Factoring out Kloosterman Sum

Asked Feb 22 '19 at 08:03

Active Feb 22 '19 at 08:03

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I'm reading Iwaniec's book and he says Kloosterman sum factors into

$S(n,n;c)=S(n\bar{q},n\bar{q};r)T(n\bar{r},n\bar{r};q)$

where $n$ is square free and $c=rq$ such that $(q,n)=(q,r)=1$(i.e. $q$ is the largest factor of $c$ coprime to $n$) and $T$ is the Salie sum. I know that

$S(n,n;c)=S(n\bar{q},n\bar{q};r)S(n\bar{r},n\bar{r};q)$

since $(q,r)=1, c=qr$ but then his claim means $S(n\bar{r},n\bar{r};q)=T(n\bar{r},n\bar{r};q)$ which means $\displaystyle\left(\frac{d}{q}\right)=1$ for all $(d,q)=1, d<q$ and I don't quite see why this is true. Does it have something to do with $n$ being square free or is this generally true?

asked Feb 22 '19 at 08:03

J.Shim

Where does he say this? – Peter Humphries Feb 22 '19 at 11:41
Page 78 of Topics in classical automorphic forms while he is estimating Fourier coefficients of cusp forms. – J.Shim Feb 22 '19 at 18:38
Be careful. The Kloosterman sum $S(m,n;c)$ is not the usual Kloosterman sum $\sum_{d \in (\mathbb{Z}/c\mathbb{Z})^{\times}} e\left(\frac{md + n\overline{d}}{c}\right)$; there is a $\vartheta$-multiplier inserted. – Peter Humphries Feb 23 '19 at 13:45

Factoring out Kloosterman Sum

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