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

Question

I am having trouble understanding parts of page 10 in Fulton and Harris.

We have thus accomplished the first two of the goals we have set for ourselves above in the case of the group $\def\S#1{\mathfrak{S}_{#1}}\S3$. First, we see from the above that *the only three irreducible representations of $\S3$ are the trivial, alternating, and standard representations $U$, $U'$ and $V$. Moreover, for an arbitrary representation $W$ on $\S3$ we can write $$ W = U^{\oplus a} \oplus U'^{\,\oplus b} \oplus V^{\oplus c}; $$ and we have a way to determine the multiplicities $a$, $b$, and $c$: $c$, for example, is the number of independent eigenvectors for $\tau$ with eigenvalue $\omega$, whereas $a+c$ is the multiplicity of $1$ as an eigenvalue of $\sigma$, and $b+c$ is the multiplicity of $-1$ as an eigenvalue of $\sigma$.

where $\tau = (123)$ and $\sigma = (12)$.

Here are some of my thoughts: I understand $c$ is the number of independent eigenvectors for $\tau$ with eigenvalue $\omega$ since in that case, $(v, \sigma(v))$ generates $V$. But I don't truly understand the rest of the argument. Is the mentioned "multiplicity" geometric or algebraic?

Answer 1

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 $b=q-c$.