Two rays for every fraction

The fraction written on every bulb read the Mandelbrot set from the inside. Every bulb on the main cardioid was traced as the region where one cycle attracts, and the fraction $p/q$ at which it hangs turned out to be its period, its rotation and, to within fourteen per cent, its size. This essay reads the same bulbs from the outside, where the set is not there and the plane is empty, and finds the fraction again — this time written in binary, as the angles of two curves that come in from infinity and land at the bulb’s root.

The curves are the external rays, and they were introduced in the shape in every picture of itself as the lines running across the escape-time bands, perpendicular to them, each labelled by an angle. They are the working tool of the subject: Douady and Hubbard used them to prove that the set is connected and to describe how its pieces join. Here they are computed and drawn, and three facts about them are measured. Two rays land at each bulb’s root, and which two is decided by the arithmetic of a straight line. They approach that root slowly, in a way that distinguishes a root from every other kind of landing point. And the regions between the pairs of rays share out the whole circle of angles, by an identity about Euler’s totient that predates the set by two centuries.

The angle of a point outside #

Far from the set, the orbit of nought under $z \mapsto z^2 + c$ escapes so fast that its behaviour is dominated by the squaring: after $n$ steps it is close to $c^{2^n}$ in a precise sense. There is a change of coordinates, the Böttcher coordinate $\Phi(c)$, which makes that exact. It is a conformal map from the outside of the set onto the outside of the unit disc, with $\Phi(c) \approx c$ far away, and in it the escape becomes plain squaring. Adrien Douady and John Hubbard proved in 1982 that this map exists and is one-to-one on the whole outside of the set, and that existence is the proof that the set is connected: the outside of a connected set can be mapped onto the outside of a disc, and the outside of a set in two pieces cannot.

The map gives every point outside the set two coordinates, the size of $\Phi(c)$, which measures how fast the orbit escapes, and its angle. Curves of constant size are the outlines of the escape-time bands. Curves of constant angle are the external rays: the ray of angle $\theta$ is the set of $c$ for which $\Phi(c)$ points in the direction $e^{2\pi i \theta}$. Each ray comes in from infinity and heads for the set, and the question is where it arrives.

Computing a ray means inverting $\Phi$, which has no formula, and the method used here is Tomoki Kawahira’s. At depth $n$ the equation $\Phi(c) = w$ is replaced by the polynomial equation that the $n$-th iterate of nought equals $w^{2^{n-1}}$, which is accurate when the iterate is large; Newton’s method solves it starting from the previous point of the ray, and the target $w$ is moved inward a fraction of a step at a time. The only delicate part is the angle. Raising $w$ to the power $2^{n-1}$ doubles its angle $n - 1$ times, and in floating point the angle would lose a binary digit at each doubling. Here every angle is a fraction with an exact integer numerator and denominator, and the doubling is done in integers, so the ray at depth 160 is computed with the angle exact. The hero figure’s rays were followed to that depth, and each ends within two hundredths of the root it was predicted to land at.

Doubling, and why rational angles cycle #

Squaring a number far from the set doubles its angle. So the dynamics of the map, seen through the Böttcher coordinate, is the doubling map on the circle of angles, $\theta \mapsto 2\theta \bmod 1$, and the doubling map is easiest to read in binary. Doubling shifts the binary digits of $\theta$ one place left and throws away the digit that crosses the point. The orbit written as a word met this as the first example of symbolic dynamics, where a map is its effect on an infinite string of digits.

A fraction with an odd denominator has a repeating binary expansion, and its orbit under doubling is a cycle. The fraction $9/31$ is $0.\overline{01001}$, repeating every five digits because $2^5 - 1 = 31$; doubling it gives $18/31 = 0.\overline{10010}$, then $5/31$, $10/31$, $20/31$, and back to $9/31$. Every fraction $m/(2^q - 1)$ has a period dividing $q$, and the rays with these angles are the periodic rays. Douady and Hubbard proved that every periodic ray lands, and that it lands at the root of a bulb or of one of the bulb’s smaller copies elsewhere in the set: at a parameter where a cycle is neutral with a multiplier that is a root of unity. Which bulb a given periodic ray lands at is the question the rest of this essay answers for the bulbs on the cardioid.

The cycle that keeps its order #

There are six cycles of period five under doubling, thirty angles in all, and the root of the two-fifths bulb receives exactly two rays. The figure below picks out which cycle the two come from.

Most cycles of the doubling map jump about the circle with no pattern: the points of a cycle visited in one order are arranged round the circle in a different one. The cycle drawn here is special. Its five angles, in increasing order, are $5/31, 9/31, 10/31, 18/31, 20/31$, and doubling carries each to the one two places further round: $9 \to 18$ is two places on, $18 \to 5$ is two places on going round past nought, and so on. The cycle is a rotation in disguise, by two places of five, which is exactly the rotation the cycle inside the two-fifths bulb performed round its fixed point. A search over all thirty period-five angles finds one cycle with this property and only one. Lisa Goldberg proved in 1992 that this always happens: for every fraction $p/q$ the doubling map has exactly one periodic cycle that keeps its cyclic order while turning $p$ places of $q$.

The two rays that land at the root are the two members of that cycle that sit side by side, $9/31$ and $10/31$, separated by the smallest possible gap, $1/31$. That is how the outside of the set knows the inside. The bulb’s cycle rotates round its fixed point by two fifths; the rays that land at its root rotate round the circle at infinity by two fifths as well, because the dynamics outside, read in Böttcher’s coordinates, is conjugate to the dynamics inside, read in the bulb.

The palindrome of a straight line #

Searching thirty angles for the one rotating cycle is easy at period five and hopeless at period fifty. The angles have a formula, and it comes from a straight line.

Draw the line of slope $2/5$ from the origin and the staircase of unit steps that stays just below it, and write $0$ for a step across and $1$ for a step up. The staircase spells $00101$, a word with two ones in five letters, and it begins with $0$ and ends with $1$. Strip those two letters and the middle, $010$, reads the same backwards. The two landing angles are $0.\overline{01001}$ and $0.\overline{01010}$: the palindrome $010$ followed by $01$, and the palindrome followed by $10$. The same holds for every fraction tried. For $1/3$ the palindrome is $0$ and the angles are $0.\overline{001} = 1/7$ and $0.\overline{010} = 2/7$; for $5/13$ the palindrome is eleven digits long and the angles are $0.\overline{0100101001001}$ and $0.\overline{0100101001010}$. The check behind the figure computes the rotating cycle by search and the palindrome from the line for every fraction with denominator up to fourteen, sixty-three of them, and finds the rule exact for all sixty-three.

The words are the ones the word a straight line spells found when a billiard ball crossed the walls of a square: a line of slope $p/q$ crossing the grid writes a balanced word, the most evenly spread arrangement of $p$ ones among $q$ letters. Its rotations are exactly the cycle that turns $p$ places of $q$ under doubling, because shifting a balanced word by one place is the same as advancing the line’s crossing point by $p/q$ of a period. The palindrome rule, worked out by Shaun Bullett and Pierrette Sentenac in 1994, picks out the two rotations whose values are adjacent integers. So a picture drawn from complex squaring has its rays labelled by the cutting sequence of a straight line on a square grid, which is about as far from complex squaring as an object in this collection gets.

Arriving at a root, and arriving at a tip #

A ray lands, but it can land fast or slow, and how fast tells what kind of point it has landed on. The fourth figure follows two rays and measures, at each depth, how far the computed ray still is from its landing point.

The ray of angle $1/7$, one of the pair at the root of the one-third bulb, is $0.215$ from the root at depth ten, $0.046$ at depth forty, $0.0114$ at depth 160 and $0.0057$ at depth 320. Doubling the depth halves the distance; on the logarithmic plot the slope is $-1.02$. The ray of angle $1/6$ lands at $c = i$, the tip of a filament, and it is $4.5 \times 10^{-3}$ away at depth ten, $7.8 \times 10^{-7}$ at depth twenty and $2.4 \times 10^{-14}$ at depth forty, the limit of double precision.

The difference is the difference between a parabolic point and a Misiurewicz point. At the root of a bulb a cycle is neutral, with multiplier exactly one in size, and an orbit near a neutral cycle escapes only gradually; the distance at which escape becomes visible shrinks like a power of the depth rather than exponentially. At $c = i$ the orbit of nought runs $0, i, -1 + i, -i, -1 + i, \ldots$ and falls after two steps onto a cycle of period two whose multiplier is $4(-1 + i)(-i)$, of size $4\sqrt 2$. Every two steps nearby points are pushed away by that factor, about $2.38$ a step, and $\log_{10} 2.38 = 0.376$, which is the measured gain of $0.38$ digits per step. A computation that needs to locate a bulb’s root from its rays therefore needs far more depth than one that locates a tip, and the reason is the bulb itself.

Most angles never reach a small bulb #

The two rays at a root fence off the bulb’s limb: every ray whose angle lies between them lands somewhere on the bulb or on what hangs from it, and every other ray lands elsewhere. The angular width of that wedge is the gap between the two angles, $1/(2^q - 1)$, and it has a meaning beyond geometry. The Böttcher map carries the uniform measure on the circle to the harmonic measure of the set, which is the chance that a random path wandering in from very far away first touches the set at a given place. So $1/(2^q - 1)$ is the probability that a random walker coming from infinity first meets the set somewhere in the $p/q$ limb.

The two quantities part company almost at once. At period seven the one-seventh bulb has size $0.0091$, half its width, and its limb takes $0.0079$ of the angles, about the same. At period twenty-four the bulb has size $2.4 \times 10^{-4}$, still easily visible in a good picture, and its limb takes $6.0 \times 10^{-8}$ of the angles. A random path coming in from far away is four thousand times less likely to land there than the bulb’s size would suggest. The reason is the shape of the set near the cusp, where the small bulbs hang: the region between them is a deep narrow fjord, and a random path that wanders into a fjord touches its walls long before it reaches the end. The area of the Mandelbrot set, from two sides used the same Böttcher map, expanded as a series, to bound the area from outside, and found that series converging slowly for the same reason: the outside’s coordinate barely sees the deepest parts of the boundary.

The limbs share out the whole circle #

Every fraction has a limb, and the limbs do not overlap, since two pairs of rays cannot cross. So their widths add up to at most one. The last figure shades the arcs of every limb up to period nine on the circle of angles, and adds up their widths period by period.

The one-half limb alone takes a third of the circle, from $1/3$ to $2/3$. With the two thirds limbs the total is $0.619$, with the quarters $0.752$, and by period twelve the limbs cover $0.9976$ of the circle. There are $\varphi(q)$ fractions of period $q$ — Euler’s totient, the count of numerators prime to $q$ — each with a limb of width $1/(2^q - 1)$, so the total is

$$\sum_{q \ge 2} \frac{\varphi(q)}{2^q - 1} = 1.$$

The identity is exact, and it is a Lambert series. Expanding each term as a geometric series, $\varphi(q)/(2^q - 1) = \sum_k \varphi(q)\, 2^{-kq}$, and collecting the powers $2^{-n}$, the coefficient of $2^{-n}$ is the sum of $\varphi(q)$ over the divisors $q$ of $n$, which is $n$ — the fact that the fractions with denominator $n$, reduced, run through every denominator dividing $n$, the same counting every fraction exactly once did with the Farey fractions. So the full sum over $q \ge 1$ is $\sum_n n\, 2^{-n} = 2$, and removing the $q = 1$ term, which is one, leaves exactly one.

The meaning is that the limbs take every angle there is. The rays that land on the cardioid’s own outline, rather than in some limb, have angles forming a set of total length nought. The cardioid is the largest piece of the set, and from far away it is invisible: a random path from infinity touches it with probability nought, and arrives instead in one of the limbs, in the $p/q$ limb with probability exactly $1/(2^q - 1)$.

What the rays cannot show #

The figures draw rays that land at the roots of bulbs on the cardioid, and they draw each one to a finite depth. Two things are therefore out of their reach. The first is the landing itself. A computed ray ends at a point a measured distance from where it is going, and that distance, at a root, falls only like one over the depth; the statement that the ray lands there is the theorem of Douady and Hubbard, not something the figure shows. The second is everything else on the boundary. Rays with irrational angles land, if they land, on points of the boundary that are not roots or tips, and whether every such ray lands at all is open: it is equivalent to the local connectivity of the set, the conjecture that the fraction written on every bulb ended on.

The angle measure has a blind spot of its own. The limbs fill the circle, so the harmonic measure assigns nothing to the cardioid’s outline, but that does not mean the outline is small in other senses — it is a smooth curve of positive length. Harmonic measure sees a boundary as a random path does, and a random path is a poor judge of the parts of a shape that lie at the bottom of narrow channels.

Still open: which rays land together elsewhere #

On the cardioid everything is settled. Each fraction has one bulb, one rotating cycle of doubling, one palindrome, one pair of rays. Away from the cardioid the bulbs have bulbs of their own, and small copies of the whole set sit in the filaments, each with roots receiving pairs of rays. Which pairs of angles land together at those roots is described by a combinatorial structure, Thurston’s quadratic minor lamination, and there is an algorithm, due to Lavaurs, that produces every pair of periodic angles landing together, period by period.

What is not known is whether that combinatorial description is the whole truth. If the set is locally connected, its boundary is exactly the circle of angles with each pair of landing angles pinched together, and the combinatorics is a complete model of the set’s shape; if it is not, some part of the boundary is wilder than any lamination records. The angles are known exactly, the pairings are known exactly, and whether pinching the circle along them gives back the boundary is a question that has stayed open since Douady and Hubbard posed it.

Two names for one fraction #

The bulb at two fifths of a turn is named, from the inside, by its cycle’s rotation round a fixed point. From the outside it is named by two angles, $9/31$ and $10/31$, that sit next to each other on the circle at infinity and form part of the one doubling cycle that rotates by two fifths. The binary digits of those angles are the staircase under a line of slope two fifths with its first and last steps rearranged, the rays they label reach the bulb’s root only slowly because a cycle there is neutral, and the wedge between them takes $1/31$ of all the angles there are. Summed over every fraction, the wedges take all of them.

The identity behind that last statement is an eighteenth-century fact about Euler’s totient, and the doubling map is an exercise in binary. Neither mentions complex numbers. The Mandelbrot set is drawn from one squaring and one addition, and the arithmetic that organises its outside — the binary digits of the doubling map, the cutting words of straight lines, and the totient’s divisor sums — is arithmetic that was complete before anyone squared a complex number.