# Center of an irreducible representation over $\mathbb{Q}$

Let $G$ be a finite group, $\rho\colon G \rightarrow \mathrm{GL}_n(\mathbb{Q})$ its irreducible representation, and $D$ the division algebra of $G$-endomorphisms of $\mathbb{Q}^n$. The division $\mathbb{Q}$-algebra $D$ is finite dimensional. What is its center?

I've heard that the center is the number field generated by the values of the character of an irreducible constituent of $\rho_{\mathbb{C}}$ but I cannot see this (and I don't know where to look this up).

• Here is a hint: the center $Z$ of $\mathbb{Q}[G]$ is the subalgebra of functions $G \rightarrow \mathbb{Q}$ constant on the conjugacy class. Because they are central, each element of $Z$ acts on a representation as an endomorphism. Finally, On a complex irreducible representation each element of $Z$ acts has an endomorphism, hence as a complex number by schur lemma... But then the number in question has to be the trace/dimension hence easily relates to the value of the character. – Simon Henry Aug 20 '15 at 17:41
• the question mathoverflow.net/questions/119658/… is related to this question – Venkataramana Oct 2 '15 at 12:25

## 1 Answer

Theorem (74.5) in Curtis' and Reiner's Methods of Representation Theory, Volume II should answer all your questions. Indeed, for an irreducible representation the center of the endomorphism algebra is the character field, i.e. the number field generated by the character values.