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The "quantum" version qTSPP of the number of totally symmetric plane partitions, contained in the cube $[0,n]^3$, is enumerated by $$f_n(q):=\prod_{j=1}^n\prod_{k=1}^j\prod_{\ell=1}^k\frac{1-q^{j+k+\ell-1}}{1-q^{j+k+\ell-2}}.$$ L'Hopital $f_n(1)=\lim_{q\rightarrow1}f_n(q)$ restores the classical version $\prod_{1\leq\ell\leq k\leq j\leq n}\frac{j+k+\ell-1}{j+k+\ell-2}$. Although $f_n(-1)=0$ trivially, when $n$ is odd, I observe the case $n$ even is decidedly striking; namely that, $$f_{2n}(-1)=\lim_{q\rightarrow -1}f_{2n}(q)=\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!},$$ the number $A_n$ of $n\times n$ Alternating Sign Matrices or $ASMs$.

QUESTION still waiting for an answer.

Is there a non-analytic (more conceptual) reason for this connection between qTSPP and ASMs?

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    $\begingroup$ Maybe it is related to this question? mathoverflow.net/questions/247965/… $\endgroup$ – Per Alexandersson Dec 6 '16 at 0:28
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    $\begingroup$ Your conjecture is a nice example of Stembridge's $q=-1$ phenomenon, not that this observation helps with a proof. $\endgroup$ – Richard Stanley Dec 6 '16 at 2:00
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    $\begingroup$ Curiosity asking: Is there even such a thing as "quantum symmetric plane partitions" known? I understand that $f_n(q)$ is a natural way to quantize the generating series, but do we know a statistic on symmetric plane partitions that the yields this series as generating function? (Sorry if this is something well-known.) $\endgroup$ – darij grinberg Dec 6 '16 at 8:09
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    $\begingroup$ Instead of $\sum_{\pi}1$, you do the weighted sum $\sum_{\pi}q^{\vert\pi\vert}$. $\endgroup$ – T. Amdeberhan Dec 6 '16 at 14:36
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    $\begingroup$ @darijgrinberg: see arxiv.org/abs/1002.4384 $\endgroup$ – Sam Hopkins Dec 6 '16 at 23:11

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