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For a finite group $G$, is the following true, where $\rho$ is a finite-dimensional complex unitary irreducible representation? $$\sum _{g \in G} \mathrm{Tr} (\rho(g)) \rho(g) = \frac{|G|}{n} \mathrm{id}_{\mathbb{C} ^n}$$ I would like a proof or a counterexample.

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2 Answers 2

up vote 2 down vote accepted

I think the correct version would be

$$\sum _{g \in G} {\chi(g)}^\ast \rho(g) = \frac{|G|}{n} \mathrm{id}_{\mathbb{C} ^n}$$

where $\chi(g) = \mathrm{Tr}(\rho(g))$ is the character of $\rho$ and $z^\ast$ denotes the complex conjugate of $z$.

Then the map $\phi = \sum_{g\in G} {\chi(g)}^\ast\, \rho(g)$ is $G$-invariant, so by Schur's lemma you get $\phi = \lambda\, \mathrm{id}_{\mathbb C^n}$, where $\lambda = \mathrm{Tr}(\phi)/n$. But now

$$\mathrm{Tr}(\phi) = \sum_{g\in G} \chi(g)^\ast \chi(g) = |G|\cdot \underbrace{ (\chi|\chi)}_{=1} = |G|$$

hence $\phi = \frac{|G|}{n}\mathrm{id}_{\mathbb C^n}$.

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A counterexample is given by $G=\mathbb Z_3$ and $\rho(n)=\mathrm e^{2\pi\mathrm in/3}$, in which case the left-hand side vanishes. Perhaps you intended to include a complex conjugation somewhere?

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