$$ \begin{array}{c|c|c|} \backslash & e & p \\ \hline e & e & p \\ \hline p & p & e \\ \hline \end{array} $$ (Original picture of the table here.)
First I want to look at the cyclic group $\mathbb{Z}_2$. In this group, the symmetric functions transform under the trivial representation ($D(e) = D(p) = 1$, so all group actions do nothing to the function), and the antisymmetric functions transform under the sign representation ($D(e) = 1, D(p) = -1$, so we need to act $D(p)$ twice to get the orignal function, acting once won't do). There are two irreducible representarion, and they all create their own subspace. All nice and dandy.
$$ \begin{array}{c|c|c|c|} \backslash & e & a & b \\ \hline e & e & a & b \\ \hline a & a & b & e \\ \hline b & b & e & a \\ \hline \end{array} $$ (Original picture of the table here.)
But what I don't understand is the group $\mathbb{Z}_3$, because it seems to me that they only create two irreducible representation also, whereas it should be three. Nothing wrong with the trivial representation subspace, but the two remaining representation:
- $D(a) = e^{2\pi i/3}$, $D(b) = e^{4\pi i/3}$
- $D(a) = e^{4\pi i/3}$, $D(b) = e^{2\pi i/3}$
seems indistinguishable. An "antisymmetric" function need $D(a)$ or $D(b)$ to act thrice to return to the original function, in both representation; and $D(a)D(b)$ or $D(b)D(a)$ also return the original function, in both representation.
So what did I misunderstand here? Are there only two irreducible representations? Thank you!
The point is that you are right if you are talking about real representations (indeed there are only two real irreducible representation of $\mathbb{Z}_3$. But the non trivial one is the $\frac{2 \cdot \pi}{3}$ rotation (which is isomorphic to the $\frac{4 \cdot \pi}{3}$ rotation) in $\mathbb{R}^2$, so its dimension is $2$).
But, since you are using complex numbers, I think you mean complex representations.
In this case the two representation (let me call them $\rho_1$ and $\rho_2$ with $\rho_i : \mathbb{Z}_3 \to GL_1(\mathbb{C}) \cong \mathbb{C}$ ) are not isomorphic because there is no $\mathbb{C}-$linear invertible function $f: \mathbb{C} \to \mathbb{C}$ (maybe your confusion is here, you seem to be looking at $\mathbb{R}-$linear functions) such that $f( \rho_1(a)z) = \rho_2(a)( f(z))$ for any $a \in \mathbb{Z}_3$ and $z \in \mathbb{C}$.
This fact is indeed true because since $f: \mathbb{C} \to \mathbb{C}$ is $\mathbb{C}-$linear and invertible, there exist $\lambda \neq 0$ such that $f(z) = \lambda z$ for all $z \in \mathbb{C}$. Let $\omega = e^{\frac{2 \pi i}{3}}$, then for $a = [1]$ we have $ \rho_1(a)(z) = \omega z$ and $\rho_2(a)(z) = \omega^2 z$. Now, the equation implies $\lambda \omega z = \lambda \omega^2 z$ for any $z \in \mathbb{C}$, which is absurd (because $\lambda \neq 0$).
Hope this is helpful.