Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 20 / 3

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.

For the unknot the three-manifolds are $S^3$, $U_0=S^1\times S^2$, and $U_n=L_n$, where $L_n$ denotes positive $n$-surgery with its surgery orientation. This fixes the orientation independently of competing conventions for the name $L(n,1)$.

Use the grading convention $\mathrm{gr}{\Theta }_{S^3}^{+}=0$ and $\mathrm{gr}{\Theta }_{S^3}^{-}=-2$. Let $R=\mathbb{Z}[U]$ and $\hat{R}=\mathbb{Z}[[U]]$. The ordinary minus groups arewhere $0\le k<n$ andThe lens-space absolute grading can be calculated from the standard torus surgery triple. For $0\le k<n$, the reverse trace structure with evaluation $2k-n$ has a small embedded holomorphic triangle with basepoint multiplicity zero and signed count $\pm 1$, taking $y_k$ to ${\Theta }^{-}$. Its degree shift is ${\Delta }_k=(1-(2k-n{)}^2/n)/4$. Thus $(d_k-2)+{\Delta }_k=-2$, which gives the boxed formula for $d_k$. Additional triangles wind around the surgery region and give the higher powers of $U$ calculated below.

Their infinity groups are obtained by inverting $U$: one Laurent tower for $S^3$ and each lens-space structure, two for the torsion structure on $S^1\times S^2$, and $\mathbb{Z}[U,U^{-1}]/(1-U^{|i|})$ in its non-torsion structures. Their plus groups are respectively ${\mathcal{T}}_0^{+}$, ${\mathcal{T}}_{1/2}^{+}\oplus {\mathcal{T}}_{-1/2}^{+}$, and ${\mathcal{T}}_{d_k}^{+}$. The hat groups are $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in each degree $\pm 1/2$, and $\mathbb{Z}$ in degree $d_k$. These follow directly by straightening the standard genus-one Heegaard diagrams: the lens space has one generator in each structure and no chain differential, while the two bigons in the torsion $S^1\times S^2$ diagram cancel with coherent signs. The sphere adjunction condition eliminates its non-torsion plus and hat groups, but not its ordinary minus group. The Non-torsion Heegaard Floer homology of the sphere-circle product calculation comes from a two-generator chain complex with chain differential $1-U^{|i|}$; its cokernel is the displayed quotient. For the triangle, complete the free minus towers by replacing $R$ with $\hat{R}$. The non-torsion summands disappear after completion, because $1-U^{|i|}$ has an inverse $1+U^{|i|}+U^{2|i|}+\cdots$ in $\hat{R}$.

For $k=0$, choose signs so that the exact triangle becomesThe zero-surgery trace has $\chi =1$, $\sigma =0$, and $c_1^2=0$ in the torsion structure; hence its absolute degree shift is $-1/2$, taking ${\Theta }^{-}$ to $x_{-}$. The auxiliary-generator map takes $x_{+}$ to $y_0$, so its effective grading shift isFor the underlying cobordism $Z_n$ in its torsion structure, $c_1^2=0$, $\chi =1$, $\sigma =0$, and its actual cobordism shift is $-1/2$. Tensoring with the canonical lens generator adds $d_0=(n-1)/4$ under the connected-sum grading convention, explaining the effective shift. For $n=1$ the auxiliary lens space is $S^3$, and this is simply the other degree-$-1/2$ surgery trace map.

The reverse trace components are explicit in every $k$. For $j=k+n\ell$, put $m_j=2j-n$. Its intersection form is $(-n)$, soIndeed $c_1^2=-m_j^2/n$, $\chi =1$, $\sigma =-1$, and matching source and target degrees gives precisely the displayed exponent. The exponent is nonnegative, and the coefficient is a unit: the infinity trace map is an isomorphism, and restriction to the free minus towers leaves the unique grading-compatible monomial.

With a compatible choice of Coherent orientations in Heegaard Floer homology, the sum $C_k$ is multiplication byFor $k=0$, pair $\ell$ with $1-\ell$: their exponents agree and their signs are opposite, so $P_0=0$, in agreement with the first triangle. For $0<k<n$ the term $\ell =0$ is the unique constant term, and every other exponent is positive. Thus $P_k=1+U(\cdots )$ is a unit in $\hat{R}$. The zero-surgery summand for such a $k$ vanishes, and the exact triangle isIt is isomorphic, as an ungraded completed-module triangle, to one with the final map the identity. The sum generally has no single absolute Heegaard Floer grading shift; each individual trace component has shift ${\Delta }_j$. This distinction and the power series explain why replacing the completed triangle by a polynomial-module triangle is not innocuous.