# A problem from Dummit Foote on Representation of finite groups over algebraic closure of $\mathbb{Q}$

Let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$.Let $\phi:G \rightarrow Gl_{m}(\bar{\mathbb{Q}})$ be an irreducible representation. Then $\phi$ can be viewed as a representation from G to $Gl_{m}(\mathbb{C})$. Prove the following:

1) $\phi$ viewed as a representation of G to $Gl_{m}(\mathbb{C})$ is irreducible.

2)Show that the set of irreducible characters over $\bar{\mathbb{Q}}$ is same as the set of irreducible characters over $\mathbb{C}$.

3) Deduce that every complex representation is equivalent to a representation over $\bar{\mathbb{Q}}$.

I did try the ways of character theory but that to my knowledge works with vector spaces over $\mathbb{C}$. I really don't know whether they work when we have vector space over algebraically closed field like $\bar{\mathbb{Q}}$. One idea that I have is the fact the root of character theory lies in Schur's lemma where we actually use for the first time the fact that we are working with $\mathbb{C}$ and that is one the main thing that make the whole theory work. If we consider algebraically closed field the Schur's lemma should still work. Maybe that is the reason that these resuls are true!!

I want to know whether this is right or else there is a more concrete reason for these results to hold. Thanks in advance!!

• My first suggestion would be to use the fact that the cyclotomic field containing $\Bbb{Q}$ and all the roots of unity of order $|G|$ is known to be a splitting field. Probably we get away with a less powerful result for this conclusion. – Jyrki Lahtonen Mar 10 '17 at 20:45

## 1 Answer

Perhaps the most important observation to make is that, for any representation $\rho : G \to Gl_n(\mathbb C)$, the character $\chi : G \to \mathbb C$ always takes values in $\bar{\mathbb Q}$.

[Hint: if $g$ is an element of order $n$ in $G$, then $\rho(g)^n = id$. What are the only possible values for the eigenvalues of $\rho(g)$? How is the character of $g$ related to these eigenvalues?]

Other hints:

(i) You need a way of identifying whether a given representation is irreducible or not. One way to do this is to use the fact that a representation is irreducible iff its character $\chi(g)$ obeys $$\frac 1 {|G|} \sum_{g \in G}\overline{\chi(g)}\chi(g) = 1.$$

(ii) You need a way to tell if two irreducible representations are equivalent. This is easy: they are equivalent iff they have the same character.

(iii) You need a way to determine whether a given collection of irreducible representations is the complete set of irreducible representations for $G$. A useful fact is that total number of irreps equals the number of conjugacy classes of $G$.

I know for sure that the results (i), (ii), (iii) above are true over $\mathbb C$. Off the top of my head, I can't think of any reason why the proofs of these results shouldn't also work over the algebraically closed field $\bar{\mathbb Q}$. But please do check whether these results do hold over $\bar{\mathbb Q}$, and please leave a comment if you find that I'm mistaken.

Anyway, I hope these hints point you in the right direction.

Edit (28/03/2017): I think the first remark about characters of representations over $\mathbb C$ taking values in $\bar{\mathbb Q}$ isn't strictly necessary to complete the argument!

One could argue like this: Let $n$ be the number of conjugacy classes in $G$. Working over $\bar{\mathbb Q}$, which is an algebraically closed field of characteristic zero, all standard results in representation theory and character theory are applicable. They tell us that there are $n$ irreps of $G$ in $Gl_n(\bar{\mathbb Q})$, with distinct characters obeying $\frac 1 {|G|} \sum_{g \in G} |\chi(g)|^2 = 1$. Now view these representations as representations in $Gl_n(\mathbb C)$. As complex representations, they are still irreducible (because $\frac 1 {|G|} \sum_{g \in G} |\chi(g)|^2 = 1$), they are still distinct (because their characters are distinct) and they still form a complete collection of irreps (because there are $n$ of them). Hence every irrep of $G$ over $\mathbb C$ is equivalent to an irrep of $G$ over $\bar{\mathbb Q}$.

• yeah I should have written in the question itself that these are the ways that I know of but the main problem is character theory holds exclusively over $\mathbb{C}$ but I dont get the idea whether it also holds for any algebraically closed field. – Riju Mar 11 '17 at 4:41
• Why do you say that character theory holds exclusively over ${\mathbb C}$? Where exactly do you use the fact that you are working over ${\mathbb C}$ rather than an arbitrary algebraically closed field of characteristic $0$. (When I first learnt representation and character theory, we were told from the beginning that the field was an algebraically closed field of characteristic $0$, of which ${\mathbb C}$ is of course an example.) – Derek Holt Mar 11 '17 at 5:23
• yeah that's what I also think that character theory holds for algebraically closed field of characterestics zero field also. – Riju Mar 11 '17 at 6:15
• @ Riju, perhaps work through the proofs of the main results in character theory. Can you see any step that might fail over $\bar{\mathbb Q}$? I can't think of any such step. [I can certainly think of a step that would fail over $\mathbb Q$: the proof of the second Schur's lemma requires the characteristic polynomial of matrices to factorise into linear factors. But over $\bar{\mathbb Q}$, we seem to be fine.] – Kenny Wong Mar 11 '17 at 9:38