Motivation for characters in Harris and Fulton

Question

At page 13, they said

Why is knowing $\sum \lambda_i^k$(is this sum over i? ) gives each eigenvalue? And how does this fact motivate the definition of character?

Answer 1

Concretely, the remark means the following: If you know $\lambda_1^k+\cdots+\lambda_n^k$ for all $1\le k\le n$, then you know $\lambda_1,\dots,\lambda_n$.

Indeed, by Newton's identities, knowing $\lambda_1^k+\cdots+\lambda_n^k$ allows you to solve for the elementary symmetric polynomials $e_k(\lambda_1,\dots,\lambda_n)$ for each $1\le k\le n$. But then $\lambda_1,\dots,\lambda_n$ are simply the roots of the polynomial $$X^n-e_1X^{n-1}+\cdots+(-1)^ne_n=0.$$

Answer 2

Consider a finite group $G$, an algebraically closed field $\mathbb{F}$, and a character $\chi : G \to \mathbb{F}$ defined as $\chi(g) \triangleq \text{Tr}(\rho(g))$ for some representation $\rho : G \to \text{GL}_d(G)$. Fix an element $g \in G$ and let $\lambda_1, \ldots , \lambda_d \in \mathbb{F}$ be the eigenvalues of $\rho(g)$. Then $\lambda_1^i, \ldots , \lambda_d^i$ are the eigenvalues of $\rho(g^i)$, which means

\begin{equation*} \chi(g^i) = \lambda_1^i + \ldots + \lambda_d^i \end{equation*}

For all $i \in \mathbb{N}_0$. Thus if you consider the generating function $f_g(x) \triangleq \sum_{i=0}^{\infty}{\chi(g^i)x^i}$, you get that

\begin{align*} f_g(x) = \sum_{i=0}^{\infty}{\chi(g^i)x^i} = \sum_{i=0}^{\infty}{\sum_{j=1}^d{\lambda_j^i x^i}} = \sum_{j=1}^d{\sum_{i=0}^{\infty}{(\lambda_j x)^i}} = \frac{1}{1 - \lambda_1 x} + \ldots + \frac{1}{1 - \lambda_d x} = \frac{p_g'(x)}{p_g(x)} \end{align*} where $p_g(x) \triangleq \prod_{j=1}^d{(1-\lambda_j x)}$. By some simple calculus, this therefore uniquely determines $p_g$ (note that $p_g(0) = 1$) given $\chi(\cdot)$ and thus uniquely determines the multiset $\{\lambda_1, \ldots , \lambda_d\}$ given $\chi(\cdot)$.