Three sines tie a knot

A sine wave is a circle seen from the side, and when two circular motions come home drove a point across a page with one sine wave and up and down with another, and found that when the two frequencies are whole numbers the point retraces a closed figure — a Lissajous figure — whose crossings can be counted in advance: $2ab - a - b$ of them for frequencies $a$ and $b$ with no common factor and a generic phase. It ended by asking what happens when a third sine drives the point in the third direction of space.

The answer is a closed curve in space, and a closed curve in space is a knot. At every crossing of the Lissajous figure seen from above, one strand passes over the other, and which one is decided by the third coordinate. So a choice of three frequencies and three phase shifts produces a knot, and the question is which knots can be produced.

The trefoil cannot. Neither can the figure-eight knot, nor most knots with up to seven crossings. The reason is a symmetry that every curve of three sines has, and the symmetry leaves a mark on the knot’s polynomial that the trefoil does not carry. This essay draws the curves, identifies the knots they tie, and checks the mark on every prime knot of seven crossings or fewer.

Five-two, from frequencies two, three and seven #

Take frequencies 2, 3 and 7 with phases $0.1$, $0.7$ and $0.3$. Seen from above, the curve is the Lissajous figure of frequencies 2 and 3, with its $2 \cdot 2 \cdot 3 - 2 - 3 = 7$ crossings. At each crossing, two different values of $\theta$ project to the same point, and the value of $\cos(7\theta + 0.3)$ at each decides which strand is higher. The figure draws the lower strand broken, as a knot diagram does.

A diagram with seven crossings is enough to compute the invariants a polynomial behind the colourings introduced. Each crossing gives a row of Alexander’s matrix; deleting one row and one column and taking the determinant gives the polynomial, here $2t - 3 + 2t^{-1}$. At $t = -1$ it evaluates to $-7$, so the determinant — the number that counts colourings — is 7. Among knots of up to seven crossings only one has this polynomial: $5_2$, the knot with five crossings that is not the cinquefoil.

The diagram itself is not the simplest picture of $5_2$, which needs only five crossings. The sine curve has seven, two of which could be undone by the moves of three moves, and what they cannot undo. But the polynomial does not care which diagram it is computed from, and neither does the knot.

A census of curves and their knots #

Most choices of frequencies and phases give the unknot, and a search has to look past many of them. The census above lists eight curves found by trying frequency triples up to 9 with random phases. Alongside the unknot and $5_2$ are $6_1$, from frequencies 2, 3 and 5 at other phases; the knot $7_4$ from 3, 7 and 2; and curves whose polynomials are those of $7_5$, of the square knot and of the ten-crossing twist knot $10_1$.

The phrasing matters there. For the smaller knots the polynomial settles the identity, because no other knot of up to seven crossings shares it. For $7_5$, the square knot and $10_1$ the census computes the polynomial and nothing more, and several larger knots share each of those polynomials. What the curve with frequencies 2, 7 and 9 has been shown to have is the polynomial of $7_5$; whether it is $7_5$ is a separate question the polynomial cannot answer.

The diagrams grow quickly. A Lissajous figure of frequencies 3 and 7 has $2 \cdot 21 - 10 = 32$ crossings, and Alexander’s matrix of that size is too large to expand by hand or even symbolically. The census computes each determinant modulo a large prime at many whole-number values of $t$ and recovers the polynomial by interpolation, and wherever a diagram is small enough to expand exactly, the two methods were run side by side and agreed.

The symmetry that half a period performs #

Replace $\theta$ by $\theta + \pi$. Each coordinate $\cos(k\theta + \varphi)$ becomes $\cos(k\theta + \varphi + k\pi)$, which is the same when $k$ is even and the negative when $k$ is odd. Since the three frequencies have no common factor, at most one of them is even.

If one frequency is even — say the first — the substitution sends $(x, y, z)$ to $(x, -y, -z)$. That is a rotation by half a turn about the $x$-axis, and it carries the curve onto itself, because $\theta + \pi$ runs over the same circle of values as $\theta$. So every Lissajous knot with an even frequency is symmetric under a half-turn about a line, and the line does not meet the knot. The figure shows it for $5_2$: points half a period apart sit opposite each other through the centre of the end-on view.

If all three frequencies are odd, the substitution sends $(x, y, z)$ to $(-x, -y, -z)$, a reflection through the origin. A reflection reverses the handedness of space, so a knot carried onto itself by one is its own mirror image. Michael Bogle, John Hearst, Vaughan Jones and Lubomir Stoilov, who introduced Lissajous knots in 1994, drew both consequences: every Lissajous knot is either symmetric under a half-turn or equal to its mirror image.

Two conditions the symmetry forces #

A symmetry of a knot constrains its invariants, and for these two kinds of symmetry the constraints have been worked out. Two were drawn in the 1990s. First, the knot’s Arf invariant vanishes — equivalently, its determinant leaves remainder 1 or 7 on division by 8. Second, its Alexander polynomial is a perfect square modulo 2: reduced to coefficients 0 and 1, it is the square of another polynomial, up to a power of $t$.

The second condition is easy to test by eye. In arithmetic modulo 2, $(a + b)^2 = a^2 + b^2$, so the square of a polynomial has only even powers of $t$. A polynomial modulo 2 is a square exactly when all its odd-power coefficients vanish (after shifting so that the powers start at nought). For $5_2$, $2t^2 - 3t + 2$ reduces to $t$ — a single term, which after shifting is 1, a square. For the trefoil, $t^2 - t + 1$ reduces to $t^2 + t + 1$, which has an odd power: not a square.

The table applies both tests to every prime knot of seven or fewer crossings. The trefoil, the figure-eight and the cinquefoil fail the square test. So do ten of the fourteen. Only four pass both: $5_2$, $6_1$, $7_4$ and $7_5$. So the trefoil and the figure-eight are not Lissajous knots, and neither are most of the small knots, however the frequencies and phases are chosen.

The four that pass have each appeared in the census, with the matching polynomial. For $5_2$, $6_1$ and $7_4$ that settles it, since the polynomial identifies them among small knots and the census curves’ own diagrams have seven, seven and 32 crossings. For $7_5$ the curve found has the right polynomial and nineteen crossings, and confirming that it is $7_5$ rather than a larger knot with the same polynomial would need a finer invariant.

What the two conditions measure #

The first condition is about the determinant, the number colours that count more than three met as a count of colourings: a knot can be coloured modulo a prime $p$ in a non-trivial way exactly when $p$ divides its determinant. A Lissajous knot must have determinant $\equiv \pm 1 \pmod 8$. The trefoil’s determinant is 3 and the figure-eight’s is 5, and both fail. The determinant of $5_2$ is 7 and of $6_1$ is 9, and both pass.

Behind that condition is the Arf invariant, a single bit attached to a knot by the surface it bounds. The surface a knot bounds built such a surface from any diagram, and the way curves on the surface link one another defines a form whose Arf invariant — nought or one — does not depend on which surface was chosen. Raymond Robertello defined it for knots in 1965, and Jerome Levine showed in 1966 that it is nought exactly when the determinant is $\pm 1$ modulo 8, which is why the test can be read off the polynomial.

The figure-eight knot is a useful case. It is its own mirror image — amphicheiral, in the language of a polynomial that tells left from right — so it has one of the two symmetries a Lissajous knot can have, at least in the weak sense that some deformation carries it to its mirror. It still fails both tests. The symmetry a Lissajous knot has is stronger: a single reflection of space through a point carries the curve onto itself, point for point, with no deformation in between, and that rigid symmetry is what the conditions require.

Turning one phase #

The frequencies fix the shadow; the phase shifts decide the knot. Turning the third phase $\varphi$ moves every strand up or down while leaving the view from above unchanged, and the knot stays the same until two strands pass through each other at a crossing. At that instant the curve meets itself, and afterwards the crossing has changed.

For frequencies 2, 3 and 7 the figure records the knot type at 180 values of $\varphi$. The curve is $5_2$ over two arcs of the circle of phases — one of them wrapping past $\varphi = 0$ — the unknot over two, and $7_4$ over two more, and the type changes six times. Between changes, $\varphi$ can be turned freely without changing the knot, which is why a census needs only one generic phase from each arc.

The shadow constrains more than the knot type. A diagram of seven crossings allows $2^7 = 128$ ways of choosing which strand is over at each, but the third sine does not choose freely: the heights along the curve are values of one smooth function, $\cos(7\theta + \varphi)$, and the choices at different crossings are linked. Only a few of the 128 diagrams are ever realised, and the symmetry of the previous section is one of the links.

Why the curve is an honest knot #

A curve of three sines could in principle pass through itself, and then it would not be a knot at all. That happens only on a thin set of phases: two parameter values $\theta_1 \ne \theta_2$ must give equal values in all three coordinates, which is three equations in two unknowns, so for a generic choice of phases it has no solution. The sweep of $\varphi$ above crosses such a set at each of its six changes, and nowhere else.

The computation treats the smooth curve as a polygon — six hundred points joined by straight segments — which is the setting of six sticks tie a trefoil, and five cannot. The polygon has the same knot type as the curve provided no segment passes through another during the straightening, and with segments this short compared with the smallest distance between strands, none does; halving the number of points, or doubling it, gave the same polynomial for every curve in the census. The Lissajous knot $5_2$, which as a smooth curve has no corners at all, is in this sense a polygon of six hundred sticks, far more than the eight that $5_2$ needs.

That comparison is itself worth a remark. A Lissajous curve is a very economical description — six numbers — but a very uneconomical shape: the diagram has more crossings than the knot needs, and the curve winds more than a minimal one would. The description is short because it is rigid, and the rigidity is what the symmetry exploits.

The same knots from a ball in a box #

A Lissajous curve has an unexpected twin. Put a ball in a cube and send it off in a straight line; each time it hits a wall it reflects. Each coordinate of the ball then moves back and forth between two walls at a steady speed — a triangle wave instead of a sine wave, but a wave with the same frequency. With the three speeds in whole-number ratios the path closes up, and a closed path in a cube is again a knot.

A bounce is a fold of the table showed why a billiard path is a straight line in disguise: unfold the table across each wall it hits and the path becomes straight. Vaughan Jones and Józef Przytycki showed in 1998 that the billiard knots in a cube are exactly the Lissajous knots: replacing each triangle wave by the cosine of the same frequency and phase changes the curve but not the knot. So the same list of knots can be generated by three sines or by a ball in a box, and the symmetry argument is the same in both — shifting by half a period reverses each odd-frequency coordinate.

That connection places the question among the questions about billiards that the word a straight line spells and its neighbours asked. A billiard path is described by the sequence of walls it hits; a Lissajous knot by three frequencies and three phase shifts. Both descriptions are finite and exact, and both produce objects whose shape is hard to predict from the description.

Still open: which knots can be drawn #

Among knots with up to seven crossings, the two known conditions and a search decide everything: four knots pass the tests and all four are Lissajous. Beyond that range, which knots are Lissajous knots is not known. Christoph Lamm and others have found Lissajous representations for many knots of eight, nine and ten crossings, and the tests rule out many more, but there are knots that pass every known test and for which no representation has been found and none has been ruled out.

The difficulty is that the tests only use the symmetry, while the realisation needs specific frequencies and phases, and nothing connects the two in general. A knot might pass both tests because it has the right symmetry, and still need frequencies larger than any search has tried — or fail to be Lissajous for a reason no invariant yet detects. The search space grows quickly: frequency triples up to 9 already give hundreds of shadows, each with dozens of crossings and several phase arcs, and the diagrams of larger frequencies run to hundreds of crossings.

Other questions surround it. Whether every knot with the half-turn symmetry and the right Arf invariant is Lissajous; whether the frequencies needed for a knot can be bounded in terms of its crossing number; whether a finer invariant than the polynomial modulo 2 separates Lissajous knots from the rest. Each is open, and each is a question about how much a knot’s symmetry determines about its shape.

What the pictures cannot show #

Each knot type here is computed exactly from a diagram: the crossings are found by exact comparison of segments of a finely sampled curve, and the polynomial by exact modular arithmetic. The sampling is fine enough that the crossing counts match the formula $2ab - a - b$ in every case, and the knot types did not change when the sampling was doubled.

The census is a sample, not a classification. It found the four small knots the tests allow, but it did not search every frequency triple or every arc of phase shifts, and it cannot show that a knot it did not find is not Lissajous. The two conditions are quoted from the literature. The table’s verdicts are computed from them; the proof that the conditions follow from the symmetry is not reproduced here, and the knots’ identities for $7_5$, the square knot and $10_1$ rest only on their polynomials.