Let $\chi_1$ be the trivial character.

Let $\chi$ be a non-trivial character such that for all $g$, $\chi(g)$ is real, and $\chi(g)\geq 0$.

Show that $\langle\chi,\chi_1\rangle$ is strictly positive, and conclude that $\chi$ is reducible.

My partial solution:

$$\langle\chi, \chi_1\rangle = \frac{1}{|G|}\sum_{g\in G} \chi(g) \overline{\chi_1(g)}$$ $$= \frac{1}{|G|} \sum_{g \in G} \chi(g)$$ $\chi(1) \geq 1$ and $\chi(g) \geq 0 \quad \forall g \in G$ so $$\chi(g) \gt 0 \quad \forall g \in G$$ so $$\langle\chi, \chi_1\rangle \gt 0$$ How do I deduce that $\chi$ is reducible?

  • $\begingroup$ Since $\chi_1$ is irreducible, $\langle \chi, \chi_1 \rangle$ is the number of copies of $\chi_1$ in $\chi$. In general, you can determine whether an irrep is a summand of a rep by looking at the inner product of their characters and noting whether it is zero or not. $\endgroup$
    – Aaron
    Nov 11, 2016 at 15:30

1 Answer 1


This is a corrolary of the Schur Orthogonality Relations. Assume $\chi$ was reducible. Then either $\chi$ is nontrivial, or $\langle \chi, \chi_1\rangle =0$. Since this is not the case, $\chi=\chi_1$, a contridiction.


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