Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 19F (Representation Theory)

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.

For a representation space $V$, let ${\chi }_S$ and ${\chi }_A$ be the characters on ${\mathrm{Sym}}^{2}V$ and ${\bigwedge }^{2}V$. If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, the respective traces are ${\sum }_{i\le j}{\lambda }_i{\lambda }_j$ and ${\sum }_{i<j}{\lambda }_i{\lambda }_j$. Since $\chi (g)=\sum {\lambda }_i$ and $\chi (g^2)=\sum {\lambda }_i^2$, this givesFinite-order matrices over $\mathbb{C}$ are diagonalizable, so this eigenvalue argument applies to every group element. Equivalently the symmetric and antisymmetric projections on $V\otimes V$ are $(I\pm \text{swap})/2$.

Use the given square-class assignments and ${\omega }^3=1$, $1+\omega +{\omega }^2=0$. The character values, in the specified class order, areFor example the fourth entry of ${\chi }_S$ is $((-{\omega }^2{)}^2-\omega )/2=0$, and that of ${\chi }_A$ is $\omega$. A class of centralizer order $c$ has size $24/c$, so a character norm is the sum of its squared moduli divided by these centralizer orders. ThereforeBoth are genuine characters by their symmetric/exterior-square construction. A character norm is the sum of squares of its irreducible multiplicities, so both characters are irreducible.