Expressions for symmetric power sums in terms of lower symmetric power sums

Question

The Newton symmetric power sums $p_k(x_1, \ldots, x_n)$, for $k \geq 1$, are given by $$ p_k(x_1, \ldots, x_n) = \sum_{i=1}^n x_i^k. $$ Do you know if it's possible to express $p_k$ in terms of (non-linear combinations) of the $p_j$'s with $j < k$ (for characteristic $0$ fields)? Is there some sort of recurrence relation involving the $p_j$'s that avoids the elementary symmetric polynomials (a.k.a. Newton's identity)?

Note in here Determinant expression for the power sum that there is such an expression if we take $x_j = j$ for each $j$. I'm wondering if something of the sort holds in general?

Answer 1

A well-known identity in the ring of symmetric functions (see, e.g., Exercise 2.9.13(a) in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, arXiv:1409.8356v5) says the following:

Let $n \in \mathbb{N}$. (Sorry, my $n$ is not your $n$. My symmetric functions are in infinitely many indeterminates.) Define a matrix $A_n = \left(a_{i,j}\right)_{i,j = 1, 2, \ldots, n}$ by $$ a_{i,j} = \begin{cases} p_{i-j+1}, & \text{if } i \geq j ; \\ i, & \text{if } i = j - 1 ; \\ 0, & \text{if } i < j - 1 \end{cases} \qquad \qquad \text{ for all } \left( i, j \right) \in \left\{ 1, 2, \ldots , n \right\}^2 . $$ This matrix $A_n$ looks as follows: $$ A_n = \left( \begin{array}[c]{cccccc} p_1 & 1 & 0 & \cdots & 0 & 0\\ p_2 & p_1 & 2 & \cdots & 0 & 0\\ p_3 & p_2 & p_1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ p_{n-1} & p_{n-2} & p_{n-3} & \cdots & p_1 & n-1\\ p_n & p_{n-1} & p_{n-2} & \cdots & p_2 & p_1 \end{array} \right) . $$ Then, $\det\left(A_n\right) = n! e_n$.

Now, let us restrict these symmetric functions to $N$ variables $x_1, x_2, \ldots, x_N$ for some $N \in \mathbb{N}$. (This $N$ is your $n$.) Then, for every $n > N$, we have $e_n = 0$, and thus $\det\left(A_n\right) = n! e_n = 0$. This is an equality involving $p_1, p_2, \ldots, p_n$, and $p_n$ only appears in it linearly with coefficient $\pm\left(n-1\right)!$, so you can solve it for $p_n$ (since your field has characteristic $0$). This expresses $p_n$ (for $n > N$) in terms of the $p_k$ with $k < n$. I don't know if this expression is as explicit as you want it to be, though...

You cannot do this for $n \leq N$ (with constant coefficients), of course, since $p_1, p_2, \ldots, p_N$ are algebraically independent.

Answer 2

Yes. The power sum symmetric functions $p_k$ satisfy the linear recurrence

$$p_k - e_1 p_{k-1} + e_2 p_{k-2} \mp \dots + (-1)^n e_n p_{k-n} = 0$$

where the $e_i$ are the elementary symmetric functions. No hypotheses on the characteristic are necessary. This comes from the identity

$$\prod_{i=1}^n (t - x_i) = t^n - e_1 t^{n-1} + e_2 t^{n-2} \mp \cdots $$

after multiplying by $t^{k-n}$, then substituting $t = x_1, x_2, \dots , x_n$ and adding up the results.