# Easy way to get real irreducible characters (reps) from complex irreducible characters?

For plenty of groups, the real irreducible characters/representations aren't the same as the complex irreducible representations. I really enjoy James Montaldi's summary of real representations, for instance.

However, what I can't find is a simple statement of going back the other way - which real representations come from a usual complex character table. (This is useful if one can compute a character table with e.g. Gap.) I assume that this is pretty nontrivial, otherwise I wouldn't have to wish I had my references for representation theory with me.

And yet... wouldn't it be nice if one could (say) in certain circumstances just say "oh yeah, if two complex non-real characters look a lot alike, just add them and you get the real character."

Question: Is there ever a time where one is justified in just "adding non-real complex irreducible characters" to get real irreducible characters?

If it turns out this is trivial and I didn't know it, let's just chalk that up to the fact that nearly all of us presumably learned our representation theory mostly over $\mathbb{C}$.

The answer is completely described by the Frobenius–Schur indicator, $$v_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi(g^2)$$ which is covered in chapter 4 of Isaacs's Character Theory of Finite Groups. In particular, we have the Frobenius–Schur theorem, as given on page 58 of Isaacs's CToFG:

Theorem: If $\chi$ is an irreducible (ordinary, complex) character of the finite group $G$, then $$v_2(\chi) = \begin{cases} 1 & \chi \text{ is the character of an irreducible real representation} \\ 0 & \chi \text{ takes complex values } \\ -1 & \text{otherwise} \end{cases}$$ If $v_2(\chi)=1$, $\chi$ is the character of a real representation. If $v_2(\chi)=0$, then $\chi + \bar \chi$ is the character of a real representation. If $v_2(\chi)=-1$, then $2\chi = \chi + \bar \chi$ is the character of a real representation.

The following is Lemma 9.18 (more or less) in Isaacs's CToFG:

Proposition: If $\chi$ is a (ordinary, complex) character of the finite group $G$, then $\chi + \overline{\chi}$ is the character of a real representation of $G$. More generally, if $K \leq F$ are fields of characteristic 0 and $\chi$ is the character of a (ordinary) $F$-representation of $G$, then for every $\sigma \in \operatorname{Gal}(F/K)$, $\chi^\sigma = g \mapsto \sigma(\chi(g))$ is the character of a (ordinary) $F$-representation of $G$, and $$\sum_{\sigma \in \operatorname{Gal}(F/K)} \chi^\sigma$$ is the character of a $K$-representation.

Proof: If $\chi$ is the character of the representation $X$, then consider the representation $X \otimes_{\mathbb{C}} \mathbb{C}_\mathbb{R} \cong X \oplus \bar X$ obtained by replacing the complex entries $a+bi$ of $X$ with the block matrices $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$. The resulting matrix has trace $\chi(g) + \bar\chi(g)$, since we replace each diagonal entry $a+bi$ with a matrix of trace $2a$. The general case is similar: just choose a $K$-basis of $F$, and write out the associated matrix representation of $f \in F$ in its action on the $K$-vector space $F$. $\square$

Since you mention GAP, I'll mention the indicators of a character table are computed using Indicator( chartable, 2) and that since every complex, ordinary representation of $G$ is a representation over the field CF(Exponent(G)), the $\sigma$ appearing in the Galois group are given by chi -> GaloisCyc(chi,k) for $k$ relatively prime to $G$. In particular, $k=-1$ is complex conjugation. To compute the matrices of the $K$-representation from the $F$-representation as described in the proof, you can use BlownUpMat(Basis(AsVectorSpace(K,F)),mat).

The function RationalizedMat can be applied to a character table to compute some smaller sums where one needs to multiple by the so called Schur index as in Geoff Robinson's answer: for example, it would take a real character and leave it real, even if its indicator were $-1$. However, for Brauer characters (where the Schur indices are always 1), it can be very handy.

At any rate, chapters 4, 9, and 10 of Isaacs's textbook are great for understanding how characters work over different fields of characteristic 0 and how that relates to the power maps of the character table.

• I was aware of the Fr-Sch indicator, but only as a way to distinguish between real and complex reps. I like that $\chi+\bar{\chi}$ "just works", though I would have been surprised if it didn't. – kcrisman Jun 3 '14 at 15:28
• And I assume it is a trivial corollary that real representations gotten via this are irreducible? – kcrisman Jun 3 '14 at 15:30
• And thanks for the GAP; I had already found (and attempted to use) Indicator, but it's nice to have it all in one place. – kcrisman Jun 3 '14 at 15:31
• Yes. If $\chi$ is irreducible and $v_2(\chi) \neq 1$, then $\chi+\bar \chi$ is irreducible (consider any possible submodule as a complex submodule to see the only possibilities are $\chi$ and $\bar \chi$ neither of which have real reps). – Jack Schmidt Jun 3 '14 at 15:34
• @kcrisman: glad to read this question (+1 to both). Please consider tagging (gap) questions where GAP plays essential part in the answer as well. – Alexander Konovalov Jun 3 '14 at 15:35

If you have a real valued irreducible (complex) character $\chi$, you can decide if it comes from a representation over the reals by considering its Frobenius-Schur indicator. If $\sum_{g \in G} \chi(g^{2}) = |G|,$ then the character is afforded by some. absolutely irreducible representation over the real field. If $\sum_{g \in G} \chi(g^{2}) = -|G|$, then the repreentation is not afforded by any real representation, but $2\chi$ may be afforded by a real representation (an example where this happens is the quaternion group of ordar $8$ and its $2$-dimensional complex irreducible representation). If $\chi$ is not real-valued, then $m_{\mathbb{Q}}(\chi)(\chi + \overline{\chi})$ may be afforded by real representation, where $m_{\mathbb{Q}}(\chi)$ denotes the Schur index, but I am not sure at the moment whether this can be read directly from the character table of $G.$

Later edit: The upshot of Jack Schmidt's comments below is that in the second case, there is no issue. If the Frobenius Schur indicator of $\chi$ is zero ( the only remaining possibility), then $\chi + \overline{\chi}$ is the character of a real representation (irreducible as real representation, but reducible as complex representation). So the characters of the real irreducible representations can indeed be read fairly directly from the complex character table.

• I think $\chi+\bar \chi$ is always afforded by a real representation, but I'm having a hard time expressing why at this late hour. I'd be interested in an example where it fails if it isn't true. – Jack Schmidt Jun 3 '14 at 2:57
• Maybe you are right. I will think about it. I don't see a reason offhand. – Geoff Robinson Jun 3 '14 at 6:56
• Figured it out: if $\chi$ is real valued, then $m_\mathbb{Q}(\chi) \chi$ is the character of a real rep, but $\chi + \bar \chi$ is always the character of the real rep that is the tensor product of the original rep with the $\mathbb{R}$-vector space $\mathbb{C}$. – Jack Schmidt Jun 3 '14 at 15:14
• I.e., the complexification, right? – kcrisman Jun 3 '14 at 15:26
• @kcrisman: it is probably ok to think of it that way. However, the complexification is tensoring with ${}_{\mathbb{R}} \mathbb{C}_{\mathbb{C}}$ while I am tensoring with ${}_{\mathbb{C}} \mathbb{C}_{\mathbb{R}}$. – Jack Schmidt Jun 3 '14 at 15:36