Representation of $S_3$ without using character theory - Fulton and Harris

We have $W=\bigoplus_{i}\mathbb{C}v_i$ with $\tau.v_i=\lambda v_i$ ,$\lambda\in\{1,\omega,\omega^2\}$

• Let's consider first the $v_i$ such that $\tau.v_i=\omega v_i$.

We have $\tau\sigma v_i=\omega^2\sigma v_i$ so $v_i$ and $\sigma(v_i)$ are indempendant.

Let's consider the family $(\mathbb{C}\{v_i,\sigma(v_i)\})_{\tau v_i=\omega v_i}$.

We can verify that $\mathbb{C}\{v_i,\sigma v_i\}$ is irreducible. Let $\lambda v_i +\alpha \sigma.v_i$, we have $$ \sigma.(\lambda v_i+\alpha \sigma v_i)=\lambda \sigma v_i+\alpha v_i=\lambda v_i+\alpha\sigma v_i \quad \Rightarrow \lambda=\alpha $$ and $$ \tau.(\lambda v_i+\alpha \sigma v_i)=\lambda \tau v_i+\alpha \tau\sigma.v_i=\lambda \omega v_i+\alpha\omega^2\sigma v_i=\lambda v_i+\alpha \sigma v_i \quad \Rightarrow \lambda\omega=\lambda\quad \mbox{and} \quad \alpha\omega^2=\alpha $$ This latter line is impossible, so the trivial representation is not in $\mathbb{C}\{v_i,\sigma v_i\}$. The same idea shows us that the alternating representation too is not in. Thus, the family $(\mathbb{C}\{v_i,\sigma(v_i)\})_{\tau v_i=\omega v_i}$ is irreducible then they are isomorphic to the standard representation $V$.

• The $v_i$ such that $\tau.v_i=\omega^2v_i$ appear all in the previous consideration as a $\sigma.v^{'}$ of a $v^{'}$.
• Let's take $v_i$ such that $\tau.v_i=v_i$,

we have $\tau\sigma.v_i=\sigma.v_i$

1. If $\sigma.v_i$ and $v_i$ are independent, the family $(V_i)=(\mathbb{C}\{v_i,\sigma.v_i\})$ are two dimensional representation. We have that, $\mathbb{C}\{\sigma.v_i+v_i\}$ and $\mathbb{C}\{\sigma.v_i+v_i\}$ are submodule of each $V_i$. Then $V_i$ are reducible and we have $$ V_i=\mathbb{C}\{\sigma.v_i+v_i\}\oplus \mathbb{C}\{\sigma.v_i-v_i\} $$ These two modules are respectively isomorphic to the trivial and alternating representation. Their basis are respectively eigenvector of $1$ and $-1$ for $\sigma$. $$ V_i=U\oplus U^{'} $$
2. If $\sigma v_i$ and $v_i$ are independent : We must have $\sigma v_i=\lambda v_i$, so $\sigma^2 v_i=\lambda^2 v_i\Rightarrow \lambda^2=1\Rightarrow \lambda\in\{-1,1\}$. Then $\sigma.v_i=v_i$ or $\sigma.v_i=-v_i$. In the first case $\mathbb{C}\{v_i\}=U$ and equal to $U'$ for the other. They are all eigenvector of $1$ and $-1$ for $\sigma$.

We can guess that all eigenvector of $1$ for $\sigma$ lead to the trivial and all eigenvector of $-1$ to the alternating. But there's still some eigenvector of $1$ and $-1$ that doesn't fit this statement.

Lemma Let $v$ such that $\tau.v=\omega v$. $$ \forall u\in \mathbb{C}\{v,\sigma.v\} \quad \sigma.u=u\Rightarrow u\in \mathbb{C}\{v+\sigma.v\} $$ and $$ \forall u\in \mathbb{C}\{v,\sigma.v\} \quad \sigma.u=-u\Rightarrow u\in \mathbb{C}\{v-\sigma.v\} $$

We have seen that the vector basis $v$ of $W$ which are eigenvector of $\omega$ for $\tau$ give the standard representation in the decomposition. The number of indempendent vector basis $v$ of $V$ which verify $\sigma.v=v$ except those presented in the lemma give the trivial. The same thing happen for the alternating. Finally, if $$ W=U^{\oplus a}\oplus U{'\oplus b} \oplus V^{\oplus c} $$

$c$ is the number independent vector basis of $W$ such that $\tau.v=\omega w$. The number of the vector said in the lemma is equal to $c$. Then if the multiplicity of $1$ is equal to $p$ and of $-1$ is $q$ then $a=p-c$ and $b=q-c$.