# On a sum taken over all irreducible characters: is $\sum\limits_{i=1}^t\chi_{V_i}(g)^2$ nonzero for all $g\in G$?

Suppose $G$ is a finite group and $\mathbb{k}$ an algebraically closed field of characteristic zero. Denote the irreducible representations of $G$ over $\mathbb{k}$ by $V_1,\ldots, V_t$. Is it necessarily true that $\sum\limits_{i=1}^t\chi_{V_i}(g)^2\neq 0$ for $g\in G$?

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Try the group of order 3. –  Dane Oct 11 '12 at 21:57

## 2 Answers

It doesn't seem to be true for a generator of $C_3$ (and $\mathbb k=\mathbb C$).

Then $\chi_{V_i}(g)$ are the three cube roots of unity, and so are their squares.

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Wow, seems obvious now. Thanks –  Bey Oct 12 '12 at 0:48

You may be thinking of $\displaystyle\sum_{i=1}^t |\chi_{V_i}(g)|^2$, which is equal to $|C_G(g)|$ and thus always positive.

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You're right, thank you –  Bey Oct 12 '12 at 0:48