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)

  • $\begingroup$ 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, ...)? $\endgroup$ Nov 6, 2015 at 14:06
  • $\begingroup$ 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/… $\endgroup$
    – vishmay
    Nov 6, 2015 at 14:10

1 Answer 1


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.

  • $\begingroup$ This is probably a stupid question. I am a physicist and not familiar with these. Does this mean that under the extension from R to C irreducibles remain irreducibles? How so? Sorry again for the possibly dumb question. $\endgroup$
    – vishmay
    Nov 6, 2015 at 15:19
  • $\begingroup$ 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. $\endgroup$
    – Jay Taylor
    Nov 6, 2015 at 16:18
  • $\begingroup$ @phys In more simple terms, it means that every simple module has a basis where the matrices of the action are rational (actually, you can make them integers), in particular, real. $\endgroup$
    – Ben Webster
    Nov 6, 2015 at 22:53
  • 6
    $\begingroup$ @phys In even more simple terms: yes, the irreducible representations of the symmetric group over $\mathbb{R}$ remain irreducible when you extend them to $\mathbb{C}$. $\endgroup$
    – Tom Church
    Nov 7, 2015 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy