I'm currently taking a course in representation theory, and I'm not entirely sure I understand the structure of the representation ring of a finite group. For a finite group $G$ with irreducible representations $(\rho_{i}, V_{i})_{i=1}^{n}$ over a field $F$, we denote

$$R_{F}[G] = \Big \{\sum_{i = 1}^{n}n_{i}V_{i} : n_{i} \in \mathbb{Z} \Big \}$$

to be the representation ring of the group $G$ with addition defined as direct addition of coefficients and multiplication defined by tensor product? I'm not entirely sure I understand the ring structure here. I understand that since we take coefficients in $\mathbb{Z}$, we clearly have an abelian group by directing taking sums of coefficients of $V_{i}$, but I'm not entirely sure I understand the multiplicative aspect of thing ring. From what I understand, it has to do with tensor products, but how is the tensor product actually taken? Do we distribute the tensor over all elements and then have to reduce the tensored representations back down to recover a ring element of this form? Would the multiplicative identity of this ring then just be the trivial representation with coefficient 1? This ring seems very odd to me, although it does seem to come up for instance in Brauer theorem.

This question is somewhat similar to this one, but reading through the question and answer here doesn't really answer the question to me.


1 Answer 1


It is better to think of $R_F[G]$ as being spanned by isomorphism classes of representations, and with the system $V_1, \ldots, V_n$ of irreducibles entering purely because they are an obvious basis.

For a finite-dimensional representation $U$ of $G$ over the field $F$, let $[U]$ denote its equivalence class up to isomorphism. Define the representation ring $R_F[G]$ to be the free abelian group with basis the set of equivalence classes, equipped with the product structure $[U][V] := [U \otimes V]$, and quotiented by the relation $[U \oplus V] = [U] + [V]$. You should check that this is well-defined (does not depend on choices of representatives from equivalence classes) and that it makes $R_F[G]$ into a commutative ring (associativity, etc).

Supposing that the representation theory of $G$ over $F$ is semisimple, and $V_1, \ldots, V_n$ is a full system of irreducibles, then it can be shown that $[V_1], \ldots, [V_n]$ is a basis for $R_F[G]$. Showing that the basis spans is easy: any representation $W$ is, by semisimplicity, isomorphic to a direct sum of the form $$W \cong V_1^{\oplus k_1} \oplus \cdots \oplus V_n^{\oplus k_n}$$ for some $k_1, \ldots, k_n \in \mathbb{N}$, and hence by the additive relation we get $$[W] = k_1 [V_1] + \cdots + k_n [V_n].$$

I think this should answer your question: if I have $[U] = [V_1] + 2[V_2]$ and $[W] = [V_3]$ then we have $$ [U \otimes W] = [U][W] = ([V_1] + 2 [V_2])[V_3] = [V_1 \otimes V_3] + 2[V_2 \otimes V_3],$$ and we need to know how to decompose $[V_1 \otimes V_3]$ and $[V_2 \otimes V_3]$ to be able to write things back in the basis of simple representations. So indeed, the trivial representation is the multiplicative identity.

A few final comments:

  • The irreducibles are just one choice of basis for $R_F[G]$, and depending on the group you study, there might be other good choices. For the symmetric group, the modules induced from trivial modules of a Young subgroup come to mind as another nice basis.
  • What I've defined above is called the split Grothendieck group. If $G$ is not semisimple with respect to $F$ then you probably want the actual Grothendieck group, which has the extra relation $[U] - [V] + [W] = 0$ whenever $0 \to U \to V \to W \to 0$ is a short exact sequence. Authors will usually clarify which one they mean.
  • $\begingroup$ Oh wow this makes a lot more sense to me actually, the structure is abundantly clear if you approach it in this way rather than the way I had previously seen. Since I've so far only dealt with finite groups, they should all be able to split in this way, but in the general case, I can see it would be a bit of a different beast, thanks! $\endgroup$
    – beeselmane
    Mar 21, 2022 at 4:31

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