# The vanishing of sum of coefficients: symmetric polynomials

Denote $$\pmb{X}_n=(x_1,x_2,\dots,x_n)$$. Consider the symmetric polynomial $$f_n(\pmb X_n)=\prod_{1\leq i Expand these in terms of elementary symmetric polynomials, say $$f_n(\pmb{X}_n)=\sum_{\mu}c_{\mu,n}\cdot e_{\mu}(\pmb{X}_n).$$

For example, \begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \\ f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}. \end{align*}

QUESTION 1. Is it true that, for integers $$n \geq 1$$, we have $$\sum_{\mu}c_{\mu,2n+1}=0?$$

POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:

QUESTION 2. Is it true that, for integers $$n \geq 1$$, we have $$\sum_{\mu}c_{\mu,2n}=(-1)^{\binom{n}2}?$$

• The proofs given so far are great. Now I'd like to point out that $f_n(\boldsymbol{X}_n)$ is the generating function for simple graphs by degree sequence. For example, with $n=4$, the coefficient of $x_1x_2^3x_3^2x_4^2$ is the number of graphs with degree sequence $1,3,2,2$. It would be fun to prove the two claims by an argument based on graph manipulation, but I don't know how to do it. – Brendan McKay Mar 4 '19 at 1:11
• That is a very interesting idea. – T. Amdeberhan Mar 4 '19 at 1:25
• @BrendanMcKay I see how it is a generating function of out-degrees of tournaments. Is it what you mean? – Fedor Petrov Mar 4 '19 at 22:22
• @FedorPetrov, ooops you are correct. My error is all the more embarrassing given that I'm working on a paper about oriented graphs (with coauthors) that uses a generalisation of this. Thanks. – Brendan McKay Mar 4 '19 at 23:54

Choose $$n$$ numbers $$x_1,\dots,x_n$$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $$f_n$$. We should get zero value for odd $$n$$. Well, what are these numbers? The roots of $$x^{n}-x^{n-1}+x^{n-2}-\ldots-1=(x^{n+1}-1)/(x+1)$$. This polynomial indeed has two roots with sum equal to 0 when $$n$$ is odd.

If $$n=2k$$ is even, we substitute the roots $$w_1,\dots,w_n$$ of the polynomial $$f(x)=x^{2k}-x^{2k-1}+\ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)\dots (x-w_n)$$. Then your claim reads as $$A:=\prod_{1\leqslant i This is done by the standard trick (and is well known itself). At first, $$|A|^2=\prod_{i=1}^n \prod_{j\ne i,1\leqslant j\leqslant n}|w_i+w_j|=2^{-n}\prod_{i=1}^n \prod_{j=1}^n|(-w_i)-w_j|=2^{-n}\prod_{i=1}^n |f(w_i)|=\\=2^{-n}\prod_{i=1}^n\left|\frac{(-w_i)^{2k+1}+1}{-w_i+1}\right|=1,$$ since $$1+(-w_i)^{2k+1}=2$$ for all $$i=1,2,\dots,n$$ and $$\prod_{i=1}^n (1-w_i)=f(1)=1$$.

At second, we need to find the argument of the complex number $$A$$. This may be done for example as follows: all pairs $$w_i+w_j$$ for which $$w_i$$ and $$w_j$$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $$w_i$$ and $$w_j$$ are complex conjugate, the sum $$w_i+w_j$$ is a real number whose sign is the sign of the real part of $$w_i$$. Therefore $$A$$ is the real number whose sign equals $$(1)^{m/2}$$, where $$m$$ is the number of $$w$$'s in the left half-plane. It is easy to see that $$m/2=[k/2]$$ and that $$(-1)^{[k/2]}=(-1)^{k(k-1)/2}$$.

• An alternative to your "standard trick" is to observe that $w_i + w_j = \dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $\prod\limits_{i<j} \left(w_i+w_j\right) = \dfrac{\prod\limits_{i<j}\left(w_i^2 - w_j^2\right)}{\prod\limits_{i<j}\left(w_i-w_j\right)}$. But the $n$ numbers $-w_1^2, -w_2^2, \ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, \ldots, w_n$, and thus $\dfrac{\prod\limits_{i<j}\left(w_i^2 - w_j^2\right)}{\prod\limits_{i<j}\left(w_i-w_j\right)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute. – darij grinberg Mar 2 '19 at 21:26
• Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument. – Fedor Petrov Mar 2 '19 at 21:38

Here is another method using more of the theory of symmetric functions. By Enumerative Combinatorics, vol. 2, Exercise 7.30, we have $$f_n(\boldsymbol{X}_n)= s_{(n-1,n-2,\dots,1)}(\boldsymbol{X}_n)$$ (a Schur function). By the dual Jacobi-Trudi identity, $$s_{(n-1,n-2,\dots,1)} = \det[e_{n-2i+j}]_{i,j=1}^{n-1},\ \ (*)$$ where $$e_0=1$$ and $$e_k=0$$ for $$k<0$$. Since $$e_k(\boldsymbol{X}_n)=0$$ for $$k>n$$, it follows that $$\sum_\mu c_{\mu,n}$$ is obtained by substituting $$e_1=e_2=\cdots=e_n=1$$ and $$e_{n+1}=e_{n+2}=\cdots=0$$ into the right-hand side of (*). When $$n$$ is odd, the two middle rows of the determinant are equal (in fact, they are all 1's), so the determinant is 0. If $$n=2m$$ then subtract row $$m-1$$ from row $$m$$, then row $$m-2$$ from row $$m-1$$, up to row 1 from row 2. Also subtract row $$m+2$$ from row $$m+1$$, row $$m+3$$ from $$m+2$$, etc. The resulting matrix $$A$$ can be transformed into a triangular matrix $$B$$ with 1's on the diagonal by row and column permutations. The permutation indexing the 1's in $$A$$ that become the diagonal elements of $$B$$ is $$1,3,5,\dots,n-1,2,4,6,\dots,n-2$$, which has $${m\choose 2}$$ inversions, and the proof follows.

• This is a really cool alternative proof. Thanks, Richard! – T. Amdeberhan Mar 3 '19 at 22:51
• It looks interesting that again we get that the answer is the sign of the permutation $x\to 2x+1$ of residues modulo $n$, before establishing the explicit value $(-1)^{m\choose 2}$. – Fedor Petrov Mar 4 '19 at 8:56

The polynomial in question is an instance of the Boolean product polynomials, which might give some extra insight. For example, I believe Lascoux have studied the Schur expansion of that exact expression.

• Thanks for the pointers. – T. Amdeberhan Mar 4 '19 at 22:50