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In Modular Forms by Toshitsune Miyake, at page 80, I am stuck at lemma $3.1.1$ Assume $W $ is a Gauss sum.

Lemma 3.1.1. Let $\chi$ be a primitive Dirichlet character mod $m$.
(1) $\sum_{a=0}^{m-1} \chi(a) e^{2π\iota ab/m} =\overline{ \chi(b)} W(x)$ for any integer $b$.
(2) $W(\chi) W(\bar{\chi})=\chi(-1)m.$
(3) $W(\chi)=\chi(-1) W(\bar{\chi})$.

While looking at the proof of (2) from (1)

$$\begin{align}W(\chi) W(\bar{\chi})&=\sum_{b=0}^{m-1} W(\chi)\overline{\chi(b)} e^{2π\iota b/m}\\&=\sum_{b=0}^{m-1} \sum_{a=0}^{m-1} {\chi(a)} e^{2π\iota ab/m} e^{2π\iota b/m}\\&=\sum_{a=0}^{m-1} {\chi(a)} \sum_{b=0}^{m-1} e^{2π\iota b(a+1)/m}\end{align}$$

Now if $a+1$ is divisible by $m$ the inner sum is $m$ and we get $W(\chi) W(\bar{\chi})=\chi(-1)m.$

My question is how we got the desired formula, this I didn't understand. Because all I can see is $$W(\chi) W(\bar{\chi})=\sum_{a=0}^{m-1} {\chi(a)}m,$$ from which I can't proceed.

From this (2), we can prove (3) easily.

Anne Bauval
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Ricci Ten
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1 Answers1

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If $a=m-1$ then $\chi(a)=\chi(-1)$, and $a+1$ is divisible by $m$ (even equal to it), so the inner sum is $m$.

If $0\le a<m-1$, the inner sum is $0$.

Anne Bauval
  • 49,005