Representation theory over $\mathbb{Q}$

Solution 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.