[proofplan]
We prove the identity inside the completed graded [tensor product](/page/Tensor%20Product) of the symmetric-function ring. The key input is the general Cauchy kernel expansion in dual bases for the Hall [inner product](/page/Inner%20Product): if $(u_\lambda)_\lambda$ and $(v_\lambda)_\lambda$ are dual homogeneous bases, then the kernel $\prod_{i,j}(1-x_i y_j)^{-1}$ expands as $\sum_\lambda u_\lambda(x)v_\lambda(y)$. We then use Schur orthonormality for the Hall inner product, so the Schur basis is dual to itself. Finally, the completion justifies summing over all partitions degree by degree.
[/proofplan]
[step:Place the Cauchy kernel in the completed graded tensor product]
Let $\Lambda_{\mathbb{Q}} = \bigoplus_{d=0}^{\infty} \Lambda_{\mathbb{Q}}^d$ be the graded ring of symmetric functions over $\mathbb{Q}$, where $\Lambda_{\mathbb{Q}}^d$ is the finite-dimensional subspace of homogeneous symmetric functions of degree $d$. Define the completed diagonal tensor product by
\begin{align*}
\widehat{\Lambda_{\mathbb{Q}} \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}}
:=
\prod_{d=0}^{\infty} \Lambda_{\mathbb{Q}}^d \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}^d.
\end{align*}
For each integer $d \geq 0$, let $\mathcal{P}_d$ denote the finite set of partitions of $d$. Since $s_\lambda \in \Lambda_{\mathbb{Q}}^{|\lambda|}$ for every partition $\lambda$, the expression
\begin{align*}
\sum_{\lambda} s_\lambda(x)s_\lambda(y)
\end{align*}
means the element whose degree-$d$ component is
\begin{align*}
\sum_{\lambda \in \mathcal{P}_d} s_\lambda(x) \otimes s_\lambda(y)
\in \Lambda_{\mathbb{Q}}^d \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}^d.
\end{align*}
Thus the right-hand side is well-defined in the completed tensor product.
Likewise, expanding each factor as the geometric series
\begin{align*}
(1-x_i y_j)^{-1}=\sum_{r=0}^{\infty} (x_i y_j)^r
\end{align*}
defines the Cauchy kernel degree by degree, because the homogeneous component of total $x$-degree and $y$-degree equal to $d$ depends only on finitely many partitions of $d$ when expressed in any homogeneous symmetric-function basis. Hence
\begin{align*}
\prod_{i,j \geq 1}(1-x_i y_j)^{-1}
\end{align*}
is an element of $\widehat{\Lambda_{\mathbb{Q}} \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}}$.
[guided]
The infinite product and the infinite sum are not ordinary finite algebraic expressions, so the first task is to specify the ambient space in which they live. The ring of symmetric functions is graded:
\begin{align*}
\Lambda_{\mathbb{Q}} = \bigoplus_{d=0}^{\infty} \Lambda_{\mathbb{Q}}^d,
\end{align*}
where $\Lambda_{\mathbb{Q}}^d$ consists of homogeneous symmetric functions of degree $d$. For the Cauchy identity, the $x$-degree and the $y$-degree always match. Therefore the correct completion is
\begin{align*}
\widehat{\Lambda_{\mathbb{Q}} \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}}
=
\prod_{d=0}^{\infty} \Lambda_{\mathbb{Q}}^d \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}^d.
\end{align*}
Now define $\mathcal{P}_d$ to be the set of partitions of $d$. This set is finite. Since $s_\lambda$ is homogeneous of degree $|\lambda|$, the degree-$d$ part of the Schur sum is
\begin{align*}
\sum_{\lambda \in \mathcal{P}_d} s_\lambda(x) \otimes s_\lambda(y),
\end{align*}
a finite sum in $\Lambda_{\mathbb{Q}}^d \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}^d$. Thus the full sum over all partitions is not an uncontrolled infinite sum; it is a sequence of finite homogeneous pieces.
The same grading controls the product
\begin{align*}
\prod_{i,j \geq 1}(1-x_i y_j)^{-1}.
\end{align*}
Each factor expands as
\begin{align*}
(1-x_i y_j)^{-1}=\sum_{r=0}^{\infty}(x_i y_j)^r.
\end{align*}
Every monomial produced has equal degree in the $x$ variables and the $y$ variables. Therefore the product defines a homogeneous degree-by-degree element of the same completed tensor product. This is the formal reason the infinite alphabets cause no convergence issue: all manipulations are algebraic and are performed one degree at a time.
[/guided]
[/step]
[step:Expand the Cauchy kernel using dual bases for the Hall inner product]
Let
\begin{align*}
(\cdot,\cdot)_{\mathrm{Hall}} : \Lambda_{\mathbb{Q}} \times \Lambda_{\mathbb{Q}} \to \mathbb{Q}
\end{align*}
denote the Hall inner product. We use the Cauchy [kernel expansion in dual homogeneous bases](/theorems/5201) for the Hall inner product: if $(u_\lambda)_\lambda$ and $(v_\lambda)_\lambda$ are homogeneous bases of $\Lambda_{\mathbb{Q}}$ satisfying
\begin{align*}
(u_\lambda,v_\mu)_{\mathrm{Hall}}=\delta_{\lambda\mu}
\end{align*}
for all partitions $\lambda,\mu$, then
\begin{align*}
\prod_{i,j \geq 1}(1-x_i y_j)^{-1}
=
\sum_{\lambda} u_\lambda(x)v_\lambda(y)
\end{align*}
in $\widehat{\Lambda_{\mathbb{Q}} \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}}$.
This is the standard reproducing-kernel property of the Hall inner product, applied degree by degree to the finite-dimensional spaces $\Lambda_{\mathbb{Q}}^d$. The required external result is not yet resolved in the wiki: (citing a result not yet in the wiki: Cauchy kernel expansion in dual bases for the Hall inner product).
[/step]
[step:Use Schur orthonormality to identify the dual bases]
For partitions $\lambda$ and $\mu$, the Schur functions satisfy the Schur orthonormality relation
\begin{align*}
(s_\lambda,s_\mu)_{\mathrm{Hall}}=\delta_{\lambda\mu}.
\end{align*}
The required external result is not yet resolved in the wiki: (citing a result not yet in the wiki: Schur orthonormality for the Hall inner product).
Therefore the Schur basis is dual to itself under the Hall inner product. In the dual-basis kernel expansion from the previous step, we may take
\begin{align*}
u_\lambda=s_\lambda,
\qquad
v_\lambda=s_\lambda
\end{align*}
for every partition $\lambda$. Substitution gives
\begin{align*}
\prod_{i,j \geq 1}(1-x_i y_j)^{-1}
=
\sum_{\lambda} s_\lambda(x)s_\lambda(y).
\end{align*}
[/step]
[step:Conclude the completed identity degree by degree]
For each $d \geq 0$, both sides have homogeneous component in
\begin{align*}
\Lambda_{\mathbb{Q}}^d \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}^d.
\end{align*}
The dual-basis kernel expansion proves equality of these degree-$d$ components. Since the completed tensor product is the direct product of the homogeneous components over all $d \geq 0$, equality in every degree implies equality in
\begin{align*}
\widehat{\Lambda_{\mathbb{Q}} \otimes_{\mathbb{Q}} \Lambda_{\mathbb{Q}}}.
\end{align*}
Hence
\begin{align*}
\prod_{i,j \geq 1}(1-x_i y_j)^{-1}
=
\sum_{\lambda} s_\lambda(x)s_\lambda(y),
\end{align*}
as claimed.
[/step]
