## Power Sum Polynomials

The power sum polynomial is defined as follows: $p_k := x_1^k + x_2^k + \ldots + x_n^k \in \Lambda_n^{(k)}.$

In this case, we do not define $p_0$, although it seems natural to set $p_0 = n.$ As before, for a partition $\lambda,$ define: $\displaystyle p_\lambda:= p_{\lambda_1} p_{\lambda_2} \ldots p_{\lambda_l}, \qquad l = l(\lambda).$

Note that we must have $l = l(\lambda)$ above since we have not defined $p_0.$ For example if $\lambda = (2, 1)$, then $p_\lambda = \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n x_i\right).$

Their generating function is given by: $\displaystyle P(t) := p_1 + p_2 t + p_3 t^2 + \ldots= \sum_{j=1}^n \frac {x_j}{1 - x_j t}.$ ## Newton’s Identities (I)

Recall that the generating function for $e_i$ is $E(t) = \prod_{i=1}^n (1 + x_i t)$. Thus we have the following relation: $\displaystyle \log E(t) = \sum_{j=1}^n \log (1 + x_j t)\stackrel{\frac d {dt}}{\implies} \frac {E'(t)}{E(t)} = \sum_{j=1}^n \frac {x_j}{1 + x_j t} = P(-t).$

From $E'(t) = E(t) P(-t)$ we take the coefficient of $t^{k-1}$ on both sides and obtain Newton’s identities: $k e_k = e_{k-1} p_1 - e_{k-2} p_2 + \ldots + (-1)^{k-1} e_0 p_k.$

for all $k\ge 1$, where $e_k = 0$ for all k>n. With that, each $p_k$ can be written as a polynomial in $e_1, e_2, \ldots, e_n$ with integer coefficients. E.g. $p_1 = e_1,\quad p_2 = e_1^2 - 2e_2,\quad p_3 = e_1^3 - 3e_1 e_2 + 3e_3, \ldots$

Conversely, we can also express each $e_k$ as a polynomial in $p_1, p_2, \ldots, p_n$ with rational coefficients. Thus we have: $\displaystyle \Lambda_n \otimes_{\mathbb{Z}} \mathbb{Q} =\mathbb{Q}[p_1, p_2, \ldots, p_n],$

where the RHS is a free $\mathbb{Q}$-algebra in $p_1, \ldots, p_n.$ In particular, $\Lambda_n^{(d)}\otimes_{\mathbb{Z}} \mathbb{Q}$ has basis given by $p_\lambda$ for all $\lambda \vdash d, \lambda_1 \le n.$

## Newton’s Identities (II)

There is another form of Newton’s identities which relate $p_k$ with the complete symmetric polynomials $h_i.$ Recall that the generating function for $h_i$ is $H(t) = \prod_{i=1}^n \frac 1 {(1 - x_i t)}.$ Hence: $\displaystyle\log H(t) = -\sum_{j=1}^n \log(1 - x_j t) \implies \frac{H'(t)}{H(t)}= \sum_{j=1}^n \frac {x_j}{1 - x_j t} = P(t).$

From $H'(t) = P(t)H(t)$, taking the coefficient of $t^{k-1}$ gives us another form of Newton’s identities: $k h_k = h_{k-1} p_1 + h_{k-2} p_2 + \ldots + h_0 p_k.$

for all $k\ge 1$. From this, we can express $p_k$ as a polynomial in $h_1, \ldots, h_n$ with integer coefficients. E.g. $p_1 = h_1, \quad p_2 = 2h_2 - h_1^2,\quad p_3 = 3h_3 - 3h_2 h_1 + h_1^3.$

As above, we can also express each $h_k$ as a polynomial in $p_1, \ldots, p _n$ with rational coefficients.

## Exercise

Using one of Newton’s identities, express $p_4 = x^4 + y^4 + z^4$ as a polynomial in: \begin{aligned} p_1 &= x+y+z,\\ p_2 &= x^2 + y^2 + z^2, \\ p_3 &= x^3 + y^3 + z^3.\end{aligned} ## Self-Duality

Recall that the involution $\omega : \Lambda_n \to\Lambda_n$ swaps $e_i \leftrightarrow h_i$ for $1\le i \le n.$ Looking at the expressions of $p_n$ in terms of $e_i$ and $h_i,$ the following is clear.

Proposition. For each $1\le k \le n$, we have $\omega(p_k) = (-1)^{k-1}p_k.$

Proof

Apply induction on k; when k=1 we have $p_1 = e_1 = h_1$ so this is clear. Suppose $1 Apply $\omega$ to the first Newton’s identity; for each $i\le n$ we have $\omega(e_i) = h_i$ and so: $k h_k = h_{k-1} \omega(p_1) - h_{k-2} \omega(p_2) + \ldots + (-1)^{k-1} h_0 \omega(p_k).$

By induction hypothesis $\omega(p_i) = (-1)^{i-1}p_i$ for $1 \le i \le k-1,$ so we get: $k h_k = h_{k-1} p_1 + h_{k-2} p_2 + \ldots + h_1 p_{k-1} + (-1)^{k-1} h_0 \omega(p_k).$

Now the second Newton’s identity gives $\omega(p_k) = (-1)^{k-1} p_k$ as desired. ♦ The above is no longer true for k>n. For example, let n=2. We get: \begin{aligned} p_1 = e_1\ &\implies\ \omega(p_1) = h_1 = p_1, \\ p_2 = e_1^2 - 2e_2\ &\implies\ \omega(p_2) = h_1^2 - 2h_2 = -p_2,\\ p_3 = e_1^3 - 3e_1 e_2 \ &\implies\ \omega(p_3) = h_1^3 - 3h_1 h_2 \ne \pm p_3.\end{aligned}

## Summary

We have seen the following symmetric polynomials: In the above diagram a → b means that b can be expressed as a polynomial in a with integer coefficients; the same holds for dotted arrows but now the polynomial may have rational coefficients.

One wonders if there is a combinatorial interpretation of the coefficients of $p_\lambda$ when expressed as a linear sum of $m_\mu.$ It turns out this has implications in the representation theory of the symmetric group $S_n.$ We will (hopefully) get to talk about this beautiful result in a later article.

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