Orthogonality relations of Characters

Question

Could somebody please help me understand the jump from Proposition 10 to Proposition 11 in the following

http://www.ms.uky.edu/~pkoester/research/charactersums.pdf

Note: The orthogonality relations of Prop 11 are the wrong way around, and one only achieves zero if the characters are different.

Thank You

Answer 1

The orthogonality relations are spelled out in a little more detail in $\S 4.3$ of Appendix B of my number theory prebook (and in many other places, of course). The missing observation from the notes you cite seems to be that if $\chi_1 \neq \chi_2$, then $\chi_1 \chi_2^{-1} = \chi_1 \overline{\chi_2}$ is not the trivial character.

Answer 2

Proposition 10 tells you that you have $\sum_{g \in G} \chi (g) = 0$ for every character $\chi \in \widehat{G}$ except the principal character.

Then proposition 11 uses proposition 10 and the fact that the characters of $G$ form a group with respect to pointwise multiplication where the inverse of $\chi (g)$ is $\chi^{-1} (g) = \overline{\chi (g)} = \chi (-g)$. In particular, $\chi (g) \overline{\psi (g)} = \varphi(g)$ for some $\varphi \in \widehat{G}$. So by proposition 10 we have $\sum_{g \in G} \chi (g) \overline{\psi (g)} = \sum_{g \in G} \varphi (g) = 0$.

On the other hand, if $\psi = \chi$ we get $\sum_{g \in G} \chi (g) \overline{\chi (g)} = \sum_{g \in G} \chi (g) \chi (-g) = \sum_{g \in G} \chi (g - g) = \sum_{g \in G} \chi (0) = \sum_{g \in G} 1 = |G|$.

Edit(In response to comment)

In this case conjugation corresponds to taking inverses. To see this, note that $\chi : G \to S^1$ because $\chi (g)^{|G|} = \chi (|G|g) = \chi (0) = 1$ hence $\chi (g)$ are roots of unity for all $g$ in $G$ and all $\chi$ in $\widehat{G}$. In particular, $|\chi (g)| = 1 = \chi (g) \cdot \overline{\chi (g)}$ hence $\overline{\chi (g)} = \chi^{-1} (g)$. On the other hand, $1 = \chi (0) = \chi (g - g) = \chi (g) \chi (-g) = \chi (g) \chi^{-1} (g)$ hence $\overline{\chi (g)} = \chi^{-1} (g) = \chi (-g)$.

Hope this helps.

Answer 3

Yes, Pete's answer gives the key point. In proposition 11, it is intended that $\chi$ and $\phi$ be (degree 1) characters of your Abelian group $G.$ Then $\chi {\bar \phi}$ is also a character of $G.$ It's either the trivial character, or it isn't. If it's the trivial character, then the sum which appears in proposition 11 is clearly $|G|$. If it isn't the trivial character, then proposition 10 tells you that the sum is zero.