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One has the nice identities $${xy\choose 1}={x\choose 1}{y\choose 1},$$ $${xy+1\choose 2}={x+1\choose 2}{y+1\choose 2}+{x\choose 2}{y\choose 2}$$ and $${xy+2\choose 3}={x+2\choose 3}{y+2\choose 3}+4{x+1\choose 3}{y+1\choose 3}+{x\choose 3}{y\choose 3}.$$

(The proof is essentially trivial by interpreting ${z\choose k}$ as a polynomial of degree $k$.)

This sequence of identities stops: There seems to be no nice expression of ${xy+k-1\choose k}$ as a linear combination of ${x+k-i\choose k}{y+k-i\choose k},i=1,\ldots,k$ for $k\geq 4$.

Is there a good reason for this breakdown? (Probably a better question is: Is there a reason for these identities to hold for $k=2$ and, especially, for $k=3$?)

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    $\begingroup$ Is it a coincidence that 1; 1,1; 1,4,1 is the Eulerian numbers sequence? $\endgroup$ Aug 3 at 14:59
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    $\begingroup$ No idea if this is coincidental. Coefficients on the right hand side should however add up to $k!$ (by comparing coefficients of $(xy)^k$ on both sides). $\endgroup$ Aug 3 at 15:30

1 Answer 1

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$\def\des{\operatorname{des}}$Let $\des(\pi)$ be the number of descents of the permutation $\pi$. Then for any permutation $\pi$ in $S_k$, we have \begin{equation*}\binom{xy+k-\des(\pi)-1}{k} =\sum_{\sigma\tau=\pi}\binom{x+k-\des(\tau)-1}{k} \binom{y+k-\des(\sigma)-1}{k}.\tag{$*$}\label{star} \end{equation*}

If $\pi$ is the identity then $\des(\pi)=0$, so the left side of \eqref{star} is $\binom{xy+k-1}{k}$ and on the right, $\tau=\sigma^{-1}$. The OP's identities correspond to the fact that if $k\le 3$ then the number of descents of $\pi$ is the same as the number of descents of $\pi^{-1}$, but this is not true for $k>3$. This also explains the occurrence of the Eulerian numbers as coefficients for $k\le3$.

As far as I know, the earliest proof of \eqref{star} is in Bogdan Mielnik and Jerzy Plebański, Combinatorial approach to Baker–Campbell–Hausdorff exponents, Annales de l’I. H. P., section A, tome 12, no 3 (1970), pp. 215–254 (see equation (11.10)); further references can be found in Jason Fulman and T. Kyle Petersen, Card shuffling and P-partitions, arXiv:2004.01659 [math.CO].

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    $\begingroup$ Very nice explanation! Thanks. $\endgroup$ Aug 3 at 16:36
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    $\begingroup$ MathJax notes: as with TeX, you can use \label+\eqref with \tag and get nice hyperlinks to boot. Unlike TeX, MathJax is not aggressive about discarding whitespace, so a preamble line $\def\des{…}$ followed by a newline will force a space in the text. Unfortunately, as ugly as it is in the source, the only way to get rid of the space is to put the closing $ directly next to the first line of the post, with no intervening whitespace. I have edited accordingly. $\endgroup$
    – LSpice
    Aug 3 at 18:56
  • $\begingroup$ is it implied that $\pi = \tau \sigma$? $\endgroup$ Aug 4 at 12:03
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    $\begingroup$ @NooneAtAll If $\sigma\tau$ is the identity then so is $\tau\sigma$ but in general $\tau\sigma\ne\sigma\tau$. $\endgroup$
    – Ira Gessel
    Aug 4 at 20:44

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