= Solution
All three displayed <transposition> <matrices> have <trace> zero. Multiplying two of them gives
$$
R(\tau_1\tau_2)=\operatorname{diag}(\alpha,\alpha^{-1}),\qquad
R(\tau_2\tau_1)=\operatorname{diag}(\alpha^{-1},\alpha),
$$
since $\alpha^3=1$. Their <traces> are $\alpha+\alpha^{-1}=-1$, whereas $\operatorname{tr}R(e)=2$. Therefore
$$
\boxed{\chi_R=(2,0,-1),\qquad
\langle\chi_R,\chi_R\rangle=\frac{4+3\cdot0+2\cdot1}{6}=1.}
$$
By <character orthogonality>, $R$ is the <irreducible representation> $\mathbf2$ from <1/c>.
There is also a direct <invariant subspace> argument. The two distinct <eigenvalues> of $R(\tau_1\tau_2)$ force any invariant line to be one of the two coordinate lines. But $R(\tau_3)$ exchanges those lines. Neither is invariant under the whole <group>, so this two-dimensional <representation> is irreducible.
Back to article page
