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Warning I am a physicist and I am not familiar with a lot of the machinary of representation theory.

I consider the regular representation of $\mathbb S_n$ over reals $\mathbb R$ ($\mathbb R \mathbb S_n$). I see that for the proof of Schur's lemma about the isormphisms $\phi:V \to V $ being the identity map times a scalar, when $V$ is an irreducible, one needs alebraically closed field.

$\mathbb R$ is not such a field. Is this lemma still true in this particular case?

I would be grateful if anyone can answer without too much sophisticated maths: If possible of course.

Kind regards.

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    $\begingroup$ The distinction between irreducible and absolutely irreducible, as in David Hill's answer, is also discussed on Wikipedia (Schur's lemma). In this case, absolutely irreducible means irreducible even after complexification (extension of scalars from real to complex numbers). $\endgroup$ – Igor Khavkine Oct 22 '14 at 11:40
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Irreducible representations of $S_n$ are absolutely irreducible, meaning that they remain irreducible after extension of scalars. Therefore, if $V$ is irreducible as an $\mathbb{R} S_n$-module and $\phi:V\longrightarrow V$ is any nonzero homomorphism, then $$\phi\otimes 1:V\otimes\mathbb{C}\longrightarrow V\otimes\mathbb{C}$$ is nonzero and $V\otimes \mathbb{C}$ is irreducible. It follows that $\phi\otimes1$ is an isomorphism, hence $\phi$ is an isomorphism.

To read more, I suggest Gordon James' book, or the James-Kerber text.

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  • $\begingroup$ Is there a way to prove absolute irreducibility without having to find all irreps? $\endgroup$ – Mariano Suárez-Álvarez Oct 22 '14 at 17:19
  • $\begingroup$ @MarianoSuárez-Alvarez I think the answer to your question is no, but that is because there is a very nice way of describing the irreducible complex reptns of $S_n$ in terms of Young diagrams, and it is a by-product of this theory that yields the result that they can all be written over the integers. $\endgroup$ – Derek Holt Oct 22 '14 at 21:29

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