Geometric Topology 6

The Goeritz Method

There is an easier way to compute the determinant of a knot.

Example:

Take the chessboarding of the Whitehead link. Label the white regions inner left 1, right 2, outside 3. Define a $3 \times 3$ matrix $G_+ = (g_{ij})$ by:

In taking the sum, the crossing should be assigned its chess-sign, so $G_+$ is symmetric and its rows and columns add up to zero. We then compute the determinant using the previous proposition. Deleting last row and column yields:

It can be shown that $| \det (G) |$ agrees with the link determinant computed from the colouring matrix $A$. What's more, the SNF of $G$ will also determine the colouring group, so typically the diagonal form of $A$ will have lots more $1$'s than that of $G$.

Pretzel Links

If $p,q,r$ are integers, the (triple) pretzel link $P(p,q,r)$ is formed by taking three twists with $p,q,r$ crossings and combining them in the form of a flat "pretzel". This provides a diagram $D$ in which the twists are shown from left to right. If $D$ is chess-boarded with white on the outside then a positive/negative integer means that all the crossings in that twist have positive/negative chess signs. With this convention, we can write the trefoil knot $R3_1$ as $P(-1,-1,-1)$.

In space, one can think of the three twists more symmetrically as dangling vertically from points of a circular light fitting. That shows that a cyclic permutation of the three integers leads to ambient isotopic links. If $p$ is even/odd then the two strands enter and leave the twist in the same/reversed order on $D$. It follows that $P(p,q,r)$ is a knot if at least two of $p,q,r$ are odd, otherwise it is a link with $2$ or $3$ components.

The Determinant of a Pretzel

Proposition: The determinant of the link $P(p,q,r)$ equals $| qr + rp + pq |$.

Justification with the Goeritz Method: Chessboarding a diagram like that of $P(-3,3,-3)$ leaves two interior white regions, one bounded by $| p | + | q |$ crossings and the other by $| q | + | r |$. The resulting Goeritz matrix:

takes account of the chess signs. Its determinant is $pq + pr + qr$. $\blacksquare$

As a corollary $P(-2,3,5)$ and $P(-3,5,7)$ have determinant one and like the unknot cannot be coloured for any modulus $n$.

Unless $p,q,r$ all have the same sign, the chess signs will also vary in sign and the usual diagram of $P(p,q,r)$ cannot be alternating. The first non-alternating knot in the tables is $8_{19}$, which happens to coincide with $P(3,3,-2)$ and is a knot that can in common with $3_1,5_1$ and $7_1$ be inscribed on a torus.

More Examples

• Recall that no knot an be $2$-coloured (because both underpasses at a crossing must be $0$ or $1$, so there can only be one colour throughout the component). So a knot always has odd determinant.
• Any link with at least $2$ components can be $2$-coloured (one knot $0$, the rest $1$). If a link possesses a diagram with $2$ components that can be separated then its determinant is zero, but the converse is false: the right link has $\Delta = 0$.