# 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?

• The answer is $\sigma_3 \sigma_2 \sigma_1 - \sigma_3^2 - \sigma_4 \sigma_1^2$, but I found this with the help of a computer. I don't know a systematic way to find it (other than guessing which linear combinations of the degree-$6$ products of symmetric polynomials seem to simplify things). Mar 24, 2014 at 18:21
• Actually, writing $\Sigma=(\sigma_2-X_1X_2-X_3X_4)(\sigma_2-X_1X_4-X_2X_3)(\sigma_2-X_2X_4-X_1X_3)$, as you've done, is a very good starting point. Now you only need to know $(X_1X_2+X_3X_4)+(X_1X_4+X_2X_3)+(X_2X_4+X_1X_3)$ (which is $\sigma_2$), $$(X_1X_2+X_3X_4)(X_1X_4+X_2X_3)+(X_1X_4+X_2X_3)(X_2X_4+X_1X_3)+(X_2X_4+X_1X_3)(X_1X_2+X_3X_4)$$ and $(X_1X_2+X_3X_4)(X_1X_4+X_2X_3)(X_2X_4+X_1X_3)$. This takes some work, but it's not too bad. Mar 24, 2014 at 19:57
• I've managed to obtain the above result, even if it was only after some tedious computations. I wonder if there is a smarter way to do this... Mar 24, 2014 at 21:55

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, $$$$\prod_{1\leq i

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 $$$$\sum_{\sigma\in S_{n}}\left( -1\right) ^{\sigma}x_{1}^{\alpha_{\sigma\left( 1\right) }}x_{2}^{\alpha_{\sigma\left( 2\right) }}\cdots x_{n}^{\alpha_{\sigma\left( n\right) }} = \det\left( \left( x_i^{\alpha_j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right)$$$$ in $$\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 $$$$s_{\lambda}\left( x_{1},x_{2},\ldots,x_{n}\right) =\dfrac{a_{\lambda+\delta }}{a_{\delta}} \label{2} \tag{2}$$$$ whenever $$\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 $$$$a_{\delta+\delta}=\det\left( \left( x_{i}^{n-j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\prod_{1\leq i (by the Vandermonde determinant formula). But $$\delta+\delta=\left( 2\left( n-1\right) ,2\left( n-2\right) ,\ldots,2\cdot0\right)$$, so that \begin{align} a_{\delta+\delta} & =\det\left( \left( x_{i}^{2\left( n-j\right) }\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) =\det\left( \left( \left( x_{i}^{2}\right) ^{n-j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) \nonumber\\ & =\prod_{1\leq i (again by the Vandermonde determinant formula). Dividing \eqref{4} by \eqref{3}, we obtain $$$$\dfrac{a_{\delta+\delta}}{a_{\delta}}=\dfrac{\prod_{1\leq i But applying \eqref{2} to $$\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. Thus, \eqref{1} is proven.

Finally, we can transform \eqref{1} into an explicit (if you allow determinants) formula for $$\prod_{1\leq i in terms of the elementary symmetric polynomials. To that aim, we let $$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 $$$$s_{\lambda^{t}}\left( x_{1},x_{2},\ldots,x_{n}\right) =\det\left( \left( e_{\lambda_{i}-i+j}\left( \overrightarrow{x}\right) \right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) ,$$$$ where $$\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$$ (this is easy to check), this simplifies to $$$$s_{\delta}\left( x_{1},x_{2},\ldots,x_{n}\right) =\det\left( \begin{array} [c]{cccc} e_{n-1}\left( \overrightarrow{x}\right) & e_{n}\left( \overrightarrow{x} \right) & \cdots & e_{2n-2}\left( \overrightarrow{x}\right) \\ e_{n-3}\left( \overrightarrow{x}\right) & e_{n-2}\left( \overrightarrow{x} \right) & \cdots & e_{2n-4}\left( \overrightarrow{x}\right) \\ \vdots & \vdots & \ddots & \vdots\\ e_{-n+1}\left( \overrightarrow{x}\right) & e_{-n+2}\left( \overrightarrow{x}\right) & \cdots & e_{0}\left( \overrightarrow{x}\right) \end{array} \right) .$$$$ Thus, \eqref{1} rewrites as $$$$\prod_{1\leq i Note that the matrix on the right hand side is not generally upper-triangular, but it has a lot of zeroes (more or less the whole lower half of its southwestern triangle consists of zeroes), so its determinant is a bit easier to compute than a general $$n\times n$$ determinant. But I don't think there is a more explicit formula.