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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.

Put $R=\mathbb{Z}[U]$, with $\mathrm{deg}U=-2$. In genus one, an intersection point is itself a generator of the Heegaard Floer chain complex. Thus the free $R$ basis isThere are infinitely many integral generators $U^{m}j$, $m\ge 0$, but these seven points are the module generators.

The obstruction to a Whitney disk class is the alpha-and-beta joining cycle in $H_1(Y)$. Identify the opposite sides of the square and straighten the two small fingers of the alpha curve. Its algebraic intersection with beta is three; the remaining intersections, one in each residue class, are represented by $1,4,5$. The four small bigons connect $1$ to $2$ to $3$, and $5$ to $6$ to $7$. A joining cycle between consecutive remaining strands is a meridian, which generates $H_1(Y)=\mathbb{Z}/3$. Consequently the equivalence classes are exactlyEquivalently, these are the three Spin-c structures represented in the Heegaard diagram.

Use the domain convention ${\partial }_{\alpha }D=y-x$. The positive bigons run from $2$ to $1$, from $2$ to $3$, from $5$ to $6$, and from $7$ to $6$. Their Maslov indices are one: their Euler measure is $1/2$, and the two endpoint averages contribute $1/4$ each. Write their basepoint multiplicities as $a,b,c,d$, respectively. The relative grading in Heegaard Floer homology is thereforeThe reproduced figure has no visible marked $z$. If $z$ is outside all four small bigons, as in the usual intended diagram, $a=b=c=d=0$. One may normalize the relative Heegaard Floer gradings asEach class has its own arbitrary additive constant, and $\mathrm{gr}(U^{m}j)=\mathrm{gr}(j)-2m$. If the basepoint is put inside one of these disjoint bigons, exactly its multiplicity is one; the preceding boxed difference formulas give the answer without an unstated basepoint assumption.