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The rank-one De Concini–Kac to Lusztig specialization map is not injective

Statement

Let ℓ>2 be odd and ζ a primitive ℓ-th root of unity. The generator assignment ϕζ:UζDK(sl2)⟶UζL(sl2),E↦E,F↦F,K±1↦K±1 is a well-defined algebra homomorphism, but it is not injective: it kills Eℓ, which is nonzero in its source.

Facts & Assumptions

Given: The rank-one forms of De Concini–Kac, Lusztig, and small quantum-group forms at roots of unity, the divided-power integral form The Lusztig divided-power integral form UA, and a primitive odd root ζ of order ℓ>2.

[F1]

The rank-one De Concini–Kac source is generated by E,F,K±1 with KEK−1=ζ2E, KFK−1=ζ−2F, and EF−FE=(K−K−1)/(ζ−ζ−1) (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, De Concini–Kac, Lusztig, and small quantum-group forms at roots of unity). An assignment satisfying these three-generator relations extends to a homomorphism from the presented rank-one source.

[F2]

In the Lusztig integral form, EF−FE=T1,0,1=(K−K−1)/(v−v−1) before specialization; the toral product defining T1,0,1 specializes by substituting v=ζ (Divided-power multiplication, toral binomials and the Ei(a)Fi(b) reordering identity, The Lusztig divided-power integral form UA).

[F3]

Eℓ is nonzero in the source by the rank-one PBW basis, and the divided power E(ℓ) is an element of the Lusztig form (The rank-one root-power quotient is a finite-dimensional Hopf algebra, The Lusztig divided-power integral form UA).

[F4]

Eℓ=[ℓ]v!E(ℓ) in the generic divided-power algebra, by the definition E(ℓ)=Eℓ/[ℓ]v! (The Lusztig divided-power integral form UA, Quantum integers, factorials, Gaussian binomials and divided powers at qi).

Proof

1.1givenF1F2constructalgebra

The denominator ζ−ζ−1 is nonzero because ℓ>2. The images of E,F,K±1 in UζL(sl2) satisfy the two toral conjugation relations by specialization of the generic defining relations. By [F2], their commutator is the specialization of T1,0,1, which equals (K−K−1)/(ζ−ζ−1). Hence they satisfy all rank-one De Concini–Kac defining relations, and the presentation's universal property gives the stated homomorphism.

1.2F4algebra

By [F4], the image of Eℓ is [ℓ]ζ!E(ℓ). Since [ℓ]ζ=(ζℓ−ζ−ℓ)/(ζ−ζ−1)=0, the product [ℓ]ζ! is zero, so ϕζ(Eℓ)=0.

2.1step 1.2F3∎

By [F3], Eℓ is nonzero in the source. Thus ϕζ has a nonzero kernel and is not injective.

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