Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 1

Label the unbounded region $r_0$ and the three lobe regions $a,b,c$ in cyclic order, and label the central region $d$. The Dehn presentation of a knot group Heegaard diagram is obtained by thickening the projection graph: take one alpha compressing curve for each of $a,b,c,d$, one beta curve for each crossing, and place the basepoint in $r_0$. Orient and order the crossing curves so that their cyclic region words areThis intersection data, together with $r_0=1$, specifies the requested pointed Heegaard diagram up to the usual isotopies and handle slides.

The generators are the alpha-compressing disks, equivalently the four bounded regions. Each beta curve supplies the relator read from its signed intersections with the alpha curves. Since the basepoint region has generator $r_0=1$, the presentation isThe relations say $d=ba=cb=ac$. Eliminating $c,d$ leaves $bab=aba$, the standard two-generator presentation of the trefoil knot group, which checks the region conventions.

The Alexander numbering of the three lobe regions is one and that of the central region is two, relative to the unbounded region numbered zero. Equivalently, abelianizing the three relators givesThus $A=B=C$ and $D=2A$. The class $A$ is the positively oriented meridian of a knot, so

Forthe Fox calculus product rule and $\partial d^{-1}/\partial d=-d^{-1}$ give, in generator order $(a,b,c,d)$,Multiplication of every row by the unit $t$ gives the equivalent Alexander matrixFor example, deleting the $a$-column gives determinant $-(t^2-t+1)$, the Alexander polynomial of a knot of the trefoil up to a unit.

Star the unbounded region $r_0$ and the adjacent lobe region $a$. The remaining columns of the row-scaled matrix are $(b,c,d)$:Choose at crossings $1,2,3$ the corners in regions $b,c,d$, respectively. Every unstarred region then contains one chosen corner, so this is a Kauffman state of a knot diagram. Circle the entriesTheir signed determinant product is $-1$, one summand inThe other two states give $t$ and $-t^2$, and the total differs from $t^2-t+1$ only by the unit $-1$.