2018
DOI: 10.2140/pjm.2018.292.305
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Noncommutative geometry of homogenized quantum 𝔰𝔩(2, ℂ)

Abstract: This paper examines the relationship between certain non-commutative analogues of projective 3space, P 3 , and the quantized enveloping algebras Uq(sl2). The relationship is mediated by certain non-commutative graded algebras S, one for each q ∈ C × , having a degree-two central element c such that S[c −1 ]0 ∼ = Uq(sl2). The non-commutative analogues of P 3 are the spaces Proj nc (S). We show how the points, fat points, lines, and quadrics, in Proj nc (S), and their incidence relations, correspond to finite di… Show more

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Cited by 7 publications
(5 citation statements)
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“…He also observed that "degenerate" versions of the 4-dimensional Sklyanin algebras lead to the quantized enveloping algebras U q (sl 2 ). A detailed examination of this degeneration procedure is carried out in [11].…”
Section: Introductionmentioning
confidence: 99%
“…He also observed that "degenerate" versions of the 4-dimensional Sklyanin algebras lead to the quantized enveloping algebras U q (sl 2 ). A detailed examination of this degeneration procedure is carried out in [11].…”
Section: Introductionmentioning
confidence: 99%
“…Or, conversely, the U q (sl n )'s are "degenerations" of Q n 2 ,n−1 (E, τ ). A detailed examination of this degeneration process for n = 2 is carried out in [CSW18].…”
Section: The Artin-van Den Bergh Theorem and The Functor γmentioning
confidence: 99%
“…Since dim k (R) = 6 = dim(P 3 × P 3 ), Γ = ∅. ) and the class of the zero locus of a non-zero element in V ⊗ V is equal to s + t. If dim(Γ) = 0, then the class of Γ is (s + t) 6 since dim(R) = 6. But (s + t) 6 = 20s 3 t 3 so the cardinality of Γ is 20 when its points are counted with multiplicity.…”
Section: The Zero Locus Of the Relations For A(α β γ)mentioning
confidence: 99%
“…In conclusion, we have Proof. Simply compose ψ and its analogue for ℓ 2 (which are double covers by Proposition 5.4) with the two-fold covers of the form (5)(6). and (α j + α k ){x j , a i } + (α j α k + 1)[x j , c i ] − (α k + 1) α j {x i , a j } + [x i , c j ] + (α j − 1) [x 0 , a k ] − {x 0 , c k } = (α 1 + α 2 + α 3 + α 1 α 2 α 3 )[x k , x 2 j ], and (α i + α k ){x i , a j } − α i (α j α k + 1)[x i , c j ] + (α i + 1) {x 0 , c k } − [x 0 , a k ] − (α k − 1) α i {x j , a i } − [x j , c i ] = − (α 1 + α 2 + α 3 + α 1 α 2 α 3 )[x k , x 2 i ] and − (α j + α k ){x 0 , c i } − α i (α j α k + 1)[x 0 , a i ] + α i (α k + 1) α j {x k , a j } − [x k , c j ] + α i (α j − 1) α k {x j , a k } + [x j , c k ] = − (α 1 + α 2 + α 3 + α 1 α 2 α 3 )[x i , x 2 0 ].…”
Section: Atmentioning
confidence: 99%