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Ben Webster
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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.

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 $\mathbb{R}$).

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.

Source Link
Ben Webster
  • 44.7k
  • 12
  • 126
  • 260

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 $\mathbb{R}$).