TITLE: Why are some q-analogues more canonical than others?
QUESTION [15 upvotes]: It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g.

the factorial and the q-Gamma function
the basic hypergeometric series (at least $_{r+1}\Phi_r$ and $_{r}\Psi_r$)
q-Pi and the q-Wallis formula

In a strict sense, q-analogs can of course not be canonical, as we might throw in almost everywhere powers of $q$ without changing the limit if $q\to1$.  

I mean canonical in the sense that these are the forms that require the least extra powers of $q$, or, more importantly, that other q-identities/theorems using them also tend to avoid such extra powers at best, making the formulae shorter.


Does it really make sense to call these (and certain other) q-analogs "canonical"? And if so, is there an explanation why some are much more canonical than others?

(Or is there a better definition of canonical?)
The canonicity of the q-binomial coefficients is obviously accounted for by their relationship with linear subspaces. (So this is not an analytical criteria using the limit $q\to1$.) 
What about the q-Gamma function? We may consider it canonical because of the (?!) q-analogue of the Bohr-Mollerup theorem proved by R. Askey, which states that for $0\lt q\lt 1$, the only logarithmically convex function satisfying $f(1)=1$ and $f(x+1)=\frac{q^x –1}{q–1}f(x)$ is the q-gamma function 
$ \Gamma_q(z)=(1-q)^{1-x}  \frac{(q\;;\; q)_{\infty}}{{(q^x;\; q) _\infty }}.$
Also note that this formula looks at least as elegant as Euler's definition of $\Gamma(z)= \dfrac{1}{z} \prod\limits_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$ 
On the other hand, one of the most basic identities, the recursion of binomial coefficients, has only an "asymmetric" q-analog and thus two of them:
${\genfrac[]00{n}{k}}_q=q^k{\genfrac[]00{n-1}{k}}_q+{\genfrac[]00{n-1}{k-1}}_q={\genfrac[]00{n-1}{k}}_q+q^{n-k}{\genfrac[]00{n-1}{k-1}}_q$.
For the classical orthogonal polynomials, it looks like there exist systematically "nice" q-analogs (see e.g. this survey), but it is not clear if there is a certain sense in which those can be considered canonical. Maybe for the Chebyshev polynomials, there is one "best" q-analog.

Is there a reasonable way of considering certain q-analogs of orthogonal polynomials "canonical", e.g. their uniqueness w.r.t. to an appropriate criteria, as for most classical orthogonal polynomials?
Has a q-analog of a polynomial identity (e.g. involving binomial coefficients) more chances of being canonical if it has a combinatorial interpretation?

For the q-derivative, there are at least two completely different approaches, both with their merits. So there is no use looking for canonicity there.
But nevertheless the next question:

Are q-analogs conceptually similar to an extension from $\mathbb R$ to $\mathbb C$?

By the latter I mean the following:
I wonder if generally speaking, the shift from an entity to its q-analog(s) can be likened, at least sometimes, to the shift of passing from $\mathbb R$ to $\mathbb C$, in the sense that some q-analogs provide a more complete picture than the entity itself (cp. for the $\mathbb R\to\mathbb C$ case the fundamental theorem of algebra or the meromorphic extension of the zeta function)?  

Many features in $\mathbb C$, e.g. the residue theorem, cannot be reduced to $\mathbb R$. Likewise for example, identities of "infinite q-polynomials" (i.e. of generating functions), cannot be taken to the limit $q\to1$.
In situations where there are several useful q-analogs, e.g. for the q-exponential function or the q-cosine, we might consider those corresponding to different panes of a Riemann surface like the one of $\sqrt{z}$ or $\ln z$.
And we can take it further:
The next step after $\mathbb R$ to $\mathbb C$ are the quaternions.
The next step after q-analogs are p,q-analogs. It looks like they haven't been studied a lot yet. 

Thank you for reading me so far.
Some of the questions may be rather subjective. As for the title, I had thought at first about "Why are most q-analogues canonical?" But then I figured that in the whole ocean of q-analogues, maybe there are just some "very" canonical islands, but for the vast majority the notion of canonicity is more or less fuzzy. Is that a feasible perception?
Even though some of these thoughts are somewhat philosopical, anyway, here goes. Looking forward to your input!

REPLY [8 votes]: As I said above I have some difficulty to denote specific $q-$analogues as canonical.
Consider as example the Catalan numbers  $\frac{1}{{n + 1}}{2n\choose n}$ . They have a simple generating function $f(z)$ which satisfies $f(z) = 1 + zf(z)^2 $ or equivalently  $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}$ and they are characterized by the fact that all Hankel determinants are 1.
Which of the following simple $q - $analogues should be called "canonical"? 
a)  The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$.  They also have very simple Hankel determinants, but there is no known formula for the polynomials themselves.
b) The polynomials  $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function and no simple Hankel determinants.
c) The $q - $Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$  is a  $q - $analogue of  $ \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$  Let $h(z)$ be the $q - $analogue of $\sqrt {1 + z}$ defined by $h(z)h(qz)=1+z$.  Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function and also simple Hankel determinants. But they are not polynomials in $q.$