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Self-dual shift normalization for a cyclotomic projective

Example

For rank one di=1 and λ≥1, let B=k[x]/xλ and P=B. The PROJECTIVE Hom dual satisfies P#≅P, so the canonical normalized projective is P with no shift. Its finite vector dual is different: DψP≅P{−2(λ−1)}. The unique vector-dual-centered shift of this regular module is P{−(λ−1)}. For λ=1 both centers agree; for λ=2, projective Hom-self-duality centers P at 0 while vector-self-duality centers it at −1. If λ=0 the block is zero and there is no indecomposable projective here. For any normalized cyclotomic projective the general relation is DψP≅P#{−2def⁡Λ(β)}; it does not redefine the lower-basis projective normalization.

Verification

Given: The rank-one cyclotomic quotient B=k[x]/xλ of The cyclotomic KLR quotient at a dominant integral weight with di=1, λ≥1, and the two dualities of Cyclotomic diagram duality and graded module duals.

[F1] The vector dual DψM=HOM⁡k(M,k)ψ reverses internal shifts and satisfies Dψ2M≅M; the projective dual P#=HOM⁡B(P,B)ψ is a finite graded projective and the symmetric theorem gives DψP≅P#{tβ} with tβ=−2def⁡Λ(β) (Cyclotomic KLR blocks are finite-dimensional graded symmetric algebras).

1.1F1

Projective Hom self-duality. Evaluation HOM⁡B(B,B)≅B gives P#=HOM⁡B(B,B)ψ≅B=P directly, the generator-fixing anti-involution producing the same left module; hence the canonical normalized projective is P with no shift.

2.1step 1.1F1

Vector duality and the shift. The functional ϵ(xa)=0 for a<λ−1 and ϵ(xλ−1)=1 is the coefficient-of-xλ−1 trace; it has degree −2(λ−1) and its anti-diagonal pairing ϵ(xaxb)=δa+b,λ−1 is perfect over every field, so DψB≅B{−2(λ−1)}. Since Dψ(P{s})=DψP{−s}, vector self-duality of P{s} solves −2(λ−1)−s=s, namely s=−(λ−1), while projective Hom self-duality solves −s=s, namely s=0.

3.1step 2.1F1∎

The general defect shift and the boundary. For a general normalized cyclotomic projective the same computation with the symmetric theorem gives DψP≅P#{−2def⁡Λ(β)}, which is a comparison of the two dualities and does not redefine the projective normalization. At λ=1 the two centers agree; at λ=2 they differ by 1 as computed; at λ=0 the defining relation is 1=0 and the block vanishes, so there is no indecomposable projective here. The trace is never halved to pick the lower-basis class.

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