# Product of $q$-analogues

### Background

Recall that the $$q$$-analogue $$[n]_q\in\mathbb Z[q]$$ of a natural number $$n\in\mathbb N$$ is defined as $$[n]_q := \frac{q^n -1}{q-1}$$ the idea being that formulas involving $$q$$ will specialize along $$\mathbb Z[q]/(q-p^n)$$ to counting formulas about finite fields, while specializing along $$\mathbb Z[q]/(q-1)$$ will yield counting formulas about finite sets.

For example, defining the $$q$$-factorial by $$[n]_q! := [1]_q [2]_q \dotsm [n]_q,$$ we have that the number of $$k$$-dimensional subspaces of $$\mathbb F_q^n$$ is $$\#\mathrm{Gr}_{n,k}(\mathbb F_q) = \binom nk_q :=\frac{[n]_q!}{[n-k]_q![k]_q!}$$ and $$\binom nk_q$$ reduces to $$\binom nk$$ when $$q\to 1$$.

### Question

Now let $$p$$ be a fixed prime. In my work, I've come across the expressions $$[p]_q^{k_1} [p^2]_q^{k_2} \dotsm [p^n]_q^{k_n}$$ with $$k_i\ge0$$, $$i=1,\dotsc, n$$, as well as funny things like $$[p^r]_{q^{p^s}} = \frac{q^{p^{r+s}}-1}{q^{p^s}-1}$$

Do these have any known combinatorial interpretation?

I will usually want to consider the above product when the values of $$n$$ and $$\sum k_i$$ are fixed, so we can consider it ranging over all partitions of $$\sum k_i$$ into $$n$$ non-negative integers.

### Motivation

Ultimately I'm interested in the specialization $$\mathbb Z[q] \to \mathbb Z_p[\![q-1]\!] \to \mathbf A_{\mathrm{inf}}(R)$$, where $$R$$ is a perfectoid ring containing a compatible choice of roots of unity $$\zeta_{p^\infty}$$. Letting $$\epsilon = (1,\zeta_p,\dots) \in R^\flat,$$ the structure map $$\mathbb Z[\![q-1]\!] \to \mathbf A_{\mathrm{inf}}$$ is given by $$q\mapsto[\epsilon]$$. Then the Fontaine map $$\tilde\theta_r\colon \mathbf A_{\mathrm{inf}}(R) \to W_r(R)$$ has kernel generated by $$[p^r]_q$$ (Example 3.16), while the map $$\theta_r = \tilde\theta_r \varphi^r$$ has kernel generated by $$[p^r]_{q^{1/p^r}}$$. You can get the product of these ideals, above, out of some $$RO(S^1)$$-graded THH calculations.

• Are the $k_i$ restricted to be less than $p$ or just have the sum fixed and no other restriction (besides non-negativity). Is $q$ an indeterminate or do you want it to be real in $(0,1)$ or complex with $|q| < 1$ or $p$-adic with $|q-1|_p < 1$? – KConrad Jun 5 at 0:23
• If you haven't come across them combinatorially, would you say how you came across them? – AHusain Jun 5 at 0:34
• @KConrad there's no restriction on the $k_i$. For the combinatorial interpretation I'd want to know what happens when $q=1$ and when $q=p^m$, but for my applications $q-1$ is going to specialize to $\mu\in \mathbf A_{\mathrm{inf}}$; see the motivation added to the question. – Yuri Sulyma Jun 5 at 1:54

$$v_p(n!) = \sum_{s=1}^\infty\left\lfloor\frac n{p^s}\right\rfloor = \sum_{r=0}^\infty a_r[r]_p$$ where $$n = \sum a_r p^r$$ is the base-$$p$$ expansion of $$n$$.
A $$q$$-analogue of this formula is provided by Lemma 4.8 of The $$p$$-completed cyclotomic trace in degree 2, which states that
$$$$\label{AClB}\tag{\ast} \large [n]_q! = u\prod_{r=1}^\infty \varphi^{r-1}([p]_q)^{\lfloor n/p^r\rfloor}$$$$
for a unit $$u\in\mathbb Z_p[\![q-1]\!]^\times$$. While my computations yielded products $$[p]_q^{k_1} [p^2]_q^{k_2} \dotsm [p^n]_q^{k_n},$$ for arbitrary $$k_i\ge0$$, the ones most relevant to my eventual application (the regular slice filtration on THH) were the principal ideal generated by the RHS of \eqref{AClB} (when $$p$$ divides $$n$$).