# What would non-zero character mean if i'm working on the complex field?

I'm solving the exercises of chapter 14 in the book Representations and Characters of groups. (Gordon James, Martin Liebeck) Always working with $\mathbb{R}$ or $\mathbb{C}$.

One of them says:

Suppose that $\chi$ is a non-zero, non trivial character of $G$, and that $\chi(g)$ is a non-negative real number for all $g$ in $G$. Prove that $\chi$ is reducible

At this stage of the book, I think that the natural procedure is to calculate $<\chi,\chi>$ and see that it is $\neq 1$

I know that non-trivial means $\exists g \in G$ such that $\chi(g)\neq 1$, but since $\chi(1_G)\in \mathbb{N}$, I don't know what they want to say with the non-zero condition.

Anyway, I've doing the next:

By definition, $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\chi(g^{-1})$

I know that there exist a base $\mathcal{B}$ where $[g]_{\mathcal{B}}$ is diagonal, then $\chi(g)$ is sum of $m$th roots of the unity (considering the $order(g)=m$) so $\chi(g^{-1})=\overline{\chi(g)}$

But $\chi(g)$ is a non-negative real number for all $g$ in $G$.

So we obtain $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)^2$

Since $\chi$ is non-trivial, $\exists g \in G$ such that $\chi(g)\neq 1$, but I'm stuck here. Maybe I'm missing something related to the non-zero condition.

Any advice would be welcome.

• Consider the inner product with the trivial character. – Qiaochu Yuan May 21 '18 at 1:37
• @QiaochuYuan Thanks for your hint. $<\chi,\chi_1>=\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\chi_1(g)=\dfrac{1}{|G|}\sum_{g\in G}\chi(g)$ And since $\chi(1_G)\geq 1$ and $\chi(g)\geq 0\ \forall g \in G$, then we have $<\chi,\chi_1> >0$. This means that the trivial character (which is irreducible) is a component of our character $\chi$. But $\chi$ is not trivial, so it must have something else as component. In conclusion, $\chi$ is reducible. – G.Jimenez May 21 '18 at 3:33