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For a torsion Spin-c structure, the relative grading of Heegaard Floer homology lifts canonically to a rational absolute grading, normalized by the bottom plus generator of $S^3$ having degree zero. Multiplication by $U$ has degree $-2$, and a Heegaard Floer cobordism map has degree $(c_1^2-2\chi -3\sigma )/4$. The correction term $d(Y,\mathfrak{s})$ is the lowest grading of the image of $H{F}^{\infty }$ in $H{F}^{+}$ for a rational homology sphere. A pure tower then has bottom plus degree $d$ and top minus degree $d-2$ in the $i<0$ minus convention. Keeping that shift explicit prevents errors of two in surgery and mixed-invariant calculations.

An admissible cut splits a four-dimensional cobordism as $W=W_1{\cup }_N{W}_2$, with $b_2^{+}(W_1),b_2^{+}(W_2)>0$, and with the Mayer-Vietoris connecting map $H^1(N;\mathbb{Z})\to H^2(W;\mathbb{Z})$ equal to zero. The last condition removes the ambiguity in gluing Spin-c structures on the two pieces. Positivity makes both infinity cobordism maps vanish, allowing passage through the Reduced Heegaard Floer homology of $N$. A useful cut is the boundary of a neighborhood of a positive-square embedded surface, chosen so that its complement still has a positive direction. The nonzero normal Euler number makes $H^1$ of the neighborhood surject onto $H^1$ of its boundary, proving the connecting-map condition.

Let $\tilde{M}=M\#{\overline{\mathbb{CP}}}^{,2}$, and let $E$ be the exceptional sphere. Extend a Spin-c structure by $c_1(\tilde{\mathfrak{s}})=c_1(\mathfrak{s})\pm \mathrm{PD}[E]$. Then ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}={\Phi }_{M,\mathfrak{s}}$. More generally, an evaluation $⟨c_1(\tilde{\mathfrak{s}}),E⟩=\pm (2\ell +1)$ introduces the factor $U^{\ell (\ell +1)/2}$. This follows by putting the blowup on one side of an admissible cut: the punctured exceptional piece is negative definite, its minus and plus tower maps are multiplication by this power of $U$, and the cobordism composition law gives the formula. In the $\pm 1$ case the exponent is zero, so nonvanishing is preserved.

A smooth oriented cobordism $W:Y_1\to Y_2$ with a Spin-c structure induces $U$-equivariant maps $F_{W,\mathfrak{s}}^{\circ }$ on the corresponding Heegaard Floer homology groups. A handle decomposition defines the maps using generator selection for one- and three-handles and holomorphic triangle counts for two-handles. Composition sums over extensions of the prescribed Spin-c structures. When the boundary determinant classes are torsion, the absolute grading shift isThe square is defined using the rational relative lift. If $b_2^{+}(W)>0$, the infinity map vanishes. Between rational homology spheres, when $b_1(W)=b_2^{+}(W)=0$, the infinity map is an isomorphism of Laurent towers.

Heegaard Floer homology assigns groups $H{F}^{\circ }(Y,\mathfrak{s})$ to a closed oriented three-manifold with a Spin-c structure. A pointed Heegaard diagram of genus $g$ gives product Lagrangian tori ${\mathbb{T}}_{\alpha },{\mathbb{T}}_{\beta }\subset {\mathrm{Sym}}^g(\Sigma )$; the chain complex is generated by their intersection points and its differential counts Whitney disk classes represented by holomorphic disks. The minus, infinity, plus and hat versions use different powers or quotients of the variable $U$. The Heegaard Floer cobordism maps and the Mixed Heegaard Floer invariant relate these groups to smooth four-dimensional topology.

For a knot $K$ in an integral homology three-sphere and $n>0$, fix a Spin-c structure ${\mathfrak{t}}_k$ on the $n$-surgery. Write ${\mathfrak{s}}_i$ for structures on zero-surgery with $⟨c_1({\mathfrak{s}}_i),[\hat{F}]⟩=2i$, using a capped Seifert surface. In compatible labels the completed-minus triangle isOne can equally use the plus theory without completing. The first map is the congruence-restricted zero-surgery trace map; the last is the sum of reverse, orientation-reversed $n$-surgery trace maps. The middle map uses a Heegaard triple with an auxiliary lens-space generator. For $n=1$ that lens space is $S^3$, giving the usual three two-ended cobordism maps. For $n>1$ it is important to retain the auxiliary boundary and its fixed Spin-c structure. The U-adic completion of Heegaard Floer homology makes the sums of maps well-defined.

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).

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.

Use ordinary integer surgery coefficients, so the positive trace has intersection form $(n)$. Let ${\hat{F}}_0\subset K_0$ be the capped Seifert surface. Label the Spin-c structures byOn the trace $X_n:S^3\to K_n$, let ${\hat{F}}_n^2=n$ and label its structures ${\mathfrak{r}}_j$ by $⟨c_1({\mathfrak{r}}_j),[{\hat{F}}_n]⟩=2j-n$. Restriction identifies ${\mathfrak{t}}_k$ with $j\equiv k(\mathrm{mod}n)$. In these compatible labels the Integer-surgery exact triangle in Heegaard Floer homology isHere boldface denotes the U-adic completion of Heegaard Floer homology. Only finitely many zero-surgery structures contribute, by the genus bound on a capped Seifert surface, so the middle direct sum causes no difficulty. The infinite sums of trace maps do require completion. The printed minus notation should be understood in this completed sense; with the polynomial version, a sum over every extension need not even have values in $\mathbb{Z}[U]$. The plus version has the same ordering and needs no completion.

The two ordinary surgery traces give $A_k$ and $C_k$. For $X_0:S^3\to K_0$, $H^2(X_0)=\mathbb{Z}$, and its Spin-c structures ${\mathfrak{q}}_i$ have determinant evaluations $2i$ on the capped Seifert surface. Their restrictions are the unique structure on $S^3$ and ${\mathfrak{s}}_i$ on $K_0$. Thus the component of $A_k$ in the $i$th summand is $F_{X_0,{\mathfrak{q}}_i}^{-}$ for $i\equiv k$. For the orientation-reversed and turned-around trace $V_n:K_n\to S^3$, $H^2(V_n)=\mathbb{Z}$ and ${\hat{F}}_n^2=-n$. Its structures ${\mathfrak{r}}_j$ still have characteristic evaluations $2j-n$, restrict to ${\mathfrak{t}}_{j\mathrm{mod}n}$ on $K_n$ and to the unique structure on $S^3$, andCharacteristic parity explains the allowed evaluations; changing a structure by the generator of $H^2$ changes its evaluation by two, and the boundary restriction reduces the label modulo $n$.

For $n>1$, $B_k$ has an auxiliary lens-space input. More precisely, take the three-boundary surgery cobordism, or equivalently a two-ended cobordism $Z_n:K_0\#L_n\to K_n$, where $L_n$ is positive $n$-surgery on the unknot. Its incoming lens-space factor is fixed to the distinguished structure ${\mathfrak{u}}_0$ and its canonical top Floer generator ${\Theta }_{L_n}$. ThenThis describes the boundary restrictions exactly, including the boundary that must not be suppressed. A concrete Kirby diagram for $Z_n$ starts with the split incoming surgeries of coefficients $0$ and $n$, then attaches a zero-framed circle linking each incoming component once. Its full linking matrix is $\left(\begin{array}{ccc}0& 0& 1\\ 0& n& 1\\ 1& 1& 0\end{array}\right)$. Quotienting cohomology by the two incoming surgery columns gives $H^2(Z_n)=\mathbb{Z}\oplus \mathbb{Z}/n$. Accordingly its Spin-c structures may be labeled $(i,q)$ with incoming restrictions $({\mathfrak{s}}_i,{\mathfrak{u}}_q)$ and outgoing restriction ${\mathfrak{t}}_{i-q}$. In particular, fixing $q=0$ gives exactly one extension for each $i$, and its outgoing class is $i\mathrm{mod}n$. Thus $B_k$ is a Heegaard Floer cobordism map with the fixed auxiliary generator, rather than an unqualified map of a trace from $K_0$ to $K_n$. When $n=1$, $L_n=S^3$ and all three arrows become the familiar sums of ordinary two-ended Heegaard Floer cobordism maps.

The geometric argument uses the Mixed Heegaard Floer invariant defined in Question 5(iii). First suppose $g(\Sigma )>0$; the genus-zero qualification is discussed below. Write $r=\Sigma \cdot \Sigma \ge 0$. Blow up $r$ distinct points of the surface. Its proper transform $\tilde{\Sigma }$ in $\tilde{M}=M\#r{\overline{\mathbb{CP}}}^2$ has classChoose the extension of the Spin-c structure withSince $E_j^2=-1$, its exceptional evaluations are $-1$, so the Blowup formula for the mixed Heegaard Floer invariant has no extra power of $U$. Thus ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}\ne 0$, andThis calculation is why the self-intersection appears with a plus sign.

It remains to bound the Chern pairing for a square-zero surface $S$. Its oriented normal bundle has Euler number zero and is trivial, so $\partial \nu S=S\times S^1$. Choose an embedded positive-square surface $T$ disjoint from $S$, with positive direction left in its complement, and cut along $N=\partial \nu T$. To see why the choice is possible, impose orthogonality to $[S]$ in the intersection form. A nonzero isotropic class uses at most one positive direction; $b_2^{+}\ge 3$ therefore leaves at least two. Choose an integral positive-square class in that complement, represent it by a surface, and remove algebraically cancelling intersections with $S$ by tubing along $S$. This alters genus but not the class. The neighborhood of $T$ supplies one positive direction and the complement supplies another. Since its normal Euler number is nonzero, $H^1(\nu T)\to H^1(N)$ is onto, proving that the cut is admissible.

Put the outgoing puncture inside $\nu S$, so that the plus Heegaard Floer cobordism map on the side containing $S$ passes through the Floer group of $\partial \nu S$. On the chain level this means splitting the cobordism along that boundary and composing its handle maps; with the fixed determinant class, every intermediate generator has restriction $\mathfrak{t}=\tilde{\mathfrak{s}}{|}_{S\times S^1}$. If the Chern pairing exceeded $2g(S)-2$, the given completed minus group would be zero. As in Question 3, this is the completed convention: for a non-torsion structure, ordinary infinity may retain $(1-U^N)$-torsion, whereas $1-U^N$ is invertible after completion. Such a pairing is positive for $g\ge 1$, so $\mathfrak{t}$ is non-torsion and the completed infinity group ${\mathbf{HF}}^{\infty }(S\times S^1,\mathfrak{t})$ is zero as well. The completed fundamental long exact sequence, whose plus group is the usual plus theory, then makes that plus group zero. The Heegaard Floer cobordism map, and consequently the Mixed Heegaard Floer invariant, would vanish, contradicting nonvanishing. ThereforeSubstituting the preceding blowup calculation provesReversing the orientation of the surface gives the usual absolute-value form of the Adjunction inequality for the mixed Heegaard Floer invariant.

The printed assumptions omit the positive-genus qualification. With the actual Floer groups, the asserted three-dimensional vanishing is false for $S=S^2$: the torsion structure on $S^2\times S^1$ has two nonzero minus towers, although its pairing is $0>-2$. Correspondingly, a small sphere embedded in a ball of a K3 surface has $[S]=0$, $S^2=0$ and Chern pairing zero, while the canonical Mixed Heegaard Floer invariant is nonzero; the desired inequality would read $0\le -2$. Thus the displayed proof establishes the intended positive-genus statement. If the supplied vanishing is instead taken literally as a formal additional axiom for every genus, the same factorization would force the inequality even for a sphere, but that axiom contradicts the actual $S^2\times S^1$ computation.

Remove two disjoint four-balls from $M$, giving $W:S^3\to S^3$. An Admissible cut for a Heegaard Floer mixed map is a separating three-manifold $N$ withThe last condition is the Mayer-Vietoris connecting-map condition. It makes a Spin-c structure on $W$ uniquely determined by its restrictions to the two pieces, avoiding an uncontrolled sum of extensions in the composition law. Such cuts exist when $b_2^{+}(M)>1$: represent an integral positive-square class by an embedded surface whose tubular neighborhood carries one positive direction and whose complement retains another. One can construct an embedded representative by resolving intersections of a representative cycle; the genus may increase without changing its class or square. The nonzero normal Euler number makes the restriction $H^1(\nu T)\to H^1(\partial \nu T)$ surjective, so the connecting-map condition holds. Puncture on either side of the cut.

Write $\mathfrak{t}=\mathfrak{s}{|}_N$ and ${\mathfrak{s}}_i=\mathfrak{s}{|}_{W_i}$. Both infinity Heegaard Floer cobordism maps vanish by positivity. Naturality with respect to the exact sequencegives $\mathrm{im}{F}_{W_1,{\mathfrak{s}}_1}^{-}\subset \ker \iota$ and $F_{W_2,{\mathfrak{s}}_2}^{+}\pi =0$. Exactness identifies $\delta$ with a degree-$-1$ isomorphism from $H{F}^{+}(N)/\mathrm{im}\pi$ onto $\ker \iota$, namely the two versions of Reduced Heegaard Floer homology. Hence the inverse on these reduced groups is defined, and one formsThis is the construction of the Mixed Heegaard Floer invariant, rather than the zero ordinary composite.

Let ${\Theta }^{-}$ be the top minus generator, of degree $-2$, and ${\Theta }^{+}$ the bottom plus generator, of degree zero. For $\zeta \in {\Lambda }^{\ast }(H_1(M)/\mathrm{torsion})$, insert its degree-one homology actions in the Heegaard Floer cobordism maps and defineThis is an integer up to the overall orientation sign. cobordism composition and associativity show independence of the admissible cut and handle choices. Equivalently, fixing the relevant orientation data removes the overall sign ambiguity.

The grading condition can be checked explicitly. Since $\chi (W)=\chi (M)-2$ and $\sigma (W)=\sigma (M)$, the sum of the ordinary cobordism shifts is $d(\mathfrak{s})+1$, where $d(\mathfrak{s})=(c_1(\mathfrak{s}{)}^2-2\chi (M)-3\sigma (M))/4$. The inverse connecting map contributes another $+1$. Therefore the output degree on $U^r{\Theta }^{-}\otimes \zeta$ is $d(\mathfrak{s})-2r-\mathrm{deg}\zeta$, and the displayed invariant vanishes unless $2r+\mathrm{deg}\zeta =d(\mathfrak{s})$.

Finally, two positive pieces are needed, so that both infinity maps vanish and the reduced-group factorization is available. For an admissible cut the connecting-map condition and signature additivity imply that the two positive subspaces inject as orthogonal subspaces of the closed intersection form; hence $b_2^{+}(M)\ge 2$. With only one positive direction no such admissible cut exists, and a chamber-dependent construction requires extra choices. With no positive direction, the ordinary infinity maps need not vanish at all. This explains the stated bound for the invariant defined here.

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.

For generators in the same Spin-c structure and a Whitney disk class $\psi$ joining them,The ambiguity is the divisibility of $c_1(\mathfrak{s})$ on $H_2(Y)$; a torsion determinant class gives an integer relative grading. Adding a sphere class changes $\mu$ by two and $n_z$ by one, so it leaves the difference unchanged. For a domain $D$, the index can be calculated as $\mu (D)=e(D)+n_{\mathbf{x}}(D)+n_{\mathbf{y}}(D)$, where $e$ is Euler measure and each $n_{\mathbf{x}}$ is the sum of averages of the four adjacent region multiplicities at the selected intersections.

Use the convention that $C{F}^{-}$ is generated by $[\mathbf{x},i]$ with $i<0$, and $U[\mathbf{x},i]=[\mathbf{x},i-1]$. The chain complexes areThe last complex is generated by $[\mathbf{x},0]$. Equivalently it is $C{F}^{-}/UC{F}^{-}$ with its absolute grading shifted up by two. The short exact sequence $0\to C{F}^{-}\to C{F}^{\infty }\to C{F}^{+}\to 0$ gives the fundamental long exact sequence. For a rational homology sphere in each Spin-c structure, $H{F}^{\infty }\cong \mathbb{Z}[U,U^{-1}]$. The plus tower is ${\mathcal{T}}_d^{+}=\mathbb{Z}[U,U^{-1}]/U\mathbb{Z}[U]$, with bottom degree $d$; its corresponding minus tower is a free $\mathbb{Z}[U]$ module with top degree $d-2$.

For intersection generators $\mathbf{x},\mathbf{y}$, a Whitney disk is a map of a disk into the symmetric product, with its two boundary semicircles mapped to the two product tori and its endpoints mapped to $\mathbf{x},\mathbf{y}$. Its homotopy class belongs to ${\pi }_2(\mathbf{x},\mathbf{y})$. Projecting it to the Heegaard surface gives an integral domain. With the surface oriented, the convention is ${\partial }_{\alpha }D=\mathbf{y}-\mathbf{x}$. The obstruction to a disk is the class $ϵ(\mathbf{x},\mathbf{y})\in H_1(Y)$ obtained by joining the points along alpha arcs and returning along beta arcs. In the Heegaard Floer setup, its vanishing is equivalent to equality of the generators' Spin-c structures; for the rational homology sphere diagrams this is equivalent to ${\pi }_2(\mathbf{x},\mathbf{y})\ne \varnothing$.