Intuition behind the choice of the trace-function in character theory

Question

I have recently been studying representation theory of finite groups and hence also character theory, so I'm quite new in the field and maybe my question is a bit silly: On the first glance, the choice of the trace-function to define the character of a representation seems pretty arbitrary, so I was wondering whether there is a deeper reason for choosing the trace to define the character. Why did the inventors of character theory pick the trace function, and why did it work out so well?

I have heard of other representation theories, for example of Lie Algebras or Lie Groups (but I don't understand those yet unfortunately), so a (perhaps) related question would be whether we have a character theory for those representation theories as well, and if so what the common properties of those characters and the character of the representation theory of finite groups are...

Answer 1

Let's recall that, concretely, the data of an $n$-dimensional representation of a finite group $G$ is a collection $\rho(g)$ of $n \times n$ matrices, one for each element $g \in G$. Classifying such things up to isomorphism amounts to classifying collections of such matrices (satisfying the group relations) up to simultaneous conjugacy.

So, what invariants of square matrices up to conjugacy are there? We can consider eigenvalues, or equivalently we can consider the coefficients of the characteristic polynomial, which include the trace as well as the determinant. These are in fact the only conjugacy invariants of a square matrix which are well-behaved in various reasonable senses: for example, every polynomial conjugacy invariant of a square matrix is a polynomial in the coefficients of the characteristic polynomial. If we want to talk about simultaneous conjugacy things are more complicated; we can consider functions of the eigenvalues of various products, e.g. $\text{tr}(AB)$ is also an invariant of $A$ and $B$ up to simultaneous conjugacy, but since for a group representation a product of two of the matrices involved is another of the matrices involved this doesn't really buy us anything.

So it's at least reasonable to consider some function of the coefficients of the characteristic polynomial. Among these, the trace is distinguished by behaving particularly nicely with respect to both direct sums and tensor products: we have both

$$\chi_{V \oplus W}(g) = \chi_V(g) + \chi_W(g)$$

and

$$\chi_{V \otimes W}(g) = \chi_V(g) \chi_W(g).$$

So that's very nice: abstractly what this implies is that taking the character of a representation defines a ring homomorphism from the representation ring $R(G)$ of $G$ to the ring of class functions $G \to \mathbb{C}$, and standard results in representation theory of finite groups implies that this ring homomorphism is in fact an isomorphism, which is great.

From this perspective it may be a little surprising that the trace captures the isomorphism class of a representation since we only consider the sum of the eigenvalues and not e.g. their product, but as an exercise you can show that for a matrix $M$ over a field of characteristic zero, knowing the traces $\text{tr}(M^k)$ of all powers of $M$ is equivalent to knowing all coefficients of the characteristic polynomial of $M$; this is an exercise in symmetric function theory. So the character of a representation tells you, at the very least, the eigenvalues of every matrix $\rho(g)$.

The character theory of Lie groups also involves taking traces, while the character theory of Lie algebras is a bit more complicated but basically involves taking traces as well.

Answer 2

Schur's lemma tells that $\dim \text{Hom}(\rho,\pi)$ is 1 if $\rho,\pi$ are isomorphic irreducible representations, and 0 if $\rho,\pi$ are non-isomorphic irreducible representations. Thus $\dim \text{Hom}(\bullet,\bullet)$ behaves somewhat as an inner product on representations, where irreducible representations form an orthogonal basis. Writing out $\dim \text{Hom}(\rho,\pi)$ differently as in here: Is it true that $\langle \psi, \omega\rangle$ is the dimension of $Hom(V,W)$? makes characters pop out completely naturally. From this "orthogonality relation" (and the implicitly-used decomposition results, like representations over a characteristic 0 algebraically closed field decompose uniquely into a direct sum of irreducibles) one can get $\rho \simeq \pi \iff \chi_\rho = \chi_\pi$.

So at least in the "finite group representation theory over a nice field" story, traces show up and are so nice because traces are very related to dimensions of projections, and (from the decomposition theorem) dimensions of projections are very related to the uniquely determined numbers of copies of a given irreducible inside a given representation.