TITLE: Realizability of a real representation using real cyclotomic coefficients
QUESTION [10 upvotes]: Let $G$ be a finite group and $\rho: G \rightarrow GL(d,\mathbb{C})$ an irreducible representation with Frobenius-Schur indicator $\frac{1}{|G|}\sum_{g\in G} \operatorname{tr} \rho(g^2) = 1$. Thus $\rho$ is a real representation.
A theorem by Brauer states that every irreducible representation over $\mathbb{C}$ can be written using coefficients taken from $\mathbb{Q}[\zeta_n]$ where $\zeta_n=\exp(\frac{2\pi i}{n})$ where $n$ is at most the exponent of the $G$.
We also know that any real representation can be written with coefficients in $\mathbb{R}$
Can we satisfy both?
For example, the irreducible representations for the dihedral groups can satisfy both conditions (real coefficients using cosines and sines of fractions of $\pi$).
I can solve the problem over the algebraic integers by computing an antilinear equivariant involution $F$ such that $F(x) = J \overline{x}$ for a suitable matrix $J$, and such that $\rho(g) F(x) = F(\rho(g) x)$, which implies $\rho(g) J = J \overline{\rho(g)}$. Then I know that $J \overline{J} = \alpha \mathbb{1}$ for a positive scalar $\alpha$, and I set $J' = J/ \sqrt{\alpha}$, so that $J'$ represents the complex conjugation. Now, $J$ can be found with cyclotomic coefficients (one iterates over a basis of the space of matrices with $\{0,1\}$ coefficients, and averages over the group). But the square root operation is not necessarily closed over the cyclotomics, this is where I am currently stuck.

REPLY [5 votes]: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$The main result of this answer will be that, if $V$ is a representation defined over $\RR$ and over $\QQ(\zeta_M)$, then there is an integer $N$ such that $V$ is defined over $\RR \cap \QQ(\zeta_{MN})$. This proof will not give us good control over the size of $N$.
We'll abbreviate $\QQ(\zeta_M)$ to $K_M$ and abbreviate $K_M \cap \RR$ to $R_M$.
Start with a $K_M$-vector space $V$ and a representation $G \to GL(V)$ which is known to descend to $\RR$. Let $V_R$ be $V$ considered as an $R_M$-vector space and let $\Delta$ be the endomorphism of $V_R$. Then $E$ contains $K_M$. Also, since $V$ descends to $\RR$, we know that $V$ is isomorphic, over $K_M$, to the complex conjugate representation $\overline{V}$. We can compose the complex conjugate map $V \to \overline{V}$ with a $K_M$-linear isomorphism $\overline{V} \to V$ to get another element $\sigma$ of $E$. We have $\sigma \alpha = \overline{\alpha} \sigma$ for all $\alpha \in K_M$. Also, $\sigma^2$ commutes with $K_M$ so, by Schur's lemma applied to $V$, we have $\sigma^2 \in K_M$, say $\sigma^2 = a$. In other words, $E$ contains the quaternion algebra $( -1, a )$ for some $a \in K_M$ and some dimension counting shows that it is this quaternion algebra. We have $V$ defined over $R_M$ if and only if this quaternion algebra splits over $R_M$: If $V$ is defined over $R_M$, it is easy to see that $E \cong \mathrm{Mat}_{2 \times 2}(R_M)$ and conversely, if $E$ splits, the image of any rank $1$ idempotent is an $R_M$-form of $V$.
So we want to show: If $E$ is a quaternion algebra over $R_M$ which splits over $\RR$ then there is an integer $N$ such that $E$ splits over $R_{MN}$.
Quaternion algebras are order $2$ elements in the Brauer group. The Brauer group injects into the direct sum of all the local Brauer groups, which are $\ZZ/2 \ZZ$ for the Archimedean places and $\QQ/\ZZ$ for the non-Archimedean. Since we assumed that $E$ splits over $\RR$, there are no contributions from the Archimedean places. If $\lambda/\kappa$ is a degree $d$ embedding of local fields, then the corresponding map on Brauer groups is multiplication by $d$ from $\mathrm{Br}(\kappa) \cong \QQ/\ZZ$ to $\mathrm{Br}(\lambda) \cong \QQ/\ZZ$. In particular, in order to split our quaternion algebra $E$, we just need to find an extension $R_{MN}$ of $R_M$ such that every local field of $K_M$ at which $E$ is not split has even degree extensions in $R_{MN}$.
Let $p_1$, $p_2$, \dots, $p_r$ be the primes of $\ZZ$ lying below the primes $R_M$ where $E$ is split. Choose $N$ relatively prime to $M$ and a quadratic nonresidue modulo every $p_j$. Then $R_M[\sqrt{N}] \subset R_{MN}$ has the desired property. $\square$.
Obviously, this is not the answer you are looking for, but it is the best I can do.