ISEGORIA / MATH ENCYCLOPEDIA
Knot theory: telling tangles apart
How to prove two tangled loops are different when no amount of pulling will tell you. Reidemeister moves and what they leave alone, colouring a diagram and the determinant behind it, and the torus knots, the one family where every invariant has a formula.
Before you begin: Algebraic topology and group theory
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Reidemeister moves
A knot is a closed curve in space; a diagram is its shadow on the page with a gap at each under-crossing. Reidemeister showed that two diagrams represent the same knot exactly when one can be turned into the other by planar wiggling and three local moves: R1 adds or removes a kink, R2 pokes one strand across another, R3 slides a strand across a crossing. Choose a move and press Play. The diagram is a real space curve, recomputed frame by frame, and the readout counts its crossings, adds up their signs (the writhe) and evaluates the determinant. Watch the writhe jump by one under R1 while the other two moves leave it alone, and watch the determinant never change: a quantity that survives all three moves is an invariant of the knot itself.
Worked example. The closed braid \(\sigma_1\sigma_2\sigma_1\sigma_2\) is a trefoil drawn with four positive crossings, so its writhe is \(4\) and its determinant is \(3\). The standard three-crossing trefoil has writhe \(\pm3\) and the same determinant. After the R1 kink the readout shows \(5\) crossings, writhe \(5\), determinant \(3\); after R2, \(6\) crossings, writhe \(4\), determinant \(3\).
Watch out. The theorem says that equal knots are connected by a chain of moves; it does not bound the length of the chain, and a diagram may have to get more complicated before it can get simpler. Invariance is only ever proved one move at a time. At the single instant of an R3 move when three strands meet, the diagram is not regular and the readout momentarily counts a triple point.
Writhe is not a knot invariant, yet it appears in the definition of the Jones polynomial. How is that repaired?
The Kauffman bracket \(\langle D\rangle\) is invariant under R2 and R3 but picks up a factor \((-A^{3})^{\mp1}\) under R1. Multiplying by \((-A^{3})^{-w(D)}\) cancels exactly that factor, because R1 changes \(w\) by the same \(\pm1\). Two non-invariants fail in the same way and their product is an invariant.
2. Colouring a knot
Cut a diagram at every under-crossing and it falls into arcs. A tricolouring paints each arc with one of three colours so that at every crossing the three arcs meeting there are all the same or all different. Painting everything one colour always works, and each Reidemeister move carries valid colourings to valid colourings one to one, so the number of tricolourings is an invariant. Choose a knot and click the arcs; the ticks show which crossings obey the rule. The rule at a crossing is the congruence 2·(over) − (in) − (out) ≡ 0 mod 3, and stacking those rows gives the colouring matrix. Any (n−1)×(n−1) minor of it has the same absolute value, the determinant of the knot, and a knot has a colouring with n colours that is not all one colour exactly when gcd(n, det) > 1. Move the slider n to test that.
Worked example. For the trefoil the matrix has rows \((2,-1,-1)\), \((-1,2,-1)\), \((-1,-1,2)\); the minor \(\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=3\) is the determinant. Over \(\mathbb F_3\) the rows are all multiples of \((1,1,1)\), so the kernel has dimension \(2\) and there are \(3^{2}=9\) tricolourings, \(3\) trivial and \(6\) that use all three colours. For the figure-eight the minor gives \(5\), \(\gcd(3,5)=1\), and only the three one-colour paintings satisfy every crossing.
Watch out. Colourability is a coarse test. The trefoil and the stevedore knot \(6_1\) both have nine tricolourings, and the figure-eight, the cinquefoil and the unknot all have three; the determinant separates some of these pairs but not all, and the knots \(6_1\) and \(9_{46}\) share the same determinant. A trefoil and its mirror image are indistinguishable by every quantity on this page. Counting colourings by brute force costs \(n^{a}\) for \(a\) arcs, which is why the readout stops at \(n=7\).
Why does every row of the colouring matrix sum to zero, and what does that say about its rank?
Each row is \(2-1-1=0\). So the all-ones vector is in the kernel over every ring: the constant colourings are always solutions, and the rank of \(M\) is at most \(n-1\). Deleting one row and one column removes exactly that redundancy, and the minor measures how much more than the constants survives; that its absolute value does not depend on which row and column are deleted is the theorem that makes the determinant well defined.
3. Torus knots T(p, q)
Wind a curve p times around the axis of a torus and q times through its hole, and close it up: that is the torus knot T(p, q), one closed curve when gcd(p, q) = 1 and a link of gcd(p, q) components otherwise. Drag the torus to rotate it, and move p and q. On the flat torus beside it the same curve is a straight line of slope q/p, which is why T(p, q) and T(q, p) are the same knot: swapping the two angles is a symmetry of the torus. For this family every hard invariant has a closed form. The crossing number is min(p(q−1), q(p−1)), the genus and the unknotting number both equal (p−1)(q−1)/2, and the Alexander polynomial is a ratio of cyclotomic products. Try (2, 3), (2, 5) and (3, 4), then a pair with a common factor.
Worked example. For \(T(3,4)\): \(\Delta(t)=\dfrac{(t^{12}-1)(t-1)}{(t^{3}-1)(t^{4}-1)}=t^{6}-t^{5}+t^{3}-t+1\), degree \(6=2g\) with \(g=3\), crossing number \(\min(9,8)=8\), and the closed 3-braid \((\sigma_1\sigma_2)^{4}\) has exactly \(8\) crossings, so that drawing is minimal. For \(T(2,3)\), \(\Delta=t^{2}-t+1\) and \(g=1\): the trefoil bounds a torus with one hole.
Watch out. The genus formula goes back to Seifert and the unknotting number is the theorem of Kronheimer and Mrowka (the Milnor conjecture); neither has an elementary proof, and the readout quotes them rather than computing them from the curve. The Alexander polynomial is defined only up to a unit \(\pm t^{k}\). For a link with \(d>1\) components the readout shows the reduced one-variable polynomial \((t-1)(t^{pq/d}-1)^{d}/((t^{p}-1)(t^{q}-1))\) and the genus of the fibre surface \(((p-1)(q-1)+1-d)/2\); the unknotting formula is stated for knots only. The drawing has \(q(p-1)\) crossings, so it is minimal only when \(p\le q\).
The unknotting number of T(p, q) equals its genus. Why is one of these inequalities easy and the other hard?
\(u\le g\) is elementary, since \((p-1)(q-1)/2\) well-chosen crossing changes of the braid diagram unknot it, while \(u\ge g\) needs a lower bound on \(u\), and lower bounds are what invariants are for. For torus knots that bound came only in 1993, from gauge theory.