# Sum on product of two charcteres, which runs on symmetric generating set

Let $$G$$ be a finite (not necessarily abelian) group and let $$S$$ be a symmetric generating set of $$G$$, i.e. if $$s\in S$$ then $$s^{-1} \in S$$.

Let $$\chi$$ be an irreducible character of $$G$$. I have managed to show that $$\sum_{g_1,g_2,g_3,g_4 \in S}\chi(g_1 g_2g_3g_4) \geq 0$$.

How? Rewrite the LHS as $$\chi((\sum_{g\in S}g)^4)$$, and use the star operator (which takes $$\sum_{i}a_ig_i$$ to $$\sum_{i}a_i^*g_i^{-1}$$), which preserves $$\sum_{g\in S}g$$ (since $$S$$ is a symmetric generating set). Then, conclude that the LHS is the same as $$\chi(((\sum_{g\in S}g)(\sum_{g\in S}g)^*)^2)$$. Finally, if $$A$$ is the corresponding matrix to $$\sum_{g\in S}g$$, the above is just $$Tr((AA^*)^2)$$, which is clearly non-negative.

I would like to generalize this result to more than a single character. For example, I would like to show that $$\sum_{g_1,g_2,g_3,g_4,g_5,g_6\in S}\chi_1(g_1g_2g_3g_4)\cdot \chi_2(g_3g_4g_5g_6) \geq 0$$ for every two irreducible characters $$\chi_1,\chi_2$$.

What can we do to show that? I tried using the same technique but didn't manage to complete the calculation.

• If $A=\sum_{s\in S} s$ then $A^\star=A$ and you're talking about $\sum_{x\in SS^{-1}} {\rm tr}_1(Ax){\rm tr}_2(Ax)$. Is there a reason to expect this is nonnegative?
– anon
Commented Jun 8, 2022 at 19:14
• @runway44, $AA*$ is a matrix with non-negative real eigenvalues, hence its trace is non-negative. Commented Jun 9, 2022 at 6:06
• Yes, I'm asking why that would mean the sum in my comment (the more complicated sum your question is about) should be nonnegative.
– anon
Commented Jun 9, 2022 at 6:34
• @runway44 I proposed an explanation below, what do you think? Commented Jun 9, 2022 at 10:23

Rewrite the sum as $$\sum_{g_3,g_4 \in S}\chi_1((\sum_{s\in S}s)^2 g_3g_4)\cdot \chi_2((\sum_{s\in S}s)^2 g_3g_4)$$. Then remember that $$\chi =\chi_1 \cdot \chi_2$$ is a character of the tensor product of the correspoding representations for $$\chi_1,\chi_2$$, hence we can rewrite $$\sum_{g_3,g_4 \in S}\chi((\sum_{s\in S}s)^2 g_3g_4) = \chi((\sum_{s\in S}s)^4)$$ and this is non-negative due to the same argument in the single character case. Can you find any incorrect argument here?