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Over $\mathbb{Z}$, holomorphic disk moduli spaces in a Heegaard Floer chain complex must be oriented compatibly with gluing: the orientation of a concatenated disk class is the product determinant-line orientation of its two factors. Boundary orientations of compactified one-dimensional moduli spaces then make broken disks cancel, yielding ${\partial }^2=0$. Isolated regular disks contribute $+1$ or $-1$. Changing the chosen orientation at an intersection generator changes the corresponding row and column signs of the differential and yields an isomorphic integral chain complex. Cobordism triangle counts require compatible orientation choices as well; an overall orientation ambiguity accounts for the usual sign ambiguity of the closed invariant.

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.

For a rational homology three-sphere and any Spin-c structure, the infinity version of Heegaard Floer homology isIt is one Laurent tower, with one integral generator in each degree of a fixed parity and none in the opposite parity. Its absolute rational grading has an overall offset determined by $(Y,\mathfrak{s})$.

To compute the Euler characteristic of the hat group, tensor with $\mathbb{Q}$, which preserves integral ranks. Put $V=H{F}^{-}(Y,\mathfrak{s};\mathbb{Q})$. It is a finitely generated graded $\mathbb{Q}[U]$ module. Localizing at $U$ gives $H{F}^{\infty }$, so the structure theorem for finitely generated modules over a principal ideal domain gives one free summand and finitely many $U$-primary torsion summands. The torsion is $U$-primary because it vanishes after localization, or directly because the reduced graded group is supported in finitely many degrees.

The exact sequence of Heegaard Floer chain complexes given by multiplication by $U$ and passage to the hat quotient identifies the hat group, as a parity-graded vector space, with an extension of $V/UV$ by $\ker (U:V\to V)$ with the latter parity reversed. The extra absolute shift by two between $C{F}^{-}/UC{F}^{-}$ and $\widehat{CF}$ is even, so it does not affect this argument. ThusThe free summand contributes one vector to $V/UV$ and none to $\ker U$. A torsion summand $\mathbb{Q}[U]/U^r$, whose top generator has parity $ϵ$, contributes one vector of parity $ϵ$ to the quotient and one of the same parity to the kernel, since $U^{r-1}$ changes degree by an even number. Their contributions cancel. Only the single free summand remains, provingInteger torsion does not enter an Euler characteristic defined by ranks.