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 ,


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.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.