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\DeclareMathOperator{\codim}{codim} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\TropB}{TropB} \DeclareMathOperator{\weight}{wt} \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\Coh}{Coh} \DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\Fuk}{Fuk} \DeclareMathOperator{\str}{star} \DeclareMathOperator{\Ob}{Ob} \DeclareMathOperator{\Coh}{Coh} \DeclareMathOperator{\CritVal}{CritV} \DeclareMathOperator{\Sing}{Sing} \DeclareMathOperator{\FS}{FS} \DeclareMathOperator{\Vect}{Vect} \DeclareMathOperator{\grad}{grad} \DeclareMathOperator{\Supp}{Supp} \DeclareMathOperator{\Bl}{Bl} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Tw}{Tw} \DeclareMathOperator{\Int}{Int} \DeclareMathOperator{\Arg}{\mathbf{M}}\begin{filecontents}{references.bib} @article{ballard2012hochschild, title={Hochschild dimensions of tilting objects}, author={Ballard, Matthew and Favero, David}, journal={International Mathematics Research Notices}, volume={2012}, 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Observe that $S^2=D^2\cup_{S^1} D^2$, so we can write $M=D^2\times S^1 \cup_{\Sigma_1} D^2\times S^1$. The Heegaard diagram $(\Sigma_1, \alpha, \beta)$ consists of a torus with two meridional cycles. \begin{figure} \label{fig:nonadmissibleHeegaardDiagram} \centering \begin{tikzpicture} \begin{scope}[] \begin{scope}[] \draw[fill=gray!20] (-2,-1.5) .. controls (-3.5,-1.5) and (-4,0) .. (-4,1) .. controls (-4,2) and (-3.5,3.5) .. (-2,3.5) (-2,1.5) .. controls (-2.5,1.5) and (-2.5,0.5) .. (-2,0.5); \end{scope} \begin{scope}[xscale=-1, shift={(-0.5,0)}]] \draw[fill=gray!20] (-2,-1.5) .. controls (-3.5,-1.5) and (-4,0) .. (-4,1) .. controls (-4,2) and (-3.5,3.5) .. (-2,3.5) (-2,1.5) .. controls (-2.5,1.5) and (-2.5,0.5) .. (-2,0.5); \end{scope} \fill[gray!20] (-2,0.5) rectangle (2.5,-1.5); \fill[gray!20] (-2,3.5) rectangle (2.5,1.5); \end{scope} \begin{scope}[] \fill[orange!20] (1,-0.5) ellipse (0.5 and 1); \fill[orange!20] (-0.5,0.5) rectangle (1,-1.5); \fill[gray!20] (-0.5,-0.5) ellipse (0.5 and 1); \end{scope} \draw (-2,-1.5) -- (2.5,-1.5) (-2,0.5) -- (2.5,0.5) (-2,1.5) -- (2.5,1.5) (-2,3.5) -- (2.5,3.5); \begin{scope}[] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[red, shift={(1.5,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[blue, shift={(3,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-0.5,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(4.5,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(4.5,3)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(0,3)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \node[left, red] at (0,-0.5) {$\beta$}; \node[right, blue] at (1.5,-0.5) {$\alpha$}; \node at (0.5,0) {$\mathcal D$}; \node[circle, fill=black, scale=.2] at (0.5,2.5) {}; \node[right] at (0.5,2.5) {$z$}; \end{tikzpicture}\caption{A non-admissible Heegaard diagram} \end{figure} If the diagram is chosen so that $\alpha, \beta$ are disjoint, then the Lagrangian intersection Floer cohomology $\HF(\alpha, \beta)$ vanishes. \begin{figure} \label{fig:admissibleHeegaardDiagram} \centering \begin{tikzpicture} \begin{scope}[] \begin{scope}[] \draw[fill=gray!20] (-2,-1.5) .. controls (-3.5,-1.5) and (-4,0) .. (-4,1) .. controls (-4,2) and (-3.5,3.5) .. (-2,3.5) (-2,1.5) .. controls (-2.5,1.5) and (-2.5,0.5) .. (-2,0.5); \end{scope} \begin{scope}[xscale=-1, shift={(-0.5,0)}]] \draw[fill=gray!20] (-2,-1.5) .. controls (-3.5,-1.5) and (-4,0) .. (-4,1) .. controls (-4,2) and (-3.5,3.5) .. (-2,3.5) (-2,1.5) .. controls (-2.5,1.5) and (-2.5,0.5) .. (-2,0.5); \end{scope} \fill[gray!20] (-2,0.5) rectangle (2.5,-1.5); \fill[gray!20] (-2,3.5) rectangle (2.5,1.5); \end{scope} \begin{scope}[] \fill[orange!20] (1,-0.5) ellipse (0.5 and 1); \fill[orange!20] (-0.5,0.5) rectangle (1,-1.5); \fill[gray!20] (-0.5,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[red, shift={(1.5,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (0.5,-1.5); \draw[fill=gray!20] (-2,0.5) .. controls (-1.5,0.5) and (-1.5,0.5) .. (-1.5,0) .. controls (-1.5,-0.5) and (0.5,0) .. (0.5,-0.5) .. controls (0.5,-1) and (-1.5,-0.5) .. (-1.5,-1) .. controls (-1.5,-1.5) and (-1.5,-1.5) .. (-2,-1.5); \clip (-0.5,0) rectangle (0.5,-1); \draw[fill=orange!20] (-1.5,0) .. controls (-1.5,-0.5) and (0.5,0) .. (0.5,-0.5) .. controls (0.5,-1) and (-1.5,-0.5) .. (-1.5,-1); \clip (-1.5,0) .. controls (-1.5,-0.5) and (0.5,0) .. (0.5,-0.5) .. controls (0.5,-1) and (-1.5,-0.5) .. (-1.5,-1); \draw[fill=gray!20] (-0.5,-0.5) ellipse (0.5 and 1); \draw (-1.5,0) .. controls (-1.5,-0.5) and (0.5,0) .. (0.5,-0.5) .. controls (0.5,-1) and (-1.5,-0.5) .. (-1.5,-1); \end{scope} \begin{scope}[blue, shift={(3,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-0.5,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(4.5,0)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(4.5,3)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \begin{scope}[shift={(0,3)}] \draw[dotted] (-2,-0.5) ellipse (0.5 and 1); \clip (-2,0.5) rectangle (-1,-1.5); \draw (-2,-0.5) ellipse (0.5 and 1); \end{scope} \node[right, red] at (2,-0.5) {$\beta'$}; \node[left, blue] at (1.5,-0.5) {$\alpha$}; \node at (0.5,0) {$\mathcal D$}; \node[circle, fill=black, scale=.2] at (0.5,2.5) {}; \node[right] at (0.5,2.5) {$z$}; \draw (-2,-1.5) -- (2.5,-1.5) (-2,0.5) -- (2.5,0.5) (-2,1.5) -- (2.5,1.5) (-2,3.5) -- (2.5,3.5); \end{tikzpicture}\caption{An admissible Heegaard diagram} \end{figure} However, if the diagram is chosen so that $\alpha, \beta'$ intersect transversely, the Lagrangian intersection Floer cohomology (with $\ZZ/2\ZZ$ coefficients) is $\ZZ/2\ZZ\oplus \ZZ/2\ZZ$. Note that $\beta'$ can be chosen so that it is Hamiltonian isotopic to $\beta$. The discrepancy between these two answers comes from the non-convergence of the homotopy between the composition of continuation maps \[f\circ g:\CF(\alpha, \beta')\to \CF(\alpha, \beta) \to \CF(\alpha, \beta')\] \[\id: \CF(\alpha, \beta')\to \CF(\alpha, \beta')\] over $\ZZ/2\ZZ$ coefficients. The presence of an annulus between $\alpha, \beta$ is the culprit for the non-convergence. One can make the quantities converge by using \underline{\href{https://jeffhicks.net/snippets/index.php?tag=def:novikovRing}{ Novikov coefficients}} instead of $\ZZ/2\ZZ$ coefficients. In that setting, the differential in \cref{fig:nonadmissibleHeegaardDiagram} will be exact unless the areas of the two strips agree --- that is, the Lagrangians $\alpha, \beta$ are Hamiltonian isotopic. \end{example} \printbibliography \end{document}