### Question About Representations of Finite Groups

#### Posted by John Baez

Here’s the first of some questions that have been bugging me. Maybe you can help!

I want to know when we can define the representations of a finite group using not the full force of the complex numbers, but only some subfield, like $\mathbb{Q}[\sqrt{5}]$ or $\mathbb{R}$. If I knew the answer to this question, it might be important for the groupoidification program, where we’re trying to replace complex vector spaces by groupoids whenever possible.

Suppose $k \subseteq \mathbb{C}$ is some subfield of the complex numbers. In what follows, ‘representation’ will mean *representation on a finite-dimensional complex vector space*. Suppose $G$ is some group with a representation $\rho$. Let’s say $\rho$ is **defined over $k$** if we can find some basis of our vector space such that the matrices corresponding to the linear transformations $\rho(g), g \in G$ all have entries lying in $k$.

Question 1.Is there a smallest subfield $k \subseteq \mathbb{C}$ such that every representation of every finite group is definable over $k$? If so, what is it?

It’s not hard to see that:

- Every representation of every finite group is definable over $k$ when $k = \overline{\mathbb{Q}}$ is the field of algebraic numbers.
- Not every representation of every finite group is definable over $k$ when $k = \mathbb{R}$. There’s an easy trick to see which ones are.
- Every representation of the symmetric group $S_n$ is definable over $k$ when $k = \mathbb{Q}$.
- Every representation of the cyclic group $\mathbb{Z}/n$ is definable over $k$ when $k = \mathbb{Q}[e^{2\pi i/n}]$ is the cyclotomic field generated by taking $\mathbb{Q}$ and throwing in a primitive $n$th root of unity.

But what I really want to know is this:

Question 2.Is every representation of every finite group definable over $k$ when $k = \mathbb{Q}^{ab}$ is the field generated by taking $\mathbb{Q}$ and throwing in all roots of unity? If not, what’s the simplest counterexample?

Here’s a pathetic shred of evidence that the answer to Question 2 is “yes”:

Theorem.Let $\rho$ be a representation of a finite group. Then for any $g \in G$, $tr(\rho(g)) \in \mathbb{Q}^{ab}.$

If this theorem were false, the answer to Question 2 would be “no”. But, that’s just a pathetic shred of evidence that the answer is “yes”. In fact, if I had to guess, I’d guess the answer is “no”!

This theorem is old. I read in here:

- Charles W. Curtis,
*Pioneers of Representation Theory: Frobenius, Burnside, Schur and Brauer*, History of Mathematics vol. 15, AMS, Providence, Rhode Island, 1999.

that it was proved by some bigshot like Frobenius or Burnside, who then went on to ponder Question 2. So, the answers to both my questions must be known by now. But, I don’t know them!

The proof of the theorem is easy. By group averaging we can find an inner product such that $\rho(g)$ is unitary for all $g$. So, for any $g$ we can find a basis in which $\rho(g)$ is diagonal. Since our group is finite, $g^n = 1$ for some $n$. So, the diagonal entries of $\rho(g)$ are roots of unity. So, $tr(\rho(g)) \in \mathbb{Q}^{ab}$.

Note: this shows that for any *one* element $g$ we can find a basis for which the entries of $\rho(g)$ lie in $\mathbb{Q}^{ab}$. But, that’s far short of finding a basis that works for all $g$ at once!

## Re: Question About Representations of Finite Groups

One way to prove it is as follows: The group ring R of G over Q

^{ab}is a semi-simple algebra, hence by Wedderburn’s theorem a product of matrix rings over finite dimensional division algebras over Q^{ab}. But the only such division algebra is Q^{ab}itself, so R must be a product of matrix rings, which implies that all representations of G are defined over Q^{ab}.Unfortunately, I don’t know an elementary proof (it follows from class field theory) for the statement about division algebras…