Evaluating elementary symmetric polynomial at $n^{th}$ powers

Question

Let $n\geq 2$ and $e_{1}, e_{2}, \ldots, e_{n}$ are elementary symmetric polynomials in $n$-variables. If I know the values of $e_{1}(a_{1}, a_{2}, \ldots, a_{n}), e_{2}(a_{1}, a_{2}, \ldots, a_{n}), \ldots, e_{n}(a_{1}, a_{2}, \ldots, a_{n})$ at some numbers $a_{1}, a_{2}, \ldots, a_{n}$, whose values I do not know. How can one go about calculating following quantities?

$$e_{1}(a_{1}^{n}, a_{2}^{n}, \ldots, a_{n}^{n}), e_{2}(a_{1}^{n}, a_{2}^{n}, \ldots, a_{n}^{n}), \ldots, e_{n}(a_{1}^{n}, a_{2}^{n}, \ldots, a_{n}^{n}).$$

My attempt: My approach was to try some examples to see whether there is any pattern. I tried some examples when $n = 2, 3$. For $n = 2$ it was fairly easy. For $n = 3$, I got that,

$$e_{1}(a_{1}^{3}, a_{2}^{3}, a_{3}^{3}) = \sum_{i=1}^{3} a_{i}^{3} = e_{1}^{3}(a_{1}, a_{2}, \ldots, a_{n}) - 3 e_{1}(a_{1}, a_{2}, \ldots, a_{n}) \cdot e_{2}(a_{1}, a_{2}, \ldots, a_{n}) + 3 e_{3}(a_{1}, a_{2}, \ldots, a_{n}).$$

However, I couldn't find a way to see if I could do the same for $$e_{2}(a_{1}^{3}, a_{2}^{3}, a_{3}^{3}) = a_{1}^{3}a_{2}^{3} + a_{2}^{3}a_{3}^{3} + a_{1}^{3}a_{2}^{3}.$$

How can one go about solving the problem for general $n$? I would appreciate any help or a nudge towards some reference/book that helps teach this.

Answer 1

Let $m$ be any integer $> 1$ and $\omega = e^{\frac{2\pi}{m}i}$ be a primitive $m^{th}$ root of unity.

For any expression $(\cdots)$, we will use $[t^\ell](\cdots)$ to denote the coefficient of $t^\ell$ appear in $(\cdots)$.

Let $\tilde{e}_k$ be the elementary symmetric polynomials for $a_k^m$. Define two polynomials $$\begin{align} f(t) &= \prod_{k=1}^n(1 + a_k t) = 1 + \sum_{k=1}^n e_k t^k\\ \tilde{f}(t) &= \prod_{k=1}^n(1 + a^m_k t) = 1 + \sum_{k=1}^n \tilde{e}_k t^k\ \end{align}$$ We have $e_k = [t^k] f(t)$, $\tilde{e}_k = [t^k] \tilde{f}(t)$ and $$ \tilde{f}(-(-t)^m) = \prod_{k=1}^n (1 - a_k^m(-t)^m) = \prod_{k=1}^n \prod_{\ell=0}^{m-1}(1 + a_k t \omega^\ell) = \prod_{\ell=0}^{m-1} f(t\omega^{\ell})\tag{*1} $$ This leads to $$\tilde{e}_k = (-1)^{(m-1)k} [t^{mk}] \prod_{\ell=0}^{m-1} f(t\omega^{\ell})$$

For any $p = (p_0, p_1, \ldots, p_{m-1}) \in \mathbb{C}^m$, let $M(p)$ be the $m \times m$ circulant matrix with entries $M(p)_{ij} = p_{{\rm mod}(j-i,m)}$. It is known that

$$\det M(p) = \prod_{\ell=0}^{p-1} g(\omega^\ell) \quad\text{ where }\quad g(t) = \sum\limits_{k=0}^{m-1} p_k t^k$$

Compare this with $(*1)$, we find $$\tilde{f}(-(-t)^m) = \det M( 1 + c_0, c_1, \ldots, c_{m-1}) \quad\text{ where }\quad c_\ell = \sum_{\substack{k=1\\k \equiv \ell \pmod m}}^n e_k t^k$$

In the special case $m = n$, this leads to

$$\tilde{e}_k = (-1)^{(n-1)k} [t^{nk}] \left|\begin{matrix} 1 + e_n t^n & e_1 t & \cdots & e_{n-2}t^{n-2} & e_{n-1}t^{n-1}\\ e_{n-1}t^{n-1} & 1 + e_nt^n & e_1 t& & e_{n-2}t^{n-2}\\ \vdots & e_{n-1} t^{n-1} & 1 + e_nt^n & \ddots & \vdots\\ e_2 t^2 & & \ddots & \ddots & e_1 t\\ e_1 t & e_2 t^2 & \ldots & e_{n-1}t^{n-1} & 1 + e_n t^n \end{matrix}\right|$$

For $n > 3$, the determinant on RHS becomes too complicated to be evaluated by hand. I have implemented following functions in maxima to compute the $\tilde{e}_k$.

Cell(n,l)   := if(l = 0) then 1 + e[n]*t^n else e[l]*t^l;
Circulant(n) := genmatrix(lambda([i,j],Cell(n,mod(j-i,n))),n);
TildeE(n)    := makelist((-1)^((n-1)*k)*coeff(expand(determinant(Circulant(n))),t,n*k),k,1,n);

For small $n$, I obtain:

$$\begin{array}{rrl} n = 2 & \tilde{e}_1 &= {e}_{1}^{2}-2\,{e}_{2},\\ & \tilde{e}_2 &= {e}_{2}^{2}\\ \\ n = 3 & \tilde{e}_1 &= {e}_{1}^{3}-3\,{e}_{1}\,{e}_{2}+3\,{e}_{3},\\ & \tilde{e}_2 &= {e}_{2}^{3}-3\,{e}_{1}\,{e}_{2}\,{e}_{3}+3\,{e}_{3}^{2},\\ & \tilde{e}_3 &= {e}_{3}^{3}\\ \\ n = 4 & \tilde{e}_1 &= {e}_{1}^{4}-4\,{e}_{1}^{2}\,{e}_{2}+2\,{e}_{2}^{2}+4\,{e}_{1}\,{e}_{3}-4\,{e}_{4},\\ & \tilde{e}_2 &= {e}_{2}^{4}-4\,{e}_{1}\,{e}_{2}^{2}\,{e}_{3}+2\,{e}_{1}^{2}\,{e}_{3}^{2}+4\,{e}_{2}\,{e}_{3}^{2}+4\,{e}_{1}^{2}\,{e}_{2}\,{e}_{4}-4\,{e}_{2}^{2}\,{e}_{4}-8\,{e}_{1}\,{e}_{3}\,{e}_{4}+6\,{e}_{4}^{2},\\ & \tilde{e}_3 &= {e}_{3}^{4}-4\,{e}_{2}\,{e}_{3}^{2}\,{e}_{4}+2\,{e}_{2}^{2}\,{e}_{4}^{2}+4\,{e}_{1}\,{e}_{3}\,{e}_{4}^{2}-4\,{e}_{4}^{3},\\ & \tilde{e}_4 &= {e}_{4}^{4} \\ \\ n = 5 & \tilde{e}_1 &= {e}_{1}^{5}-5\,{e}_{1}^{3}\,{e}_{2}+5\,{e}_{1}\,{e}_{2}^{2}+5\,{e}_{1}^{2}\,{e}_{3}-5\,{e}_{2}\,{e}_{3}-5\,{e}_{1}\,{e}_{4}+5\,{e}_{5},\\ & \tilde{e}_2 &= {e}_{2}^{5}-5\,{e}_{1}\,{e}_{2}^{3}\,{e}_{3}+5\,{e}_{1}^{2}\,{e}_{2}\,{e}_{3}^{2}+5\,{e}_{2}^{2}\,{e}_{3}^{2}-5\,{e}_{1}\,{e}_{3}^{3}+5\,{e}_{1}^{2}\,{e}_{2}^{2}\,{e}_{4}-5\,{e}_{2}^{3}\,{e}_{4}\\ &&\quad -5\,{e}_{1}^{3}\,{e}_{3}\,{e}_{4}-5\,{e}_{1}\,{e}_{2}\,{e}_{3}\,{e}_{4}+5\,{e}_{3}^{2}\,{e}_{4}+5\,{e}_{1}^{2}\,{e}_{4}^{2}+5\,{e}_{2}\,{e}_{4}^{2}-5\,{e}_{1}^{3}\,{e}_{2}\,{e}_{5}\\ &&\quad +10\,{e}_{1}\,{e}_{2}^{2}\,{e}_{5}+10\,{e}_{1}^{2}\,{e}_{3}\,{e}_{5}-15\,{e}_{2}\,{e}_{3}\,{e}_{5}-15\,{e}_{1}\,{e}_{4}\,{e}_{5}+10\,{e}_{5}^{2},\\ & \tilde{e}_3 &= {e}_{3}^{5}-5\,{e}_{2}\,{e}_{3}^{3}\,{e}_{4}+5\,{e}_{2}^{2}\,{e}_{3}\,{e}_{4}^{2}+5\,{e}_{1}\,{e}_{3}^{2}\,{e}_{4}^{2}-5\,{e}_{1}\,{e}_{2}\,{e}_{4}^{3}-5\,{e}_{3}\,{e}_{4}^{3}+5\,{e}_{2}^{2}\,{e}_{3}^{2}\,{e}_{5}\\ &&\quad -5\,{e}_{1}\,{e}_{3}^{3}\,{e}_{5}-5\,{e}_{2}^{3}\,{e}_{4}\,{e}_{5}-5\,{e}_{1}\,{e}_{2}\,{e}_{3}\,{e}_{4}\,{e}_{5}+10\,{e}_{3}^{2}\,{e}_{4}\,{e}_{5}+5\,{e}_{1}^{2}\,{e}_{4}^{2}\,{e}_{5}+10\,{e}_{2}\,{e}_{4}^{2}\,{e}_{5}\\ &&\quad +5\,{e}_{1}\,{e}_{2}^{2}\,{e}_{5}^{2}+5\,{e}_{1}^{2}\,{e}_{3}\,{e}_{5}^{2}-15\,{e}_{2}\,{e}_{3}\,{e}_{5}^{2}-15\,{e}_{1}\,{e}_{4}\,{e}_{5}^{2}+10\,{e}_{5}^{3}\\ & \tilde{e}_4 &= {e}_{4}^{5}-5\,{e}_{3}\,{e}_{4}^{3}\,{e}_{5}+5\,{e}_{3}^{2}\,{e}_{4}\,{e}_{5}^{2}+5\,{e}_{2}\,{e}_{4}^{2}\,{e}_{5}^{2}-5\,{e}_{2}\,{e}_{3}\,{e}_{5}^{3}-5\,{e}_{1}\,{e}_{4}\,{e}_{5}^{3}+5\,{e}_{5}^{4},\\ & \tilde{e}_5 &= {e}_{5}^{5} \end{array}$$