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Let $\chi$ be a primitive Dirichlet character of conductor $q$. I want to compute numerically $$G(k)=\sum_{n\bmod q}\chi(n)e^{2\pi i n(n-k)/(2q)}$$ for all $k$ with $0\le k<2q$ with $k\equiv q\pmod2$ (thanks to this last condition the sum $G(k)$ is well defined). I have two questions:

  1. For now I compute these values in a naive way, so using $q*q$ steps (of course precomputing the $2q$th roots of unity). Is there a better method (even reducing to $q^2/2$ steps would be nice)? Imagine $q=10^5$ or $q=10^6$.

  2. When $q$ is not prime, $G(k)$ is often equal to $0$. For instance if $p$ is a prime congruent to $3$ mod 4 dividing $q$ and $k$, it seems that $G(k)=0$, at least for quadratic characters (but I am interested in all characters). My question is: give a necessary and sufficient condition for $G(k)=0$ (sufficient would already be nice).

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    $\begingroup$ Do you want to quickly evaluate for a single $k$, or all $k$ at once? If the latter, this is an evaluation of a polynomial at the the roots of unity, so you could use FFT. $\endgroup$
    – Aurel
    Jan 24 at 13:47
  • $\begingroup$ @Aurel: yes all at once, so FFT is conceivable. $\endgroup$ Jan 24 at 15:58
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    $\begingroup$ I realised afterwards that a complexity of $O(q^2)$ could only mean computing all the $G(k)$. Then the FFT version should be $O(q\log q)$. $\endgroup$
    – Aurel
    Jan 24 at 16:03

2 Answers 2

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For the second question, if $p$ is an odd prime dividing $q$, $p$ divides $k$ with at least the multiplicity with which it divides $q$, and $\chi_p(-1)=-1$, where $\chi_p$ is the $p$-adic part of $\chi$, then $G(k)=0$.

This generalizes what you wrote in the $\chi$ quadratic case. In that case, the multiplicity which which $p$ divides $q$ is always one, and $\chi(-1)=-1$ if and only if $p$ is congruent to $3$ mod $4$.

Indeed, the Chinese remainder theorem gives

$$G(k)=\frac{1}{2} \sum_{n\bmod 2q}\chi(n)e^{2\pi i n(n-k)/(2q)} = \prod_{ p \mid 2q} \sum_{n \bmod p^{v_p(2q)}} \chi_p (n) e^{ 2 \pi i \lambda_p n (n-k)/ p^{v_p(2q)}}$$

where $v_p(2q)$ denotes the $p$-adic valuation of $2q$, $\lambda_p$ is the inverse of $q/ p^{v_p(2q)}$ modulo $p^{v_p(2q)}$, and $\chi_p$ is the $p$-adic part of $\chi$.

Fixing on an odd prime $p$, we see that if $k$ is a multiple of $p^{v_p(2q)}=p^{v_p(q)}$, the factor at $p$ simplifies to

$$ \sum_{n \bmod p^{v_p(2q)}} \chi_p (n) e^{ 2 \pi i \lambda_p n^2/ p^{v_p(2q)}}$$

and if $\chi_p(-1)= -1$ then the terms for $n$ and $-n$ in the sum always cancel, giving a value of $0$.

If $\chi(-1)= 1$ then we get a sum of two Gauss sums and I don't see any reason it should be zero.

In general, if $v_p(2q)=1$, the local factor is a nice complete exponential sum. For fixed $k \neq 0$ mod $p$, Katz's equidistribution theory will tell us that the sum is nonzero for a density 1 subset of characters $\chi$. This probably can be proven also for fixed $\chi$ and a density one set of values of $k$, but might be harder to prove (at least by this method - there may be a clever congruence that proves nonvanishing).

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  • $\begingroup$ Thanks Will, this pretty much answers my second question (and thanks for correcting my typo in the case of quadratic characters). $\endgroup$ Jan 24 at 19:13
  • $\begingroup$ I accepted Will's answer, but am still interested in question 1, in addition to Aurel's suggestion of using FFT $\endgroup$ Jan 24 at 22:15
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I might be missing something, but I believe the below gives $O(q \log q)$ to compute $G(k)$ for all $k$.

For a FFT-type solution to the first question, we write $k = 2k' + (q\bmod 2)$ (where we in particular mean the residue of $q\bmod 2$ in $\{0,1\}$), and note that

$$\exp\left(-\frac{2\pi i n(2k'+(q\bmod 2))}{2q}\right) = \exp\left(-\frac{2\pi i nk'}{q}\right)\exp\left(-\frac{2\pi i n(q\bmod 2)}{2q}\right) = \zeta_q^{-nk'}\zeta_{2q}^{-n(q\bmod 2)}.$$

We can therefore write

$$G(2k'+(q\bmod 2)) = \sum_{n\bmod q} \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}\zeta_q^{-nk'}$$

Let $a_n = \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}$. Then computing $G(k)$ for all $k\in [0, 2q)$ such that $k\equiv q\bmod 2$ reduces to evaluating the polynomial

$$A(x) = \sum_{n = 0}^{q-1}a_nx^n$$

at the points $\zeta_q^{-k'}$ for $k'\in [0,q)$. This can be done with the standard complex DFT. To see this, we use the exposition of the complex DFT in terms of matrix multiplications, namely that evaluating $A(x)$ on $q$ points $x_0,\dots, x_{q-1}$ is equivalent to multiplying the vector $(a_0,\dots, a_{q-1})$ by the Vandermonde matrix associated with those points, i.e. computing the product

$$ \begin{pmatrix} 1 & 1 & 1&\dots & 1\\ 1 & \zeta_q^{-1} & \zeta_q^{-2} & \dots & \zeta_q^{-(q-1)}\\ 1 & \zeta_q^{-2} & \zeta_q^{-4}&\dots & \zeta_q^{-2(q-1)} \\ \vdots & & & \ddots&\vdots\\ 1 & \zeta_q^{-(q-1)} & \zeta_q^{-2(q-1)} & \dots & \zeta_q^{-(q-1)(q-1)} \end{pmatrix}\begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_{q-1} \end{pmatrix} $$

This product can be computed in $O(q\log q)$ time using standard FFT techniques. Note that the $i$th coordinate of the output is precisely $\sum_{n = 0}^{q-1} a_n \zeta_q^{-in} = \sum_{n = 0}^{q-1} \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}\zeta_q^{-ni} = \sum_{n = 0}^{q-1} \chi(n)\zeta_{2q}^{n(n-2i-(q\bmod 2))}$, i.e. is precisely $G(2i+(q\bmod 2)) = G(k)$ for some $k\in[0, 2q)$ with $k\equiv q\bmod 2$.

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  • $\begingroup$ Thank you, this is the detailed explanation of what Aurel suggested. $\endgroup$ Jan 25 at 8:33

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