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Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely related characterization is that this subspace is the kernel of $\Delta^{d+1}$, where $\Delta : \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ is the difference operator defined by $(\Delta P)(x) = P(x) - P(x-1)$. Roughly stated, my question asks for a $q$-analogue of either of these results.

To make this more concrete, use the standard notation, so $[n]_q = 1+q+\cdots + q^{n-1}$, $[n]_q! = [n]_q[n-1]_q\ldots [1]_q$, and $\binom{n}{d}_q = \frac{[n]_q!}{[d]_q![n-d]_q!}$. Fix $N \in \mathbb{N}$ and for each $m \in \mathbb{N}_0$ let

$$u^{(m)}_q = \bigl( \binom{0}{m}_q, \binom{1}{m}_q, \ldots, \binom{N-1}{m}_q \bigr) \in \mathbb{C}[q]^{N}$$

By the first paragraph, for $P \in \mathbb{C}[x]$, we have

$$\bigl( P(0), P(1), \ldots, P(N-1) \bigr) \in \bigl\langle u^{(0)}_1, u^{(1)}_1, \ldots, u^{(d)}_1 \bigr\rangle$$

if and only if $\mathrm{deg} P \le d$. In a current research problem, it's useful that the same holds for $( P(N-1), \ldots, P(1), P(0))$; in fact the evaluation points $0$, $1, \ldots, N-1$ can be varied by an arbitrary affine transformation.

Is there an analogous characterization of the span $\langle u_q^{(0)}, u_q^{(1)}, \ldots, u_q^{(d)} \rangle$?

As a follow up, what transformations preserve this space? In particular, is it invariant under a $q$-analogue of the affine transformation just mentioned? Despite some thought I have not found any reasonable answer to these questions, but I find it hard to believe that there is nothing to be said.

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We have $$\binom{n}d_q=\frac{(q^n-1)\ldots(q^n-q^{d-1})}{(q^d-1)\ldots(q^d-q^{d-1})}=f_d(q^n)=g_d([n]_q) $$ where $f_d$ and $g_d$ are polynomials of degree $d$ (depending on $q$ of course). Therefore $$\bigl( P([0]_q), P([1]_q), \ldots, P([N-1]_q) \bigr) \in \bigl\langle u^{(0)}_q, u^{(1)}_q, \ldots, u^{(d)}_q \bigr\rangle$$ if and only if $\deg P\leqslant d$.

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  • $\begingroup$ See also: mathoverflow.net/questions/218696/q-integer-valued-polynomials $\endgroup$
    – Sam Hopkins
    Commented Jul 14, 2020 at 17:17
  • $\begingroup$ Thank you. I think the critical point is that the binomial coefficients $\binom{x}{d}$ are the minimal degree polynomials that interpolate the functions $0 \mapsto 0, 1 \mapsto 0, \ldots, d-1 \mapsto 0, d \mapsto 1$. Thus $(v_0, \ldots, v_{N-1})$ is in the span of $u_1^{(0)}, \ldots, u_1^{(d)}$ if and only if there is a polynomial $P$ of degree $\le d$ such that $P(j) = v_j$ for each $j$. This generalizes to the $q$-binomial coefficients replacing $j \mapsto 0$, $d \mapsto 1$ with $[j] \mapsto 0$ and $[d]_q \mapsto 1$ and $P(j) = v_j$ with $P([j]_q) = v_j$. $\endgroup$
    – Mark Wildon
    Commented Jul 16, 2020 at 10:21
  • $\begingroup$ Hence if $P(ax+b) = Q(x)$ then $(P([0]_q), \ldots, P([N-1]_q)) \in \langle u_q^{(0)}, \ldots, u_q^{(d)} \rangle$ if and only if $(Q([0]_q), \ldots, Q([N-1]_q)) \in \langle u_q^{(0}), \ldots, u_q^{(d)} \rangle$. For binomial coefficients this is useful, because such affine changes of variable move the evaluation points in a 'natural way'. (E.g. they can be reversed.) But for $q$-binomial coefficients, it's not so useful: e.g, there is no affine transformation sending $[0]_1, [1]_q, [2]_q$ to $[2]_q, [1]_q, [0]_q$ when $q\not= 1$. $\endgroup$
    – Mark Wildon
    Commented Jul 16, 2020 at 10:26
  • $\begingroup$ Anyway, thank you for the very clear answer, which I think is definitive and shows that the generalization (I now realise) I really wanted is not feasible. $\endgroup$
    – Mark Wildon
    Commented Jul 16, 2020 at 10:31

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