There are three elements in a $C_3$ group and there are three classes in this group cause it is an Abelian group.

\begin{array}{c|ccc|cc} \hline C_3 & E & C_3 & C_3^2 & & \varepsilon = \exp(2\pi\mathrm{i}/3) \\ \hline \mathrm{A} & 1 & 1 & 1 & z, R_z & x^2+y^2, z^2 \\ \mathrm{E} & \left\{ \begin{aligned}1 \\ 1\end{aligned} \right. & \begin{aligned}\varepsilon \\ \varepsilon^* \end{aligned} & \left. \begin{aligned}\varepsilon^* \\ \varepsilon \end{aligned}\right\} & (x, y), (R_x, R_y) & (x^2-y^2,xy), (xz, yz) \\ \hline \end{array}

According to group representation theory, the number of classes of a group should be the same as the number of irreducible representations of the group, which means there should be 3 irreducible representations for $C_3$ group.

In a character table, this means there should be 3 rows.

Further more, according to $$\sum_{i}l_{j}^2= h$$ in which $l_{j}$ stands for the dimension of the $j-th$ irreducible representation, and $h$ stands for the order of the group.

All of the three irreducible representation of $C_3$ group should be of 1 dimension.

But in the link above, for $C_3$ group, there are only two irreducible representations, and one of the irreducible representation is of 2 dimension, so it seems it violates the basic results of group representation theory.

So, what I have misunderstood?


1 Answer 1


You got everything right. It is just that two of these representations would use complex numbers ($\varepsilon=e^{2\pi i\over3}$ and $\varepsilon^2$), and we don't like that, so we stick them together into one representation of dimension 2, which makes the imaginary part vanish. Afterwards we treat this representation as "kinda irreducible". In fact it can be reduced, but we just don't want to go that way.

The same applies to other cyclic groups.

  • $\begingroup$ Thank you! So actually the representation labeled $E$ in the character table is actually a reducible representation? I observe that the inner product $1+\epsilon \epsilon^* +\epsilon^*\epsilon =3 $. $\endgroup$ Commented Nov 2, 2017 at 12:23
  • $\begingroup$ Well, let's put it this way: it is reducible if you allow complex numbers, and irreducible otherwise. $\endgroup$ Commented Nov 2, 2017 at 12:26

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