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Let $A_n = \{\sigma \in S_n : \mathrm{sign}\;\sigma=1\}$ be alternating group.

Let $\rho: A_n \to \mathcal{L}(\mathbb{C}^n) $ stand for standart complex representation of $A_n$ defined by $$ \rho(\sigma)(e_i) = e_{\sigma(i)} $$ where $e_i$ is a standart basis of $\mathbb{C}^n$.

It is well known fact that simmilar representation for $S_n$ decomposes to sum of two irreducible ones for subspaces $$ U_1 = \mathbb{C}\sum^n_{i=1} e_i $$ $$ U_0 = \left\{ v \in \mathbb{C}^n \bigg| \sum^n_{i=1} v_i = 0 \right\} $$

But will $U_0$ still be irreducible for $A_n$?

Are there any invariant irreducible subspaces of $U_0$ which exist only for $A_n$?

Clearly $A_2$ is trivial and $A_3$ can't have any irreducible representations of degree 2 as $|A_3| = 3 < 4 = 2^2$. This indicates that there must be one-dimensional invariant subspace of $U_0$ But I don't know how to find equation for this subspace.

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  • $\begingroup$ One-dimensional subspaces are eigenspaces, so they at least come from linear algebra. $\endgroup$ – anon Mar 31 '18 at 15:19
  • $\begingroup$ I understand that $U_0$ always must be invariant, but I can't comprehend how it is irreducible for $A_3$ which should not hav irreducible representations of degree bigger than 1. $\endgroup$ – Nik Pronko Mar 31 '18 at 15:19
  • $\begingroup$ @anon I think I got what you meant $\endgroup$ – Nik Pronko Mar 31 '18 at 15:24
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    $\begingroup$ If $n > 3$ then $U_0$ is irreducible under the action of $A_n$. Note that $A_n$ acts $2$-transitively when $n>3$ and there is a general result that the permutation representation over ${\mathbb C}$ of any $2$-transitive group is the sum of the trivial representation and an irreducible module. $\endgroup$ – Derek Holt Mar 31 '18 at 15:35
  • $\begingroup$ @DerekHolt Thanks, this is useful. $\endgroup$ – Nik Pronko Mar 31 '18 at 15:49
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You are right, when $n=3$ there is an invariant subspace of $U_0$ of dimension $1$. It is generated by $(1,j,j^2)$.

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  • $\begingroup$ Yes I think it comes from the intersection of eigenspaces of $\rho(123)$ and $\rho(132)$ as @anon suggested. $\endgroup$ – Nik Pronko Mar 31 '18 at 15:26
  • $\begingroup$ I will add that $j = - \frac{1}{2} + \frac{\sqrt{3}i}{2}$ $\endgroup$ – Nik Pronko Mar 31 '18 at 17:30

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