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I am trying to understand the $3$-dimensional irreducible representation in the alternation group $A_4$.

$(\rho, \mathbb{C^4})$ is a $4$-dimensional representation (reducible) where for any element $g \in A_4$, $\rho(g)$ acts on a vector in $\mathbb{C}^4$ by permuting the basis elements. e.g. if $v\in \mathbb{C^4}$ is represented as $v = v_1 \alpha_1 + v_2 \alpha_2 + v_3 \alpha_3 + v_4 \alpha_4$ , for some basis $\mathcal{B}=\{ \alpha_1,\alpha_2,\alpha_3,\alpha_4 \}$ ,then for example $\rho((123))v= v_1 \alpha_3 + v_2 \alpha_1 + v_3 \alpha_2 + v_4 \alpha_4$. it is reducible, because it has a one dimensional invariant subspace, which is $$ W_1 = \langle \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \rangle \equiv \langle \alpha \rangle $$ and every $\rho(g)$ acts trivially on this one dimensional subspace, in other words, $(\rho_4,W_1)$ is the trivial representation of $A_4$. Now how/why can I find the three-dimensional complement $W$ of $W_1$ in $\mathbb{C}^4$, i.e. $$ \mathbb{C}^4=W_1 \bigoplus W $$ such that $W$ be also invariant under the action of $\rho_4(g)$, for all $g \in A_4$ ? If it is possible to find such $W$, what is its basis?

Thank you ,

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Let me write $V=\Bbb{C}^4$. $W$ is the space of vectors whose coordinates sum to zero. This is codimension 1 in $V$ so 3-dimensional, and its clearly invariant. It has trivial intersection with $W_1$ and so you're done. As for its basis... Just pick any three independent vectors!

As for the "why", note that $V/W_1$ is a 3-dimensional quotient, which is isomorphic to $W$. But by Maschke's theorem $V$ is semi simple and so every quotient arises as a subrepresentation.

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