Columns: the circle \(\mathbb R/\mathbb Z\) cut into \(Q\) cells \([\tfrac jQ,\tfrac{j+1}Q]\), drawn from \(-\tfrac12\) to \(\tfrac12\).
Rows: for \(q\mid Q\), the row \(R^Q_{q,u}\) has a \(1\) in each cell \(u+aQ/q\) with \(\gcd(a,q)=1\). It is legal when \(\max\{|u|,|u+1|\}<Q/\sqrt{\operatorname{lcm}(N,q^2)}\): the cell \(u\) lies under the semicircle of the \(c_q\)-twist.
Theorem: for some \(Q\) the legal rows span \(\mathbb Q^{\mathbb Z/Q\mathbb Z}\) if and only if \(16\nmid N\) and \(p^2\nmid N\) for every odd prime \(p\).
levels that span
excluded levels
rows with q = 1, 2 (a single entry)rows with q ≥ 3 (one entry per a/q)columns carrying the obstruction x
Reading the picture
Each row is one twisted functional equation used on one cell: it controls the sum of \(f\) over the marked cells.
Rows with \(q=1,2\) have a single \(1\) (teal); rows with \(q\ge3\) have one \(1\) for each fraction \(a/q\) (purple).
Each \(q\)-block is a band of parallel diagonals, one for each \(a\).
The bar under the matrix is the rank of the rows so far; the rows span when it reaches \(Q\).
The strip at the top shows the circle covered by the semicircles with \(q\mid Q\): teal for \(q\le2\), purple where only \(q\ge3\) reach.
Levels that span
The admissible \(N\le70\) from the paper's table whose smallest \(Q\) is at most \(1596\), each drawn at that smallest \(Q\).
Spanning means every value of \(f\) near the line is a combination of values higher up.
Excluded levels
If \(s\) is an odd prime with \(s^2\mid N\), or \(s=4\) with \(16\mid N\), take \(Q\) divisible by \(s\).
The vector with \(\sin(2\pi m/s)\) in the cell \(mQ/s\) (amber columns) and \(0\) elsewhere is orthogonal to every legal row.
So the rows never span, for any \(Q\). This is the imaginary part of the vector in the paper's proof of necessity.
Ranks computed modulo the prime \(2^{61}-1\) (full rank modulo a prime implies full rank over \(\mathbb Q\)). For every level that spans, the rank is \(Q\) at the table's \(Q\) and below \(Q'\) at \(Q'=Q/p\) for each prime \(p\mid Q\), as the table says. For the excluded levels the orthogonality of the amber vector to all legal rows was checked numerically (largest pairing below \(10^{-15}\)).