A polynomial that tells left from right

Tie a trefoil in a rope, and then tie another while watching the first in a mirror. The two knots look alike in every way a description in words can capture — three crossings, one loop through another, the same threefold symmetry — and they differ in the way a left glove differs from a right one. Whether one can be deformed into the other, without cutting the rope, is a question about the knot and not about the picture, and the picture gives no help with it.

Every invariant met so far is blind to the difference. Three colours colour both trefoils in exactly the same number of ways. Colourings modulo a prime are counted by the determinant, and reflecting a diagram does not change a single row of the system that computes it. The Alexander polynomial is symmetric — it reads the same with $t$ replaced by $1/t$ — and reflection does exactly that replacement, so the mirror image has the same polynomial. Even the surface a knot bounds is no help, since the mirror of a surface has the same number of handles.

Max Dehn proved in 1914 that the two trefoils are different knots, and the proof was a hard argument about the symmetries of the knot’s group. Half a century later the signature, a number read from how curves on a Seifert surface link one another, gave a second proof, still by way of a surface and a matrix. Then, in 1984, Vaughan Jones found a polynomial by a route through operator algebras that had nothing to do with knots on its surface, and it separated the two trefoils in a line. Three years later Louis Kauffman showed that the same polynomial could be computed by a child with a pencil: smooth every crossing, count circles, add.

Two ways to remove a crossing #

A crossing is two strands, one passing over the other, and there are exactly two ways to cut both strands at the crossing and reconnect the four loose ends without making a new crossing. Seifert’s construction chose between them using arrows on the strands. Kauffman’s chooses between them using nothing but which strand is on top, so it needs no arrows at all.

Stand at the crossing and turn the upper strand counter-clockwise, about the crossing point, until it lies along the lower one. It sweeps through two of the four corners around the crossing, opposite each other, and those are labelled A; the other two are B. The A smoothing cuts the crossing open so that the two A corners become one region; the B smoothing does the same for the two B corners. Nothing about this depends on which way the strands run or on how the crossing is turned on the page, since turning the whole picture turns the upper strand with it.

The rule is then written as an equation about a symbol $\langle D \rangle$, the bracket of a diagram $D$:

$$\langle \text{crossing} \rangle = A\,\langle \text{A smoothing} \rangle + A^{-1}\,\langle \text{B smoothing} \rangle .$$

Here $A$ is a variable, and the equation is applied at one crossing while the rest of the diagram is left alone. Each application replaces one diagram by two diagrams with one fewer crossing, so after $c$ applications a diagram with $c$ crossings has become $2^c$ diagrams with none. A diagram with no crossings is a collection of disjoint circles, and the last rule says what those are worth: each circle beyond the first contributes a factor $d$, where $d$ is a fixed expression in $A$ that the next section will be forced to choose.

Eight pictures of the trefoil, and what each is worth #

Each of the $2^c$ end results is called a state: a choice, at every crossing, of the A smoothing or the B smoothing. A state with $a$ crossings smoothed the A way and $b$ the B way leaves some number of circles, $|s|$, and contributes the term $A^{a-b}\,d^{|s|-1}$. The bracket is the sum over all states:

$$\langle D \rangle = \sum_{\text{states}} A^{a-b}\, d^{\,|s|-1}.$$

That formula is the whole definition. It says nothing about knots being the same; it is a sum over pictures, and whether it means anything is a separate question.

The trefoil has three crossings and so eight states, and the figure draws them all. The pattern in the circle counts is worth reading before any algebra. Smoothing all three crossings the A way leaves two circles, one inside the other. Switching any single crossing to B merges them into one. Switching a second splits them again, and switching all three leaves three separate circles. Each change of one smoothing changes the number of circles by exactly one, up or down, because it either joins two circles into one or cuts one into two; it can do nothing else, since it only reconnects the four ends at a single point.

The numbers of states with three, two, one and no A smoothings are $1, 3, 3, 1$ — the row of Pascal’s triangle that counts the ways of choosing which crossings to switch. For a diagram with $c$ crossings the states are the subsets of the crossings, and the bracket is a sum over all $2^c$ subsets. That is also why nobody computes it this way for large knots: a knot with forty crossings has more than a million million states, and no general shortcut is known.

Why a circle is worth $-A^2 - A^{-2}$ #

The bracket is to be an invariant, so it must survive the three Reidemeister moves, and the second move decides $d$ on its own. The second move lays one strand over another to create two crossings, or pulls them apart to remove them. Expand the bracket of the two-crossing picture at both crossings. There are four combinations. One of them — A at one crossing and B at the other — leaves the two strands pulled apart, with coefficient $A \cdot A^{-1} = 1$: this is the picture the move should produce. The other three all leave the same different picture, the two strands turned back on themselves: both smoothings A, with coefficient $A^2$; both B, with coefficient $A^{-2}$; and the other mixed choice, which also traps a small closed circle between the strands and so carries coefficient $d$. Together they contribute $(A^2 + A^{-2} + d)$ times the unwanted picture.

For the bracket to be unchanged by the move, the unwanted picture must drop out, and the only way is

$$d = -A^2 - A^{-2}.$$

There is no freedom here: the value of a circle is not a convention but the one number that makes the second move invisible. With it the third move follows too, by expanding one crossing of the triangle and applying the second move to both halves. Two of the three moves are paid for by one choice.

The table above does the addition for the trefoil. The single all-A state contributes $A^3 d = -A^5 - A$. The three states with two A smoothings leave one circle each and contribute $3A$. The three with one A smoothing leave two circles each and contribute $3A^{-1}d = -3A - 3A^{-3}$. The all-B state leaves three circles and contributes $A^{-3}d^2 = A + 2A^{-3} + A^{-7}$. Every power of $A$ in the middle of the range is attacked from several rows at once, and the coefficients of $A^1$ — namely $-1$, $+3$, $-3$ and $+1$ — cancel completely, as an alternating sum along a row of Pascal’s triangle must. What survives is

$$\langle \text{trefoil} \rangle = A^{-7} - A^{-3} - A^{5}.$$

Of the fourteen terms written out, three are left. That pattern of heavy cancellation is typical, and it is where the information is: the bracket is small because the states conspire, and the way they conspire is a property of the knot.

The first move, and a correction that costs one number #

The first Reidemeister move removes a small twist — a kink — from a strand, and here the bracket fails. Expanding the kink’s single crossing gives two pictures: one smoothing leaves the strand straight, and the other leaves the strand straight with a small circle beside it. Which is which depends on the direction of the twist, and for one direction the result is

$$\langle \text{kink} \rangle = A + A^{-1} d = A - A - A^{-3} = -A^{-3},$$

times the bracket of the straightened strand. A kink twisted the other way gives $-A^{3}$ instead.

So the bracket is not an invariant: it multiplies by $-A^{\pm 3}$ every time a twist is added or removed. The repair needs one more number read off the diagram. Give the knot a direction, and at each crossing record $+1$ or $-1$ according to whether the lower strand passes from right to left or from left to right beneath the upper one, looking along the upper strand. The sum of these signs over all crossings is the writhe $w(D)$. The second and third moves create or destroy crossings only in cancelling pairs, or rearrange them without changing their signs, so they leave the writhe alone; the first move changes it by exactly $\pm 1$, matching the $-A^{\pm3}$ it does to the bracket. A consistent choice of sign is what orientation always buys, and here it buys a correction factor:

$$f(D) = (-A^3)^{-w(D)}\,\langle D \rangle$$

is unchanged by all three moves, and so is an invariant of the knot. Reversing the knot’s direction flips both strands at every crossing, which leaves every sign as it was, so the choice of direction does not matter either.

The twisted circle in the figure checks the repair. Its single crossing has sign $-1$, so its writhe is $-1$, and $(-A^3)^{1} \cdot (-A^{-3}) = 1$ — exactly the value an untwisted circle has. The correction has not been fitted to this example; it was fixed by the general kink, and it lands on the right answer for a picture it was never shown.

For the trefoil drawn above the writhe is $+3$, and multiplying by $(-A^3)^{-3} = -A^{-9}$ gives $-A^{-16} + A^{-12} + A^{-4}$. Every exponent is a multiple of four, which is true of every knot and is why the customary substitution $t = A^{-4}$ loses nothing. After it, the result is the Jones polynomial:

$$V_{\text{trefoil}}(t) = t + t^3 - t^4.$$

The mirror image, and the reason it differs #

Reflecting a diagram swaps which strand is on top at every crossing. That turns every A corner into a B corner, so every state with $a$ A smoothings becomes the state with $a$ B smoothings, and the bracket of the mirror image is the bracket of the original with $A$ replaced by $A^{-1}$. Reflection also reverses the sign of every crossing, so the writhe changes sign. The two changes together replace $t$ by $1/t$:

$$V_{\text{mirror}}(t) = V(1/t) = t^{-1} + t^{-3} - t^{-4}.$$

The two polynomials are different, so the two trefoils are different knots, and the figure puts them side by side with the Alexander polynomial $t - 1 + t^{-1}$ that both share. What took Dehn a paper about groups is a calculation that fits on a napkin. The point is not that the arithmetic is shorter but that the invariant has a direction built into it: the A and B corners are defined by turning counter-clockwise, and a reflection is exactly what swaps counter-clockwise for clockwise. Every invariant met before this one was built from rules that are the same in a mirror.

The consequence runs in one direction only. A knot that can be deformed into its own mirror image — an amphichiral knot — must have a Jones polynomial unchanged by $t \mapsto 1/t$. The figure-eight knot, whose diagram has writhe $0$, has $t^{-2} - t^{-1} + 1 - t + t^{2}$, symmetric as required, and it is in fact amphichiral: a figure-eight tied in rope can be turned into its mirror image by a sequence of moves that can be followed by hand. The cinquefoil and the seven-crossing torus knot have polynomials that are visibly lopsided, so neither is its own mirror, and the table settles that as quickly as it settled the trefoil. The symmetries of a square include reflections; the symmetries of most knots do not, and a lopsided polynomial is the certificate.

Changing one crossing at a time #

The bracket’s defining rule also gives a way to compute without listing states. Take a diagram and look at one crossing. Call the diagram $L_+$ when that crossing is positive and $L_-$ when it is switched to negative, with everything else unchanged, and call the two smoothings of that crossing $L_0$ (the one that respects the strands’ directions) and $L_\infty$ (the one that does not). Switching the crossing swaps which smoothing is the A one, so

$$\langle L_+ \rangle = A\langle L_0\rangle + A^{-1}\langle L_\infty\rangle, \qquad \langle L_- \rangle = A^{-1}\langle L_0\rangle + A\langle L_\infty\rangle .$$

Multiply the first by $A$, the second by $A^{-1}$, and subtract: the unoriented smoothing $L_\infty$ disappears, leaving $A\langle L_+\rangle - A^{-1}\langle L_-\rangle = (A^2 - A^{-2})\langle L_0\rangle$. Folding in the writhe, which differs by one between $L_+$ and $L_0$ and by one the other way for $L_-$, turns this into Jones’s skein relation:

$$t^{-1}V(L_+) - t\,V(L_-) = \left(t^{1/2} - t^{-1/2}\right)V(L_0).$$

The smoothing $L_0$ is usually a link of two loops rather than a knot, which is where the half-powers of $t$ come from; for knots they never survive. The relation has the shape of the rule that computes a graph’s colourings by deleting and contracting one edge at a time, and like that rule it turns one hard computation into two easier ones: switching crossings until the knot falls apart into unknots, and smoothing them until there are none left. It is how the polynomial is usually computed by hand, since a switch and a smoothing at a well-chosen crossing can reduce a knot to unknots in a handful of steps where the states would number in the hundreds.

Eight pictures that prove nothing on their own #

The states figure shows eight pictures, and none of them shows invariance. The bracket is defined from one diagram, and a diagram is not a knot: the same trefoil has infinitely many diagrams, with any number of crossings, and each one has its own $2^c$ states with their own circles. Nothing visible in the eight pictures says that a diagram of the trefoil with thirty crossings, and a billion states, will produce $t + t^3 - t^4$ as well. That fact lives entirely in the algebra of the two moves above — the value of $d$ that makes the second move vanish, and the writhe that absorbs the first — and it is the reason the definition is interesting rather than merely computable.

Equally, the hero figure shows two trefoils that look different only because one is drawn as the other’s reflection, and a reader can stare at them indefinitely without learning whether some clever deformation would turn one into the other. The eye is no help with that question, and never has been; it took the whole of Dehn’s argument, and now takes a polynomial.

The polynomial also has limits, and they are stated rather than hidden. A symmetric Jones polynomial does not prove a knot is amphichiral: the knot listed as $9_{42}$ in the standard tables is not its own mirror image, and its Jones polynomial cannot tell. And whether the Jones polynomial detects the unknot — whether $V = 1$ forces the knot to be unknotted — is the open question the Alexander polynomial left behind, still unanswered after checking every knot up to twenty-four crossings. There is a sharper version of the polynomial that does detect it. In 2000 Mikhail Khovanov built, from the same states and the same circles, a sequence of groups whose alternating count of sizes recovers the Jones polynomial, and in 2011 Peter Kronheimer and Tomasz Mrowka proved that these groups are trivial only for the unknot. The information was in the states all along; the polynomial adds it up too early.

The same sum in a physics paper and in a colouring count #

The bracket’s shape — a sum over all ways of making a local choice at every site, each weighted by a power of a variable and by a factor for every connected piece — is the shape of a partition function in statistical mechanics, where the sites are atoms, the choices are spins, and the sum is over every configuration of the material. The resemblance is not an accident of notation. The algebra at the centre of Jones’s construction had appeared more than a decade earlier, in Temperley and Lieb’s work on the Potts model of a magnet, which is a sum of exactly this kind.

The resemblance became a theorem in 1987, when Morwen Thistlethwaite showed that the Jones polynomial of an alternating knot is a value of the Tutte polynomial of a graph drawn from its diagram — shade the regions of the diagram like a chessboard and join the shaded regions through the crossings. The Tutte polynomial is a two-variable polynomial of a graph, defined by deleting and contracting edges, and it contains the chromatic polynomial that counts a map’s colourings as a different value of the same two variables. So one polynomial of one graph answers two questions that seem to have nothing to do with each other: how many ways a map can be coloured with $k$ colours, and whether a knot is its own mirror image. Both are evaluations, at different points, of a count over the same pictures.

That kinship also explains the $2^c$. Computing the Tutte polynomial of a general graph at most points is as hard as any counting problem is believed to be, and the Jones polynomial inherits that hardness: no method is known that does fundamentally better than summing something exponential. There is one use of the bracket where that does not matter, because only the two extreme states — all A and all B — are needed, and it settles a question about crossing numbers that Tait asked in the 1880s and that stood for a century.