Where one functional equation stops, and what the Ramanujan-sum twists add
q =bar under the axis: the smallest q whose semicircles cover the pointnot covered, or two semicircles only touchfundamental domain (q = 1 only)
One functional equation relates \(f(z)\) and \(f(-1/(Nz))\): a half-turn about \(i/\sqrt N\).
Inside the semicircles \(|z-n|<1/\sqrt N\) it moves points up (a reflection in the semicircle, then in a vertical line).
For \(N\le4\) the semicircles cover the line.
At \(N=5\) a gap opens at \(\tfrac12\).
The fundamental domain reaches the line: its area is infinite.
Over the gap there is no control: for \(N\ge5\) the solutions form an infinite-dimensional space.
The Euler product gives a functional equation for every Ramanujan-sum twist \(c_q\), of conductor \(\operatorname{lcm}(N,q^2)\).
For \(q=2\) only \(z+\tfrac12\) occurs: one more half-turn, about \(\tfrac12+i/\sqrt{\operatorname{lcm}(N,4)}\).
It closes the gap.
With \(q\le2\) the line stays covered up to \(N=8\).
At \(N=9\) the semicircles only touch, at \(\tfrac13\) and \(\tfrac23\).
These are the two cusps of \(\Gamma_0(9)\) with denominator \(3\).
At \(N=11\) gaps open that no single half-turn reaches.
\(N=10\) and \(N=12\) are still covered with \(q\le2\); from \(N=13\) on, never.
Twists with \(q\ge3\) reach the gaps.
But each controls only the sum of \(f\) over \(z+a/q\), \(\gcd(a,q)=1\): the hollow dots turn together.
Spanning theorem: the sums can be untangled exactly when \(16\nmid N\) and \(p^2\nmid N\) for odd \(p\).
At \(N=9\) they cannot be untangled.
Only one semicircle contains \(\tfrac13\): the \(q=3\) one centred there; all others are at best tangent.
It sees only \(f(z+\tfrac13)+f(z+\tfrac23)\), where \(+1\) near \(\tfrac13\) and \(-1\) near \(\tfrac23\) cancel.
Choose \(N\) and the largest \(q\) with the sliders.
Hover (or touch) to see a point reflected in its semicircle, then in a vertical line.
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One functional equation
For a sequence \((f_n)\) put \(f(z)=\sum_{n\ge1} f_n n^{\frac{k-1}{2}}e^{2\pi i n z}\). The functional equation of \(L_f(s)\) says
$$f\!\left(-\frac{1}{Nz}\right)=\epsilon_f\,(-i\sqrt N z)^k f(z).$$
\(z\mapsto-1/(Nz)\) is a half-turn of the hyperbolic plane about \(i/\sqrt N\).
\(\operatorname{Im}(-1/(Nz))=\operatorname{Im}(z)/(N|z|^2)\), so it raises \(z\) exactly when \(|z|<1/\sqrt N\).
With \(f(z+1)=f(z)\): values near the line are determined by values higher up wherever the semicircles \(|z-n|<1/\sqrt N\) reach, which is everywhere iff \(N\le4\).
For \(N\le4\) the region \(0\le\operatorname{Re}z\le1\), \(|z|\ge1/\sqrt N\), \(|z-1|\ge1/\sqrt N\) is a fundamental domain of area \(2\arcsin(\sqrt N/2)\): \(\pi/3,\pi/2,2\pi/3,\pi\).
Twists by Ramanujan sums
\(c_q(n)=\sum_{\gcd(a,q)=1}e^{2\pi i an/q}\).
If \(L_f\) has an Euler product, its \(c_q\)-twist has a functional equation of conductor \(\ell_N(q)=\operatorname{lcm}(N,q^2)\).
In terms of \(f\), it relates \(\sum_{\gcd(a,q)=1}f(z+\tfrac aq)\) to the same sum at \(-1/(\ell_N(q)z)\): one half-turn applied to all the translates at once.
It raises points inside the semicircles of radius \(1/\sqrt{\ell_N(q)}\) centred at the fractions \(a/q\).
The spanning theorem (no Euler factors)
For \(q\mid Q\) and \(u\in\mathbb Z\): \(R^Q_{q,u}=\sum_{\gcd(a,q)=1}e_{u+aQ/q}\), legal if \(\max\{|u|,|u+1|\}<Q/\sqrt{\ell_N(q)}\).
A legal row is one relation used on the cell \([\tfrac uQ,\tfrac{u+1}{Q}]\), shifted to each \(a/q\): the cell must lie under the semicircle.
Some \(Q\) makes the legal rows span \(\mathbb Q^{\mathbb Z/Q\mathbb Z}\) iff \(16\nmid N\) and \(p^2\nmid N\) for every odd prime \(p\).
Spanning means every value of \(f\) near the line is a combination of values strictly higher up.
At an excluded level the points \(a/s\) (\(s=p\) with \(p^2\mid N\), or \(s=4\) with \(16\mid N\)) lie inside exactly one semicircle each, the one with \(q=s\) centred there. This was checked in exact arithmetic for all \(q\le200\) at \(N=9,16,18,25,27,36\).