# Plane partitions as sums of determinants

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i The number of plane partitions in an $$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 $$V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$$.

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 $$\det\left[\binom{x_j}i\right] =\frac1{\prod_{i results in integer entries, and thus integer determinant $$\frac{V(\mathbf{J})}{V(\mathbf{I})}$$. $$\qquad\square$$

• Related (also unanswered!): math.stackexchange.com/questions/3367836/… Apr 28 at 23:30
• Hm, looks like Binet-Cauchy. This $V(\mathbf{J})/V(\mathbf{J})$ is a determinant of an $m\times m$ matrix $(f_{k-1}(j_i))_{k, i\in [m]}$, where $f_k(x)$ is a polynomial of degree $k$ with leading coefficient $1/k!$, for example $f_k(x)={x+c_k\choose k}$ for some $c_k$. If I am not mistaken, this gives a determinant ${i+j+m+n\choose m+n-1}_{0\le i, j\le m-1}$. That has some LGV interpretation but not which immediately gives these 3d Young diagrams in a box. Apr 30 at 11:45

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.
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 $$\mathbf{J}=\{j_1 are equal to $$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).