# $$q$$-Pochhammers

q_analogs q_pochhammers q_binomial_coefficients q_vandermonde_chu_identity

## $$q$$-Pochhammers

By [{gasper-rahman04}], for $$n \in \mathbb Z$$ define

$(a)_n := (a; q)_n := \begin{cases} (1 - a) (1 - a q) ... (1 - a q^{n - 1}) & n > 0 \\ 1 & n = 0 \\ {1 \over (1 - a q^{-1}) (1 - a q^{-2}) ... (1 - a q^{-n})} & n < 0 \end{cases}$

Thus for $$n < 0$$

$(q)_n = \infty \qquad (1)$

$$\phi(q) := (q)_\infty$$ is called Euler's function.

## $$q$$-binomial coefficients

For $$n \ge 0$$ define the $$q$$-binomial coefficients $${n \choose k}_q$$ by

${n \choose k}_q = {(n)_q \over (k)_q (n - k)_q} \qquad (2)$

Thus by (1) when $$k < 0$$ or $$k > n$$, $${n \choose k}_q = 0$$.

Claim 1. $$\sum_{\lambda \subset n^k} q^{|\lambda|} = {n + k \choose k}_q$$

Proof. Denote the LHS by $$F(n, k)$$, then

$F(n, k) = F(n - 1, k) + q^n F(n, k - 1)$

with $$F(0, k) = F(n, 0) = 1$$. This recursive relation with the boundary conditions agree with those of the binomial coefficients. $$\square$$

Remark. The Young diagrams confined in the $$k \times n$$ rectangle correspond to walks from $$(0, 0)$$ with $$n$$ ups and $$k$$ downs. Setting $$q = 1$$ in (2) this degenerates to the binomial coefficients as enumerations of such walks.

## Properties

1. $$(x; q^{-1})_n = (- 1)^n x^n q^{-{n \choose 2}} (x^{-1}; q)_n$$
2. $$(q^{-n}; q)_k = {(n)_q \over (n - k)_q} (-1)^k q^{{k \choose 2} - n k}$$
3. $${n \choose k}_{q^{-1}} = q^{- n (n - k)} {n \choose k}_q$$
4. $$(a)_{n - k} = {(a)_n \over (a^{-1} q^{1 - n})_k} (- q a^{-1})^k q^{{k \choose 2} - n k}$$
5. $$(a^{-1} q^{1 - n})_n = (a)_n (-a)^{- n} q^{- {n \choose 2}}$$

## Cauchy's $$q$$-binomial series

Theorem.

${}_1\phi_0(a;;q;z) = \sum_{k \ge 0} {(a)_k \over (q)_k} z^k = {(a z)_\infty \over (z)_\infty}. \qquad |q| < 1, |z| < 1. \qquad (3)$

Proof.

\begin{align} F(a, z) - F(a, q z) &= (1 - a) z F(a q, z)\\ F(a, z) - F(a q, z) &= - a z F(a q, z) \end{align}

Eliminating $$F(a q, z)$$ we have

$F(a, z) = {1 - a z \over 1 - z} F(a, q z).$

Applying this recursively we have

$F(a, z) = {(a z)_n \over (z)_n} F(a, q^n z).$

Let $$n \to \infty$$ we have

$F(a, z) = {(a z)_\infty \over (z)_\infty} F(a, 0) = {(a z)_\infty \over (z)_\infty}.$

$$\square$$

## $$q$$-deformations of binomial theorem

### Version 1

Let $$x, y$$ be the generators of a $$q$$-Grassmann algebra, namely they satisfy

$y x = q x y$

Theorem. $${n \choose k}_q$$ satisfies

$(x + y)^n = \sum_k {n \choose k}_q x^k y^{n - k}$

This is also called $$q$$-binomial theorem.

Claim.

${n \choose k}_q = \sum_{S \in {[n] \choose k}} q^{\sum_{a_i \in S} (a_i - i)} = \sum_{S \in {[n] \choose k}} q^{(\sum_{a_i \in S} a_i) - {k + 1 \choose 2}}$

Pf. We know by the previous thm $${n \choose k}_q$$ gathers the products with $$k$$ $$x$$'s and $$n - k$$ $$y$$'s, and each of such product corresponds to $$S = \{a_1 \le a_2 ... \le a_k\} \in {[n] \choose k}$$ in a obvious way, e.g. when $$n = 8$$ and $$k = 2$$, $$yyxyxyyy \leftrightarrow \{3, 5\}$$. Shifting all the $$x$$'s to the left corresponds $$\sum_i (a_i - i)$$ shifts. $$\square$$

### Version 2

Another $$q$$-binomial theorem (a.k.a. Gauss binomial theorem) is

$\sum_k q^{{k \choose 2}} \alpha^k {n \choose k}_q = (- \alpha; q)_n. \qquad (4)$

Proof 1. Set $$a := q^{- n}$$ and $$z = - \alpha q^n$$ in (3) and simplify. $$\square$$

Proof 2. The RHS is

$(1 + \alpha) (1 + \alpha q) ... (1 + \alpha q^{n - 1})$

The coefficient of $$\alpha^k$$ enumerates the partitions of $$k$$ distinct parts (with the possibility that the last part is $$0$$) and first part $$\le n - 1$$, and that by removing $$0$$ from the last part, $$1$$ from the second last part, ... $$k - 1$$ from the first part, we have partitions with length $$\le k$$ and first part $$\le n - k$$. Finally applying Claim 1 we arrive at the conclusion:

\begin{align} (1 + \alpha) &(1 + \alpha q) ... (1 + \alpha q^{n - 1}) \\ &= \sum_k \sum_{\lambda: \ell(\lambda) = k, \lambda_1 \le n - 1, \lambda_1 > \lambda_2 > ... > \lambda_k \ge 0} \alpha^k q^{|\lambda|}\\ &= \sum_k \alpha^k q^{{k \choose 2}} \sum_{\mu: \ell(\mu) \le k, \mu_1 \le n - k} \\ &= \sum_k \alpha^k q^{{k \choose 2}} {n \choose k}_q. \end{align}

$$\square$$

### Version 3

Yet a third $$q$$-deformation of the binomial theorem is as follows:

$\sum_k b^k (a; q)_k (b; q)_{n - k} {n \choose k}_q = (a b; q)_n \qquad (5)$

which when $$q \to 1$$ reads

$\sum_k (b - a b)^k (1 - b)^{n - k} {n \choose k} = (1 - a b)^n.$

see Exercise 1.3 on pp25 of [{gasper-rahman04}].

Proof. By (3)

$\hg{1}{\phi}{0} (a; ; q, z) \hg{1}{\phi}{0} (b; ; q, a z) = \hg{1}{\phi}{0} (a b; ; q, z).$

Expanding both sides and compare coefficients of $$z^n$$ we are done. $$\square$$

This is also called a $$q$$-Vandermonde-Chu identiy. It is used to define the $$\phi_{q, \xi, \eta}$$ distribution in q_hahn_process. As in the derivation of (4) using (3), (5) becomes (6) with certain parameters, see binomial_distribution.

Remark. Using Item 4 in Properties the LHS of (5) can be written as

$(b)_n \hg{2}{\phi}{1} (a, q^{- n}; b^{-1} q^{1 - n}; q, q)$

and (5) can be derived using

$\hg{2}{\phi}{1}(q^{- n}, b; c; q, q) = {(c / b)_n \over (c)_n} b^n. (5.5)$

and Item 5 in Properties.

### Version 4

$\sum_{k = 0}^n \alpha^k (\alpha; q)_{n - k} {n \choose k}_q = 1$

When $$q \to 1$$ it reduces to the usual binomial theorem

$\alpha^k (1 - \alpha)^{n - k} {n \choose k} = 1$

## $$q$$-Vandermonde-Chu identity

$\sum_s q^{(m_1 - s) (k - s)} {m_1 \choose s}_q {m_2 \choose k - s}_q = {m_1 + m_2 \choose k}_q. \qquad (6)$

Proof 1.

$(- x)_{m_1 + m_2} = (- x)_{m_1} (- x q^{m_1})_{m_2}.$

Applying (4) to both sides of the above formula and gather the coefficients of $$x^k$$ we are done. $$\square$$

Proof 2. Applying (5.5) with $$n = m_1, b = q^{- k}, c = q^{m_2 - k + 1}$$. $$\square$$

## Classical limits of $$q$$-Pochhammers

In the following let $$q = e^{- \epsilon}$$.

### quantum dilogarithm

For $$t > 0$$

$(q^t; q)_\infty = \Gamma(t)^{-1} \exp(A(\epsilon) + (1 - t) \log \epsilon + O(\epsilon))$

where $$A(\epsilon)$$ is defined as

$A(\epsilon) = - {\pi^2 \over 6 \epsilon} - {1 \over 2} \log {\epsilon \over 2\pi}$

Specifically, when $$t = 0$$ we have asymptotics for Euler's function:

$\phi(q) = \exp(A(\epsilon) + O(\epsilon))$

This can be derived for example from Thm3.2 of [{banerjee-wilkerson16}].

Hence

$(n)_q = \Gamma(n + 1)^{-1} \epsilon^{- n} \exp(O(\epsilon))$

### Another classical limit

in Lemma 3.1 in [{gerasimov-lebedev-oblezin12}]:

Lemma 3.1. For $$\alpha \ge 1$$,

$(\alpha \epsilon^{-1} \log \epsilon^{-1} + \epsilon^{-1} y)_q = \begin{cases} e^{A(\epsilon) + e^{- y} + O(\epsilon)} & \alpha = 1 \\ e^{A(\epsilon) + O(\epsilon)} & \alpha > 1 \end{cases}$

## References

• [banerjee-wilkerson16] Lambert series and q-functions near q= 1, , arXiv preprint arXiv:1602.01085 2016.
• [gasper-rahman04] Basic hypergeometric series, , , Vol. 96 2004.
• [gerasimov-lebedev-oblezin12] On a classical limit of $q$-deformed Whittaker functions, , Letters in Mathematical Physics, Vol. 100, p.279–290 2012.