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

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.

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.

For the shaded bigon, a holomorphic strip is a conformal parametrization of its interior. Send its two corners to the two ends of $\mathbb{R}\times [0,1]$. The Riemann map is unique up to translation, so quotienting the moduli space by translations leaves a single point. The embedded strip is regular, and a coherent orientation assigns it sign $+1$ or $-1$. Thus the signed count is $\pm 1$. The other three small bigons have the same argument.

For the shaded rectangle, a disk in ${\mathrm{Sym}}^2(\Sigma )$ is represented by a two-sheeted surface over the strip, with its projection to $\Sigma$ covering the rectangle once. There is one simple branch point. The four alternating boundary arcs and the conformal modulus of the rectangular region determine the branched cover up to strip translation: cutting along an arc from the branch point produces two simply connected strips, whose conformal maps are fixed by their marked corners, and gluing fixes the relative translation. The index calculation in part (ii) gives a zero-dimensional quotient; regularity of this embedded empty rectangle makes its one solution an isolated regular point. Hence its signed count is also $\pm 1$. The same construction applies to the lower rectangle. This is the empty bigon and rectangle counts in Heegaard Floer homology calculation, including the geometric reason for the counts.

The listed bigons and rectangles give all positive index-one domains in these diagrams. One can check this by following the alpha and beta boundary arcs: the four two-corner regions in Figure 2a and the two empty four-corner regions in Figure 2b are the possibilities with convex corners; a domain reaching around a handle has an additional corner or a negative region coefficient. There is no nonzero basepoint-normalized periodic domain, because the displayed homology presentations have nonzero determinant. Adding the whole surface increases the Maslov index by two, so does not give another index-one representative. Thus no extra chain differential term is hidden in the calculations below.

For Figure 2a with the basepoint outside the four small bigons, coherent signs can be absorbed into the choices of generator orientations, yielding a chain complex of the formThe first summand has homology $R⟨[1]⟩$, with $[3]=-[1]$. The singleton $4$ contributes $R$. In the last summand, $6$ is a boundary and the cycles are $R⟨5-7⟩$. Consequentlyone free tower in each Spin-c structure. With the normalizations in part (i), their top relative Heegaard Floer gradings are $0,0,1$.

The absent visible basepoint in Figure 2a does not change this homology answer. For the multiplicities of part (i), replace the three nonzero chain differential formulas by $\partial 2=\pm U^{a}1\pm U^{b}3$, $\partial 5=\pm U^{c}6$, $\partial 7=\pm U^{d}6$. At most one of $a,b,c,d$ equals one, and the others are zero, so each two-term relation contains a unit. Elementary elimination again leaves one free $R$ tower in each class; its relative Heegaard Floer grading is read from the multiplicity-dependent grading formulas.

For Figure 2b, write $A=(1,a)$, $B=(2,b)$, $C=(3,c)$. The two rectangles give, after choosing orientations,and both singleton generators have zero chain differential. The first summand has cycles $RA\oplus RC$ modulo the primitive relation $A+C$, hence is $R$. Thereforeagain with one free tower in each Spin-c structure. All three can have top relative Heegaard Floer grading zero in the independent normalizations of part (ii).