Derivation and application of Newton's identity

Question

How is the following identity derived? $$\sum_{\ell =0}^{n-1}(-1)^\ell e_\ell s_{n-\ell}+(-1)^nne_n=0$$ Is there an example demonstrating the context in which this might be applied?

Answer 1

Vieta's formulas state

$$p(T)=\prod_{k=1}^n(T-x_k)=\sum_{k=0}^n(-1)^{n-k}e_{n-k}(x_1,\cdots,x_n)T^k.$$

Clearly $p(x_i)=0$ for each $i=1,\cdots,n$. Evaluating and summing,

$$0=\sum_{i=1}^n p(x_i)=\sum_{i=1}^n\sum_{k=0}^n(-1)^{n-k}e_{n-k}x_i^k=\sum_{k=0}^n(-1)^{n-k}e_{n-k}\left(\sum_{i=1}^n x_i^k\right)=\sum_{k=0}^n(-1)^{n-k}e_{n-k}p_k.$$

As $n$ varies, these are known as Newton(-Girard)'s identities.

You've used to tag (representation-theory), which suggests you'd appreciate an application from that particular context. Rewriting the identities allows us $e_n$ in terms of $e_i$ for $i<n$ and power-sums:

$$e_n=\frac{1}{n}\sum_{k=0}^{n-1}(-1)^ke_{n-k}p_k=\cdots=\frac{1}{n!}\sum_{\sigma\in S_n}{\rm sgn}(\sigma)p_1^{c_1(\sigma)}\cdots p_n^{c_n(\sigma)}.$$

Note $p_0=n$, and $c_i(\sigma)$ stands for the number of cycles of length $i$ in $\sigma$'s disjoint cycle representation.

By iterating the first equality for each $e_i$, we end up with the last one. The general linear group ${\rm GL}(V)$ acts on the alternating power $\Lambda^n(V)$ by $(\Lambda^nA)(v_1\wedge\cdots\wedge v_n)=Av_1\wedge\cdots\wedge Av_n$, and we can form an eigenbasis for $\Lambda^n(V)$ out of an eigenbasis for $V$ to show the trace of $\Lambda^nA$ is the elementary symmetric polynomial in the eigenvalues of $A$ acting on $V$. Plugging into $(*)$, we get the trace formula

$${\rm tr}\Lambda^nA=\frac{1}{n!}\sum_{\sigma\in S_n}{\rm sgn}(\sigma)({\rm tr}A^1)^{c_1(\sigma)}\cdots({\rm tr}A^n)^{c_n(\sigma)}.$$

This generalizes to trace formulae for Schur functors.