In Fulton, Harris, "Representation Theory. A first Course" there's the following paragraph which I don't really understand:
The representation ring $R(G)$ of a group $G$ is easy to define. First, as a group we just take $R(G)$ to be the free abelian group generated by all (isomorphism classes of) representations of $G$, and mod out the subgroup generated by elements of the form $V+W-(V\oplus W)$. Equivalently, given the statement of complete reducibility, we can just take all integral linear combinations $\sum a_i\cdot V_i$ of the irreducible representations $V_i$ of $G$; elements of $R(G)$ are corrrespondingly called virtual representations. The ring structure is then given simply by tensor product, defined on the generators of $R(G)$ and extended by linearity.
OK, let's start with the "free abelian group generated by … representations of $G$". Now from what I understand, a free group is generated from a set of groups by formally multiplying elements of different groups together. Here, the multiplication is obviously replaced by addition.
However, here's my first problem: While representations are of course groups under element multiplication, they are in general not abelian; also, it wouldn't work well with the addition notation; I think something different is meant. But then, the only notion of addition for representations I can see is the direct sum (point-wise adding the matrix representations will not give another representation, and is not even well-defined if the dimension of the representations is different). But the direct sum doesn't give rise to a group structure because there's no additive inverse. Also the fact that it is explicitly used alongside the "group addition" in the "mod out the subgroup" part indicates that it is not what is meant.
So either I have an incorrect understanding of what the "free abelian group" means, or I'm missing a way how to define sums of representations, and especially the additive inverse $-V$ of a representation.
So can someone please enlighten me?
Best Answer
"Now from what I understand, a free group is generated from a set of groups by formally multiplying elements of different groups together. Here, the multiplication is obviously replaced by addition."
This is your first problem: that's really not the definition of a free group, even approximately. (Maybe you're thinking of free products instead.) A free group and a free abelian group are both associated to a perfectly naked set $S$. The free abelian group on $S$ is easier to understand: you can look it up here. One can consider the free abelian group on $S$ as the set of all finite formal combinations $\sum_{s \in S} a_s [s]$ with $a_s \in \mathbb{Z}$ and $a_s = 0$ for all but finitely many $s$. The group operation is $\sum_{s \in S} a_s [s] + \sum_{s \in S} b_s [s] = \sum_{s \in S} (a_s + b_s) [s]$. (If $S$ is finite, this is just the direct product of $\# S$ copies of the infinite cyclic group $\mathbb{Z}$.)
Since this definition is new to you, you should probably soak it in for a while before trying to understand the definition of the representation ring $R(G)$.