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Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$

Definition

Let $(I, A, D, P, P^{\lor}, q)$ be a symmetrizable Cartan datum for a quantum group (Symmetrizable Cartan data for quantum groups) and put $q_{i} = q^{d_{i}}$ . For $m \in N$ , define the $q_{i}$ -integer

$[m]_{i} : = \frac{q_{i}^{m} - q_{i}^{- m}}{q_{i} - q_{i}^{- 1}}, [0]_{i} : = 0,$

and the $q_{i}$ -factorial

$[m]_{i}! : = \prod_{k = 1}^{m} [k]_{i}, [0]_{i}! : = 1.$

For $0 \leq r \leq m$ , define the Gaussian binomial ${(\binom{m}{r})}_{i} : = [m]_{i}! / ([r]_{i}! [m - r]_{i}!)$ , and set it to $0$ when $r < 0$ or $r > m$ . In the published one-parameter convention, write ${(\binom{m}{r})}_{t} : = {(\binom{m}{r, m - r})}_{t}$ for the two-part $q$ -multinomial coefficient of The $q$ -integer, $q$ -factorial and $q$ -multinomial coefficients.

For an element $x$ of a unital $Q (q)$ -algebra and $m \geq 0$ , define its divided power at $q_{i}$ by $x^{(m)} : = x^{m} / [m]_{i}!$ , so $x^{(0)} = 1$ and $x^{(1)} = x$ .

The symmetric convention and the published asymmetric convention are related, for $m \geq 0$ and $0 \leq r \leq m$ , by

$[m]_{i} = q_{i}^{- (m - 1)} [m]_{q_{i}^{2}}, {(\binom{m}{r})}_{i} = q_{i}^{- r (m - r)} {(\binom{m}{r})}_{q_{i}^{2}} .$

In particular $[m]_{i}$ is invariant under $q_{i} \mapsto q_{i}^{- 1}$ . These quantities depend on the symmetrizer only through $q_{i} = q^{d_{i}}$ .

Facts & Assumptions

Given: A symmetrizable Cartan datum with $q$ indeterminate over $Q$ , and an element $x$ of a unital $Q (q)$ -algebra.

[F1]

The datum has $q_{i} = q^{d_{i}}$ with positive integer $d_{i}$ (Symmetrizable Cartan data for quantum groups).

[F2]

The asymmetric $q$ -integer and $q$ -factorial are $[m]_{t} = 1 + t + \dots + t^{m - 1}$ and $[m]_{t}! = \prod_{j = 1}^{m} [j]_{t}$ , with $[0]_{t} = 0$ and $[0]_{t}! = 1$ ; the $q$ -multinomial is the factorial quotient (The $q$ -integer, $q$ -factorial and $q$ -multinomial coefficients).

Verification

technique · Expand the symmetric $q_i$-integer and multiply the resulting finite product

1.1givenF1F2algebra

For $m \geq 1$ , cancel the nonzero factors $q_{i} - q_{i}^{- 1} = q_{i}^{- 1} (q_{i}^{2} - 1)$ to obtain $[m]_{i} = q_{i}^{- (m - 1)} (q_{i}^{2 m} - 1) / (q_{i}^{2} - 1) = q_{i}^{- (m - 1)} \sum_{j = 0}^{m - 1} q_{i}^{2 j} = q_{i}^{- (m - 1)} [m]_{q_{i}^{2}}$ . At $m = 0$ the same identity holds by the zero convention. For $m \geq 1$ the quotient is nonzero because its numerator and denominator are nonzero in $Q (q)$ by [F1].

2.1step 1.1F1F2algebra

Multiplying the identity of step 1.1 for $j = 1, \dots, m$ gives $[m]_{i}! = q_{i}^{- m (m - 1) / 2} [m]_{q_{i}^{2}}!$ ; for $m = 0$ this is the equality of empty products. Hence for $0 \leq r \leq m$ , division by the nonzero factorials is valid and ${(\binom{m}{r})}_{i} = q_{i}^{- m (m - 1) / 2 + r (r - 1) / 2 + (m - r) (m - r - 1) / 2} {(\binom{m}{r})}_{q_{i}^{2}} = q_{i}^{- r (m - r)} {(\binom{m}{r})}_{q_{i}^{2}}$ , since $- m (m - 1) + r (r - 1) + (m - r) (m - r - 1) = - 2 r (m - r)$ .

3.1step 1.1step 2.1F1algebra∎

Since every $[j]_{i}$ for $j \geq 1$ is nonzero, $[m]_{i}!$ is a nonzero scalar and therefore invertible in $Q (q)$ ; this makes $x^{(m)}$ well-defined. Replacing $q_{i}$ by $q_{i}^{- 1}$ negates numerator and denominator in $[m]_{i}$ , so $[m]_{i}$ is invariant, as are its factorials and Gaussian quotients.

Depends on

Used by

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Sources

Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)
Benjamin Enriquez, PBW and Duality Theorems for Quantum Groups and Quantum Current Algebras, Journal of Lie Theory 13 (2003), 21–64 (standard reference, not scraped)

Quantum integers, factorials, Gaussian binomials and divided powers at qiqi​