8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $|q-1|_p < 1$? $\endgroup$
    – KConrad
    Commented Jun 5, 2019 at 0:23
  • $\begingroup$ If you haven't come across them combinatorially, would you say how you came across them? $\endgroup$
    – AHusain
    Commented Jun 5, 2019 at 0:34
  • $\begingroup$ @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. $\endgroup$ Commented Jun 5, 2019 at 1:54

1 Answer 1

2
$\begingroup$

Recall Legendre's formula

$$ v_p(n!) = \sum_{s=1}^\infty\left\lfloor\frac n{p^s}\right\rfloor = \sum_{r=0}^\infty a_r[r]_p $$ 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

\begin{equation}\label{AClB}\tag{$\ast$} \large [n]_q! = u\prod_{r=1}^\infty \varphi^{r-1}([p]_q)^{\lfloor n/p^r\rfloor} \end{equation}

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 $n$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .