Symmetric quantum binomial coefficients #
This file defines the symmetric quantum binomial coefficient [n choose k]_q in a field by the
Pascal recursion
[n+1 choose k+1]_q = q^(n-k) [n choose k]_q + q^(-(k+1)) [n choose k+1]_q,
with [n choose 0]_q = 1 and [0 choose k+1]_q = 0, and proves the noncommutative
q-binomial theorem: if q ≠ 0 and two elements of an algebra satisfy y x = q² x y, then
(x + y)^n = ∑_k q^(k(n-k)) [n choose k]_q x^k y^(n-k).
The symmetric normalization makes [n choose k]_q invariant under q ↦ q⁻¹, and under
k ↦ n - k for k ≤ n. It gives [n choose 1]_q = [n]_q whenever q - q⁻¹ ≠ 0, so that the
quantum integer [n]_q is defined by a nonzero denominator, and it specializes to Nat.choose
at q = 1. The q-binomial theorem computes the coproducts of the powers of the root
generators of U_q(sl₂).
Negating the parameter gives [n choose k]_{-q} = (-1)^{k(n-k)} [n choose k]_q.
This identity uses no division by quantum integers, so it includes roots of unity and
q = ±1; at q = -1 it gives a signed ordinary binomial coefficient. These signs enter
comparisons of quantum divided-power relations at opposite parameters.
Main definitions #
TauCeti.quantumBinomial q n k: the symmetric quantum binomial coefficient[n choose k]_q.
Main results #
TauCeti.quantumBinomial_succ_succandTauCeti.quantumBinomial_succ_succ': the two Pascal rules, obtained from each other byq ↦ q⁻¹.TauCeti.quantumBinomial_succ_right_eq: the absorption identity[n choose k + 1]_q [k + 1]_q = [n choose k]_q [n - k]_q.TauCeti.quantumBinomial_invandTauCeti.quantumBinomial_symm: invariance underq ↦ q⁻¹, and underk ↦ n - kfork ≤ n.TauCeti.quantumBinomial_neg_left: the sign change[n choose k]_{-q} = (-1)^{k(n-k)} [n choose k]_q.TauCeti.quantumBinomial_one_right:[n choose 1]_q = [n]_qwhenq - q⁻¹ ≠ 0.TauCeti.quantumBinomial_one: the classical specialization[n choose k]_1 = n.choose k.TauCeti.add_pow_of_mul_eq_smul_mul: theq-binomial theorem fory x = q² x ywithq ≠ 0.TauCeti.coeff_prod_range_one_add_C_mul_X: Cauchy's binomial theorem in coefficient form, the coefficient ofX ^ kin∏_{j < n} (1 + q^(2j) X)isq^(k(n-1)) [n choose k]_q.
References #
- C. Kassel, Quantum Groups, Graduate Texts in Mathematics 155, Section IV.2, for Gaussian
binomial coefficients and the
q-binomial formula fory x = q x y. - J. C. Jantzen, Lectures on Quantum Groups, Graduate Studies in Mathematics 6, Chapter 0, for the symmetric normalization.
- G. Lusztig, Introduction to Quantum Groups, Sections 1.3 and 33.2, for the symmetric normalization and sign comparisons.
- V. Kac and P. Cheung, Quantum Calculus, Universitext, Chapter 5, for Gauss's (Cauchy's) binomial formula.
The symmetric quantum binomial coefficient [n choose k]_q, defined by [n choose 0]_q = 1,
[0 choose k+1]_q = 0 and the Pascal rule
[n+1 choose k+1]_q = q^(n-k) [n choose k]_q + q^(-(k+1)) [n choose k+1]_q.
Instances For
[n choose 0]_q = 1.
[0 choose k + 1]_q = 0.
The defining Pascal rule
[n+1 choose k+1]_q = q^(n-k) [n choose k]_q + q^(-(k+1)) [n choose k+1]_q.
[n choose k]_q = 0 for n < k.
[n choose n]_q = 1.
The defining Pascal rule with integer exponents. This form holds for all indices,
including k > n, because the corresponding quantum binomial coefficients vanish.
Over a semifield, the quantum binomial coefficients at q = 1 are the ordinary
binomial coefficients.
Cauchy's binomial theorem, in coefficient form: for q ≠ 0, the coefficient of X ^ k
in ∏_{j < n} (1 + q^(2j) X) is q^(k(n-1)) [n choose k]_q.
The absorption identity [n choose k + 1]_q [k + 1]_q = [n choose k]_q [n - k]_q. At a
generic parameter it determines [n choose k]_q from [n choose 0]_q = 1.
The second Pascal rule
[n+1 choose k+1]_q = q^(-(n-k)) [n choose k]_q + q^(k+1) [n choose k+1]_q, the image of the
defining rule under q ↦ q⁻¹.
The quantum binomial coefficients are invariant under q ↦ q⁻¹.
Negating the parameter multiplies the symmetric quantum binomial by
(-1)^(k * (n - k)). This includes zero parameters and indices above the diagonal.
The symmetry [n choose n - k]_q = [n choose k]_q for k ≤ n.
The q-binomial theorem: if y x = q² x y for a nonzero scalar q, then
(x + y)^n = ∑_k q^(k(n-k)) [n choose k]_q x^k y^(n-k).