Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 20 / 2 / ii / Solution

A generator selects one point from each alpha circle and each beta circle. Here the points $1,3$ lie on ${\alpha }_1\cap {\beta }_2$, $2$ on ${\alpha }_1\cap {\beta }_1$, $a,c$ on ${\alpha }_2\cap {\beta }_1$, and $b$ on ${\alpha }_2\cap {\beta }_2$. Hence, with $R=\mathbb{Z}[U]$, the Heegaard Floer chain complex has free basisA pair such as $(2,a)$ is invalid because it uses beta-one twice.

Orient the attaching circles so that their algebraic-intersection matrix is $\left(\begin{array}{cc}1& -2\\ -2& 1\end{array}\right)$. The presentation of $H_1(Y)$ is then $m_1-2m_2=0$, $m_2-2m_1=0$, giving $H_1(Y)=\mathbb{Z}/3$ with $m_2=-m_1$. The upper rectangular domain connects $(2,b)$ to $(3,c)$; the lower one connects $(2,b)$ to $(1,a)$. Thus these three generators have zero joining-cycle obstruction. Changing just the point on the second alpha circle from $a$ to $c$, or just the point on the first from $1$ to $3$, gives the two opposite nonzero meridian classes. This distinguishes the other two generators. The equivalence classes of Whitney disk classes are thereforeThe dot in the central unshaded region represents the basepoint. Each displayed rectangle has $n_z=0$, Euler measure zero and four corner averages $1/4$, hence Maslov index one. The relative grading in Heegaard Floer homology satisfiesThus a convenient normalization is degree zero for $(1,a),(3,c)$ and degree one for $(2,b)$; either singleton can independently be put in degree zero. As before, multiplication by $U^m$ subtracts $2m$ from every degree.