Geometric Topology 5

Geometry of Diagrams

Let $D$ be a diagram of a link $L$ with $c$ crossings. We shall assume that the diagram is connected and contains no closed arc. Then its shadow is a planar graph with $c$ vertices and $c+2$ regions, including "outside". We state without proof the fairly obvious fact that the shadow can be "chess-boarded":

Chess Lemma: Black and white can be assigned to the regions of $D$ in such a way that the same colours only meet at a vertex and not along an edge.

We shall use this result to show that:

1. A chess-boarded shadow distinguishes an alternating diagram
2. A colouring matrix can be defined so that the row vectors sum to zero
3. There is a quick way to compute the determinant this way.

Alternating knots

Let $S$ be the shadow of a knot diagram with a chess-boarding. Orient $S$ and start anywhere. At the first crossing one reaches, make the strand an overpass if black is on the right as you arrive, otherwise make it an underpass. Since black and white alternate, this rule will determine an alternating diagram with the property that (if, as usual, we regard an overpass as the $x$-axis) quadrants $Q_1$ and $Q_3$ are shaded black.

Corollary: The shadow of any knot diagram has two alternating diagrams associated to it, otherwise (up to ambient isotopy) to a knot $K$ and its mirror image $mK$.

It is convenient, whenever possible, to chess-board a shadow so that white is on the outside. This then distinguishes one of $K$ and $mK$. For example, making the the three leaves of a trefoil knot shadow black determines $R3_1$.

Any knot up to and excluding $8_{19}$ can be represented by an alternating diagram. Such a knot is therefore determined by its shadow up to reflection. The absence of over and underpasses in a knot diagram assumes it is alternating. If $8_{19}$'s shadow is rendered alternating it becomes a diagram of $8_{18}$.

Reduced Diagrams

For future reference, it will help to restrict to diagrams that have no twists that can easily be eliminaed either by an $R_1$ move on a single arc or by a 'macro move'.

Definition: The diagram of a link is reduced if no crossing is an isthmus, one that is bounded by $3$ (rather than $4$) regions.

An isthmus arises from a 'macro' twist contained in some region, and defines a closed curve $f(S^1) \subset \mathbb{R}^2$.

By Schoenflie's theorem (a strengthening of Jordan's), $f(S^1)$ is unknotted and there is a homeomorphism $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that the image of $F \circ f$ is a circle, the boundary of a disk that can be rotated in space to eliminate the twist. This is the idea behind:

Theorem: Any link is isotopic to one with a reduced diagram.

Return to Colouring Matrix

Let $D$ be a reduced diagram of a knot or link $L$. In addition, suppose that $D$ is connected and contains no closed arc. The colouring equations are then encoded in a matrix $A_+$ of size $c \times c$.

Given a chess-boarding of $D$, assign a chess-sign (also called a Goeritz index) $+1$ to a crossing if (with $x$-axis as the overpass) quadrants $Q_1,Q_3$ are black, otherwise $-1$. For $+1$ use the entries $2,-1,-1$ in the row of the colouring matrix, otherwise $-2,1,1$.

Lemma: In this setup, the integers in each column of $A_+$ (defined by an arc) sum to $0$.

Proof: Consider 'subarcs' of $D$, equivalently edges $x_i$ of the shadow $S$. Each colouring equation involves exactly four of them:

where $x_i$ has a positive sign $\iff$ it approaches the crossing with black on its right. The overpass will consist of either $\{x_1,x_3\}$ or $\{x_2,x_4\}$. Each subarc will be assigned opposite signs from its two crossings, and all the crossing equations sum to zero. $\blacksquare$

The Colouring Group

Let $A_+$ be the colouring matrix determined by a chess-boarding of $D$ as before.

Definition: The determinant of $D$ is $| \det(A) |$ where $A$ is a submatrix of $A_+$ obtained by deleting any one row and one column.

The previous results guarantee that this non-negative quantity does not depend on $A$. Since we have obtained it in a systematic way from the system of colouring equations, which is unchanged by $R$-moves, $| \det (A) |$ depends only on the ambient isotopy class of the link $L$. We may therefore talk about the determinant of a link or knot.

Definition: Having chosen $A$, the colouring group $C$ of $D$ is the abelian group generated by the symbols $x_1,...,x_{c-1}$ subject to the relations given by the rows of $A$.

It can be shown that the isomorphism class of $C$ likewise only depends on the ambient isotopy class of the link, and that $| C | = | \det (A) |$, provided $A$ is invertible.

Example: It follows from the matrix calculation that the knot $4_1$ has $C \cong \mathbb{Z}_5 = \mathbb{Z}/5 \mathbb{Z}$, a cyclic group of order $5$. We shall revisit this example.

Diagonalization of Integer Matrices

Let $D$ be the diagram of $4_1$ with:

Then:

where the $x_i$ should now be thought of as mere symbols, not integers. More abstractly $C$ is the quotient $\mathbb{Z}/(A^T \mathbb{Z}^3)$ of the set of all integer column vectors by the subgroup generated by columns of $A^T$. By applying elementary row operations to $A$ we can make it upper-triangular and deduce that:

which coincides with what we discovered at the start.

Lemma: (Smith Normal Form, SNF) Suppose that $A$ is a square matrix with integer entries. There exist integer matrices $P,Q$ with determinant $1$ such that $PAQ$ is diagonal with entries $d_1,d_2,...$

The proof of this lemma (which crops up in the matrix presentation of modules) uses row and column operations. We omit the details.

The Colouring Theorem

Theorem: Let $D$ be a connected reduced diagram of some knot or link $L$. Let $\Delta = | \det (A) |$ denote its determinant.

1. If $\Delta = 0$ then $L$ is $n$-colourable for any $n \geq 2$.
2. If $\Delta \neq 0$ then $L$ is $n$-colourable $\iff$ $gcd (\Delta, n) > 1$.

Proof: We can assume that $A$ is obtained from $A_+$ by deleting the last row and column. Use SNF to write $PAQ = B$ where $B = diag(d_1,...,d_{c-1})$. Allowing $P$ to have determinant $\pm 1$, we can assume that each $d_i$ in the SNF is a non-negative integer. Set $\underline{y} = Q^{-1}\underline{x}$. The system of equations for $n$-colourability can be written:

so $d_iy_i = 0 \ \mod \ n$ for $1 \leq i \leq c-1$. We are assuming that $x_c = 0$.

1. If $\Delta = 0$ then $d_i = 0$ for some $i$ and we can take $y_i = 1$ and all other $y_j = 0$.
2. If $gcd(\Delta, n) > 1$ then $n$ has a prime factor $p > 1$ in commmon with some $d_i$. Thus $d_i = pe$ and $n = pf$, and we can take $y_i = f$ and $y_j = 0$ if $j \neq i$. Conversely if there is a solution with $y_i \neq 0 \ \mod \ n$ then $gcd(d_i,n) > 1$. $\blacksquare$

Finite Abelian Groups

Any finite abelian group is known to be the product of cyclic groups. We are using SNF to establish this theorem in our situation, although we are working with additive notation. The proof of the colouring theorem actually determines the structure of the set of solutions to the colouring equations, and so the colouring group $C$. The latter is an abelian group with addition. By converting (changing basis) from the symbols $x_j$ to $y_i$, we see that:

where $\mathbb{Z}_d = \mathbb{Z}/d\mathbb{Z}$ is a cyclic group of order $d$, except that $\mathbb{Z}_0 = \mathbb{Z}$ denotes the infinite cyclic group. Note that $Z_1 = \{0\}$ is the identity group, so if $d_i = 1$ there is effectively no contribution to the direct sum above.

Corollary: If $\Delta \neq 0$ then $| C | = \Delta$. But if $\Delta = 0$ then $C$ is infinite.

There are different ways of decomposing a finite abelian group $C = \bigoplus \mathbb{Z}_{d_i}$ as a direct sum of cyclic groups. One can always arrange that $d_i | d_{i+1}$ to get a unique description. Or (since $\mathbb{Z}_{pq} \cong \mathbb{Z}_p \oplus \mathbb{Z}_q$ if $gcd(p,q) = 1$), one can take each $d_i$ to be a prime power.