I am trying to show that a representation of the symmetric group $S_3$ is irreducible.
I am viewing $S_3$ as the symmetry group of the regular triangle in $\mathbb{C}$ with vertices $\mu_3=\{1,\omega, \omega^2\}$, generated by rotation through $\frac{2\pi}{3}$ and reflection $(x,y)\mapsto (x,-y)$. Then $a$ is the matrix \begin{bmatrix}\cos\frac{2\pi}{3}&-\sin\frac{2\pi}{3}\\\sin\frac{2\pi}{3}&\cos\frac{2\pi}{3}\end{bmatrix} and b is \begin{bmatrix}1&0\\0&-1\end{bmatrix}
Then by this construction, $S_3=\langle a,b \rangle$ is a matrix group in GL$(2,\mathbb{C})$, so it has a given representation on $V=\mathbb{C}^2$.
I am reading from lecture notes, and have stumbled onto something that is confusing me. In these, the argument that this representation is irreducible goes like this:
I claim that this representation is irreducible over $\mathbb{C}$. For a non-trivial $\mathbb{C}G$-submodule $U \subset V$ must be 1-dimensional, so a simultaneous eigenspace of $a$ and $b$. But the eigenspaces of $b$ are the $x$ and $y$ axes.
So far this makes sense to me: the eigenvalues of $b$ are $1,-1$, so it follows that the eigenspaces of $b$ are so. The argument continues:
But the eigenspaces of $a$ are $\mathbb{C} \cdot (1,\omega)$ and $\mathbb{C} \cdot (1,\omega^2)$.
I am stuck here. The eigenvalues of $a$ are $\omega$ and $\omega^2$, so the corresponding eigenvectors are $(1,i)$ and $(1,-i)$. Thus, the eigenspaces of $a$ are $\mathbb{C} \cdot (1,i)$ and $\mathbb{C} \cdot (1,-i)$. Have I misunderstood how to calculate eigenspaces (the space of $0$ union with multiples of the eigenvector)? Or is there a mistake in the notes I am reading from?
The argument then finishes nicely:
As these eigenspaces of $a$ and $b$ do not coincide, there cannot be a 1-dimensional representation.
Thus, the given representation is irreducible.
Any insight on my confusion surrounding the issue of eigenspaces would be appreciated!
You are right. It must be a typo.
We have more generally for any multiple of a rotation matrix : $$\begin{pmatrix}a & -b \\ b & a\end{pmatrix} \cdot \begin{pmatrix}1 \\ \pm i\end{pmatrix} = \begin{pmatrix}a\mp bi \\ b\pm ai\end{pmatrix} = (a \mp bi)\begin{pmatrix}1\\ \pm i\end{pmatrix}$$ so the eigenvectors must be $\begin{pmatrix}1\\ \pm i\end{pmatrix}$.
Of course, this doesn't change the rest of the argument, since these eigenvectors are still different than the axes.