## Edit: yes

After doing a bit of work, I can now say, yes, it's always possible
to realise $\rho$ over $\mathbb{Q}(\zeta_n)\cap\mathbb{R}$, see
https://arxiv.org/abs/2107.03452

The main ingredient in the latter is Serre's induction theorem,
from Jean-Pierre Serre. Conducteurs d’Artin des caractères réels.Invent.Math., 14:173–183, 1971
(also in Charles W. Curtis and Irving Reiner.Methods of representation theory.Vol. II. Pure and Applied Mathematics (New York). John Wiley, NY, 1987). It's an analog of Brauer's induction thereom, used by Brauer to show that $\mathbb{Q}(\zeta_n)$ is a splitting field, but it works over $\mathbb{Q}(\zeta_n)\cap\mathbb{R}$.

## Splitting field of a real representation - summary

This summarises the state of the art on the question, at least IMHO. In particular, there is an exact algorithm to decide whether the field needs to be extended (and do the base change), but no example showing that extending might be needed in some cases.

The paper [1] studies a closely related question of minimising the degree of the number field needed to write down the representation, and in fact the "yes" answer can be distilled from there with a bit of extra work. In fact, we basically need to modify the proof of Thm 4.19 (Frobenius-Schur) in [3] (which probably goes back to the original paper by Frobenius and Schur) to show that the splitting field $\mathbb{F}$ need not be extended.
The proof of Thm 4.19 starts with the elementary fact that if $Q$ is the transformation making $\rho$ real then $Q^{-1}\rho Q=\overline{Q^{-1}\rho Q}$, thus $\overline{Q}Q^{-1}\rho=\overline{\rho Q}Q^{-1}$, and $P:=\overline{Q}Q^{-1}$ transforms $\rho$ to $\overline{\rho}$, i.e. $P^{-1}\overline{\rho}P=\rho$ (such $P$ must exist as the characters of $\rho$ and $\overline{\rho}$ are equal).

Now we have the equation
$$PQ=\overline{Q}, \qquad \det Q\neq 0\tag{1}$$ implying $\overline{P}PQ=\overline{PQ}=Q$, i.e. $\overline{P}P=I$. The latter is an extra restriction, as $P$ can be multiplied by any $\lambda\neq 0$, and so
$$P\overline{Pg}=Pg\overline{P}=\overline{g}P\overline{P},$$
i.e. $P\overline{P}$ centralises an irreducible representation $\overline{\rho}$, and so $P\overline{P}=\alpha I$, for some $\alpha>0$.

As $\rho$ is real, it affords a nonzero $G$-invariant symmetric linear form $M$, i.e. $M^\top=M$, $g^\top Mg=M$ for all $g\in\rho$ - this is well-known (cf. e.g. Thm 4.14 in [loc.cit.]). As $M$ can be found in the trivial sub-respresentation of the symmetric square of $\rho$, $M\in M_d(\mathbb{F})$. As well, $\det M\neq 0$, as the kernel of $M$ would give rise a sub-representation of $\rho$, contradicting irredudicibility of $\rho$.

Let ${\Sigma}:=\sum_{h\in\rho}h^\top \overline{h}$. Note that ${\Sigma}$ is a Hermitian positive definite, in particular $\det{\Sigma}> 0$.

Choose $P$ as $P:=\Sigma^{-1}M$. Let's check that $P^{-1}\overline{\rho}P=\rho$ (we use $\det M\neq 0$ here).
Let $g\in\rho$. Then, as $(\overline{g}\Sigma^{-1}g^\top)^{-1}=(g^\top)^{-1}\Sigma (\overline{g}^\top)^{-1}=\Sigma$,
$$\Sigma^{-1}Mg=\overline{g}\Sigma^{-1}g^\top Mg=\overline{g}\Sigma^{-1} M,$$
as required.

Let $\alpha=\lambda\overline{\lambda}$. Then we can scale $P$ by $1/\lambda$ to ensure $P\overline{P}=I$.
It follows immediately from Thm 3 of [1] that $\lambda$ may be chosen in $\mathbb{F}$ is and only if there exists $A\in GL_d(\mathbb{F})$ s.t. $A^{-1}\rho A\in GL_d(\mathbb{R})$. (See Conclusion below for discussion).

It remains to solve (1) and so that $Q$ belongs to the splitting field of $\rho$.
Note that the solution of (1) in [3] assumes that $\rho$ is unitary; i.e. $\Sigma=I$; so in this case $P^\top=P$, and an explicit formula for $Q$ provided - which however does not work for us, as it involves square roots of eigenvalues of $P$.
Fortunately, in [2], Prop. 1.3, there is an algorithmic proof of existence of the required solution of $P\overline{Y}=Y$ (which is the same, just set $Y=\overline{Q}$). In [2] it is done for finite fields (and in bigger generality, for a field automorphism $\sigma$ of finite order, referring to this result as a generalisation of *Hilbert Theorem 90*), and in [3] it was noted that it works for number fields as well.

[2], Prop. 1.3 is a deterministic linear algebra way to solve (1), and they also give a much quicker to describe probailistic one. Namely, let $Y\in M_d(\mathbb{F})$ be chosen randomly; then $Q=\overline{Y}+\overline{P}Y$ satisfies $PQ=P\overline{Y}+Y=\overline{Q}$, and with high probability $\det Q\neq 0$.

**Conclusion.** We are left here at mercy of number theory; e.g., if it happened that $\mathbb{F}=\mathbb{Q}[i]$ (i.e. cyclotomic field of degree 4) and $\mu=2$, we'd been good, as $(1-i)(1+i)=2$, but if $\mu=3$ then $\mu$ cannot be expressed as a norm, and we need to extend the field, to degree 12 cyclotomics, in fact, as $\sqrt{3}=-\zeta^7+\zeta^{11}$, with $\zeta$ a 12th primitive root of 1.

Whether there are examples showing that $\mathbb{F}$ might need to be extended in some case remains to be seen, apparently. While in the literature (see refs. in [1]) one may find examples of representations with the minimal splitting field not being $\mathbb{F}$, all these examples are not real representations.

**References.**

[1]: C. Fieker Minimizing representations over number fields, Journal of Symbolic Computation,
Volume 38, Issue 1, 2004,833-842, DOI:
https://doi.org/10.1016/j.jsc.2004.03.001.

[2]: S. P. Glasby & R. B. Howlett (1997) Writing representations over minimalfields, Communications in Algebra, 25:6, 1703-1711, DOI: https://doi.org/10.1080/00927879708825947

[3]: M. Isaacs, Character theory of finite groups

## older notes

- There is always a way to write down a real $2d$-dimensional representation $\hat\rho$ given a $d$-dimensional complex $\rho$, using the usual represenation of $\mathbb{C}\subset M_2(\mathbb{R})$
given by $a+ib\mapsto \begin{pmatrix}a &-b\\ b& a\end{pmatrix}$.
As cyclotomics are closed under complex conjugation, this can be carried over cyclotomics.

For $\mathbb{R}$-irreducible representations $\hat\rho$ this is how these kinds of real irreducible representations can be constructed - by the classification on p 108 of Serre's "Linear Representations of Finite Groups" this is if and only if $\rho$ cannot be written over $\mathbb{R}$.

If on the other hand $\hat\rho$ is reducible then its character is twice the character of a real irreducible (more generally, its the character plus its conjugate), and perhaps there is an easy way to find a proper $G$-invariant (cyclotomic) subspace, which would solve our problem. E.g. look at the symmetric square of $\hat\rho$ - this is the action on symmetric bilinear forms, and so one should have 2-dimensional subspace of fixed points there. Up to scaling, two forms there are rank $d$, so finding any of these will solve the problem.

Note that one can simultaneously permute row and columns of $\hat\rho$ so that each $\hat\rho(g)=\begin{pmatrix}A &-B\\ B& A\end{pmatrix}$, with $A=A_g,B=B_g\in M_d(\mathbb{R})$.

- Some conditions may be derived following the proof of Thm 4.19 in M.Isaacs "Character theory of finite groups", from the elementary fact that if $Q$ is the transformation making $\rho$ real then $Q^{-1}\rho Q=\overline{Q^{-1}\rho Q}$, thus $\overline{Q}Q^{-1}\rho=\overline{\rho Q}Q^{-1}$, and $P:=\overline{Q}Q^{-1}$ transforms $\rho$ to $\overline{\rho}$, i.e. $P^{-1}\overline{\rho}P=\rho$ (such $P$ must exist as the characters of $\rho$ and $\overline{\rho}$ are equal).

Now we have the equation $PQ=\overline{Q}$ implying $\overline{P}PQ=\overline{PQ}=Q$, i.e. $\overline{P}P=I$.

Now, if $\rho$ is unitary, it may be shown (as in the proof of Thm 4.19) that in this case $P^\top=P$, and an explicit formula for $Q$ provided. Even in this case it's not clear if no further field extension is needed, as, while $Q$ is in the same field as $P$, it's not clear how to guarantee $\overline{P}P=I$ (we can get, as a consequence of Schur lemma, that $\overline{P}P=\alpha I$, and $\alpha>0$ by linear algebra, but we need $\sqrt{\alpha}$ to belong to the original field).

can'tdo is just say that if it's defined over $L$ and defined over $K$ then it's defined over $L \cap K$, see this question. So you're going to need to try to mimic the proof of the original result rather than just applying the original result. That said, I think once you get through the notation, the result from Serre I mention tells you pretty clearly where to look. $\endgroup$12more comments