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Let $q \geq 2$. What does the expression $(q^n-1)(q^n-q)(q^n-q^2)(q^n-q^3)\ldots(q^n-q^{n-1})/n!$ count? If $q$ is a prime power, then this is the number of bases of an $n$-dimensional vector space over a field with $q$ elements.

edited Jun 28 2010 at 15:07

asked Jun 26 2010 at 12:43

Martin Erickson
767●2●6

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@Darji, the "fact" (that the expression in question counts the number of bases of an $n$-dimensional vector space over a field with $q$ elements) only holds when $q$ is a prime power. I believe the OP wants a more general fact that works for all integer $q \geq 2$. – Doug Chatham Jun 26 2010 at 15:24

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@rdchat: if you follow carefully Darji's comment, you'll find that this is exactly what Darji is interested in. Yes, the intrigue is to give a counting interpretation for $q$ an arbitrary integer $>1$. A preliminary exercise: prove that the number is an integer. (Only elementary number theory is required!) – Wadim Zudilin Jun 26 2010 at 15:58

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@Wadim: I think this is a perfectly respectable question to ask, and it is likely to lead to some interesting combinatorics. For example, the polynomial 1/n sum_{d | n} mu(d) q^{n/d} counts, when q is a prime power, the number of irreducible polynomials over F_q. If one asks what this polynomial counts for general q, one gets several interesting answers, e.g. the number of Lyndon words of length n over an alphabet of q letters, the dimension of the nth graded part of the free Lie algebra on q generators... – Qiaochu Yuan Jun 26 2010 at 15:59

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@Martin: probably you already know this, but (1+q)(1+q+q^2)...(1+q+...+q^{n-1}) is the generating function for permutations with respect to inversion number. Unfortunately I can't think of an obvious way to get a free action of S_n on any combinatorial quantity related to this. – Qiaochu Yuan Jun 26 2010 at 16:36

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Hi, Wadim, see meta.mathoverflow.net/discussion/474 – Will Jagy Jun 27 2010 at 1:49

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Wadim Zudilin · Answer 1 · 2010-06-27 11:39:06Z UTC

This is a partial answer...perhaps someone will improve on it!

Historically, most of the $q$-analog formulae (beginning from Euler) were derived based on the assumption that $|q|< 1$ (to ensure series convergence) or $q=p^k$ for a prime $p$. John Baez in one of his weekly finds (week184) discusses the geometric interpretation of $q=1$ (counting over $\mathbb CP^n$), $q=-1$ (counting over $\mathbb RP^n$) and $q=$a prime power (counting over PG($\mathbb F_q$)). There is no discussion for other values of $q$.

However, in Gasper and Rahman's Basic Hypergeometric Series, there is an inversion identity on page 4 which can be used when $|q| > 1$:

$(a; q)_n = (a^{-1}; p)_n (-a)^n p^{-n(n-1)/2} $ where $p=1/q$.

This returns a new expression in base $|1/q| < 1$. You can see some examples of the identity being applied in Gasper's Lecture Notes on q-series (Exercise 1.1, Exercise 2.3, pg 14)

I have no idea how to interpret the result geometrically. Could it be relevant to Buildings, Buekenhout geometry or $p$-adic geometry?

What does the expression count?

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