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:](https://mathoverflow.net/a/412397/165920)


$$\widehat{t\_{x}}(\rho) := \mathbf{1}\_{d\_{\rho}} \exp( \frac{1}{d\_{\rho}} \sum\_{s \in S}\chi\_{\rho}(s) x\_s )$$


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.



Is $\sum_{\rho \text{ irred. }} \deg(\rho) \chi_{\rho}(g)=0$ for every group element $1 \neq g \in G$ of the finite group $G$?
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
$$
\mathrm{tr}(L\_g)=\sum\_{y\in G}\langle L\_g\delta\_y,\delta\_y\rangle=\sum\_{y\in G}\langle \delta\_{gy},\delta\_y\rangle=0.
$$


