Geometric Topology 9

Tangles and Mutants

The notion of a tangle was introduced by John Conway around 1970 as a means to construct knots. A $2$-tangle is part of a diagram contained in a disk, which exits the boundary at $4$ points, which here could be labelled NE,SE,SW,NW. Rational tangles have a sequence of twists inside the disk, as in a pretzel knot, and (remarkably) can be characterised by an associated continued fraction.

Given a knot $K$ incorporating a tangle, a mutant can be formed by (for example) rotating in space only that part in the disk. The two are distinguished by the least genus of the oriented surfaces they bound, or by the fundamental group of their complements in $\mathbb{R}^3$.

The Alexander Polynomial

The first polynomial to represent the ambient isotopy class of a link $L$ was discovered by J.W. Alexander in 1923. It was (and often still is) introduced by labelling regions of a diagram and examining colouring-type equations:

where $(x_2,x_4)$ is an oriented overpass, and $\Delta_L(t)$ plays the role of the determinant. The Alexander polynomial $\Delta(L) = \Delta_L(t)$ is no longer an integer, but a Laurent polynomial in a variable $t$, well-defined up to multiplying by an integer power $t^k$. The determinant of $L$ is recovered by taking $t=-1$ (so what we previously called $\Delta$ is now $\Delta_L(-1)$).

Conway showed in 1969 that the Alexander polynomial satisfies a simpler version of the skein relation, namely:

Although neither $V(L)$ nor $\Delta(L)$ can be obtained from the other, a $2$-variable 'HOMFLY' polynomial defined by an enhanced skein relation generalises both of them.

Knot DT Codes

Let $D$ be an oriented diagram with $c$ crossings. Choose a starting point so that the first crossing encountered is an underpass, and chess-board the diagram so that white is on the right at the start. Number the crossings $1,2,3,...,c,c+1,c+2,...,2c$ as they are encountered in succession around the circuit. If a crossing is labelled with an odd number the first time it is encountered then it will have an even label the next time and vice-versa. This is an application of the existence of chess-boarding. Thus, this set of crossings determines a bijective mapping $f: \{1,3,5,...\} \rightarrow \{2,4,6,...\}$, which we abbreviate by the the ordered list:

If $D$ is alternating, the overpasses will all be labelled by the even integers $2,4,...,2c$. In general one adds a minus sign to the even integer if it labels an underpass.

The resulting list of even integers is called the Dowker-Thistlethwaite code or DT code of the knot. Of the $2^cc!$ possibilities, many will give rise to an unknot and few will correspond to a knot $K$ whose crossing number $cr(K)$ equals $c$.

Example: The Figure Eight Diagram With an orientation on the figure-eight diagram for $4_1$, it does not matter before which underpass we start. We always return to the initial crossing after $5$ more subarcs and the DT-code is $6,8,2,4$. With the opposite orientation it is $4,6,8,2$. The code $2,4,6,8$ gives the unknot.

Example: $8_{19}$

This is the first non-alternating knot in the Rolfsen table. It is formed by linking together two open trefoil knots and then closing them. It is clear from the diagram of the knot that $8_{19}$ is the pretzel knot $P(3,-2,3)$. Thus it has determinant $|-6+9-6|=3$ and is therefore $3$-colourable. It is sometimes called the "true lovers knot". With a given orientation its DT code is:

Minus signs record that circuit markers $10$ and $2$ occur at underpasses.

Reconstructing a Diagram

If we are given a DT code we can reconstruct a knot diagram (and its mirror) up to ambient isotopy. We can construct a knot diagram given a DT code, however it is known that the DT code of a prime knot will uniquely specify the ambient isotopy class of the knot up to reflection. However if we consider the DT code $4,6,2,10,12,8$. The alternating diagrams $L3_1 \# R3_1$ and $R3_1 \# R3_1$ both have this code, however they are not mirror images. The composite nature of the knots can be recognised by observing that the DT code pairs numbers within the two subsets $\{1,2,3,4,5,6\}$ and $\{7,8,9,10,11,12\}$ of crossing labels. The example then exploits the ambiguity in reconstructing a knot and its mirror image.

Surfaces from Knots

Let $D$ be a knot diagram with $c$ crossings and give it a chess boarding. This defines a 'cloth' surface $\mathscr{M}$ in $\mathbb{R}^3$ by interpreting each crossing vertex as hiding a 180-degree twist in the cloth. In practice, one could manufacture such a surface by first taking disjoint pieces of cloth to match the black regions, and then joining them with a slim ribbon twisted by 180 degrees for each crossing. If the ribbon is attached to one region, its free end should be twisted anti-clockwise if and only if the crossing has chess-sign $+1$. Provided the twist matches the strands of the crossing in this way, the boundary of $\mathscr{M}$ will be the knot we started with.

Orientation

The surface $\mathscr{M}$ is called two-sided in space if one can consistently distinguish two sides throughout, so that if one side is coloured black and the other grey, the two colours only meet at the surface's boundary if there is one. This is possible if all circuits from one region back to itself pass through an even number of crossings.

Two-sided surfaces are orientable, meaning there is a well-defined notion of clockwise rotation at any point. Any surface without boundary in $\mathbb{R}^3$ (such as the sphere or torus) has this property because it has an inside and an outside. But orientability is a characteristic of the abstract surface, not just the way it sits in space.

THe trefoil diagram chessboarded with three black regions gives a one-sided surface. It is in fact a $3$-twisted Mobius band in $\mathbb{R}^3$, though it is homeomorphic to the usual $1$-twisted Mobius band. One-sided surfaces are non-orientable: it is possible to carry around the surface a small set of Cartesian axes so that $y$ points are to the left of $x$ at start, but to the right after a path returning to the start.

Euler Characteristic

Suppose that a surface $\mathscr{M}$ has been subdivided using a total of $V$ vertices, $E$ edges (homeomorphic to finite intervales), and $F$ faces (these could be triangles or other regions homeomorphic to a disk).

Theorem: $\chi = V - E + F$ is independent of the choice of subdivision and is therefore a topological invariant of $\mathscr{M}$. It is called the Euler characteristic or number.

A version of the theorem applies to a planar graph. In this case $F$ is the number of regions including the outside and $\chi = 2$, which corresponds to the Euler characteristic of the sphere $S$.

For $S$ itself, one can choose any vertex and take its complement to be its face, to yield $(V,E,F) = (1,0,1)$. A less economic subdivision would consist of $2$ faces (north and south hemispheres), $2$ vertices (e.g. in Iquitos and Singapore) and the $2$ half-equators, so $(2,2,2)$. Any subdivision gives $\chi = 2$.

The Torus

A torus of revolution in $\mathbb{R}^3$ can be constructed by identifying the ends of a flexible cylinder. Topologically a cylinder is equivalent to a sphere minus two disks; each time a disk is removed, $\chi$ drops by $1$, so $\chi(cylinder) = 2 - 1 - 1 = 0$. This is consistent with the model of cylinder obtained from a rectangle by identifying the edges top and bottom, which leaves $(V,E,F) = (2,3,1)$, and again $\chi = 2$.

The closed edges left and right of the rectangle correspond to the boundary circles of the cylinder that we need to further identify to obtain the torus. All $4$ vertices of the rectangle have now become one, so $\chi = 1-2+1=0$.

If we choose the dimensions correctly, we can map the rectangle continuously onto the surface of revolution in $\mathbb{R}^3$ so that the united edges map to the two orthogonal circles intersecting in a single vertex.

A Preview of Classification

Recall that a subset of $\mathbb{R}^3$ is compact if and only if it is closed and bounded. A compact surfaced (being closed) must include its boundary if it has one. The boundary (being compact) must be a link consisting of $r \geq 0$ knots, trivial or otherwise.

We shall see that there are three quantities that can be used to distinguish a compact surface $\mathscr{M}$ topologically:

1. Whether or not it is orientable (Yes/No)
2. The Euler characteristic is an integer $\chi \leq 2$ (Yes/No)
3. The number $r \geq 0$ of boundary components it has.

The precise result will be formulated combinatorially in a later article.

If $\mathscr{M}$ is an orientable surface without boundary then it is homeomorphic to a 'torus with $g$ holes', equivalently a sphere with $g$ handles attached to it. To attach a handle one must first remove two disks (reducing $\chi$ by 2) and then add a curved cylinder (leaving $\chi$ unchanged). So $\chi(\mathscr{M}) = 2 - 2g$, where $g \geq 0$ is the so-called genus.

Seifert's Algorithm

For a cloth surface, $\chi(\mathscr{M}) = B - c$ where $B$ is the number of black regions. This is because each black region is a face, and each crossing ribbon amounts to subtracting an edge from each face and adding a rectangle, so $\chi \mapsto \chi - 2 + 1 = \chi -1$ reduces by 1. The twist is irrelevant, $\chi$ depends only on the shadow of the diagram. The standard diagram of $4_1$ gives a non-orientable surface with $\chi = 3-4=-1$. However:

Theorem: Any knot is the boundary of some orientable surface in $\mathbb{R}^3$.

Proof: Choose an oriented diagram for the knot, and define a "Seifert state" $\sigma$ to split each crossing as in the right-hand side of the skein relation. We are left with a certain number $|\sigma |$ of oriented circles, each of which is the boundary of a disk in the plane. Paste in a 180-degree-twisted ribbon between pairs of them to correctly interpret each crossing. (To do this, stack the disks vertically if one is inside another). This will ensure that the boundary of the resulting surface is faithfully encoded by the diagram. Each disk acquires an orientation from its boundary circle, and the ribbons will consistently propagate the orientation from disk to disk. $\blacksquare$

Example: Trefoil Knot

The $3$-leaved diagram $D$ of $R3_1$ has writhe $+3$. If $D$ is chess-boarded with white on the outside, the chess-signs are all positive. That implies that $\sigma = s_+$ and $|\sigma| = W = 2$. The resulting Seifert surface $\mathscr{S}$ is formed from a disk for each circle and a twisted band to interpret each crossing, so that the boundary of $\mathscr{S}$ is a trefoil knot. $\mathscr{S}$ is orientable, because it is $2$-sided in $\mathbb{R}^3$.

$\chi(\mathscr{S}) = 12 - 18 + 5 = -1$ since there are $6$ vertices on each disk, add $6$ edges to cap off the strips. Or more simply $2-3$ ($2$ disks and $3$ ribbons).

The classification of surfaces will tell us that $\mathscr{S}$ is homeomorphic to a torus minus a disk. If, on the other hand, the ribbons are not twisted, we merely have a sphere with $3$ disks removed, yet again $\chi = 2-3=1$.

Example: Figure-Eight Knot

Orient a diagram of $4_1$. There is then only one way to split each crossing preserving the orientation. This gives a state $\sigma$ consisting of $3$ circles, one nested inside another, leading to two images of the resulting Seifert surface, the second deformed from the disks and ribbons of the first.

It is easy to compute the Euler characteristic of the surface $\mathscr{S}$ produced by Seifert's algorithm in general. Each disk contributes $+1$ to $\chi$, and each ribbon $-1$. So $\chi(\mathscr{S}) = |\sigma | - c$, and the examples previously mentioned again have $\chi = -1$.