I am looking for books or papers which tell me something about representation theory of finite groups over $\mathbb{Q}$ (or finite extensions thereof which are not splitting fields of the group algebra).

To be more precise, I'd like to learn of theorems which, for example, provide me with the following information, given a finite group $G$: how many irreducible representations with coefficients of $\mathbb{Q}$ are there, how can I compute their characters and can this information somehow be obtained from the "usual" character table of the group.

I have some background in ordinary representation theory over $\mathbb{C}$.


1 Answer 1


This is a well-established theory, which is very nicely presented in the second volume of the two-volume work of Curtis and Reiner. Here is the gist of it:

Since a rational representation is also a complex representation, you still have character theory to help you. In particular, a rational representation is still uniquely determined by its character, which, of course, only take values in $\mathbb{Q}$.

So suppose that you wanted to do the converse: start with the knowledge of all the complex representations (including the full character table), and construct all the irreducible rational ones. The absolute Galois group of $\mathbb{Q}$ acts on the set of complex representations by acting on each entry in each matrix, and so also acts on the set of characters. If $\chi$ is the character of an irreducible rational representation, then it must be invariant under the Galois action. In particular, if $\phi$ is an irreducible complex character sitting inside $\chi$, then every Galois conjugate $\phi^\sigma$ also has to sit in $\chi$ with the same multiplicity. So the first step is to take an irreducible complex character $\phi$ and to "rationalise" it by $\chi = \sum_{\sigma\in \text{Gal}}\phi^\sigma$, with the sum running over the distinct Galois conjugates of $\phi$.

So now you have a $\mathbb{Q}$-valued character, but it does not mean that the corresponding representation can be realised over $\mathbb{Q}$ (as an example, think of the standard representation of the quaternion group $Q_8$). However, there is a unique minimal integer $m(\chi)$ such that $m(\chi)\chi$ can be realised over $\mathbb{Q}$, and this representation is in fact irreducible over $\mathbb{Q}$. This $m(\chi)$ is called the Schur index of $\chi$, and is also nicely treated in Curtis and Reiner, but also in Isaacs for example. It is now easy to see that all irreducible rational representations arise in this way. If you are interested in general number fields, then you only have to average over the Galois conjugates over that field, but you may still have a Schur index flying around.

The answer to your question about the number of irreducible rational representations is really neat: it is equal to the number of conjugacy classes of cyclic subgroups of $G$ (as opposed to conj classes of elements, like in the complex case). I seem to remember that this is proven, among other places, in Serre's book on representation theory. This is one of the ways of stating Artin's induction theorem.

  • $\begingroup$ This is a really great answer, thank you very much for it. I'll certainly take a look at the book by Curtis and Reiner. That result about the number of irreducible rational representations is really interesting. I wonder if it generalizes to arbitrary number fields somehow. Do you happen to know anything about that, too? (I haven't had the opportunity to check out Serre's book yet...) $\endgroup$ Feb 14, 2014 at 16:17
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    $\begingroup$ @OliverBraun: yes, the same averaging (over the Galois group) and Schur index trick works. See the question math.stackexchange.com/questions/175117/… where a link to a paper is given. $\endgroup$ Feb 14, 2014 at 16:24
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    $\begingroup$ @OliverBraun: glad you like the answer. Regarding generalisations about numbers of irreducible representations, as Jack says, and as is expanded in the linked post, there is a compatible action of the Galois group not only on the characters of a group, but also on its elements. The general result then says that the number of irreducible representations of a finite group over a number field $K$ is equal to the number of conjugacy classes of orbits of elements under this compatible Galois action. Over $\mathbb{Q}$, to be in one orbit is exactly the same as to generate the same cyclic subgroup. $\endgroup$
    – Alex B.
    Feb 14, 2014 at 16:28
  • $\begingroup$ This is also very helpful. Thanks a lot to the both of you. $\endgroup$ Feb 14, 2014 at 16:31
  • $\begingroup$ @AlexB. I can guess that (with $n$ the maximal divisor of $|G|$ not divisible by ${\rm char}\,K$) the Galois group ${\rm Gal}(K(\zeta_n)/K)$ acts on $G$ by $[\zeta\mapsto\zeta^r]x=x^r$. What does it mean that this action is "compatible"? $\endgroup$
    – anon
    Feb 15, 2014 at 7:53

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