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.

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    $\begingroup$ I think one should be able to answer this question with the techniques of Serre's Linear Representations of Finite Groups Section 12.6 (the generalization of Brauer's theorem to representations defined over particular fields) applied to the real numbers. But I don't immediately see how to do it (or which way the answer will go!) without doing some calculations. $\endgroup$ Mar 17, 2021 at 2:58
  • $\begingroup$ Ouch, that's above my pay grade, though I appreciate Serre's little book immensely. I'm posting this to assess the difficulty of the question as well -- because then the next step is to find an algorithm to compute the images. $\endgroup$ Mar 17, 2021 at 13:57
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    $\begingroup$ What you can't do 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. $\endgroup$ Mar 17, 2021 at 15:12
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    $\begingroup$ The most interesting example I've found is the 4-dimensional representation of $Q_8 \rtimes S_3$ where $S_3$ acts by an inner automorphism. I think that twice this representation has a $\mathbb{Q}$ form with quaternion algebra $(-1,3)_\mathbb{Q}$. Here the exponent is $12$, and indeed the original irrep is defined over the real subfield $\mathbb{Q}(\sqrt{3}) \subset \mathbb{Q}(\zeta_{12})$. But on the other hand, the rep is defined over the cyclotomic field $\mathbb{Q}(i)$ but not over its real subfield $\mathbb{Q}$. (Possible error, maybe $i$ should be $\sqrt{-3}$.) $\endgroup$ Mar 22, 2021 at 16:51
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    $\begingroup$ Is it true that if you have a monomial representation over $\mathbb{Q}(\zeta_n)$ and its induction is defined over some field $K$ then the induction is also defined over $K \cap \mathbb{Q}(\zeta_n)$? Or more generally if $V$ is defined over $L$ and $\mathrm{Ind}_H^G V$ is defined over $K$ is $\mathrm{Ind}_H^G V$ defined over $L \cap K$? $\endgroup$ Mar 22, 2021 at 21:14

3 Answers 3


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.


[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

  1. 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 $A=A_g,B=B_g\in M_d(\mathbb{R})$.

  1. 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).


$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$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.


(Edit: the original claim was much more ambitious).

It is possible to transform the representation into a representation which is slightly less general, without extending the field. Namely, this can be done if the transformation $M=PU$ which was used to make $\rho$ non-real, with $U$ Hermitean p.s.d. and $U$ unitary, satisfies $U=I$ (the decomposition $M=PU$, a polar decomposition, always exists).

Let $\mu:G\to GL_d(\mathbb{R})$ s.t. $\rho(G)=M\mu(G)M^{-1}$ for $G\in GL_d(\mathbb{C})$. W.l.o.g. $\mu$ is orthogonal, i.e. $hh^\top=I$ for any $h\in\mu(G)$. Thus $$|G|I=\sum_{h\in\mu(G)} hh^\top=\sum_{h\in\mu(G)} hh^*= \sum_{g\in\rho(G)} M^{-1}gM (M^{-1}gM)^*=\sum_{g\in\rho(G)} M^{-1}gM M^*g^*(M^*)^{-1}.$$ Multiplying on the right by $M^*$ and on the left by $M$ one obtains $$|G|MM^*=\sum_{g\in\rho(G)} gM M^*g^*.$$

Up to scaling by a positive real, there is unique, by irredicibility of $\rho$, Hermitean p.s.d. matrix $H$ providing $G$-invariant sesquilinear scalar product, i.e. $gHg^*=H$ for any $g\in\rho(G)$. Therefore $H=MM^*$. Note that $H$ only records the info about $M$ up to a unitary factor. Namely, $M=PU$, with $U$ unitary and $P$ Hermitean positive definite ($PU$ is called left polar decomposition of $M$). Thus $H=MM^*=PUU^*P^*=PP^*=P^2$.

Note that $H$ affords a (unique, in fact), $LDL^*$ decomposition, with $D=diag(D_1,\dots,D_d)$, and $D_k>0$ for all $1\leq k\leq d$.

It is well-known that $L$ and $D$ can be computed by solving a system of linear equations, without extending the field (e.g. as on p.43 of our preprint (=MSc thesis of my student K.Hymabaccus)).

Set $M=LD^{1/2}U$. Then $$\mu(G)=M^{-1}\rho(G)M=U^*D^{-1/2}L^{-1}\rho(G)LD^{1/2}U.$$ Thus we can perform the conjugation by $L$, and so reduce to the case of $M=D^{1/2}U$.

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    $\begingroup$ I don't understand the step where you take a square root of $D$. The entries of $D$ are in $\mathbb{Q}(\zeta_n) \cap \mathbb{R}_{>0}$, but their square roots need not be. (I also think there is something wrong where you say "set $M = L D^{1/2}$; both $M$ and $L D^{1/2}$ have predefined meanings.) $\endgroup$ Mar 25, 2021 at 14:24
  • $\begingroup$ this is not a bug, as I don't use $D$ - the bug is bigger: if $M$ is unitary it cannot be recovered, as $MM^*=I$. I probably just reduce from the general $M$ to $M$ unitary... $\endgroup$ Mar 25, 2021 at 15:31
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    $\begingroup$ @DimaPasechnik Thanks for giving it a try! I was hesitant on posting that question on MathOverflow (vs. the Math StackExchange)... and now I see we're facing an interesting problem! $\endgroup$ Mar 29, 2021 at 13:28
  • $\begingroup$ there are also 2 deleted answers here as well. $\endgroup$ Mar 29, 2021 at 22:28
  • $\begingroup$ @MartyIsaacs might know the answer here... $\endgroup$ Mar 29, 2021 at 22:31

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