Prove that a representation have a base and it's irreductible

43 Views Asked by At

I'm quite new in representations and I'm trying to do next problem: (It's supposed that I don't know anything about characters theory)

We want to study $S_3=(\tau=(123),\sigma=(1,2)\,|\, \sigma\tau=\tau^2\sigma)$ We have also $\omega=e^{{2i\pi/3}}$. I have to prove that we can find a representation $V$ with a base $(v,w)$ satifying $$\rho(\tau)v=\omega v\quad \rho(\sigma)v=w \quad \rho(\tau)w=\omega^2w\quad \rho(\sigma)w=v $$ and prove that $V$ is irreductible.

I have a solution of a classmate but I don't understand, it talks about eigenvalues of $\tau$. I will be very thankfull if anybody can explain this exercise.

1

There are 1 best solutions below

1
On BEST ANSWER

I'll give some strategy how one could prove this. There are two problems here, first proving that you can define such a representation, second proving it is irreducible.

For the first part: The easiest two-dimensional vector space I can think of is $\mathbb{C}^2$ with its standard basis, let's denote this basis as $(v,w)$.
By definition giving $\mathbb{C}^2$ the structure of a representation is the same as defining a group homomorphism $\rho:S_3\to \operatorname{GL}(\mathbb{C}^2)$. Now, $\rho(\tau)$ should be a linear map sending $v$ to $\omega v$ and $w$ to $\omega^2 w$. This linear map is represented by the matrix $\rho(\tau)=\begin{pmatrix}\omega&0\\0&\omega^2\end{pmatrix}$. Finding a similar matrix for $\sigma$ should not be that difficult. Since $\sigma$ and $\tau$ generate $S_3$, you can now define a map $S_3\to \operatorname{GL}(\mathbb{C}^2)$ by extending this map multiplicatively, e.g. $\rho(\sigma\tau)=\rho(\sigma)\rho(\tau)$.
You have to check this is well-defined, it could be that by this definition $\tau^3=1$, but $\rho(\tau)^3\neq 1$ if we had done a mistake. An easy way to check this is to have a presentation of your group by generators and relations, e.g. for $S_3$ you have $S_3=\langle \sigma,\tau | \sigma^2=1, \tau^3=1, \sigma \tau=\tau^2\sigma\rangle$. Then for checking this map is well-defined, you just have to prove $\rho(\sigma)^2=1$, $\rho(\tau)^3=1$ and $\rho(\sigma\tau)=\rho(\tau^2\sigma)$. This is defining the representation.

For checking it is irreducible, by definition you have to prove it has no proper non-zero subrepresentation. Equivalently, you can prove that every vector in the representation generates the whole space. So, take a non-zero vector $u=av+bw$ with $a,b\in \mathbb{C}$. You have to prove that the one-dimensional subspace spanned by $u$ cannot be invariant under the action of $S_3$. Applying $\rho(\sigma)$ to this $u$ gives $\rho(u)=bv+aw$. Thus, if you assume that $u$ spans a subrepresentation you get $\rho(u)=\lambda u$, hence $bv+aw=\lambda av+\lambda bw$. Using it is a basis, you get $b=\lambda a$ and $a=\lambda b$. As a next step you can try to get some more conditions applying $\rho(\tau)$ and combining them should give you the only possibility $a=b=0$, a contradiction.