Geometric Topology 4

Colouring Matrices

We label each of the crossings in a figure-eight diagram with $1,2,3,4$ left to right. We label its arcs by their overcrossings, which gives the colouring matrix:

A_+ = \begin{pmatrix} 2 & -1 & 0 & -1 \\ -1 & 2 & -1 & 0 \\ -1 & 0 & 2 & -1 \\ 0 & -1 & -1 & 2 \end{pmatrix}

where rows represent crossings and columns arcs. If the arcs are coloured $x_1,x_2,x_3,x_4$ $\mod n$ then we need:

A_+ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix} n

for some integers $x_i,b_j$ . Since $A_+ (1,1,1,1)^T = 0$ , we can take $x_4 = 0$ , but the other $x_i$ cannot all be $0$ .

Notice that the column vectors of $A$ and the row vectors of $A$ add up to $0$ .

Cofactors

Since the fourth row is a linear combination of the others, it suffices to consider the top $3 \times 3$ block $A$ of $A_+$ and solve:

\begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = n \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}

or:

A \underline{x} = n \underline{b}

Now:

\underline{x} = n A^{-1}\underline{b} = \frac{n}{\det A} \tilde{A} \underline{b} \\

= \frac{n}{5} \begin{pmatrix} 4 & 2 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}

We now see that $n=5$ possible colourings are generated by the columns of $\tilde{A}$ , the transpose of the matrix of cofactors of $A$ . In fact $\tilde{A}$ has rank $1$ modulo $5$ , so there are $4$ possible colourings with $x_4 = 0$ , generated by taking $\underline{b} = (1,0,0)^T$ , namely:

\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}, \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}.

The Determinant First, suppose $L$ is an alternating link, that is one that possesses some diagram $D$ that is alternating. Define the colouring matrix $A_+$ of $D$ as before by assigning the overpassing arc to each crossing and setting $2$ down the diagonal. Beware that if a given arc passes through the same crossing more than once, the numbers $2,-1,-1$ might become $1,-1$ or $2,-2$ . Each arc contains exactly one overpass. So not only do the entries of each row (crossing) sum to zero, but also the entries of each column (arc) sum to zero. In particular, the row vectors (or crossing equations) sum to zero.

Proposition: Let $M$ be a $c \times c$ matrix whose row and column vectors both sum to zero. Then all the first minors (determinants of submatrices of size $(c-1)\times (c-1)$ ) are equal up to sign.

We shall see that we can always define a colouring matrix so as to use the proposition for any link $L$ , alternating or not. The determinant of $L$ will then be defined to equal the absolute value of the first minor $A$ (equivalently, cofactor) of $A_+$ .

Minors of $M$ Proof of previous proposition:

Let $M'$ be the matrix obtained from $M$ by adding $1$ to every entry of $M$ . Fix $(i,j)$ and let $M''$ be the matrix obtained from $M'$ by adding to row $i$ of $M'$ to all its other rows, and then adding to the new column $j$ all its other columns. The entries in row $i$ or column $j$ of $M''$ are then $n$ except that $(M'')_{i,j} = n^2$ . Let $M'''$ be the matrix obtained from $M''$ by subtracting $\frac{1}{n}$ times row $i$ from all the other rows.

Let $M_{i,j}$ be the submatrix of $M$ formed by deleting row $i$ and column $j$ . Row $i$ of $M''$ equals $(n,...,n^2,...,n)$ so $M'''$ coincides with $M$ everywhere except in row $i$ and column $j$ (where all the entries are $0$ except for the $n^2$ ). So if we can compute its determinant by expanding along column $j$ we obtain:

\det (M''') = (-1)^{i+j}n^2 \det(M_{i,j}).

The rules for determinants tell us that $\det (M''') = \det (M')$ and the right hand side is independent of $(i,j)$ . $\blacksquare$

A list of determinants

After introducing some technicalities, we shall show that the determinant of a general link is well-defined, and see that $L$ is $n$ -colourable $\iff$ $n$ and $\det (A)$ have a non-trivial common factor. It follows that: If $\det (A) = 0$ then $L$ is colourable for any $n$ (impossible for a knot, whose determinant must be odd. Possible for a link). If $| \det (A) | = 1$ then $L$ is colourable for no $n$ (possible even for a non-trivial knot).