Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 / 3

Give each bounded region $R_i$ a generator $a_i$ and give the unbounded region $R_0$ the identity generator. At every crossing, read the four incident regions cyclically as $a,b,c,d$ and imposeUsing the opposite cyclic convention inverts all such relators and gives the same group. One crossing relation is redundant, leaving $n$ generators and $n-1$ relators; this is the Dehn presentation of a knot group.

Orient the diagram and assign its regions an Alexander numbering with $R_0$ numbered zero. The abelianization $\varphi :{\pi }_1(E_K)\to ⟨t⟩$ sends $a_i$ to $t^{m_i}$, where $m_i$ is the number of $R_i$. Since $R_1$ is adjacent to $R_0$, $m_1=\pm 1$.

Form the square matrixfrom the Fox derivatives with respect to $a_i$ for $i>1$. This is the Alexander matrix with the $a_1$ column deleted. The Fox identity implies that its maximal minors differ by the factors $\varphi (a_i)-1$, and the standard presentation of the Alexander module therefore givesbecause $m_1=\pm 1$. Thus ${\Delta }_K(t)$ is $\det A_1$ up to a unit $\pm t^r$.

Expand the determinant of the matrix $A_1$ from part (b) by the Leibniz formula for determinants. A matrix entry is a signed sum of monomials arising from the possible corners at its crossing. Choosing one summand in every row chooses one corner at every crossing, while choosing distinct columns puts exactly one chosen corner in every region $R_i$ with $i>1$ and none in the two deleted regions $R_0,R_1$. The surviving determinant terms are therefore in bijection with the Kauffman states $s\in \mathcal{S}(D)$.

The sign of a permutation in the determinant together with the corner signs gives $(-1{)}^{ϵ(s)}$, and multiplying the corner monomials gives $t^{\delta (s)}$. Since part (b) identifies this determinant with the Alexander polynomial of a knot up to a unit,