# Reference Request: Irreducibles of the regular representation of the permutation group is absolutely irreducible

I am writing a paper(physics) where I am using the fact that the irreducible's of the regular representations of the permutation group are absolutely irreducible in the following sense.

If $V$ is an irreducible of the regular representation of $S_n$ over $R$ (real numbers) then it remain irreducible under the extension of the field to complex numbers $C$.

In particular,I only need the fact that they remain irreducible when we go from $R$ to $C$.

Q1) Firstly , is this TRUE?

Q2) Can anyone please provide a reference where this is explicitly mentioned? (if possible also where in the text it appears)

• Maybe he means an infinite permutation group (of which there are many possibilities: all permutations on an infinite set, or those moving only finitely many, ...)? Nov 6, 2015 at 14:06
• What I meant was, if I take the regular representation of $S_n$ it can be decomposed into irreducible corresponding to Young diagrams. My question is whether they remain irreducible under field extensions, in particular $R \to C$. See this related question and answer (I need a reference for this.)mathoverflow.net/questions/185072/… Nov 6, 2015 at 14:10

Actually, all of these representations are defined over $\mathbb{Q}$ (i.e., they are absolutely indecomposable). This is stated in the second sentence of the Encyclopedia of Math article on Representation of the symmetric groups. This follows immediately from the fact that the Young symmetrizer is defined over $\mathbb{Q}$ (and thus also $\mathbb{R}$).
If you have a simple representation $V$ of an arbitrary group over the reals, and you extend it to the complexes, it might not stay irreducible. But if that is the case, then the smaller representations can't be defined over the reals. If we have an isomorphism $\mathbb{C}\otimes_{\mathbb{R}}V\cong V_1\oplus V_2$, then we must have that $V$ is isomorphic to $V_1$, thought of as a real vector space of twice the dimension. If $V_1$ had a real form, then thinking of it as a real vector space would give two copies of the same real representation, not an irreducible.
• It means that for any field $K$ of characteristic 0 a simple $K\mathfrak{S}_n$-module is of the form $M \otimes_{\mathbb{Q}} K$ where $M$ is a simple $\mathbb{Q}\mathfrak{S}_n$-module. Moreover the map $M \mapsto M\otimes_{\mathbb{Q}} K$ is a bijection between the simple modules. From this one easily gets that for any field extension $K \subseteq L$, with $K$ still of characteristic 0, we have the map $M \mapsto M\otimes_K L$ is a bijection between simple $K\mathfrak{S}_n$ and $L\mathfrak{S}_n$ modules. Nov 6, 2015 at 16:18
• @phys In even more simple terms: yes, the irreducible representations of the symmetric group over $\mathbb{R}$ remain irreducible when you extend them to $\mathbb{C}$. Nov 7, 2015 at 0:37