Six sticks tie a trefoil, and five cannot

A knot drawn on paper is a smooth curve, and every argument about it so far has been about its crossings. Build one instead from rigid sticks — straight segments joined end to end, the last back to the first — and a new question appears. The string can no longer bend wherever it likes; it bends only at the corners. What is the fewest number of sticks from which a given knot can be made? That number is the knot’s stick number.

Three sticks make a triangle, which lies in a plane and is unknotted. Any closed chain of sticks can be deformed into many others without passing through itself, so the stick number is a property of the knot rather than of a particular chain, and it is at least three for everything. The questions are how many the simplest knots need, how to prove that fewer will not do, and how the count relates to the crossings that three moves, and what they cannot undo used to tell knots apart.

The answers for the two simplest knots are six for the trefoil and seven for the figure-eight. Both halves of each answer — that the number is enough, and that one fewer is not — can be seen in the figures below, and the second half is the more interesting one.

Six sticks, with whole-number corners #

The six corners are whole-number points in a box ten units on a side. Seen from above, the six sticks cross three times, and at each crossing the stick that is higher at that point is drawn unbroken. The knot is identified the way a polynomial behind the colourings identified knots: from the crossings come the arcs, from the arcs the colouring matrix, and its minor is the determinant, 3. Alexander’s matrix built from the same crossings gives the polynomial $t^2 - t + 1$, the trefoil’s.

A single picture could mislead, so the computation is repeated with the hexagon turned to three other directions before projecting. The crossings differ each time — a turned hexagon casts a different shadow — and the polynomial comes out as $t^2 - t + 1$ every time, as it must for an invariant.

Six is therefore enough. It is also, as it turns out, necessary: no chain of five sticks is knotted at all. The reason is not obvious, and the simplest way to it starts from counting crossings.

Why count sticks at all #

A smooth curve is an idealisation. Anything that is actually knotted — a molecule, a strand of DNA, a chain of linked rods in a mechanism — is built from pieces that are straight or nearly so, joined at angles. In a ring-shaped molecule the sticks are chemical bonds of roughly fixed length, and whether the ring can be knotted depends on how many atoms it has. Jean-Pierre Sauvage and his colleagues made the first synthetic molecular trefoil in 1989 by threading strands round metal ions, and the question of how few units a knotted ring can have is, for chemists, a question about the stick number with extra constraints.

For topology the interest is different. The stick number is a measure of complexity that comes from geometry rather than from a picture. The crossing number of the crossings an alternating knot cannot lose is read off the best diagram; the unknotting number is read off the best sequence of changes; the stick number is read off the best way of placing the knot in space with straight pieces. None of the three determines the others, and comparing them is one of the ways the subject learns what each one sees.

The hexagon above shows three crossings when viewed from above, the fewest a trefoil can show. Turned to other directions it shows more. A hexagon’s picture could in principle have nine crossings, one for each pair of sticks that share no corner; this one, viewed from 100,000 random directions, shows three crossings from about two directions in five, six from about one in three, and never more than seven. A single chain of sticks casts many shadows, and the knot is what they have in common.

Four sticks cannot cross often enough #

In a picture of a chain of sticks, two sticks can cross at most once, since two straight segments meet at most once, and two sticks that share a corner cannot cross at all. With $n$ sticks there are $n(n-3)/2$ pairs that share no corner, so a picture has at most that many crossings: two for four sticks, five for five, nine for six.

Every knot other than the unknot needs at least three crossings in every picture — the trefoil’s three are the fewest any knot has, and a diagram with one or two crossings can always be undone by the moves three moves, and what they cannot undo described. So four sticks cannot tie a knot: whatever direction they are seen from, they show at most two crossings.

The count also gives a lower bound for any knot, and it is instructive how weak it is. A knot whose every picture has at least $c$ crossings needs enough sticks that $n(n-3)/2 \ge c$. For the trefoil, $c = 3$, that asks for five sticks, one short of the truth. For the figure-eight, $c = 4$, it again asks for five, two short. For a knot with a hundred crossings it asks for sixteen, while the best general construction known guarantees only about a hundred and fifty. The count uses only the fact that straight sticks cross at most once, and a knot made of sticks has to satisfy far more than that: its crossings cannot be arranged freely, because the sticks that produce them are rigid and each one takes part in many crossings at once.

The same count tells nothing about five sticks. Five pairs share no corner, so a pentagon can show five crossings, which is as many as the cinquefoil, the five-crossing torus knot, needs. The counting argument stops here, and something with more geometry in it is needed.

A star with five points #

A pentagon that shows five crossings has every pair of non-adjacent sticks crossing, and the only way that happens is a five-pointed star. The star is a familiar knot diagram: two strands wound round each other five times. If every one of the five twists turns the same way, the diagram is a cinquefoil. If four turn one way and one the other, the odd twist cancels one of the others and three remain — a trefoil. If three turn one way and two the other, four cancel in pairs and one twist is left, which is no knot at all.

So the question for five sticks is which of those a star made of straight sticks can be, and the figure’s star is the third kind. The sign of a crossing is the one two loops and one number used to count linking: turn the lower strand’s direction onto the upper’s the short way round, and the crossing is positive if the turn is anticlockwise. Along the star, a twist one way and a twist the other have opposite signs, so the sum of the five signs is the net number of twists. The twists are read from the crossings’ signs, computed from the directions of the two sticks and which is higher; they run $+ - + + -$, net $+1$.

That is not an accident of this pentagon. Among 20,000 random pentagons, 533 show a five-pointed star from above, and in every one of them the twists are three one way and two the other. Straight sticks cannot wind the star tightly enough to make all five twist together — the geometry of five straight segments forbids it — and the crossing count, which could not see the difference, has been overruled by a finer count, the signed one.

Folding a corner flat #

The star pictures are a special case. A proof that no pentagon is knotted has to handle every pentagon, and the cleanest one works in space rather than in a picture.

Take any corner of the pentagon and the triangle it makes with its two neighbours. If no stick passes through that triangle, the corner can be pushed straight across it, dragging its two sticks along, until they lie on the line between the neighbours and become a single stick. Nothing passes through anything during the push, so the knot is unchanged — and the pentagon has become a quadrilateral, which, having only four sticks, is unknotted. The push is not an improvised trick. It is the triangle move, which Reidemeister took as the very definition of deforming a polygonal knot: replace two sides of an empty triangle by the third, or the reverse. Every deformation of a knot made of sticks is a sequence of such moves.

The only stick that can pass through a corner’s triangle is the one that shares no corner with it — the other four sticks all touch the triangle at a vertex, and a straight stick through a vertex of a triangle cannot also pass through its interior unless it lies in the triangle’s plane. So each of the five triangles has exactly one possible intruder. For the argument to work, one of the five triangles must be free, and in the census it always is: 10,745 of the pentagons have all five triangles free and the other 9,255 have exactly one pierced. Not one pentagon has two.

That is the whole argument, and it shows that five sticks cannot knot: some corner can always be folded flat, and four sticks are too few. The right-hand panel shows what the argument looked past. Crossing counts from 0 to 5 all occur, and a picture of a five-stick chain can look as tangled as a cinquefoil’s, but a picture is a shadow, and the corners fold flat in space where the shadow cannot see.

How often six or seven sticks knot #

With six sticks knots appear, and they are rare. Among 20,000 random hexagons with corners in a cube, 102 are trefoils, about one in two hundred. None is anything else. Richard Randell proved in 1994 that none could be: a hexagon is either unknotted or a trefoil, of one handedness or the other.

Seven sticks make both knots more common — 284 trefoils in 20,000 — and allow the figure-eight for the first time, in 12 of them, about one heptagon in 1,700. The determinant used to sort them is the one colours that count more than three met as the number that decides which colourings a knot admits: three colours work for a knot of determinant 3 and five for one of determinant 5. Every polygon in the census has determinant 1, 3 or 5, and those three values sort it completely, because for chains of seven or fewer sticks no other knot can occur — Jorge Calvo proved in 2001 that a heptagon is an unknot, a trefoil or a figure-eight — and those three knots have different determinants.

The 102 trefoils are of two kinds that the census cannot tell apart. A trefoil and its mirror image are different knots — a polynomial that tells left from right needed Jones’s polynomial to prove it — but they have the same determinant and the same Alexander polynomial, which are all the census computes. Reflecting the cube through a plane turns every random hexagon into another equally likely one and every left-handed trefoil into a right-handed one, so the two kinds must occur equally often on average, about fifty of each here.

The rarity is worth noticing. A trefoil can be made from six sticks, but random corners almost never make one: the chain has to wind in a particular way, and a random hexagon usually does not. The stick number is the smallest number that can work, not a number that typically does. How many changes undo a knot drew the same distinction for crossing changes: the minimum is a property of the best possible arrangement, and random arrangements are nowhere near the best.

Seven for the figure-eight #

The figure-eight knot has four crossings in its simplest diagram, one more than the trefoil, and its stick number is seven, one more than the trefoil’s. The seven corners shown are whole-number points in a box twelve units wide, found by searching random heptagons for one whose polynomial is the figure-eight’s, $-t + 3 - t^{-1}$, from every direction tried. Randell’s theorem supplies the other half: a hexagon cannot be a figure-eight, so seven is the fewest.

For larger knots the pattern of “one more crossing, one more stick” does not continue in any simple way. The cinquefoil and the knot $5_2$, with five crossings each, need eight sticks, and so do several knots with six crossings. The counting bound runs the other way: $n$ sticks show at most $n(n-3)/2$ crossings, so the number of sticks grows at least like the square root of the crossing number. Seiya Negami showed in 1991 that twice the crossing number is always enough, and Youngsik Huh and Seungsang Oh improved that in 2011 to one and a half times the crossing number, plus one and a half.

Between a square root and a linear bound there is a great deal of room, and for one family it has been closed exactly. Gyo Taek Jin proved in 1997 that the torus knot that winds $p$ times one way and $q$ times the other, with $p < q < 2p$, has stick number exactly $2q$. The trefoil is the torus knot with $p = 2$ and $q = 3$, and its six sticks are $2 \times 3$.

Still open: whether sticks of equal length need more #

For most knots the stick number is not known. Tables of knots list it exactly for the smallest ones and give only a range — a lower bound from some argument and an upper bound from a chain someone has built — for many knots of nine or ten crossings. The lower bounds are the hard part, as they were for the unknotting number: a construction proves that a number of sticks is enough, and nothing about one construction proves that fewer will not do.

A sharper version of the question asks for sticks that are all the same length, as in a molecule whose bonds are equal. The fewest such sticks is the equilateral stick number, and it is at least the ordinary one, since an equal-length chain is in particular a chain. For the trefoil the two agree: six equal sticks suffice. Whether they agree for every knot is open. No knot is known for which equal lengths cost an extra stick, and no argument shows that they never do — the freedom to choose lengths might be exactly what some knot needs, and there is no invariant yet that can see the difference between a chain that could be made equilateral and one that could not.

What the pictures do not settle #

Every knot type in these figures is computed from crossings, arcs and matrices, and every one is checked from four directions, so an error in reading one picture would have shown up as a disagreement. The census counts are random samples with a fixed seed, and the rates they report — one hexagon in two hundred knotted, one heptagon in 1,700 a figure-eight — are estimates, with an uncertainty of a few percent for the trefoils and much more for the figure-eights.

The claim that no pentagon has two pierced triangles is observed in 20,000 cases, not proved here; the argument needs only one free triangle, and the census never failed to find one. Randell’s theorem about hexagons, and the bounds of Negami, Huh and Oh, and Jin, are quoted. The link to six points in space and a pair that must link runs the other way: there, straight segments forced linking to appear; here, straight segments prevent a knot until there are six of them.