Plane partitions as sums of determinants

Question

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).

Proof. Replacing $\mathbf{x}=(x_1,\dots,x_m)$ by $\mathbf{J}=(j_1,\dots,j_m)$ into $$\det\left[\binom{x_j}i\right] =\frac1{\prod_{i<j}(j-i)}\det\left[\prod_{k=0}^{i-1}(x_j-k)\right] =\frac{\det\left[x_j^{i-1}\right]}{\prod_{i<j}(j-i)} =\prod_{1\leq i \lt j\leq m} \frac{x_j-x_i}{j-i}.$$ results in integer entries, and thus integer determinant $\frac{V(\mathbf{J})}{V(\mathbf{I})}$. $\qquad\square$

Answer 1

I haven't worked out the details, but $V(\mathbf{J})/V(\mathbf{I})$ is the principal specialization of a Schur function. Then $\left( \frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2$ corresponds to a pair of SSYT (semistandard Young tableaux), which can be merged into a plane partition as in EC2, proof of Theorem 7.20.1. Summing over $\mathbf{J}$ should give all plane partitions fitting in an $n\times m\times m$ box.

Answer 2

Well, Binet–Cauchy plus Lindström–Gessel–Viennot work indeed.

Let $i,j$ vary from 0 to $m-1$, and $k$ vary from $0$ to $m+n-1$. Denote also $k^*=m+n-1-k$, so $k^*$ also varies from 0 to $m+n-1$. Consider the $m\times (m+n)$ matrices $A=(a_{i,k})$ and $B=(b_{j,k})$ defined by $$a_{i,k}=(-1)^k{-i-1\choose k}={i+k\choose i}\\ b_{j,k}=(-1)^{k^*}{-j-1\choose k^*}={j+k^*\choose j}.$$ Note that for fixed $i$, $a_{i,k}$ as a function of $k$ is a polynomial of degree $i$ with leading coefficient $1/i!$; and so is $(-1)^ib_{i,k}$. Thus the minors of $A$, $B$ indexed by $\mathbf{J}=\{j_1<j_2<\ldots<j_m\}$ are equal to $V(\mathbf{J})/V(\mathbf{I})$ and $(-1)^{m\choose 2}V(\mathbf{J})/V(\mathbf{I})$ respectively. Thus by Binet–Cauchy we have \begin{align} \sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2=(-1)^{m\choose 2}\det AB^t. \end{align} As for $AB^t$, its elements $c_{ij}$ are equal to \begin{align} c_{ij}&=\sum_{k=0}^{m+n-1}a_{i,k}b_{j,k}=(-1)^{m+n-1}\sum_{k=0}^{m+n-1} {-i-1\choose k}{-j-1\choose k^*}\\ &=(-1)^{m+n-1}{-i-j-2\choose m+n-1}= {i+j+m+n\choose m+n-1} \\ &=P((0,-i),(m+n-1,j+1)), \end{align} where, for $u,v\in \mathbb{Z}^2$, $P(u,v)$ is the number of lattice paths from $u$ to $v$ (paths go up and right). So, $\det AB^t$ has Lindström–Gessel–Viennot interpretation. The collection of $m$ non-intersecting lattice paths from the points $u_i:=(0,-i)$ to the points $v_j:=(m+n-1,j+1)$ of course must join $u_i$ with $v_{m-1-i}$. This, in particular, gets $(-1)^{m\choose 2}$ factor out. Also, the path from $v_i$ must start with a horizontal part of length at least $i$; analogously for finishing paths. So, counting these paths is equivalent to counting the non-intersecting paths which go from $(i,-i)$ to $(n+i, m-i)$. These paths correspond to Young diagrams in $m\times n$ boards, and the condition of them being non-intersecting is equivalent to corresponding diagrams be $m$ consecutive sections of a 3d Young diagram in the $m\times m\times n$ box. (This last argument is a standard proof of MacMahon formula).

Since Binet–Cauchy itself has interpretation in terms of Lindström–Gessel–Viennot, you may rephrase this argument more combinatorially, if you prefer. I expect that what is obtained is exactly the answer outlined by Richard Stanley.