# Why is introductory representation theory over C instead of a closure of Q?

In an introductory representation theory course (finite groups and associative initial algebras), we used commonly used an algebraically closed field of characteristic 0, and most often this was just taken to be $\mathbb{C}$, instead of $\bar{\mathbb{Q}}$. Is this just for student ease, or are there results in say, finite group representation theory which use the extra analytic structure of $\mathbb{C}$?

For finite groups, it does not make any difference. For instance, any representation of a finite group over $\mathbb{C}$ is obtained by taking one over $\overline{\mathbb{Q}}$ and just tensoring up to $\mathbb{C}$ (or concretely, you can always choose a basis so that all your matrices have entries in $\overline{\mathbb{Q}}$).
For more general representation theory, however, the analytic structure of $\mathbb{C}$ is important. For instance, an important generalization of the representation theory of finite groups is the representation theory of compact topological groups, where you require the representations to be continuous. The representation theory of compact groups over $\mathbb{C}$ is formally quite similar to the representation theory of finite groups (which are of course a special case), but over $\overline{\mathbb{Q}}$ there is no good topology to get a similar theory.