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I try to understand a number theoretical identity used by Jan-Christoph Schlage-Puchta in this answer.

He defined the function

$$S(\alpha)=\sum_{n\leq N}\Lambda(n) e(n\alpha)$$

where $\Lambda(n)$ is the Mangoldt function and $e(x)$ the exponential $e(x)=e^{2\pi i x}$.

Assume $\alpha=\frac{p}{q}$ is rational, and $p:= p_1 p_2...p_n$ and $q:= q_1 ...p_m$ are coprime positive integers where $p_i$ and $q_j$ are primes such that every pair $p_i, q_j$ is pair wise different.

Why the following identity is true:

$$\sum_{n\leq N}\Lambda(n) e(n\alpha) = \sum_{(a,q)=1} e(\frac{ap}{q})\underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n)$$

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  • $\begingroup$ It should follow from grouping the sum over $n\leq N$ into residue classes mod $q,$ and for each residue class representative $a$ and just use the geometric series applied to the sum of the Fourier coefficients $\endgroup$
    – kodlu
    Commented Dec 22, 2020 at 23:18
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    $\begingroup$ so I guess I am wondering what exactly broke down in this approach (if you tried it)? otherwise worth trying it. $\endgroup$
    – kodlu
    Commented Dec 22, 2020 at 23:22
  • $\begingroup$ @kodlu: Thank you I tried to apply your approach but there is a problem. We grouping them as you suggested $\sum_{n\leq N}\Lambda(n) e(n\alpha) = \sum_{j=1}^q \sum_{l \in \mathbb{N}: \ j+l \cdot q \leq N}^{l_{j,N}} \Lambda(j+l \cdot q) \cdot e(\alpha(j+l \cdot q))$ $\endgroup$
    – user267839
    Commented Dec 23, 2020 at 0:11
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    $\begingroup$ Now $e(a+b)= e(a)a(b)$ and since $\alpha =p/q$, $e(\alpha \cdot q \cdot l) =1$, so $e(\alpha(j+l \cdot q))= e(\alpha \cdot j)$. Thus $$...= \sum_{j=1}^q e(\alpha \cdot j) \sum_{l \in \mathbb{N}: \ j+l \cdot q \leq N}^{l_{j,N}} \Lambda(j+l \cdot q)$$ $\endgroup$
    – user267839
    Commented Dec 23, 2020 at 0:13
  • $\begingroup$ That looks that we are almost done, but now how get rid of the $j \in {1,2,..., q}$ in the first sum with $(j,q) \neq 1$? $\endgroup$
    – user267839
    Commented Dec 23, 2020 at 0:13

1 Answer 1

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$$\sum_{n \leq N} e(n\alpha) \Lambda(n) =\left(\sum_{p \leq N} \log(p) \sum_{r=1}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)\right)$$

But, the right hand side is gives, $$\left(\sum_{(a,q)=1 ,a<q} e(a\frac{p}{q})\sum_{\substack{n \leq N \\ n \equiv a (\text{modulo q})}} \Lambda(n) \right)=\sum_{\substack{p \leq N \\ p \nmid q}} \log(p) \sum_{r=0}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)$$.

[ This is because $q_i \nmid n, \text{where} n \equiv a (\text{modulo q}), (a,q)=1$ ].

$$\chi(q)=\sum_{\substack q_i} \log(q_i) \sum_{r=0}^{b_i: {q_i}^{b_i} \leq N} e({q_i}^r\alpha)$$.

($q_i$ s are prime factors of $q$)

This, $\chi(q)$ obviously depends on the $N$ and isn't zero generally.

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  • $\begingroup$ could you elaborate a bit more extensive why the identity $$\left(\sum_{(a,q)=1 ,a<q} e(a\frac{p}{q})\sum_{\substack{n \leq N \\ n \equiv a (\text{modulo q})}} \Lambda(n) \right)=\sum_{\substack{p \leq N \\ p \nmid q}} \log(p) \sum_{r=0}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)$$. is true. I not understand. $\endgroup$
    – user267839
    Commented Dec 24, 2020 at 1:41
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    $\begingroup$ Because the Mangoldt function $\Lambda(n)=log(p),$ when $n=p^{a} \rightarrow \text{perfect power of some primes}$. Now, $n$ can't be any power of the primes $q_i$ as these divides $q$, but doesn't divide $a$ as $(a,q)=1$. For $n$ which isn't a perfect power of prime Mangoldt function is zero. $\endgroup$
    – Alapan Das
    Commented Dec 24, 2020 at 2:07

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