# Field of definition of representations of symmetric groups

Can one show in an elementary way, without recourse to Young tableaux etc., that the complex representations of symmetric groups are realisable over $\mathbb{R}$? It is easy to show that they are all self-dual, since the conjugacy classes of symmetric groups are self-dual, so one just has to exclude the possibility of quaternionic representations. Surely, there must be a similarly elementary argument? If there is, it is escaping me at the moment.

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There is a non-theorem which, if it were a theorem, would resolve this problem. It was observed at mathoverflow.net/questions/42646/… and mathoverflow.net/questions/53126/… that for a group with no complex representations, it often happens that the F-S indicator is a central character. Since the symmetric groups $S_n, n \ge 3$ have no center, the F-S indicator has to be $1$. Unfortunately, this is not always true, but frequently... – Qiaochu Yuan May 15 '11 at 4:07
@Qiaochu Thank you. It was me, who asked the first question you link to, so I was aware of this :-) In the particular case of symmetric groups though, I am still hoping for an argument that is true and ideally more elementary than Noah's. – Alex B. May 15 '11 at 4:15
There's a theorem (Frobenius-Schur) which, among other things, characterizes those real-valued irreducible characters that are not afforded by a real representation. Given a group $G$ and an element $g \in G$, denote by $f(g)$ the number of square roots of $g$ in $G$. Then $\langle f, \chi \rangle$ is $0, 1$ or $-1$ according as $\chi$ is not real-valued, real and afforded by a real representation, or real and not-afforded by one. Now $S_n$ certainly doesn't allow the first case to occur; perhaps there's a simple reason why the last case doesn't occur either? – Alon Amit May 15 '11 at 4:56
(PS: the theorem can be found in ch. 4 of Isaacs' book "Character Theory of Finite Groups".) – Alon Amit May 15 '11 at 4:57
@Alon If you can think of such a reason, please let me know - I haven't been able to get this to work. The Frobenius-Schur indicator is defined as $\sum_g\chi(g^2)$. Now, the map (of sets) $g\mapsto g^2$ is onto $A_n$ (that's quite easy to see). So the above sum is $\langle \text{Res}_{S_n/A_n}\chi,1\rangle$ plus further terms. The inner product is of course 1 if $\chi$ is a linear char. and 0 otherwise. But I haven't been able to express the remaining terms in any sensible way (some elements of $A_n$ get hit many times as $g$ ranges over $G$). – Alex B. May 15 '11 at 5:12

Some thoughts. Let $r_2(g)$ denote the number of square roots of $g$. Then $\langle r_2, \chi \rangle$ is the Frobenius-Schur indicator of $\chi$, and we want to show that this is equal to $1$ for all irreducibles $\chi$. This would follow if we could directly construct a representation of $S_n$ with character $r_2$, since then we would immediately have $\langle r_2, \chi \rangle \ge 0$.
The representation associated to this character would have dimension $r_2(e)$, or the number of involutions of $S_n$. And, indeed, there is a natural permutation representation of $S_n$ on the set of involutions (by conjugation), but it has the wrong character. In fact $r_2$ cannot in general be the character of a permutation representation: since it contains the trivial representation only once, it must be transitive, but its degree does not divide $n!$ in general.
But it still might be possible to construct this representation in a reasonably elementary way without going through the full construction of the irreducible representations of $S_n$.
Interesting thought. Maybe the first attempt should be to find any representation that contains all irreducible reps as summands, except for the obvious regular rep? Note also that the dimension of $r_2$ is not quite the number of involutions, but that number +1, since you have to count $e$ as its own square root, too. So the obvious permutation representation of the right dim has at least two orbits, one of them being $\{e\}$. But as soon as $n>3$, $S_n$ has involutions of different cycle types, so in fact there are then even more orbits, namely $\lfloor \frac{n}{2}\rfloor+1$ of them. – Alex B. May 15 '11 at 8:51