Sectors that shrink with the primes

Which way a prime’s two squares point drew every prime $p = a^2 + b^2$ as the point $(a, b)$, with $a > b > 0$, and asked about its angle $\theta = \arctan(b/a)$, which lies between $0°$ and $45°$. Erich Hecke proved in 1920 that the angles are evenly spread: the share of primes up to $X$ whose angle falls in a fixed sector tends to the sector’s share of the $45°$. That essay showed fixed sectors filling on schedule and thin ones filling late, and it ended on the refinement that is not settled. How thin may a sector be, as a function of the size of the primes, and still be guaranteed its share?

The conjectured answer is: any width $X^{-1/2 + \varepsilon}$, for any $\varepsilon > 0$. The proved answers are far wider. This essay does not prove anything about the question; it measures it. Take the primes between $X/2$ and $X$ with $X$ forty million, cut the $45°$ range into sectors of width $X^{-\alpha}$, and watch what happens to the counts as $\alpha$ grows from small values, where Hecke’s method works, past $\tfrac12$, where the conjecture stops.

At width $1/\sqrt X$ the picture looks like a band of noise around the line, which is what Hecke’s theorem would predict for wide sectors and what the conjecture predicts for these. Two things are visible on a closer look. The band is narrower than independent random counts would make it. And at the two ends, near $0°$ and $45°$, the counts fall away and a couple of sectors are empty. Both turn out to be properties of the lattice that the primes live on, not of the primes.

Finding every prime, and cutting the range #

The list of angles is complete. Every pair $a > b > 0$ with $a^2 + b^2$ between twenty and forty million was visited, and $a^2 + b^2$ was looked up in a sieve; since a prime that leaves remainder 1 on division by 4 is a sum of two squares in exactly one way with $a > b > 0$, as two squares and a lattice explained, each prime appears once. There are 581,517 of them.

For each $\alpha$ the $45°$ range is cut into $K = \tfrac{\pi}{4} X^{\alpha}$ equal sectors, so that each has width about $X^{-\alpha}$ radians, and the primes are counted in each. At $\alpha = 0.2$ there are 26 sectors with over twenty thousand primes apiece; at $\alpha = 0.5$, 4,967 sectors with 117 on average; at $\alpha = 0.75$, nearly four hundred thousand sectors with fewer than two each. Three statistics describe the counts: how much they vary relative to their mean, how many sectors are empty, and where the empty ones are.

The lattice points come first #

Before the primes there are the lattice points they are drawn from, and those are spread almost perfectly. The number of points $(a, b)$ with $a^2 + b^2 \le X$ in a sector of width $w$ is the sector’s area, $wX/2$, plus an error that depends only on the sector’s boundary — the arc and the two straight sides — and is much smaller than the area for any width above about $1/\sqrt X$. The dots a circle catches measured the arc’s share of that error; the straight sides contribute in a way that depends on the slope of the side, small for a generic slope and large for a slope of small height, which is the effect the empty sectors will show.

The primes are a thin subset of those points. About one point in $\log X$ is a prime, and the sums of two squares themselves are already rare, as almost no number is one showed. So the share a sector should hold is its share of the lattice points times the density of primes among them, and the question of thin sectors splits in two: whether the lattice points are evenly spread at that width, which is geometry, and whether the primes among them are, which is arithmetic. At widths above $1/\sqrt X$ the geometry is settled and only the arithmetic is in doubt. Below it, the geometry fails first.

Steadier than chance in wide sectors #

If the angles were independent random points, the count in a sector would vary with variance equal to its mean — the Poisson rule — and the ratio of the two would be 1 at every width.

The ratio is not 1. For the widest sectors it is 0.38: the counts in 26 sectors of about $1.7°$ each fluctuate with less than two-fifths of the variance random points would give. As the sectors narrow the ratio climbs — 0.48, 0.69, 0.77, 0.87 — and from about $\alpha = 0.55$ on it sits within a few per cent of 1. The primes’ angles are more regular than random at large scales and indistinguishable from random at small ones.

That shape is what Ze’ev Rudnick and Ezra Waxman predicted in 2019 from a random-matrix model of Hecke’s L-functions, the model the previous essay mentioned as its open question. The regularity at large scales has a plain source. The count in a wide sector is controlled by the low-frequency averages of the angles, which Hecke’s L-functions govern and which are close to their limits; the fluctuations of independent points would need those averages to be noisier than they are. At scales below $1/\sqrt X$ the low frequencies no longer matter and what remains looks like independent noise. The measurements at four million and forty million agree in shape, which is evidence the transition is a property of the primes rather than of one size.

More empty sectors than chance allows #

The variance is an average over all sectors. The empty sectors are where the average hides something.

Below $\alpha = 0.5$ no sector is empty, as chance would also predict. From $\alpha = 0.5$ on there are empty sectors that chance cannot account for: 2 at $\alpha = 0.5$, 10 at 0.55, 65 at 0.6 — where the expected number, with about twenty primes per sector, is effectively nought — and 431 at 0.65, where chance expects fourteen. Only by $\alpha = 0.7$, where sectors hold three or four primes on average and many are empty by chance, does the count of empty sectors fall back to the random prediction.

Some of the excess is forced. Near the axis, a point $(a, b)$ with $b \ge 1$ and $a$ at most $\sqrt X$ has angle at least about $1/\sqrt X$, so the first few sectors narrower than that contain no lattice point at all; the same happens near $45°$, where $a - b \ge 1$. Those forced sectors are the third line in the figure, and they are a minority of the empty ones. At $\alpha = 0.6$, nine of the 65 are forced. The other 56 are empty for some other reason.

Where the empty sectors are #

Sorting the empty sectors by position finds the reason.

The empty sectors are not scattered through the range. Almost every one sits right beside a direction of small rational slope: 16 beside $1/3$, 12 beside $1/1$, 7 at the axis, 6 beside $1/7$, 5 beside $1/2$, and the rest beside $1/5$, $1/4$, $3/5$, $3/11$, $7/11$, $2/3$. Near any rational direction the points of the integer lattice line up in parallel rows along that direction, the way they line up in rows $b = 1, 2, 3, \ldots$ along the axis, and the rows are spaced by $1/\sqrt{q^2 + r^2}$ for the slope $q/r$. A sector thin enough to fit between two rows catches nothing, whatever the primes are doing. The same rows are what make the fractions with small denominators stand out among all fractions, the effect how evenly the fractions spread measured on the interval. The slopes that matter are the ones every fraction, exactly once produced first by adding tops and bottoms: $0/1$ and $1/1$, then $1/2$, then $1/3$ and $2/3$. A slope of small height has widely spaced rows, and the further down that tree of mediants a slope lies, the closer its rows and the narrower the sector needed to fall between them.

Rows of this kind are a familiar way for structure to hide in apparent randomness. A rotation that hides the lattice described the pairs of outputs of a congruential random-number generator, which look scattered and lie on a few parallel lines when viewed along the right direction. The lattice points here do the same: from far away they fill the plane evenly, and seen along a direction of small slope they are lines with gaps between them.

And the slopes with both terms odd — $1/1$, $1/3$, $1/5$, $1/7$, $3/5$, $3/11$, $7/11$ — collect two-thirds of the empty sectors. The reason is parity. A prime $a^2 + b^2$ other than 2 is odd, so $a$ and $b$ have opposite parity. Along a direction of slope $q/r$ with $q$ and $r$ both odd, the rows of lattice points alternate: on every other row, $a$ and $b$ have the same parity, $a^2 + b^2$ is even, and no prime can lie there. The usable rows are twice as far apart as the lattice’s own, and the gaps a thin sector can fall into are twice as wide. For a slope with one even term every row contains points of both parities, and the effect is weaker. The slope $1/1$ is the $45°$ end of the range itself, where the rows $a - b = 1, 2, 3, \ldots$ run along the boundary, so some of its empty sectors are among the forced ones; set it aside and the odd-odd slopes still hold the largest share of what is left, with $1/3$ alone holding sixteen.

Sector by sector #

The swings near a rational direction can be seen one sector at a time.

Beside the axis the first five sectors are empty, then one holds 74 — a row of lattice points, $b = 1$ or $b = 2$, running along it — and the counts swing between nothing and several times the share before settling. Beside the slope $1/3$ the counts are ordinary until the sectors come within a few widths of the direction; then they swing the same way, and one sector holds 142, seven times its share, because it runs along a row, while sectors either side of it hold none. Averaged over a wide sector, the feast and the famine cancel. In a thin one they are the whole story. The same swings would appear with every lattice point counted instead of only the primes; the primes inherit them, at about one in fourteen of the lattice points in this range, because they can only sit where lattice points are.

This is why the variance ratio climbed to 1 rather than above it. The excess empty sectors are paired with overfull ones beside the same directions, and over the whole range both are a small part of the total. They are not a failure of equidistribution; they are equidistribution at a scale where the lattice is visible.

The barrier at the axis #

The axis is where the lattice’s rows are farthest apart, and where the conjecture’s exponent $-\tfrac12$ comes from.

The prime closest to the axis is always one with $b = 1$, a prime of the form $a^2 + 1$, and its angle is about $1/a$, which is about $1/\sqrt X$. No prime, and no lattice point with $b \ge 1$, can do better. So a sector of width narrower than $1/\sqrt X$ placed at the axis is empty by geometry, and the conjecture cannot ask for anything thinner than $X^{-1/2}$. At exactly that width, a sector at the axis contains primes only if there are primes $a^2 + 1$ near $X$, and whether there are infinitely many such primes is Landau’s problem — the open question the same bell on thinner sets met from the side of prime factors, where the count of primes $n^2 + 1$ up to two million matched Hardy and Littlewood’s prediction to within a tenth of a per cent. The conjecture for thin sectors therefore contains Landau’s problem as its extreme case, which is one way to see why it is hard.

What is proved, and what is measured #

Hecke’s theorem gives the even spread for fixed sectors. Making it quantitative, with sectors shrinking at a rate, requires bounds on the error terms of all of Hecke’s L-functions at once. A sector of width $w$ needs about $1/w$ of the L-functions to describe it, each contributing its own error, and even the Riemann hypothesis for every one of them, used in the standard way, gives the share only in sectors far wider than $X^{-1/2}$; unconditionally, the known widths are wider still. The difficulty is that a thin sector is a sharp edge in angle, and a sharp edge needs many frequencies, each of which must be controlled at once. Proved results of a different kind, from sieve methods, place infinitely many primes — though not their full share — in sectors as thin as $X^{-0.38}$. The measurements here run from $X^{-0.2}$ to $X^{-0.75}$, across all of those thresholds and past the conjectured one.

What they show is consistent with the conjecture and with Rudnick and Waxman’s model: the counts are close to their shares at every width down to about $X^{-1/2}$, with fluctuations smaller than chance, and the exceptions that appear at thinner widths are explained by the lattice. They cannot show that the conjecture is true. At forty million, $X^{-1/2}$ and $X^{-0.6}$ differ by a factor of six, and a statement about exponents is a statement about how such factors behave as $X$ grows without bound.

Still open: thin sectors away from rational directions #

Whether every sector of width $X^{-1/2 + \varepsilon}$ eventually holds its share of primes is not known. The measurements suggest a sharper form of the question. The lattice forces empty sectors beside rational directions at widths a little below $1/\sqrt X$, and doubles the effect beside slopes with both terms odd. A natural refinement asks for the share in every sector of width $X^{-\alpha}$ that keeps a distance from rational directions of small height, for some range of $\alpha$ beyond $\tfrac12$, and how that range depends on the height allowed. The measurements raise it rather than answer it, and the experiment here is too small to suggest an exponent.

The parity effect also makes a prediction that a larger computation could test. Beside a slope whose two terms are both odd, only every other row can hold a prime, so at a given width the empty sectors there should behave like those beside an even-odd slope whose rows are twice as far apart. If that matching held across many widths and sizes, it would show that the rows and their parity account for all of the excess, and that away from rational directions the thin sectors behave like chance all the way down — which is the content the conjecture would need.

The second open matter is Rudnick and Waxman’s prediction itself, which is a conjecture about the variance at every scale, derived from a model rather than from the primes. The two sizes measured here agree with its shape, and confirming its precise form would need sizes far beyond forty million and many more sector widths.

The lattice behind the primes #

Hecke’s theorem is about primes, and its refinement is still about primes, but at the widths where the refinement is hard, the lattice underneath shows through. Wide shrinking sectors hold their share more steadily than chance would; thin ones fluctuate like chance; and the thin ones that go empty sit beside the directions where the lattice lines up in rows, most of all where parity removes every other row. The barrier at $1/\sqrt X$ is the lattice’s barrier, and at the axis it becomes Landau’s problem. What the primes do away from those directions, at widths below $1/\sqrt X$, is the question the measurements point to and do not answer.