Parts a and b are trivial so those are of no concern. My conclusion for part c is that a proper q analog is the q binomial coefficient, counting the number of subspaces of dimension k in a vector space of dimension n whereas dimension n is over a finite field of q elements. However, this just seems like something "that is" and not something I can think of how to prove..
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Use the q-Pochhammer and then it’s regular pochhammer symbol gamma function definition to find the q gamma function like this. Also this looks like a problem statement question, so please add more detail. – Тyma Gaidash Oct 20 '21 at 19:41
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@TymaGaidash Thank you very much for the response. I am not familiar with the q-Pochhammer. What would I apply it to? – MathMan2021 Oct 20 '21 at 20:02
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You can derive many other “q functions” from it and it is useful for many products/sums: Q-Pochhammer Symbol – Тyma Gaidash Oct 20 '21 at 20:03
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@TymaGaidash Awesome. So I am to take the original function and apply the q-Pochhammer? I am very lost... – MathMan2021 Oct 20 '21 at 20:14
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I may have explained it a bit weirdly, so please just see equation (1) and (3) in this article. – Тyma Gaidash Oct 20 '21 at 20:16
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Let $N$ be an $\mathbb F_q$-vector space with dimension $n$.
$\binom{n}{m}_q\binom{m}k_q$ enumerates pairs of subspaces $(M,K)$ so that $K\subseteq M\subseteq N$, $\dim M=m$ and $\dim K=k$.
On the other hand, in order to choose $(M,K)$, you could first choose $K$ in $\binom{n}k_q$ ways, and then count the number of ways to choose $M$ for that given $K$. Subspaces of $N$ containing $K$ are in bijection with subspaces of $N/K$ (fourth isomorphism theorem). Therefore, the number of subspaces of $N$ containing $K$ with dimension $m$ is equal to the number of subspaces of $N/K$ with dimension $m-k$, or $\binom{n-k}{m-k}_q$.
Putting this altogether, we see that the original identity still holds when you stick little $q$-subscripts on all of the binomial coefficients.
Mike Earnest
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