The loops on a torus that never cross themselves

A torus can be made from a square by gluing its opposite edges: the right edge to the left, the top to the bottom. A path that runs off one side comes back in on the other. A loop on the torus then crosses the vertical edge some number of times and the horizontal edge some number of times, counted with a sign for the direction, and those two counts classify it: two loops can be deformed into one another exactly when their counts agree. The loops of the torus, up to deformation, are the pairs of whole numbers $(p, q)$ — the group computed by cutting the torus into pieces is exactly the pairs, added coordinate by coordinate.

The classification says nothing about what the loops look like. In particular it does not say which classes contain a loop that never crosses itself — a simple closed curve, a loop of string lying on the surface without touching itself anywhere. The class $(0, 0)$ does: a small circle. Does $(2, 0)$, a loop going twice round the same way? Does $(2, 3)$?

The answer is a piece of arithmetic that has nothing to do with topology on its face. A class $(p, q)$ other than $(0, 0)$ contains a loop that never crosses itself exactly when $p$ and $q$ have no common factor. And the same two numbers, taken from two loops at once, say how many times those loops must meet.

Straight loops of every slope #

On the square the natural loop of class $(p, q)$ is a straight line: start anywhere, head off with slope $q/p$, and follow the line through each edge until it returns to where it began. It crosses the vertical edge $p$ times and the horizontal edge $q$ times, so it is in the class $(p, q)$.

For each of the four classes drawn, $p$ and $q$ share no factor, and each straight loop closes up after a single pass and never crosses itself. The segments it leaves on the square are parallel, evenly spaced, and never meet. Seen on a doughnut rather than a square, the loop of class $(1, 0)$ runs once round the tube, $(0, 1)$ once round the hole, and $(2, 1)$ twice round the tube while going once round the hole, in a single unbroken strand.

The reason the straight loop never meets itself is Bézout’s identity. Two moments $t$ and $t'$ on the line land on the same point of the torus exactly when the difference between the positions, $(t - t')(p, q)$, is a pair of whole numbers — that is what gluing the edges means. If $p$ and $q$ have no common factor, there are whole numbers $a$ and $b$ with $ap + bq = 1$, found by running Euclid’s algorithm backwards, and then

$$t - t' = a\,(t - t')p + b\,(t - t')q,$$

which is a whole number. So the only times the line revisits a point are whole numbers of passes apart, which is the loop closing up — not crossing itself. The argument is the same one that makes two coprime clocks visit every pair of readings exactly once, and the walk drawn there is a straight line on a torus of this kind.

Steeper slopes, still one strand #

The loops in the first figure are short. The same argument covers every slope in lowest terms, however many times the loop has to wind before it closes, and the drawings stay just as clean.

The loop of class $(2, 5)$ leaves seven pieces on the square, one for each time it passes through an edge, and they are all parts of one strand: follow any piece off an edge and it continues as the next. No piece meets another. The spacing between neighbouring pieces is the same everywhere — a loop of class $(p, q)$ cuts every horizontal line of the square into $q$ equal parts and every vertical line into $p$ — because the pieces are the translates of one line by the lattice, and in lowest terms the translates that land on the square are evenly spread.

That even spacing is also how to read the class off a drawing without following the strand. Count the pieces crossing any vertical line of the square and that is $p$; count those crossing any horizontal line and that is $q$. For a straight loop the count is the same wherever the line is drawn. For a loop with a common factor drawn straight, as a single track run over $d$ times, the count along a line would still be $p$ — but the pieces would lie in only $p/d$ distinct places, each carrying $d$ passes on top of one another, and a loop of string cannot do that without touching itself.

A common factor forces a crossing #

Now take $(2, 0)$. The straight line of slope 0 closes after going round once, and to be in the class $(2, 0)$ it must go round twice — along exactly the same track, which is a loop lying on top of itself rather than a loop that avoids itself. Push the second pass slightly off the first, and at some point it has to come back to where it started, which means crossing the first pass.

The figure draws each class with a common factor $d$ as $d$ passes of the loop in lowest terms, each shifted a little to the side of the one before, with a single return at the end that crosses the other passes to get back to the start. The return crosses $d - 1$ passes: 1 crossing for $(2, 0)$ and $(2, 2)$ and $(4, 2)$, where $d = 2$, and 2 for $(3, 0)$, where $d = 3$.

No drawing does better. The fewest self-crossings of any loop in the class $(p, q)$ is $\gcd(p, q) - 1$, so a common factor always forces a crossing. The figure shows that $d - 1$ is enough; that it is necessary is a theorem, and for the case that matters here — that $d \geq 2$ forces at least one crossing — the argument is short enough to give.

Why a simple loop has coprime counts #

Suppose a loop lies on the torus without crossing itself, and cut the torus along it. Two things can happen.

The cut may split the torus into two pieces. Then the loop is the boundary of one of them, and any other loop that crosses it must cross an even number of times, going in and coming out — or rather, with signs, it must cross it a net zero times. But the net number of crossings with the loop of class $(1, 0)$ is $\pm q$ and with $(0, 1)$ it is $\pm p$, as the section below explains, so both counts are zero and the loop is in the class $(0, 0)$: it bounds a disc and can be shrunk away.

Or the cut may leave the torus in one piece. Then there is a path in the cut surface from one side of the cut to the other, and closing it up across the cut gives a loop that crosses the original loop exactly once. If the original is in class $(p, q)$ and the new loop in class $(r, s)$, their net crossing number is $\pm(ps - qr)$, so $ps - qr = \pm 1$. Any common factor of $p$ and $q$ would divide $ps - qr$, and nothing but 1 divides 1.

So a loop that never crosses itself is either shrinkable or has coprime counts. Together with the straight lines, which supply a simple loop for every coprime class, that is the whole classification: the simple loops on the torus, up to deformation, are the class $(0, 0)$ and the classes $(p, q)$ with no common factor.

How often two loops must meet #

The argument used the fact that two loops of classes $(p, q)$ and $(r, s)$ cross a net $ps - qr$ times. That is worth seeing, because it is the second half of what the two counts know.

Each crossing of two oriented loops can be given a sign, according to whether the second crosses the first from left to right or right to left. The total of the signs does not change when either loop is deformed, since crossings appear and disappear in cancelling pairs, and for straight loops every crossing has the same sign, so the total is the count. On the square it can be computed directly: the determinant $ps - qr$, the signed area of the parallelogram the two classes span. The figure’s counts are $|1 \cdot 1 - 0 \cdot 0| = 1$, $|1 \cdot 3 - 0 \cdot 1| = 3$, $|2 \cdot 2 - 1 \cdot 1| = 3$ and $|3 \cdot 1 - 2 \cdot 2| = 1$.

Because the signed total is an invariant, no deformation of two loops makes them meet fewer than $|ps - qr|$ times, and the straight loops meet exactly that often, so it is the minimum. Two simple loops can be pulled apart entirely only when $ps = qr$, which for coprime classes means they are the same class or opposite ones — parallel strands. Two simple loops that meet exactly once, like the first and last pairs in the figure, can be cut along to open the torus into a square with the two loops as its edges, and so either pair could have been the pair of edges the square started with.

Slopes in lowest terms #

The simple loops are thus the fractions. Each coprime class $(p, q)$, with its opposite $(-p, -q)$ describing the same loop run backwards, is a slope $q/p$ in lowest terms, including $1/0$ for the vertical loop. The simple loops on the torus are the rational numbers, together with infinity.

In the lattice of all classes, the coprime ones are the points visible from the origin: those with no other lattice point on the segment back to it, since a common factor $d$ would put the point $(p/d, q/d)$ in the way. The figure marks them in a square of side 17. They are most of the points, and as the square grows their share tends to $6/\pi^2$, about 61 per cent — the probability that two whole numbers chosen at random share no factor, which is a statement about the primes, since each prime independently divides both with probability $1/p^2$.

The relation of meeting once has an arithmetic meaning too. Two slopes $q/p$ and $s/r$ whose loops meet exactly once satisfy $ps - qr = \pm 1$, and that is precisely the condition for two fractions to be neighbours in the Farey sequences — the pairs whose mediant is the next fraction to appear between them. The tree of fractions built by mediants is, read on the torus, a tree of simple loops in which each loop meets its parent and its neighbours exactly once.

Loops on a doughnut in space #

The torus of the square is an abstract surface, but the standard doughnut in space is one too, and on it a simple loop of class $(p, q)$ is a closed curve winding $p$ times round the tube and $q$ times round the hole. Such a curve is a torus knot. The class $(2, 3)$ is the trefoil; $(2, 5)$ is the cinquefoil, whose spanning surface is two discs and five bands. A class with a common factor gives no knot at all, because it cannot be drawn without crossing itself — which is the theorem above, visible in space as the reason there is no “$(2, 4)$ torus knot”. The nearest thing is two separate loops of class $(1, 2)$ side by side, a link of two components, and that is what the class $(2, 4)$ becomes if it is allowed to fall apart into pieces rather than cross itself. The classes $(p, q)$ and $(q, p)$ give the same knot, since the doughnut can be turned inside out to exchange the tube and the hole; that symmetry is invisible on the square, where the two are simply different slopes.

The same coprimality turned up, in a different disguise, in a point driven round by two frequencies: a ratio in lowest terms closes up after one period, and the curve it draws is a projection of a straight loop on a torus.

What the pictures cannot show #

The crossings counted are the drawing’s, not the class’s. Each figure counts intersections in the segments it draws, and a drawing can only show that a number of crossings is achievable. That fewer is impossible is the invariance argument for two loops and the cutting argument for one; for the exact minimum $\gcd(p, q) - 1$ in the class of a loop going $d$ times round, the lower bound beyond one crossing is quoted rather than proved. The figures draw one representative in each class and cannot show that no cleverer drawing exists.

The torus is flat in every figure. The square with glued edges is the torus up to deformation, and nothing here depends on how it sits in space; the torus knots are named, not drawn.

And the density $6/\pi^2$ is a limit. The share in the drawn square is a finite count close to it, and the limit itself is Dirichlet’s, quoted from the arithmetic of the primes. Nothing about the torus itself depends on it: it is a count of how common the simple classes are, not a fact about any one of them.

Where the twists come in #

A simple loop of any slope can be carried to a simple loop of any other slope by a symmetry of the torus — not a rigid motion, but a cut, a twist and a regluing. Those twists act on the classes by two-by-two matrices, and the matrices they generate are all the matrices of whole numbers with determinant one, which is the same group whose two generators build the tree of fractions. That essay draws the twist and follows it.

The point of that construction is that the classification here, for all its arithmetic, has a single orbit. Every loop that never crosses itself and does not shrink away is carried to every other one by some symmetry of the torus. What distinguishes $(2, 5)$ from $(1, 0)$ is not anything about the loop but its position relative to the edges of the square, and the edges were a choice made when the torus was built. Change the choice of edges and the counts change with it, by exactly the matrices the twists produce — so the arithmetic of this essay is, underneath, a statement about which changes of edges are allowed.

Two numbers, and a loop that stays clear of itself #

The classes of loops on a torus are the pairs of whole numbers, and the classes that hold a loop without crossings are the pairs with no common factor — slopes in lowest terms. A common factor forces a crossing; coprime counts allow a straight loop that never meets itself, by Bézout’s identity. The determinant of two classes counts the fewest times their loops can meet, and meeting once is the Farey relation between fractions.

The whole of it rested on two facts about the square: that gluing turns a straight line into a closed loop exactly when its slope is rational, and that a line of rational slope revisits its own points only as often as the numerator and denominator allow. Everything else — the forced crossing, the meeting number, the visibility of lattice points — followed from those by arithmetic, and none of it needed a picture of anything more complicated than a straight line folded into a square.

When a topological question reduces to whole numbers, check whether it has reduced to divisibility — here the shape of a loop was decided by a greatest common divisor, and the meeting of two loops by a determinant.