= Solution
Write $\mathbf1$ for the <trivial representation>, $\varepsilon$ for the <sign representation>, and $\mathbf2$ for the given two-dimensional <group representation>. Their <matrix representations> on the <generators of a group> are
$$
\begin{array}{c|cc}
&r&s\\\hline
\mathbf1&1&1\\
\varepsilon&1&-1\\
\mathbf2&\operatorname{diag}(\alpha,\alpha^{-1})&\begin{pmatrix}0&1\\1&0\end{pmatrix}
\end{array}
$$
with $\alpha=e^{2\pi i/3}$. The two <eigenvalues> of $r$ in $\mathbf2$ are distinct, and $s$ exchanges its two <eigenlines>; no line is invariant under both. Thus $\mathbf2$ is an <irreducible representation>. The dimensions already exhaust the <sum of squares of irreducible representation dimensions>, so this is the complete list.
Since $\alpha+\alpha^{-1}=-1$, the <character table> is
$$
\boxed{
\begin{array}{c|rrr}
\text{class}&I&\{r,r^2\}&\{s,rs,r^2s\}\\
\text{class size}&1&2&3\\\hline
\chi_{\mathbf1}&1&1&1\\
\chi_{\varepsilon}&1&1&-1\\
\chi_{\mathbf2}&2&-1&0
\end{array}.}
$$
Multiplying <characters> and using <character orthogonality> gives every <tensor product of group representations>:
$$
\boxed{
\begin{array}{c|ccc}
\otimes&\mathbf1&\varepsilon&\mathbf2\\\hline
\mathbf1&\mathbf1&\varepsilon&\mathbf2\\
\varepsilon&\varepsilon&\mathbf1&\mathbf2\\
\mathbf2&\mathbf2&\mathbf2&\mathbf1\oplus\varepsilon\oplus\mathbf2
\end{array}.}
$$
For example $\chi_{\mathbf2}^2=(4,1,0)=\chi_{\mathbf1}+\chi_\varepsilon+\chi_{\mathbf2}$. Explicit <Clebsch-Gordan coefficients> can be read from two doublets $x=(x_1,x_2)^T$, $y=(y_1,y_2)^T$:
$$
\mathbf1:\ x_1y_2+x_2y_1,\qquad
\varepsilon:\ x_1y_2-x_2y_1,\qquad
\mathbf2:\ \begin{pmatrix}x_2y_2\\x_1y_1\end{pmatrix}.
$$
The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet <matrices>.
Back to article page
