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.
Field of definition of representations of symmetric groups
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representation-theory
symmetric-groups
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1There 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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1@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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0@Alex: ha. Didn't notice that... – 2011-05-15
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2There'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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1(PS: the theorem can be found in ch. 4 of Isaacs' book "Character Theory of Finite Groups".) – 2011-05-15
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1@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