Characters add on direct sums, multiply on tensor products, and conjugate on duals

Writing the direct-sum action in the concatenated basis of [A1], its trace is the sum of the traces of the two diagonal blocks, so ${\chi }_{V\oplus W}(g)=\mathrm{tr}({\rho }_V(g)\oplus {\rho }_W(g))=\mathrm{tr}{\rho }_V(g)+\mathrm{tr}{\rho }_W(g)={\chi }_V(g)+{\chi }_W(g)$ by [F1]. This is claim 1.

By [F3] and [A3], the matrix of the tensor-product action in the basis $(v_i\otimes w_j)$ has entry ${\rho }_V(g{)}_{i{i}^{\mathrm{\prime }}}{\rho }_W(g{)}_{j{j}^{\mathrm{\prime }}}$ at the position $(i,j)\mapsto (i^{\mathrm{\prime }},j^{\mathrm{\prime }})$, because $g\cdot (v_{i^{\mathrm{\prime }}}\otimes w_{j^{\mathrm{\prime }}})={\sum }_{i,j}{\rho }_V(g{)}_{i{i}^{\mathrm{\prime }}}{v}_i\otimes {\rho }_W(g{)}_{j{j}^{\mathrm{\prime }}}{w}_j$.

With dual bases $(v_i)$ of $V$ and $(v_i^{\ast })$ of $V^{\ast }$, the matrix of the dual action of [F2] is the transpose of the matrix of ${\rho }_V(g^{-1})$: writing $g^{-1}\cdot v_j={\sum }_i{\rho }_V(g^{-1}{)}_{ij}{v}_i$ gives $(g\cdot v_i^{\ast })(v_j)=v_i^{\ast }(g^{-1}\cdot v_j)={\rho }_V(g^{-1}{)}_{ij}$.

Its trace is the sum of its diagonal entries: ${\sum }_{i,j}{\rho }_V(g{)}_{ii}{\rho }_W(g{)}_{jj}=({\sum }_i{\rho }_V(g{)}_{ii}\left)\right({\sum }_j{\rho }_W(g{)}_{jj})$. By [F1] the two factors are ${\chi }_V(g)$ and ${\chi }_W(g)$, so ${\chi }_{V\otimes W}(g)={\chi }_V(g){\chi }_W(g)$, which is claim 2.

A matrix and its transpose have the same diagonal entries, hence the same trace, so ${\chi }_{V^{\ast }}(g)=\mathrm{tr}{\rho }_V(g^{-1})={\chi }_V(g^{-1})$. By [A4], ${\chi }_V(g^{-1})=\overline{{\chi }_V(g)}$, which is claim 3.