The crossings an alternating knot cannot lose

The crossing number of a knot is the fewest crossings any diagram of it can have, and it is the first number anybody writes beside a knot in a table. It is also, on the face of it, impossible to compute. Any diagram gives an upper bound, since the knot can certainly be drawn with that many crossings. A lower bound is a claim about every diagram of the knot at once — infinitely many, of every size and shape — and the three moves that connect them are allowed to pass through diagrams with far more crossings on the way to one with fewer. No search over diagrams can ever finish.

Peter Guthrie Tait began tabulating knots in the 1870s, for a reason that sounds strange now: Lord Kelvin had proposed that atoms were knotted vortices in the aether, and a table of knots would be a table of the chemical elements. The physics came to nothing. The tables survived, and in building them Tait formed three conjectures that he could not prove, of which the first was the most useful. It says that a diagram whose crossings alternate — over, under, over, under, all the way round — and which is reduced, with no crossing that a half-turn of part of the diagram would undo, already has the fewest crossings its knot allows.

A hundred years later, in 1987, Louis Kauffman, Kunio Murasugi and Morwen Thistlethwaite proved it, independently and within months of each other. All three used the bracket that Kauffman had just written down, and the argument needs only two of its $2^c$ states.

Two states that fix the top and the bottom #

Every state of a diagram contributes a term $A^{a-b} d^{\,|s|-1}$ to the bracket, where $a$ and $b$ count the crossings smoothed the A way and the B way and $|s|$ counts the circles left over. Since $d = -A^2 - A^{-2}$, the largest power of $A$ that one state can produce is $a - b + 2(|s| - 1)$.

Start from the all-A state, where $a = c$ and $b = 0$, and write $|s_A|$ for its circles. Its top power is $c + 2|s_A| - 2$. Any other state is reached by switching some number $k$ of crossings to B, one at a time, and each switch lowers $a - b$ by two while changing the number of circles by exactly one, up or down. At best, every switch adds a circle, which raises $2(|s| - 1)$ by two and exactly repays the loss. So no state can reach a power above

$$c + 2|s_A| - 2,$$

and by the same argument from the other end, no state reaches a power below $-c - 2|s_B| + 2$, where $|s_B|$ counts the circles of the all-B state. The bracket lives inside that window, and its width is

$$\bigl(c + 2|s_A| - 2\bigr) - \bigl(-c - 2|s_B| + 2\bigr) = 2c + 2\bigl(|s_A| + |s_B|\bigr) - 4.$$

For the trefoil in the figure, $|s_A| = 2$ and $|s_B| = 3$, so the window runs from $-7$ to $5$ and is twelve wide — and the bracket $A^{-7} - A^{-3} - A^{5}$ reaches both ends of it.

When the extreme term cannot cancel #

The window is a bound, and a bound can be slack: the all-A state puts a term at the top of the window, but other states might put terms there too and cancel it. The first switch away from all-A decides whether that can happen.

Switching one crossing changes the circles only where that crossing is. In the all-A state, the two arcs of the A smoothing at that crossing belong either to two different circles or to the same circle passing by twice. If they belong to two different circles, the switch joins them into one — the count goes down. Every later switch can raise it by at most one again. A state with $k$ switches then has at most $|s_A| + k - 2$ circles, and its top power is at least four below the top of the window. Nothing else reaches the top, the all-A term $\pm A^{c + 2|s_A| - 2}$ stands alone there, and it survives. A diagram in which that holds at every crossing, at both ends, is called adequate.

The argument says more than that the extreme terms survive: it says what they are. The top coefficient comes from the all-A state alone, so it is $\pm 1$, and the same is true at the bottom. Every bracket computed so far bears this out — the trefoil’s begins at $A^{-7}$ and ends at $-A^{5}$, the figure-eight’s runs from $A^{-8}$ to $A^{8}$, both with coefficient one — and it carries over to the Jones polynomial, which for every reduced alternating knot begins and ends with a coefficient of $\pm 1$. A knot whose Jones polynomial begins with a two cannot be alternating, whatever diagram it is drawn with, and that is a test for alternation that looks at no diagram at all.

When one of the all-A circles does meet itself at a crossing, the first switch can cut it in two and keep pace with the all-A term, and the top term may cancel. That is exactly the picture of a crossing that can be untwisted: a small loop hanging off the rest of the diagram, attached at a single crossing, where one extreme state’s circle runs through the crossing twice. For an alternating diagram the two conditions are the same — adequate means reduced — and that is the first half of the proof.

Euler’s formula counts the circles #

The second half is a count, and it is the one part of the argument that is about pictures rather than algebra.

A knot’s shadow — the diagram with over and under forgotten — is a closed curve in the plane that crosses itself $c$ times. The regions it cuts the plane into can be shaded like a chessboard, with regions sharing an edge always coloured differently: at every crossing four corners meet, and they alternate shaded, unshaded, shaded, unshaded. In an alternating diagram, the A corners at every crossing are always the shaded ones, or always the unshaded ones: alternation is exactly what keeps the upper strand turning the same way into the same colour from one crossing to the next. Say they are the shaded ones. Then in the all-A state every crossing opens a channel between two shaded corners, the shaded regions all run together through those channels, and each unshaded region is left enclosed by a circle of its own — its own boundary, with the crossings smoothed off its corners. The all-A circles are the unshaded regions, one for one, and by the same reasoning the all-B circles are the shaded ones. The trefoil above has two unshaded regions, the centre and the outside, and three shaded lobes: two circles and three.

So for an alternating diagram, $|s_A| + |s_B|$ is the number of regions. And the number of regions is fixed by Euler’s formula: the shadow is a map on the sphere with $c$ vertices, each of degree four, so $2c$ edges, and $V - E + F = 2$ gives

$$F = 2 - c + 2c = c + 2.$$

The first figure shows it for the figure-eight knot. Its diagram has four crossings and so six regions, and the all-A state leaves three circles and the all-B state three, one circle for each region, and the bracket $A^{-8} - A^{-4} + 1 - A^{4} + A^{8}$ reaches from one end of the window to the other with nothing to spare.

Put that into the window’s width: $2c + 2(c + 2) - 4 = 4c$. For a reduced alternating diagram the bracket runs from one end of a window $4c$ wide to the other, and so its span is exactly $4c$. For any diagram at all, alternating or not, the circles of the two extreme states number at most $c + 2$ — the same count, run as an inequality — and the span is at most $4c$.

A surface that measures the distance from alternating #

The inequality for a general diagram has a proof as pleasing as the equality, and it turns the circle count into a shape. Take the all-A circles and float them slightly above the plane, and the all-B circles slightly below. At every crossing the A smoothing above and the B smoothing below are the two ways of resolving the same four ends, and a small saddle — a piece of surface shaped like a mountain pass — joins them, its upper edge following the one and its lower edge the other. Cap every circle with a disc on the outside of the sandwich. The result, introduced by Vladimir Turaev in 1987, is a closed surface with no edge, and the diagram lies on it.

Its Euler characteristic is the number of discs less the number of saddles, $|s_A| + |s_B| - c$, and a closed surface with $g$ handles has characteristic $2 - 2g$. So the surface has

$$g = \frac{c + 2 - |s_A| - |s_B|}{2}$$

handles, and since a number of handles cannot be negative, $|s_A| + |s_B| \le c + 2$ for every diagram. The surface is a sphere exactly when the diagram alternates. Every diagram with one crossing switched in the figures above has $|s_A| + |s_B| = c$, and so lives on a torus: one handle is the price of a single strand passing over twice in a row.

Why no other diagram can do better #

Now the invariance does the rest. The bracket of a diagram changes, under the three moves, only by a factor $-A^{\pm 3}$ — and multiplying by a single power of $A$ shifts every exponent by the same amount, which leaves the span untouched. So the span of the bracket is a property of the knot, the same for every one of its infinitely many diagrams.

Suppose a knot has a reduced alternating diagram $D$ with $c$ crossings. Its span is $4c$. Any other diagram $D'$ of the same knot, with $c'$ crossings, has the same span, and the inequality gives $4c \le 4c'$. So $c' \ge c$: no diagram of the knot has fewer crossings than $D$. That is Tait’s conjecture, and the proof is complete — the search over every diagram has been replaced by one inequality that every diagram satisfies.

The same inequality gives a lower bound for every knot, alternating or not: whatever diagram is in hand, the knot needs at least a quarter of the span of its bracket in crossings. The bound is exact precisely when the knot has a reduced alternating diagram, and for the knots in these figures — trefoil, figure-eight, cinquefoil, the seven-crossing torus knot — it confirms the crossing numbers 3, 4, 5 and 7 that the tables have always given them, now with a proof attached to each.

It is worth seeing the bound fail to be exact, because the failure is informative. The smallest non-alternating knot in the tables, listed as $8_{19}$, is the knot a strand makes winding three times round a torus while it passes four times through the hole. Its Jones polynomial is $t^{3} + t^{5} - t^{8}$, a span of five in $t$ and so twenty in $A$, and the bound says the knot needs at least five crossings. It needs eight. The inequality is true and the gap of three is real: no diagram of $8_{19}$ alternates, so no diagram brings its two extreme states up to $c + 2$ circles, and the bracket’s span always falls short of four times its crossings.

A switched crossing, and a twist left in #

The figure below takes the trefoil and switches one crossing, so that one strand passes over twice in a row and the diagram stops alternating.

The state circles no longer follow the checkerboard. The all-A state leaves one circle and the all-B state two, three between them where alternation would have given five, and the window is only eight wide. Inside it the terms cancel almost to nothing: the bracket is $-A^{3}$, a single power, with span zero. The diagram has three crossings and the invariant says it needs none, and it is right — this is a diagram of the unknot, and a strand that passes over twice in a row can be lifted clear.

The cinquefoil with one crossing switched is a more instructive failure, because it does not collapse. Its span is 12, so the knot needs at least three crossings; its Jones polynomial is the trefoil’s; and the knot is in fact the trefoil, since switching one crossing of a twisted band of five half-twists leaves a band of three. Here both kinds of loss are visible. The two extreme states hold five circles instead of seven, which narrows the window from 20 to 16, and then the all-A term at the top of the window is cancelled by other states, because this diagram is not adequate at that end. What remains is exactly the trefoil’s span, found in a diagram with two surplus crossings.

The last kind of failure keeps alternation and loses reduction. A circle with one twist alternates trivially — it has one crossing — but the crossing is undone by untwisting, the two extreme states hold one and two circles, and the bracket is a single term. The table below lists all three failures beside the diagrams that succeed.

What a diagram cannot say about the others #

No picture in this essay shows that a knot cannot be drawn with fewer crossings. A figure shows one diagram and its states, and the claim is about every diagram the knot has, most of which nobody will ever draw. The whole weight rests on two facts that are not visible in any single picture: that the span is the same for all of them, which comes from the algebra of the three moves, and that it never exceeds four times the crossings, which comes from Euler’s count run as an inequality. The figures are evidence that the bound is tight on the diagrams drawn; the theorem is what makes the tightness mean something.

The argument also stops at alternation, and the limits are sharp. A non-alternating knot has a span below four times its crossing number, so for most knots the bracket gives a lower bound that is not the answer. The smallest knots that cannot be drawn alternating have eight crossings, and for them the crossing number had to be found another way. And one of the simplest-sounding questions remains open: whether the crossing numbers of two knots add when the knots are tied one after the other in the same string. For alternating knots the spans add, so the crossing numbers do; in general nobody knows.

Tait’s other conjectures, and the colouring he was chasing #

Tait made two further conjectures. The second says that two reduced alternating diagrams of the same knot have the same writhe — the sum of their crossing signs, which on a general diagram changes with every twist. It fell within a year, to Murasugi and Thistlethwaite, by the same circle of ideas. The third says that any two reduced alternating diagrams of one knot are related by a sequence of flypes, a move that turns over a tangle hanging between two strands and carries a crossing from one side of it to the other. William Menasco and Thistlethwaite announced a proof in 1991, and it gives an algorithm that decides whether two alternating diagrams show the same knot, something no general method yet does quickly for arbitrary knots.

The old tables had errors, and the most famous one shows how weak the methods were outside alternation. Two ten-crossing knots appeared as separate entries in Charles Little’s table of 1899 and in the tables that followed it, and in 1974 Kenneth Perko, a lawyer working on knots by hand, showed they were the same knot. Both are non-alternating, the region the conjectures never covered.

The checkerboard used above has one more history. Tait shaded knot diagrams because he was working, at the same time, on the four colour problem, and in 1880 he recast it as a question about colouring the edges of a map with three colours. The graph that joins a diagram’s shaded regions through its crossings is still called the Tait graph, and it carries the older invariants too. For an alternating knot, the determinant that counts colourings is the number of spanning trees of the Tait graph: the trefoil’s graph is a triangle, which has three spanning trees, and its determinant is three. The same diagrams are the ones on which Seifert’s surface is already as small as a surface can be, so alternating knots give up their genus, their determinant and their crossing number to a single diagram, where other knots hide all three. Through it the knot’s bracket becomes a value of the same two-variable polynomial of a graph that counts a map’s colourings at another value — so the man who drew the first knot tables and the man who tried to colour every map were one person, working on two problems that turned out, long after he died, to share a polynomial.