Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 20 / 5

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.

The map is a homogeneous Laurent polynomial ring homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree isConsequently the general graded-module description isIf the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called ${\Phi }^{\infty }$ in the question.

The standard topological statements for Heegaard Floer cobordism maps, which require no proof in this part, areIn the second case $a=\pm 1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational homology spheres, the rational intersection form is nondegenerate, so $b_2^{+}=0$ means that it is negative definite, allowing also the rank-zero form.

The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite cobordism. A concrete check is $W=(S^1\times S^3)\setminus (B^4\bigsqcup B^4):S^3\to S^3$. It has $b_2^{+}=0$ and $b_1(W)=1$. A one-handle followed by a three-handle gives $e\mapsto e\otimes {\theta }^{+}$ and then kills the ${\theta }^{+}$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual cobordism, not just its boundary Betti numbers.

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.