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I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the conditions that $S_q(0,k)=\delta_{0,k}$ and $S_q(n,0)=\delta_{n,0}$.

How is this recurrence arrived at? Even if this recurrence is taken as definition, there must be some motivation for it. Thank you.

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  • $\begingroup$ @anon: hmm, I don't want to make a broad discussion about that, but I think to remember there are different approaches to define q-analogues for the Stirling-numbers. So I think it should be not only legitime but also interesting to ask: "how did they arrive at that form of definition" which implies a small hint to the focus at "why did they prefer this specific definition", and not just "they are defined" $\endgroup$ Feb 11, 2012 at 8:58

2 Answers 2

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The following justification is based on material in Johann Cigler’s lecture notes, Chapters 1 and 3.

The ordinary Stirling numbers of the second kind are characterized by the identity $$(xD)^n=\sum_k\left\{\matrix{n\\k}\right\}x^kD^k\;,\tag{1}$$ where $D$ is the ordinary differentiation operator. Thus, one approach to defining a $q$-analogue $S_q(n,k)$ is to require that it satisfy the analogue of $(1)$ with $D$ replaced by $D_q$, defined by $$D_qf(x)=\frac{f(qx)-f(x)}{qx-x}$$ and satisfying $$D_qx^n=\frac{(qx)^n-x^n}{(q-1)x}=\frac{q^n-1}{q-1}x^{n-1}=(n)_qx^{n-1}$$ and $$D_qx=qxD_q\;.$$

In other words, one might reasonably define $S_q(n,k)$ to satisfy

$$(xD_q)^n=\sum_{k=0}^nS_q(n,k)x^kD_q^k\;.\tag{2}$$

Assume that $(2)$ holds for some $n$; then

$$\begin{align*} xD_q(xD_q)^n&=\sum_kS_q(n,k)xD_qx^kD_q^k\\ &=\sum_kS_q(n,k)x\Big(q^kx^kD_q+(k)_qx^{k-1}\Big)D_q^k\\ &=\sum_kS_q(n,k)q^kx^{k+1}D_q^{k+1}+\sum_k(k)_qS_q(n,k)x^kD_q^k\\ &=\sum_k\left(S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\right)x^kD_q^k\;, \end{align*}$$

so if we want $(2)$ to hold for $n+1$, we must set

$$S_q(n+1,k)=S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\;.\tag{3}$$

$(2)$ clearly requires that $S_q(n,0)=\delta_{n,0}$; it imposes no constraint on $S_q(0,k)$ for $k>0$, but setting it to $0$ is the natural thing to do and is compatible with $(3)$.

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  • $\begingroup$ Dear Brian, thanks for writing this up! I saw it just now. $\endgroup$ Mar 10, 2012 at 22:21
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If you can read German you find an answer in my lecture notes at University of Vienna, Elementare q-Identitäten, Chapter 3.

Your $S_q(n,k)$ is the same as my $q^{\binom{k}{2}} S[n,k]$

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  • $\begingroup$ Thanks for these notes, Johann. $\endgroup$ Mar 10, 2012 at 22:22

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