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Divided powers and the Lusztig form in quantized sl2

Example

Work in the rank-one subalgebra $A_{i}$ generated by $E = E_{i}$ , $F = F_{i}$ , $K^{\pm 1} = K_{i}^{\pm 1}$ inside $U_{q} (g)$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules), with $q : = q_{i}$ and $[m] : = [m]_{i}$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ), and let $x^{(a)} : = x^{a} / [a]!$ denote divided powers.

(a) Divided powers multiply. $E^{(2)} E^{(1)} = [3] E^{(3)}$ and $F^{(2)} F^{(2)} = {(\binom{4}{2})}_{i} F^{(4)} = [4]! / ([2]!)^{2} F^{(4)}$ , since $x^{(a)} x^{(b)} = {(\binom{a + b}{a})}_{i} x^{(a + b)}$ ; numerically ${(\binom{4}{2})}_{i} = q^{- 4} + q^{- 2} + 2 + q^{2} + q^{4}$ , a nonzero element of $A = Z [q, q^{- 1}]$ because $q$ is transcendental.

(b) The commutator is the first toral binomial. $E^{(1)} F^{(1)} - F^{(1)} E^{(1)} = [E, F] = \frac{K - K^{- 1}}{q - q^{- 1}} = T_{i, 0, 1}$ , and for $b \geq 1$ the published reordering gives $E F^{(b)} = F^{(b)} E + F^{(b - 1)} \frac{1}{q^{b} - q^{- b}} \sum_{s = 0}^{b - 1} (q^{- 2 s} K - q^{2 s} K^{- 1})$ ; for $b = 2$ the toral factor is $\frac{K - K^{- 1} + q^{- 2} K - q^{2} K^{- 1}}{q^{2} - q^{- 2}}$ , a combination of the integral toral binomials $T_{i, c, t}$ .

(c) Reordering at $a = b = 2$ . $E^{(2)} F^{(2)} = F^{(2)} E^{(2)} + F^{(1)} T_{i, - 2, 1} E^{(1)} + T_{i, 0, 2}$ , where $T_{i, - 2, 1} = \frac{K q^{- 2} - K^{- 1} q^{2}}{q - q^{- 1}}$ and $T_{i, 0, 2} = \frac{(K - K^{- 1}) (K q^{- 1} - K^{- 1} q)}{(q - q^{- 1}) (q^{2} - q^{- 2})}$ ; all three toral binomials are elements of the integral form $U_{A}$ (The Lusztig divided-power integral form $U_{A}$ ), and on a weight vector of weight $μ$ their $q = 1$ values are $μ (h_{i})$ , $μ (h_{i}) - 2$ and $(\binom{μ (h_{i})}{2})$ (The q=1 toral specialization identifies the classical Cartan generators).

Facts & Assumptions

Given: The rank-one algebra $A_{i}$ with its presentation, the divided powers and the toral binomials.

[F1]

$[a]! \neq 0$ in $Q (q)$ and $x^{(a)} x^{(b)} = {(\binom{a + b}{a})}_{i} x^{(a + b)}$ for all $a, b \geq 0$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ , Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules (ii)).

[F2]

$E F - F E = (K - K^{- 1}) / (q - q^{- 1})$ and, for $b \geq 1$ , $E F^{(b)} = F^{(b)} E + \frac{F^{(b - 1)}}{q^{b} - q^{- b}} \sum_{s = 0}^{b - 1} (q^{- 2 s} K - q^{2 s} K^{- 1})$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules (ii)).

[F3]

The general divided-power reordering identity $E^{(a)} F^{(b)} = \sum_{t = 0}^{\min (a, b)} F^{(b - t)} T_{i, 2 t - a - b, t} E^{(a - t)}$ with $T_{i, c, t} = \prod_{s = 1}^{t} \frac{K q^{c - s + 1} - K^{- 1} q^{- c + s - 1}}{q^{s} - q^{- s}}$ (Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity (iii), The Lusztig divided-power integral form $U_{A}$ ).

[F4]

The toral binomials lie in $U_{A}$ , commute with one another and with the $K_{h}$ , and on a weight vector of weight $μ$ act by the Gaussian value ${(\binom{μ (h_{i}) + c}{t})}_{i}$ , whose value at $q = 1$ is $(\binom{μ (h_{i}) + c}{t})$ (The Lusztig divided-power integral form $U_{A}$ , The q=1 toral specialization identifies the classical Cartan generators (b)).

[F5]

${(\binom{4}{2})}_{i} = [4] [3] / [2] = q^{- 4} + q^{- 2} + 2 + q^{2} + q^{4}$ , a nonzero Laurent polynomial, and $q$ is transcendental, so this element of $A$ is not zero (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

Verification

technique · expand the divided powers by definition, apply the published rank-one commutation and reordering formulas at a=b=2, and evaluate the toral binomials on weights

1.1F1algebra

Divided powers multiply. By definition $E^{(2)} E^{(1)} = \frac{E^{2}}{[2]!} \cdot E = \frac{E^{3}}{[2]!} = \frac{[3]!}{[2]!} E^{(3)} = [3] E^{(3)}$ and $F^{(2)} F^{(2)} = \frac{F^{4}}{([2]!)^{2}} = \frac{[4]!}{([2]!)^{2}} F^{(4)} = {(\binom{4}{2})}_{i} F^{(4)}$ , using $[a]! \neq 0$ and $x^{(a)} = x^{a} / [a]!$ from [F1]; the general identity $x^{(a)} x^{(b)} = {(\binom{a + b}{a})}_{i} x^{(a + b)}$ is the same computation.

1.2F2F3algebra

The commutator and the first binomial. By [F2] and [F3], $E F - F E = \frac{K - K^{- 1}}{q - q^{- 1}}$ , which is exactly $T_{i, 0, 1}$ by the definition of the toral binomial; more generally the displayed formula for $E F^{(b)}$ is [F2], and at $b = 2$ the toral factor is $\frac{(K - K^{- 1}) + (q^{- 2} K - q^{2} K^{- 1})}{q^{2} - q^{- 2}}$ , which equals $T_{i, - 1, 1} = (K q^{- 1} - K^{- 1} q) / (q - q^{- 1})$ : use $1 + q^{- 2} = q^{- 1} [2]$ and $1 + q^{2} = q [2]$ with $q^{2} - q^{- 2} = [2] (q - q^{- 1})$ . It is an integral toral binomial.

1.3F3algebra

Reordering at $a = b = 2$ . Setting $a = b = 2$ in the general identity of [F3] gives $E^{(2)} F^{(2)} = \sum_{t = 0}^{2} F^{(2 - t)} T_{i, 2 t - 4, t} E^{(2 - t)} = F^{(2)} E^{(2)} + F^{(1)} T_{i, - 2, 1} E^{(1)} + T_{i, 0, 2}$ with $T_{i, - 2, 1} = \frac{K q^{- 2} - K^{- 1} q^{2}}{q - q^{- 1}}$ and $T_{i, 0, 2} = \frac{(K - K^{- 1}) (K q^{- 1} - K^{- 1} q)}{(q - q^{- 1}) (q^{2} - q^{- 2})}$ by the product definition. The nontrivial toral factors $T_{i, - 2, 1}$ and $T_{i, 0, 2}$ are generators of the integral form, hence lie in $U_{A}$ .

2.1F3F4F5step 1.1step 1.3algebra∎

Values on weights and integrality. On a weight vector of weight $μ$ , $T_{i, 0, 1}$ acts by $[μ (h_{i})]_{i}$ , $T_{i, - 2, 1}$ by $[μ (h_{i}) - 2]_{i}$ and $T_{i, 0, 2}$ by ${(\binom{μ (h_{i})}{2})}_{i}$ by [F4]; at $q = 1$ these are the integers $μ (h_{i})$ , $μ (h_{i}) - 2$ and $(\binom{μ (h_{i})}{2})$ . The numerical value ${(\binom{4}{2})}_{i} = q^{- 4} + q^{- 2} + 2 + q^{2} + q^{4}$ is nonzero in $A$ by [F5], so the product formula of step 1.1 exhibits a nonzero integral coefficient, and no division by a vanishing denominator occurs: every expression used is an element of $U_{A}$ or a polynomial identity valid in $Q (q)$ . No choice is used; the computation is a finite expansion in $A_{i}$ , with the $a = b = 2$ identity justified by the locally proved reordering supplier.

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Sources

Math 263C Quantum Groups (Stanford, 2022), lecture slides 'Lusztig's quantum group. Tilting modules. The Fusion Category in Action' (standard reference, not scraped)
Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)
George Lusztig, Introduction to Quantum Groups (Progress in Mathematics 110, Birkhauser, 1993; second printing 1994; author-hosted scan) (standard reference, not scraped)