Fix an integer vector $\mathbf m\in \mathbb Z^k$. Let $q$ be a positive integer.

Is there a "good" upper bound in terms of $q,\bf m$ for the exponential sum: \[\sum_{\mathbf n} e\left(\frac{\langle m, n\rangle}{q}\right).\] Here $\langle \cdot, \cdot \rangle$ is usual inner product and $e(z) = e^{2\pi i z}$. The sum is taken over all ${\mathbf n=(n_1,\cdots,n_k)}$ such that $1\le n_i\le q$ and $gcd(n_1,\cdots,n_k,q)=1$.

When $k=1$, this is the famous Ramanujan sum. I have tried searching relevant results, but could not find any useful reference for $k\ge 2$. Any answers or references are appreciated!

  • $\begingroup$ Don't you need to assume some additional coprimality condition between each $n_i$ and $p$? For $k = 1$, for example, the usual Ramanujan sum includes the condition $(n,p) = 1$, for otherwise the sum is zero. $\endgroup$ – Peter Humphries Feb 7 '17 at 0:54
  • $\begingroup$ Yes, you are right! I definitely forgot to add that. $\endgroup$ – Changguang Feb 7 '17 at 1:46
  • $\begingroup$ Doesn't $\sum e((m_1n_1+\cdots+m_kn_k)/p)$ just factor as $\sum e(m_1n_1/p)\times\cdots\times\sum e(m_kn_k/p)$? $\endgroup$ – Gerry Myerson Feb 7 '17 at 2:16
  • $\begingroup$ Yes, but the coprimality condition doesn't quite factor in a trivial way. $\endgroup$ – Peter Humphries Feb 7 '17 at 2:16

Call this sum $c_q(m_1,\ldots,m_k)$. The indicator function of the condition $(n_1,\ldots,n_k,q) = 1$ can be written as the sum $\sum_{cd = (n_1,\ldots,n_k,q)} \mu(c)$, so that $d \mid q$, $c = \frac{q}{d}$, and $n_j \equiv 0 \pmod{c}$ for all $j \in \{1,\ldots,k\}$. It follows that $c_q(m_1,\ldots,m_k)$ may be written as \[\sum_{cd = (n_1, \ldots, n_k,q)} \mu(c) \prod_{j = 1}^{k} \sum_{n_j = 1}^{q} e\left(\frac{n_j m_j}{q}\right)= \sum_{d \mid q} \mu\left(\frac{q}{d}\right) \prod_{j = 1}^{k} \sum_{\substack{n_j = 1 \\ n_j \equiv 0 \pmod{\frac{q}{d}}}}^{q} e\left(\frac{n_j m_j}{q}\right). \] Writing $n_j = \frac{q}{d} \ell_j$, the inner sum is \[\sum_{\ell_j = 1}^{d} e\left(\frac{\ell_j m_j}{d}\right) = \begin{cases} d & \text{if $d \mid m_j$,} \\ 0 & \text{otherwise.} \end{cases}\] So \[c_q(m_1,\ldots,m_k) = \sum_{d \mid (m_1, \ldots, m_k, q)} \mu\left(\frac{q}{d}\right) d^k.\] This is multiplicative as a function of $q$, so that \[c_q(m_1,\ldots,m_k) = \prod_{p^r \parallel q} c_{p^r}(m_1,\ldots,m_k).\] So we need only determine $c_q(m_1,\ldots,m_k)$ when $q = p^r$ is a prime power. Since \[\mu(p^r) = \begin{cases} 1 & \text{if $r = 0$,} \\ -1 & \text{if $r = 1$,} \\ 0 & \text{if $r \geq 2$,} \end{cases}\] we have that for $q = p$, \[c_p(m_1,\ldots,m_k) = \begin{cases} p^k - 1 & \text{if $p \mid (m_1,\ldots,m_k)$,} \\ -1 & \text{if $p \nmid (m_1,\ldots,m_k)$,} \end{cases}\] while for $q = p^r$ with $r \geq 2$, \[c_{p^r}(m_1,\ldots,m_k) = \begin{cases} p^{(r - 1)k} (p^k - 1) & \text{if $p^r \mid (m_1,\ldots,m_k)$,} \\ -p^{(r - 1)k} & \text{if $p^{r - 1} \parallel (m_1,\ldots,m_k)$,} \\ -1 & \text{if $p^{r - 1} \nmid (m_1,\ldots,m_k)$.} \end{cases}\]

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  • $\begingroup$ Peter, thank you very much for your answer. I think it's correct, but could you please double check your formulas? For example, line 4 with line 2; and last two formulas (not consistent). $\endgroup$ – Changguang Feb 7 '17 at 2:57
  • $\begingroup$ @Changguang, it should be fixed now. $\endgroup$ – Peter Humphries Feb 7 '17 at 3:19
  • $\begingroup$ For the last two formulas, when $r=1$, I think that should be $p^k-1$, no? $\endgroup$ – Changguang Feb 7 '17 at 3:21

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