Character of a representation and dimensione of $G$-invariant space

Question

Let $G$ be a finite group. Let $V$ be a finite dimensional vectorial space on an algebraically closed field and let $\rho: G \to Gl(V)$ an irreducible linear representation of $G$ in $V$.

Since $V$ is irreducible, so the same holds for the representation $\rho^*:G \to Gl(V^*)$. If we denote $\chi_{V}$ the character of $\rho$, we know that $\overline{\chi_V}$ is the character of $\rho^*$.

Since $V$ and $V^*$ are irreducible, we have $$(\overline{\chi_V},\chi_V)\in \{0,1\}$$ and with the first orthogonality relation $$dim(V^* \otimes V^*)^G=dim Hom_G(V^*,V)= (\overline{\chi_V},\chi_V).$$ At this point, in my notes I have the equality: $$(\overline{\chi_V},\chi_V)=(\chi_0,\chi_V^2)=(\chi_0,\chi_{V \otimes V})$$ where $\chi_0$ is the character of the trivial representation of $G$. I don’t understand how I can prove this equality, it’s clear to me that is since $dim(V^* \otimes V^*)^G=dim Hom_G(V^*,V)$, but I don’t why, formally, $dim(V^* \otimes V^*)^G=(\chi_0,\chi^2_V)$

Answer 1

It seems like the step that's giving you trouble is proving that $(\overline{\chi_V},\chi_V)=(\chi_0,\chi_V^2)$.

The calculation is:

\begin{align*} (\overline{\chi_V},\chi_V) &= \frac{1}{|G|}\sum_{g \in G} \overline{\chi_V}(g)\chi_V(g)^\star \\ &= \frac{1}{|G|}\sum_{g \in G} \chi_V(g)^\star\chi_V(g)^\star \\ &= \frac{1}{|G|}\sum_{g \in G} 1 \times\left(\chi_V(g)^2\right)^\star \\ &= \frac{1}{|G|}\sum_{g \in G} \chi_0(g)\left(\chi_V(g)^2\right)^\star \\ &= (\chi_0, \chi_V^2). \end{align*}

To spell it out:

• The first equality follows from the definition of the inner product on characters.
• The second equality follows from the fact that $\overline{\chi_V}(g) = \chi_V(g)^\star$ for all $g \in G$.
• The third equality is merely an algebraic rearrangement.
• The fourth equality follows from the fact that $\chi_0(g) = 1$ for all $g \in G$.
• The fifth equality, again, follows from the definition of the inner product on characters.