# What happens to representation theory if we replace $\Bbb C$ with $\Bbb Q$?

The question is restricted to finite groups. The definition of representation is obvious if we replace $\operatorname{GL}(n, \Bbb C)$ with $\operatorname{GL}(n, \Bbb Q)$. The definitions of irreducibility stay the same, but starting with Schur's lemma something definitely goes wrong since its proof relies on the existence of eigenvectors, something which is not guaranteed in $\operatorname{GL}(n, \Bbb Q)$. To see what happens I started with cyclic groups:

• The cyclic group $C_3$: If we look at the representation generated by the matrix $M = \left(\begin{matrix}0 & -1\\1 & -1\end{matrix}\right)$ which satisfies $x^3 -1 = 0$. Searching for intertwining operators (using the method described here using Kronecker products) I found that they form a two dimensional vector space with vectors of the form $J = \left(\begin{matrix}a + b & - a\\a & b\end{matrix}\right)$. Indeed we have $JM = MJ$. We also have that $\det(J) = a^2+ab+b^2$ which is only zero iff $a=0, b=0$. So a hint for a replacement of Schur's lemma would be to state that every intertwining operator is either zero or has maximal rank (the intertwining operators form a division ring?).
• The cyclic group $C_5$: Here we take $M = \left(\begin{matrix}0 & 0 & 0 & -1\\1 & 0 & 0 & -1\\0 & 1 & 0 & -1\\0 & 0 & 1 & -1\end{matrix}\right)$ and the intertwining operators form a four dimension vector space : $J = \left(\begin{matrix}a + b + c + d & - a & - b & - c\\a + b + c & b + c + d & - a - b & - b - c\\a + b & b + c & c + d & - a - b - c\\a & b & c & d\end{matrix}\right)$ but I can't prove that $\det(J)$ is strictly positive unless $a=b=c=d=0$.
• Yeah, the intertwining operators from an irreducible representation to itself form a division ring. To prove this, try checking what would happen if one of them was neither zero nor invertible (or look up Schur's lemma in an advanced text that works over non-algebraically closed space). Commented May 1, 2017 at 15:44

We still have Schur's lemma. In this context it states that $\text{Hom}_G(V,W)$ vanishes if $V$ and $W$ are non-isomorphic irreducible $\Bbb Q G$-modules and that $\text{Hom}_G(V,V)$ is a finite-dimensional division algebra over $\Bbb Q$ if $V$ is irreducible. This can be non-commutative, e.g., a quaternionic algebra. Then $\Bbb QG$ is isomorphic to a direct sum of matrix algebras over these division algebras (this is the Wedderburn decomposition).
For cyclic groups $C_n$ we get an irreducible representation for each divisor $d\mid n$. The endomorphism algebra will then be the $d$-th cyclotomic field.
• I suppose the proof of the first statement relies on the fact that if $AB = BA$ then $A$ preserves the kernel and the image of $B$ (and vice versa)? Commented May 3, 2017 at 15:05