i2 and ψ1y2e(i)=y1ψ1e(i) for i1≠i2, and the quadratic relation ψ12e(i)=(y2−y1)e(i) for i1→i2, are explicit identities in the six-dimensional algebra F3S3.

Verification

technique · direct

Given: The group algebra F3S3 with the transpositions

c=(12), s=(23), a=(13), the additive Jucys–Murphy elements

x1=0, x2=c, x3=(13)+(23), and the level-one conventions of the

Statement.

1.1givenalgebra

The two residue projectors and the y-values. Put C=c+a+s. Group multiplication in F3S3 gives C2=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)/2 and e−=(1−c)/2; since c2=1 and 2 is invertible in F3, 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 x3=C−c one computes y1=y2=0 and y3=C: indeed (c−1)e+=(c−1)(1+c)/2=0, (c−2)e−=(c+1)(1−c)/2=0, and x3−i3 equals C on both projectors. All other residue words vanish, and both projectors are nonzero by the group basis.

1.2givenalgebra

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 F3S3 is a single block and the two projectors combine as eα=e++e−=1.

2.1step 1.1step 1.2algebra∎

The ψ-generators and the explicit identities. For the explicit source choice of the power series, ψ1=0. On e+ one has p2=−1+C and q2=−1; on e− one has p2=1+C and q2=1. Thus ψ2e+=(1−s−C)e+ and ψ2e−=(s+1+C)e−. Direct group multiplication using c2=s2=1 and csc=scs gives ψ2e+=e−ψ2, ψ2e−=e+ψ2, ψ22e+=Ce+, ψ22e−=−Ce−, and Cψ2=ψ2C=0. These are exactly the claimed mixed and quadratic identities with y3=C and y2=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.

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