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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 http://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory and http://mathoverflow.net/questions/53126/finite-groups-in-which-every-character-has-real-values-grading-the-representatio 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...2011-05-15
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    @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.2011-05-15
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    @Alex: ha. Didn't notice that...2011-05-15
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    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?2011-05-15
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    (PS: the theorem can be found in ch. 4 of Isaacs' book "Character Theory of Finite Groups".)2011-05-15
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    @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 $G$).2011-05-15

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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 $S_n$.

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    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.2011-05-15
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    It looks like a lot of (non-regular) transitive permutation reps involve every character. The multiplicities in the rep on the cosets of Alt(3) (n≥5) might be easy to express combinatorially, but I don't see how to use this. If this was modular rep theory, it sounds to me like you are asking for an injective cogenerator or a projective progenerator. r2 is asking for a minimal one, and the resulting algebra is called the basic algebra. Looking for reps that just involve every (or most) characters using perm reps and Hecke algebras is called condensation.2011-05-15