Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is well-known that the Gaussian polynomial (or Gaussian coefficient, q-binomial coefficient) $\binom{n}{k}_q$ counts the number of $k$-dimensional subspaces of an $n$-dimensional vector space over $GF(q)$.

A generalization of $\binom{n}{k}_q$ are the so-called $p,q$-binomial coefficients,

$\binom{n}{k}_{p,q}=\frac{{[n]}!}{[k]![n-k]!}$,

where $[n]=\frac{p^n-q^n}{p-q}$ and $[n]!=[n][n-1]\cdots[2][1]$.

The $p,q$-binomials equal the $q$-binomials when $q=1$.

Question 1: Is there a vector space combinatorial interpretation for the $p,q$-binomials? If there is, how does the underlying two-parameter field look like? (There is a combinatorial interpretation in terms of tableaux and lattice paths but I'm more interested with the subspace interpretation.)

Question 2 (somewhat related): A number of mathematicians have talked about the so-called $q$-disease (the widespread (at least those among working in $q$-series) interest in extending classical results to $q$-analogues.) Is there a $p,q$-disease?

PS I'm supposed to write $[n]_{p,q}$ but it doesn't seem to work with \frac.

share|improve this question
    
btw, you may wish to add the "q-analogs" tag. –  Pietro Majer Jul 6 '12 at 10:03
    
Thanks. I actually tried the British spelling (analogue) but there weren't such a tag. –  Ken Gonzales Jul 6 '12 at 10:09
2  
The variable $p$ is just making these polynomials homogeneous in an obvious way, so I would not be hopeful about a similar interpretation in a more general context. The "right" reason Gaussian polynomials count subspaces is the cell decomposition of the Grassmannian, and from that point of view the lack of homogeneity is crucial. –  Vladimir Dotsenko Jul 6 '12 at 10:27
1  
Amplifying part of Vladimir Dostenko's comment: These $p,q$-binomial coefficients are just what you get from the ordinary $q$-binomial coefficients if you replace $q$ by $q/p$ and then multiply by appropriate powers of $p$. So I doubt that there will be any serious combinatorics here that isn't already in the $q$-binomial coefficients. –  Andreas Blass Jul 6 '12 at 16:18
1  
I apologize for misspelling "Dotsenko" in my previouis comment. –  Andreas Blass Jul 6 '12 at 17:43
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.