Consider the determinantal function $$F(a,b,c):=\det\left[\binom{i+j+a+b}{i+a}\right]_{i,j=0}^{c-1}.$$ I would like to ask:

QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry $$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$ without actually computing the its values? I know how to justify this based on direct evaluation.

Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or $$\det\begin{pmatrix} 10&15&21&28\\20&35&56&84\\35&70&126&210\\56&126&252&462 \end{pmatrix}= \det\begin{pmatrix} 15&35&70\\21&56&126\\28&84&210 \end{pmatrix}=\det\begin{pmatrix} 35&56\\70&126 \end{pmatrix}.$$

POSTSCRIPT. To add more examples to David Speyer's "mincing" in his reply to a related MO question, consider $$G(a,b,c)=\det\left[\binom{a+b}{a+j-i}\right]_{i,j=0}^{c-1}.$$ Then, we have \begin{align*} F(a,b,c)&=G(a,b,c) \\ G(a,b,c)&=G(c,a,b)=G(b,c,a)=G(a,c,b)=G(c,b,a)=G(b,a,c). \end{align*}

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    $\begingroup$ Did you try to express this as the number of tuples of non-intersecting lattice paths? $\endgroup$ – Martin Rubey Feb 10 '19 at 19:05
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    $\begingroup$ The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b \in \{1,\ldots, 10\}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma? $\endgroup$ – Mark Wildon Feb 10 '19 at 20:44
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    $\begingroup$ Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10. $\endgroup$ – Gerhard Paseman Feb 10 '19 at 20:47

Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East). Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $a\times b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $a\times b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $a\times b\times c$ box. Thus the symmetry.


Just for the record:

In view of Fedor's nice reply, the interpretation gives away $$\det\left[\binom{i+j+a+b}{i+a}\right]_{i,j=0}^{c-1} =\prod_{i=0}^{a-1}\prod_{j=0}^{b-1}\prod_{k=0}^{c-1} \frac{i+j+k+2}{i+j+k+1}.\tag1$$ An apparent symmetry!

As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.

Let $(q)_k=(1-q)(1-q^2)\cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by $$\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}.$$ Then, we may generalize the above identity (1) as $$\det\left[q^{-i^2-ai-bj}\binom{i+j+a+b}{i+a}_q\right]_{i,j=0}^{c-1} =\prod_{i=0}^{a-1}\prod_{j=0}^{b-1}\prod_{k=0}^{c-1}\frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$

  • $\begingroup$ How does one prove that this first identity holds? $\endgroup$ – Carl Schildkraut Feb 11 '19 at 7:13
  • $\begingroup$ The first identity follows from Fedor's interpretation of the determinant as counting the number of 3D partitions in the $a \times b \times c$ box, or equivalently, the number of plane partitions in the $a \times b$ box with largest part at most $c$. The product formula counting these objects is a famous result of MacMahon: see e.g. (1) in math.mit.edu/~rstan/pubs/pubfiles/75.pdf. The first identity follows from the second by specializing $q$ to $1$: then $\binom{n}{m}_q$ becomes $\binom{n}{m}$ and $(1-q^n)/(1-q^m)$ becomes $n/m$. But I would like to see a proof of the second identity. $\endgroup$ – Mark Wildon Feb 11 '19 at 15:10
  • $\begingroup$ I think there is something wrong with the powers of $q$. Take $c=2$ and $a=b=1$. Then your matrix becomes $\left( \begin{matrix} \binom{2}{1}_q & q\binom{3}{1}_q \\ q^2\binom{3}{1}_q & q^3\binom{4}{2}_q \end{matrix}\right) = \left(\begin{matrix} 1 + q & q(1+q+q^2) \\ q^2(1+q+q^2) & q^3(1+q^2)(1+q+q^2) \end{matrix} \right)$ which has determinant $q^6(1+q+q^2)$. But the right hand side is the generating function for plane partitions in the $1\times 1 \times 2$-box, namely $1+q+q^2$. (And clearly the right-hand side always has constant term 1.) $\endgroup$ – Mark Wildon Feb 13 '19 at 21:13
  • $\begingroup$ @MarkWildon: You're right. Thanks. There was a minor type which is now fixed: $q^{-i^2-ai-bj}$. $\endgroup$ – T. Amdeberhan Feb 14 '19 at 22:37
  • $\begingroup$ Thank you! We can take out a factor of $q^{-i^2-ai}$ from row $i$ and $q^{-bj}$ from column $j$ to replace the left-hand side with $q^{-1^2 - \cdots - (c-1)^2} q^{-(a+b)\binom{c}{2}} \det\Bigl[ \binom{i+j+a+b}{i+a}_q \Bigr]_{i,j=0}^{c-1}$. Curiously when I tried to get your identity from the Jacobi–Trudi formula, I ended up with $q^{1^2 + \cdots + (c-1)^2} \det\Bigl[ q^{-ij} \binom{i+j+a+b}{i+a}_q \Bigr]_{i,j=0}^{c-1}$, which seems to be essentially different. I'll post the details if I have time. $\endgroup$ – Mark Wildon Feb 15 '19 at 11:42

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