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


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

Thank You

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3 Answers 3

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.

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Thanks for your comment pete! –  rk101 Mar 22 '12 at 15:42

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.

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Thanks for helping Geoff! –  rk101 Mar 22 '12 at 15:43

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.

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Thanks Matt, but doesn't the bar above a character correspond to complex conjugation, rather than inverse? (Unless the character has exponential form) This is the bit I'm confused about. –  rk101 Mar 22 '12 at 14:17
@rk101 Yes, that's right, the bar corresponds to complex conjugation. In this case conjugation corresponds to taking inverses. I've added the explanation in the answer. Hope this helps. Otherwise don't hesitate to ask. –  Matt N. Mar 22 '12 at 14:55
Ah, I see! You've been a great help, Thanks! –  rk101 Mar 22 '12 at 15:41
@rk101 Glad I could help : ) –  Matt N. Mar 22 '12 at 15:54

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