# The “ring of characters” of a finite group and its automorphisms

Let $G$ be a finite group and let $C(G)$ denote the set of characters of $G$ (in my representation theory course the values these characters take are in $\mathbb{C}$, but this is one point I'd like to ask about later). Then $C(G)$ has quite a lot of algebraic structure; if $\rho$ and $\sigma$ are two representations of $G$ with corresponding characters $\chi_\rho$ and $\chi_\sigma$ then $\chi_\rho +\chi_\sigma$ is again a character of $G$ corresponding to the representation $\rho \oplus \sigma$. Further, $\chi_\rho \cdot \chi_\sigma$ is a character of $G$ corresponding to the representation $\operatorname{Hom}(V^*, W)$, where $V$ and $W$ are the carrier spaces of $\rho$ and $\sigma$ respectively.

So we are allowed to add and multiply characters together within $C(G)$, giving two operations on $C(G)$ which I'd assume interact nicely and satisfy the ring axioms except for the fact that the additive structure is not complete - additive inverses don't exist! So I suppose the natural thing to do is define $\mathbb{Z}C(G)$ as the free abelian group on $C(G)$; then $\mathbb{Z} C(G)$ is a ring.

Now I'm aware of one automorphism of this ring; namely, complex conjugation of characters: if $\chi$ is a character of some representation $\rho$ with carrier space $V$ then $\bar{\chi}$ is the character of the representation $\rho^*: G\rightarrow GL(V^*)$ defined by $\rho^* (\phi) (v) = \phi(\rho(g^{-1})(v))$. So my questions are:

1. What is sort of object is $\mathbb{Z} C(G)$? Are there special cases where it is a field?
2. I've written down one automorphism of $\mathbb{Z} C(G)$, namely complex conjugation of characters. Are there other automorphisms coming from different modifications to representations, perhaps in the case that we consider our characters to take values in some general field $K$ rather than just $\mathbb{C}$? Is there a "Galois theory" of these automorphisms?
• You should not take the free abelian group of $C(G)$, as that will not work. You either consider the subring of the ring of all functions $G\to\mathbb G$ generated by $C(G)$, or you do something like the Grothendieck construction, which in this case gives the same thing, – Mariano Suárez-Álvarez Feb 16 '16 at 16:16

$\mathbb{Z}C(G)$ is called the representation ring of $G$, it is an example of a Grothendieck ring of a tensor category. In addition to the ring operations you mentioned, it comes equipped with additional operations coming from taking exterior or symmetric powers of representations, making it into a so called $\lambda$-ring.
As for your question about if it can be a field, the answer is always no. A quick way to see this is to consider the character $\chi_{reg}$ corresponding to the regular representation of $G$, it's an easy lemma to show this satisfies $\chi_{reg} \cdot \chi = Dim(\chi)\chi_{reg}$ for any character $\chi$. In particular, $\chi_{reg}$ is never invertible unless $G$ is trivial but then $\mathbb{Z}C(G)= \mathbb{Z}$ and it's still not a field.
Indeed any field isomorphism of $\mathbb{C}$ will give an isomorphism of $\mathbb{Z}C(G)$, complex conjugation is just one nice example. Really, all of the representation theory of $G$ takes place over a much smaller field than $\mathbb{C}$. What you could do is take the smallest field $k_G$ of characteristic zero over which all irreducible complex representations are defined and look at the action of $Gal(k_G,\mathbb{Q})$ on $\mathbb{Z}C(G)$.
• Thank you for a lovely answer! The connection with Grothendieck rings of monoidal categories is very interesting, and I'm pleased there is a general framework here. So does the same thing go through when we look at representations on vector spaces over an arbitrary field $K$? i.e. does an automorphism of $K$ necessarily induce an automorphism of the $K$-representation ring? – Alex Saad Feb 16 '16 at 16:36
• You need to be a bit careful when the characteristic of $K$ divides the size of $G$, in that case representations don't always split as a direct sum of irreducibles and the usual notion of characters doesn't behave quite as well as it does in characteristic zero. However it is still true that you can twist representations of $G$ by an automorphism of $K$, and this will induce an automorphism of the Grothendieck ring once defined more carefully. – Nate Feb 16 '16 at 16:43