# Product of Characters

Let $\rho:G\rightarrow\text{GL}_n(\mathbb{C})$ be a representation of a finite group and let $\chi_\rho$ be the corresponding character. If $\chi(e)>1$, then I want to show that $\chi_\rho\cdot\overline{\chi_\rho}$ is not an irreducible character. (Here, $\overline{\chi_\rho}$ denotes the pointwise conjugate of $\chi_\rho$ and $\cdot$ denotes the pointwise product.) I see that $\chi_\rho\cdot\overline{\chi_\rho}$ is self conjugate so it must be a real valued function, though I can't see how to use this.

A character is irreducible iff the inner product is $1$, so I want to show that $\langle \chi_\rho\cdot\overline{\chi_\rho},\chi_\rho\cdot\overline{\chi_\rho}\rangle\neq1$. Then I have the following: $$\langle \chi_\rho\cdot\overline{\chi_\rho},\chi_\rho\cdot\overline{\chi_\rho}\rangle=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_\rho(g)\overline{\chi_\rho(g)}}\chi_\rho(g)\overline{\chi_\rho(g)}$$

$$=\frac{1}{|G|}\sum_{g\in G}\chi_\rho(g)\overline{\chi_\rho(g)}\chi_\rho(g)\overline{\chi_\rho(g)}=\frac{1}{|G|}\sum_{g\in G}|\chi_\rho(g)|^4$$ But I'm not sure how to continue from here. I tried considering the character as the sum of roots of unity: $$\chi_\rho(g)\overline{\chi_\rho(g)}=(w_1+\dots+w_n)(\frac{1}{w_1}+\dots+\frac{1}{w_n})$$ for $w_i$ an (order of $g$)th root of unity and $n=\chi_\rho(e)$. Then I can rewrite this as: $$n+\sum_{i<j}^n\frac{w_i}{w_j}+\frac{w_j}{w_i}=n+2\sum_{i<j}^n\text{Re}\left(\frac{w_i}{w_j}\right)$$ but I can't see any way to proceed.

• $\chi_\rho\cdot\bar{\chi}_\rho$ is the character of $\rho \otimes \bar{\rho}$, which is the product of two representations of dimension>1 and so it is reducible. – fqq Apr 19 '16 at 15:31
• Thank you. We haven't covered the tensor product of representations yet, so I think the problem should have a purely character theoretic solution. – Tachyon Apr 19 '16 at 15:55
• Try to show that $[\chi\overline{\chi},1_G]=[\chi,\chi]>0$ for any character $\chi$. Since $\chi\overline{\chi}$ has dimension greater than $1$, you are done. – Levent Apr 19 '16 at 19:04
• Thanks for your hint. I have attempted a solution below. – Tachyon Apr 19 '16 at 22:59

## 1 Answer

Following Levent's comment:

Let $\chi=\sum_i n_i\chi_i$ be an irreducible decomposition of $\chi$. Then $$(\chi,\chi)=\left(\sum n_i\chi_i,\sum n_j\chi_j\right)=\sum_{i,j}n_in_j(\chi_i,\chi_j)=\sum_{i,j}n_in_j\delta_{i,j}=\sum_i n_i^2>0$$

Then $(\chi\overline{\chi},1_g)>0$. If $(\chi\overline{\chi},1_g)>1$, then $\chi\overline{\chi}$ is reducible. If $(\chi\overline{\chi},1_g)=1$, then there must be some other irreducible character appearing in the decomposition since $\chi\overline{\chi}(e)=\chi(e)^2>1=1_g(e)$.