= Solution
Star the adjacent regions $o,e$. Choose the <Kauffman state of a knot diagram> putting the marker at $A$ in region $d$, at $B$ in region $a$, at $C$ in region $b$, and at $D$ in region $c$. Every unstarred region is used exactly once, and every crossing has exactly one marker.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-141-dehn.png]
{title=The figure-eight graph, regions and Heegaard state}
In the full <Alexander matrix>, circle the entries $(A,d),(B,a),(C,b),(D,c)$, namely
$$
\begin{pmatrix}
\boxed{-1}&-t&1&0&0\\
-t&0&\boxed{1}&-1&t\\
0&t&-t&\boxed{1}&-1\\
-t&-1&0&0&\boxed{1}
\end{pmatrix}.
$$
In the reduced <Alexander matrix> $B$, these are the diagonal entries in column order $(d,a,b,c)$. The associated <permutation> is the identity, with <sign of a permutation> $+1$, so this <Kauffman state of a knot diagram> contributes
$$
\boxed{(+1)(-1)(1)(1)(1)=-1}
$$
to $\det B=-t^2+3t-1$. The other four nonzero permutation terms are $t,t,t,-t^2$. The drawing circles the same four entries literally; the boxes above identify them in the typeset matrix. Different row or generator conventions multiply the <Alexander polynomial of a knot> and its state weights by the corresponding overall unit.
Back to article page
