Is $\sum_{\rho \text{ irred. }} \deg(\rho) \chi_{\rho}(g)=0$ for every group element $1 \neq g \in G$ of the finite group $G$?

Question

Is $\sum_{\rho \text{ irred. }} \deg(\rho) \chi_{\rho}(g)=0$ for every froup element $1\neq g \in G$ of the finite group $G$?

I have searched for but not found a proof to this. Probably it is not so difficult, but has as application that:

$$\det(T_G) = 1$$

where $T_G = (t_{gh^{-1}})_{g,h \in G}$ is the group matrix defined for the functions defined in this answer:

$$\widehat{t_{x}}(\rho) := \mathbf{1}_{d_{\rho}} \exp( \frac{1}{d_{\rho}} \sum_{s \in S}\chi_{\rho}(s) x_s )$$

where $S$ (with $1 \notin S$) generates the finite group $\left< S \right > = G$.

From this we get, since we know by Frobenius, the factorization of the group determinant :

$$\det(T_G) = \prod_{\rho \text{ irred.}} \det( \sum_{g \in G} t_g(x) \rho(g) )^{d_{\rho}} = \prod_{\rho \text{ irred.}} \det( \widehat{t_x}(\rho))^{d_{\rho}} = \prod_{\rho \text{ irred.}} \det( \mathbf{1}_{\rho} \exp \left ( \frac{1}{d_{\rho}} \sum_{s \in S} \chi_{\rho}(s) x_s \right ) )^{d_{\rho}} $$ $$= \prod_{\rho \text{ irred. }} \exp( \sum_{s \in S} \chi_{\rho}(s) x_s)^{\deg(\rho)}$$ $$ =\exp\left( \sum_{\rho \text{ irred.}} \deg(\rho) \sum_{s\in S} \chi_{\rho}(s) x_s \right)$$

and which is equal to:

$$=\exp(\sum_{s \in S} x_s \cdot \left ( \sum_{\rho} \deg(\rho) \chi_{\rho}(s) \right ) ) =^? \exp(0)=1$$

So if the question answers positive, then the determinant should be equal to $1$ .

Thanks for your help.

Answer 1

The assertion holds with deg$(\rho)^2$ replaced with deg$(\rho)$. You also need the extra requirement that the group element $g$ is non-trivial. Then the Plancherel Theorem implies that the right hand side equals the trace of the left translation $L_g$ on $\ell^2(G)$. Then you compute this trace as $$ \mathrm{tr}(L_g)=\sum_{y\in G}\langle L_g\delta_y,\delta_y\rangle=\sum_{y\in G}\langle \delta_{gy},\delta_y\rangle=0. $$