Hi everyone!

Mandatory orz to Elegia whose blog introduced me to these concepts.

Today, I would like to write about MacMahon's master theorem (MMT). It is a nice result on the intersection of combinatorics and linear algebra that provides an easy prove to some particularly obscure combinatorial identities, in particular the Dixon's identity:

To begin with, let's formulate MacMahon's master theorem.

MMT. Let $$$\mathbf A = [a_{ij}]_{n\times n}$$$, $$$\mathbf X = \operatorname{diag}(x_1,\dots,x_n)$$$ and $$$\mathbf k = (k_1,\dots,k_n) \geq 0$$$. Then,

With MMT, the Dixon's identity is proven as elegantly as this:

Variable-less MMT

Now, let's look into MMT and try to understand what does it actually stand for? First of all, we can multiply both sides by $$$\mathbf x^\mathbf k$$$ and sum them up over all $$$\mathbf k \geq 0$$$. In this case, the RHS, by definition, will turn into $$$\det(\mathbf I - \mathbf X \mathbf A)^{-1}$$$, while the LHS will turn into

We can now do the substitution $$$\mathbf B = \mathbf X \mathbf A$$$ and use $$$b_{ij} = a_{ij} x_i$$$ to reformulate MMT in the equivalent form

This form is more convenient for us to work with, as it doesn't depend on any variables in RHS.

LHS as the sum of permanents

Now, let's take a closer look on the individual summands of the LHS. What do they enumerate?

where $$$(i_1,\dots,i_k)$$$ is a tuple such that $$$1 \leq i_1 \leq \dots \leq i_k \leq n$$$ and each $$$i$$$ occurs exactly $$$k_i$$$ times in it. Correspondingly, $$$(j_1,\dots,j_k)$$$ are all possible rearrangements of such tuple. Wait a second. We go over all rearrangements of a certain set of indices and sum up the products of $$$b_{i_s j_s}$$$ over them. Doesn't it sound familiar?

If we consider all permutations of $$$(1,\dots,k)$$$ rather than direct rearrangements of $$$(i_1,\dots,i_k)$$$, this will just be a permanent of a $$$(k_1+\dots+k_n) \times (k_1+\dots+k_n)$$$ matrix, in which the element $$$b_{ij}$$$ occurs as a block of size $$$k_i \times k_j$$$.

This concept usually occurs in literature defined even for two distinct vectors $$$\mathbf k \geq 0$$$ and $$$\mathbf l \geq 0$$$, so that $$$b_{ij}$$$ occurs in a block of size $$$k_i \times l_j$$$. The resulting matrix is typically denoted as $$$\mathbf B^{(\mathbf k, \mathbf l)}$$$. So, e.g. for $$$n = 3$$$, $$$\mathbf k = (3, 2, 1)$$$ and $$$\mathbf l = (1, 2, 3)$$$, $$$\mathbf B^{(\mathbf k, \mathbf l)}$$$ would look like

Note that to go back to rearrangements of $$$(i_1,\dots,i_k)$$$ rather than permutations of $$$(1,\dots,k)$$$ we should divide the permanent by $$$k_1 ! \dots k_n!$$$, which is commonly denoted in literature as $$$\mathbf k!$$$. In this terms, MMT rewrites concisely as

Or, in terms of the original matrix $$$\mathbf A$$$, as

Sums of permanents as traces of symmetric powers

Now, let's denote $$$|\mathbf k| = k_1+\dots+k_n$$$. We can show that

where $$$\operatorname{S}^k(\mathbf B)$$$ is the so-called symmetric power of $$$\mathbf B$$$.

Let's properly define it to better understand what it means. Before doing so, we should define tensor products. Let $$$V$$$ and $$$W$$$ be vector spaces with bases $$$B_V$$$ and $$$B_W$$$ correspondingly. Then, their tensor product $$$V \otimes W$$$ is the vector space associated with the tensor product operation $$$\otimes : V \times W \to V \otimes W$$$, such that the set $$$\{v \otimes w : v \in B_V, w \in B_W\}$$$ forms a basis of $$$V \otimes W$$$.

Computationally, if we represent vectors in coordinate form, one possible way to represent tensor products is via Kronecker product operation applied to coordinate arrays. When tensor product is defined on vectors, it is also implicitly defined on linear maps. So, if $$$A : V_1 \to V_2$$$ and $$$B: W_1 \to W_2$$$ are linear maps, then their tensor product is the linear map $$$A \otimes B : V_1 \otimes W_1 \to V_2 \otimes W_2$$$, such that $$$(A \otimes B)(v \otimes w) = (Av)\otimes (Bw)$$$. If $$$A$$$ and $$$B$$$ are represented by matrices, $$$A \otimes B$$$ can be represented as their Kronecker product.

Now, to define symmetric powers, we should consider $$$V \otimes V$$$, a tensor product of $$$V$$$ with itself.