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In this paper, there's an identity that I can't prove to my satisfaction, (there's a similar statement in here) which is that, given a permanent of a $(\mathbf{k,l})$-replicated matrix $A$, (written $A^{(\mathbf{k,l})}$), $$\sum_{\mathbf{k,l}\geq 0}\mathrm{per}(A^{(\mathbf{k,l})})\frac{\mathbf{x}^{\mathbf{k}}\mathbf{y}^{\mathbf{l}}}{\mathbf{k}!\mathbf{l}!}=\exp{\mathbf{x^{T}}A\mathbf{y}}$$

Notation: $\mathbf{x^{k}}=\prod_{i=1}^{n}x_{i}^{k_{i}}$, $\mathbf{k}!=\prod_{i=1}^{n}k_{i}!$, and $A^{(\mathbf{k,l})}$ is the block matrix with the entry $a_{i,j}$ repeated $k_{i}\times l_{j}$ times.

This is related to MacMahon's Master Theorem where $$ \sum_{\mathbf{k}\geq 0}\mathrm{per}(A^{(\mathbf{k,k})})\frac{\mathbf{x}^{\mathbf{k}}}{\mathbf{k}!}=\det (1-XA)^{-1} $$ with $X_{ij}=x_{i}\delta_{ij}$ and $A=A^{(\mathbf{1,1})}$.

The Amplitwist
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TeeJay
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1 Answers1

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Here is a double-counting argument. The coefficient of $\mathbf{x}^{\mathbf{k}}\mathbf{y}^{\mathbf{l}}$ on either side can be written as $$\frac{1}{m!^2}\sum_{K,L,\pi} a_{K(1),L(\pi(1))}\dots a_{K(m),L(\pi(m))}$$ where $m=\sum k_i=\sum l_j$, and the sum is over:

  • $K:[m]\to [n]$ with $|K^{-1}(\{i\})|=k_i$ for each $i$
  • $L:[m]\to [n]$ with $|L^{-1}(\{j\})|=l_j$ for each $j$
  • permutations $\pi:[m]\to[m]$

To see this is equal to the left-hand-side coefficient, note that the sum over $\pi$ for any fixed $(K,L)$ is $\mathrm{per}(A^{(\mathbf{k,l})})$; the $\sum_K$ gives a factor of $m!/\mathbf k!$ and likewise the $\sum_L$ gives a factor of $m!/\mathbf l!$. To see this is equal to the right-hand-side coefficient, recognise the sum over $(K,L)$ for any fixed $\pi$ as the coefficient of $\mathbf{x}^{\mathbf{k}}\mathbf{y}^{\mathbf{l}}$ in $(\sum_{i,j=1}^n a_{i,j}x_iy_j)^m$, with the $\sum_\pi$ giving a factor of $m!$.

darij grinberg
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Dap
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