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Let $G$ be a finite group and $K$ a field with $\mathbb{Q} \subseteq K \subseteq \mathbb{C}$.

For $K=\mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.

For $K=\mathbb{R}$ the number of irreducible representations of $KG$ is equal to $\frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.

For $K=\mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.

(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)

Question:

Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $\mathbb{Q}$.

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    $\begingroup$ To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $\mathbb{Q}$ and $\mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.) $\endgroup$
    – Jim Humphreys
    Commented Nov 4, 2018 at 14:30
  • $\begingroup$ @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven. $\endgroup$
    – Mare
    Commented Nov 4, 2018 at 15:49

1 Answer 1

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There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $\zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(\zeta)/K)$. We can identify $H$ with a subgroup of $(\mathbb Z/n\mathbb Z)^*$. Call $a,b\in G$ $K$-conjugate if $b^j=gag^{-1}$ for some $g\in G$ and $j\in H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.

In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.

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    $\begingroup$ Are there convenient references? $\endgroup$
    – Jim Humphreys
    Commented Nov 4, 2018 at 14:38
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    $\begingroup$ The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 4, 2018 at 17:40
  • $\begingroup$ Im traveling now so don't have the books handy. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 4, 2018 at 17:43
  • $\begingroup$ In particular, this says that every irrep can be defined over $\mathbb{Q}(\zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem. $\endgroup$
    – Kevin Casto
    Commented Nov 4, 2018 at 17:58
  • $\begingroup$ I believe the induction theorem is used in the proof. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 4, 2018 at 18:13

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