Express symmetric polynomial $\prod_{i < j} (X_i+X_j)$ in terms of elementary symmetric functions Exercise: Define a polynomial $\Sigma(X_1,\ldots,X_n)$ as
\begin{align*}
\Sigma(X_1,\ldots,X_n) = \prod_{i < j} (X_i+X_j)
\end{align*}
This is a symmetric polynomial, quite clearly. I want to express $\Sigma$ as a polynomial in the elementary symmetric functions, for $n=2,3,4$. Case $n=2$ is trivial: $\Sigma(X_1,X_2)= \sigma_1 = X_1+X_2$. Case $n=3$ is easy. I write $\Sigma$ as follows:
\begin{align*}
\Sigma(X_1,X_2,X_3) = (\sigma_1 - X_1)(\sigma_1 - X_2)(\sigma_1 - X_3),
\end{align*}
then apply Vieta's formulas formally, obtaining
\begin{align*}
\Sigma(X_1,X_2,X_3) = \sigma_1^3 - \sigma_1 \sigma_1^2 + \sigma_2 \sigma_1 - \sigma_3 = \sigma_2 \sigma_1 - \sigma_3.
\end{align*}
The case $n=4$ seems harder to me, and I'm unable to work it out. I hope there is some trick to avoid unnecessary computations. My attempt is to write down $\Sigma$ completely:
\begin{align*}
\Sigma = (X_1+X_2)(X_3+X_4)(X_1+X_4)(X_2+X_3)(X_2+X_4)(X_1+X_3)
\end{align*}
and then multiply factors in groups of two, obtaining:
\begin{align*}
\Sigma = (\sigma_2 - X_1X_2-X_3X_4)(\sigma_2 - X_1X_4 - X_2X_3)(\sigma_2 - X_2X_4-X_1X_3),
\end{align*}
but it doesn't seem to help much. How can I solve the exercise?
 A: Here is an answer for any number $n$ of variables.
For any partition
$\lambda$, we let $s_{\lambda}$ denote the Schur
function
corresponding to $\lambda$. Let $\delta$ be the partition $\left(
n-1,n-2,\ldots,1\right)  $ of $n\left(  n-1\right)  /2$. Then,
\begin{equation}
\prod_{1\leq i<j\leq n}\left(  x_{i}+x_{j}\right)
= s_{\delta}\left( x_{1},x_{2},\ldots,x_{n}\right)  .
\label{1}
\tag{1}
\end{equation}
This well-known formula (which, I think, is due to Jacobi) can easily be
derived from the definition of Schur polynomials through alternants. Let me explain:
For every $n$-tuple $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in\mathbb{N}^{n}$ of nonnegative integers, we define the
alternant $a_{\alpha}$ to be the polynomial
\begin{equation}
\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}x_{1}^{\alpha_{\sigma\left( 1\right)  }}x_{2}^{\alpha_{\sigma\left(  2\right)  }}\cdots x_{n}^{\alpha_{\sigma\left( n\right)  }}
= \det\left(  \left(  x_i^{\alpha_j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}\right)
\end{equation}
in $\mathbb{Z}\left[  x_{1},x_{2},\ldots,x_{n}\right]  $. If $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $ and $\beta=\left(  \beta
_{1},\beta_{2},\ldots,\beta_{n}\right)  $ are two $n$-tuples in $\mathbb{N}
^{n}$, then the $n$-tuple $\alpha+\beta$ is defined to be $\left(  \alpha
_{1}+\beta_{1},\alpha_{2}+\beta_{2},\ldots,\alpha_{n}+\beta_{n}\right)  $.
Any partition $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{k}\right)  $ of length $k\leq n$ will be identified with the $n$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k},\underbrace{0,0,\ldots
,0}_{n-k\text{ times}}\right)  \in\mathbb{N}^{n}$. Notice that $\delta$ is
thus identified with the $n$-tuple $\left(  n-1,n-2,\ldots,1,0\right)  $.
Then, the "alternant formula" for the Schur polynomial $s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{n}\right)  $ says that
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\dfrac{a_{\lambda+\delta
}}{a_{\delta}}
\label{2}
\tag{2}
\end{equation}
whenever $\lambda$ is a partition of length $\leq n$. This is the historically
first definition of Schur polynomials, long before the modern definitions via
Young tableaux or the Jacobi-Trudi formulas were discovered; its main
disadvantages are that it only defines $s_{\lambda}$ in finitely many
variables and that it requires proof that $\dfrac{a_{\lambda+\delta}
}{a_{\delta}}$ is indeed a polynomial. But this is fine for us.
(For a proof of the fact that this definition of $s_\lambda$ is equivalent
to the modern combinatorial definition, you can consult Corollary 2.6.6 in Darij Grinberg and Victor Reiner, Hopf algebras in combinatorics, arXiv:1409.8356v5. In the same chapter you will find an exercise proving the Jacobi-Trudi identity, which is yet another popular definition of the Schur polynomials.)
We can now prove \eqref{1}. Indeed, $\delta=\left(  n-1,n-2,\ldots,0\right)
\in\mathbb{N}^{n}$, so that the definition of $a_{\delta}$ yields
\begin{equation}
a_{\delta+\delta}=\det\left(  \left(  x_{i}^{n-j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right)  =\prod_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)
\label{3}
\tag{3}
\end{equation}
(by the Vandermonde determinant formula). But $\delta+\delta=\left(  2\left(
n-1\right)  ,2\left(  n-2\right)  ,\ldots,2\cdot0\right)  $, so that
\begin{align}
a_{\delta+\delta}  & =\det\left(  \left(  x_{i}^{2\left(  n-j\right)
}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  =\det\left(  \left(
\left(  x_{i}^{2}\right)  ^{n-j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\right)  \nonumber\\
& =\prod_{1\leq i<j\leq n}\left(  x_{i}^{2}-x_{j}^{2}\right)
\label{4}
\tag{4}
\end{align}
(again by the Vandermonde determinant formula). Dividing \eqref{4} by
\eqref{3}, we obtain
\begin{equation}
\dfrac{a_{\delta+\delta}}{a_{\delta}}=\dfrac{\prod_{1\leq i<j\leq n}\left(
x_{i}^{2}-x_{j}^{2}\right)  }{\prod_{1\leq i<j\leq n}\left(  x_{i}
-x_{j}\right)  }=\prod_{1\leq i<j\leq n}\underbrace{\dfrac{x_{i}^{2}-x_{j}
^{2}}{x_{i}-x_{j}}}_{=x_{i}+x_{j}}=\prod_{1\leq i<j\leq n}\left(  x_{i}
+x_{j}\right)  .
\end{equation}
But applying \eqref{2} to $\lambda=\delta$, we obtain $s_{\delta}\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\dfrac{a_{\delta+\delta}}{a_{\delta}}
=\prod_{1\leq i<j\leq n}\left(  x_{i}+x_{j}\right)  $. Thus, \eqref{1} is proven.
Finally, we can transform \eqref{1} into an explicit (if you allow
determinants) formula for $\prod_{1\leq i<j\leq n}\left(  x_{i}+x_{j}\right)
$ in terms of the elementary symmetric polynomials. To that aim, we let
$e_{k}\left(  \overrightarrow{x}\right)  $ denote the $k$-th elementary
symmetric polynomial in $x_{1},x_{2},\ldots,x_{n}$. Notice that $e_{k}\left(
\overrightarrow{x}\right)  =0$ whenever $k<0$ and also whenever $k>n$.
Recall that one of the two Jacobi-Trudi formulas says that if $\lambda$ is a
partition of length $\leq n$, then
\begin{equation}
s_{\lambda^{t}}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\det\left(  \left(
e_{\lambda_{i}-i+j}\left(  \overrightarrow{x}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right)  ,
\end{equation}
where $\lambda^{t}$ denotes the conjugate partition of $\lambda$ (this is the
partition whose $i$-th entry is the number of entries of $\lambda$ that are
$\geq i$, for each $i\in\left\{  1,2,3,\ldots\right\}  $). Applying this to
$\lambda=\delta$, we obtain
\begin{align*}
s_{\delta^{t}}\left(  x_{1},x_{2},\ldots,x_{n}\right)    & =\det\left(
\left(  e_{\left(  n-i\right)  -i+j}\left(  \overrightarrow{x}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  \\
& =\det\left(  \left(  e_{n-2i+j}\left(  \overrightarrow{x}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}\right)  \\
& =\det\left(
\begin{array}
[c]{cccc}
e_{n-1}\left(  \overrightarrow{x}\right)   & e_{n}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-2}\left(  \overrightarrow{x}\right)  \\
e_{n-3}\left(  \overrightarrow{x}\right)   & e_{n-2}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-4}\left(  \overrightarrow{x}\right)  \\
\vdots & \vdots & \ddots & \vdots\\
e_{-n+1}\left(  \overrightarrow{x}\right)   & e_{-n+2}\left(
\overrightarrow{x}\right)   & \cdots & e_{0}\left(  \overrightarrow{x}\right)
\end{array}
\right)  .
\end{align*}
Since $\delta^{t}=\delta$ (this is easy to check), this simplifies to
\begin{equation}
s_{\delta}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\det\left(
\begin{array}
[c]{cccc}
e_{n-1}\left(  \overrightarrow{x}\right)   & e_{n}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-2}\left(  \overrightarrow{x}\right)  \\
e_{n-3}\left(  \overrightarrow{x}\right)   & e_{n-2}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-4}\left(  \overrightarrow{x}\right)  \\
\vdots & \vdots & \ddots & \vdots\\
e_{-n+1}\left(  \overrightarrow{x}\right)   & e_{-n+2}\left(
\overrightarrow{x}\right)   & \cdots & e_{0}\left(  \overrightarrow{x}\right)
\end{array}
\right)  .
\end{equation}
Thus, \eqref{1} rewrites as
\begin{equation}
\prod_{1\leq i<j\leq n}\left(  x_{i}+x_{j}\right)
=\det\left(
\begin{array}
[c]{cccc}
e_{n-1}\left(  \overrightarrow{x}\right)   & e_{n}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-2}\left(  \overrightarrow{x}\right)  \\
e_{n-3}\left(  \overrightarrow{x}\right)   & e_{n-2}\left(  \overrightarrow{x}
\right)   & \cdots & e_{2n-4}\left(  \overrightarrow{x}\right)  \\
\vdots & \vdots & \ddots & \vdots\\
e_{-n+1}\left(  \overrightarrow{x}\right)   & e_{-n+2}\left(
\overrightarrow{x}\right)   & \cdots & e_{0}\left(  \overrightarrow{x}\right)
\end{array}
\right)  .
\end{equation}
Note that the matrix on the right hand side is not generally upper-triangular,
but it has a lot of zeroes (more or less the whole lower half of its
southwestern triangle consists of zeroes), so its determinant is a bit easier
to compute than a general $n\times n$ determinant. But I don't think there is
a more explicit formula.
