Invariant subspaces to the permuting action of the alternating group. This question arose while researching Galois groups of random reciprocal polynomials. Throughout $A_n$ is the alternating group.
Let us define the following group action of $A_n$ on the vector space  $\mathbb{C}^n$
$$\sigma \circlearrowright (x_1,\dots,x_n)=(x_{\sigma^{-1}(1)},\dots,x_{\sigma^{-1}(n)})$$
My interest is to classify all the subspaces invariant to this action. My idea was to classify them in terms of linear systems. the invariant spaces I found are the following ones:

*

*$\{0\}\subseteq \mathbb{C}^n$ which corresponds to $x_1 = \dots = x_n = 0$,

*$V_1=\operatorname{span}\{(1,\dots, 1)\}\subseteq \mathbb{C}^n$ which corresponds to $x_1 = \dots = x_n$,

*$V_2\subseteq \mathbb{C}^n$ which corresponds to $\sum_{i=1}^nx_i=0$,

*$\mathbb{C}^n \space$, the whole space.

In case we extend the action to the whole symmetric group $S_n$ it can be shown that those truly are all the invariant subspaces, however, for $A_n$ I suspect I missed a few.
Any Help would be much appreciated!
 A: If $n=3$ then there might be more subspaces, for example the span of $(1,\omega,\omega^2)$ where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$.
If $n>3$ then the subspaces you listed are indeed all the subspaces. Let $e_1,...,e_n$ be the standard basis vectors of $\mathbb{C^n}$. We can define an action of $A_n$ on $\{e_1,...,e_n\}$ by $\sigma.e_i=e_{\sigma(i)}$. Just like any other action, it induces a representation of $A_n$. Formally, the representation is defined as $\pi:A_n\to GL(\mathbb{C^n})$ where $\pi(\sigma)(a_1e_1+...+a_ne_n)=a_1e_{\sigma(1)}+...+a_ne_{\sigma(n)}$. So we are actually looking for the invariant subspaces of this representation.
We can obviously define the invariant subspaces $V_1$ and $V_2$ like you did in the question. The subspace $V_1$ is one dimensional, thus irreducible. Also, since $n>3$ the action of $A_n$ on $\{e_1,...,e_n\}$ is $2$-transitive. (can you prove it?). This implies that if $\chi$ is the character of $\pi$ and $\chi_1$ is the character of the trivial representation then $\chi-\chi_1$ is an irreducible character. Since $\mathbb{C^n}=V_1\oplus V_2$ it follows that $\chi-\chi_1$ is exactly the character of $V_2$, so $V_2$ is also irreducible.
Now the result follows. Take a non-trivial invariant subspace $U$. It has an invariant complement $W$, which also has to be non-trivial. By Maschke's theorem you can decompose both into a direct sum of irreducible subspaces. But such a decomposition of $\mathbb{C^n}$ is unique, it is exactly $V_1\oplus V_2$. So we have no choice, $U$ must be either $V_1$ or $V_2$.
A: Suppose first that $n \geq 4$. You’re looking at the so-called permutation representation with respect to the standard action of $A_n$ on $n$ letters. Recall that such a representation always decomposes as a direct sum of the forms (2) and (3) (where by (2) and (3) I mean the invariant subspaces 2. and 3. you already give in your list). (3) is irreducible if and only if the action of the group is $2$-transitive (see theorem 7.2.11 in http://users.metu.edu.tr/sozkap/513-2013/Steinberg.pdf#page85). But I’m sure you can prove the action of $A_n$ is $2$-transtive for $n \geq 4$. So $\mathbb{C}^n$ decomposes as a direct sum of irreducible represenations
$$\mathbb{C}^n=V_1 \oplus V_2$$
Let $W \subset \mathbb{C}^n$ be an irreducible invariant subspace. Then by unicity of decomposition of irreducible represenations (in its strongest form), $W$ is equal to either $V_1$ or $V_2$, and we’re done. The cases $n=1$ and $n=2$ are trivial. If $n=3$, $\mathbb{C}^3$ decomposes as
$$\text{span}\{(1,1,1)\} \oplus \text{span}\{1, \zeta, \zeta^2\} \oplus \text{span}\{1,\zeta^2, \zeta\}$$
where $\zeta$ is a primitive third root of unity. One easily sees the representations occuring in the above decomposition are irreducible and inequivalent, so now any invariant subspace is the direct sum of any of the $2^3=8$ choices of subsets of
$$\{ \text{span}\{(1,1,1)\}, \text{span}\{1, \zeta, \zeta^2\} , \text{span}\{1,\zeta^2, \zeta\} \}. $$
