Sarkar-Wang Algorithm for Heegaard-Floer

Tags: #floer #reading_projects #my_talks #active_projects #papers #definitions Refs: Floer homology

Sarkar Wang Paper Talk

References

Paper: Sarkar-Wang, An algorithm for computing some Heegaard-Floer homologies

Annals version of the paper

Intro/Overview

  • Goals:
    • Given \(Y\) a closed oriented smooth 3-manifold, compute the Heegaard-Floer homology \(\widehat{\operatorname{HF}}(Y; {\mathbb{F}}_2) \in {\mathsf{gr}\,}{\mathsf{Ab}}\).
    • Given \(K \hookrightarrow Y\) a knot, compute the knot Floer homology \(\widehat{\operatorname{HFK}}(Y, K; {\mathbb{F}}_2)\)
  • Strategy: turn admissible diagrams into nice diagrams using isotopies and handleslides
    • Note: stabilization not used in this paper.
  • Main theorem:

figures/image_2021-05-05-12-26-55.png

Definitions

Use a cylindrical reformulation of \(\widehat{\operatorname{HF}}\) due to Lipshitz.

Given a pointed Heegaard diagram \begin{align*} ( \Sigma, \vec \alpha, \vec \beta, w) ,\end{align*} the generators of \(\operatorname{CF}\) are formal sums of points \(\mathbf{x} = \sum_{i=1}^{g-k-1} x_i\) where each \(\alpha\) curve contains some \(x_i\) and each \(\beta\) curve contains some \(x_j\), and \(k\) is the number of basepoints in \(\vec w\).

A region is a connected component of the complement of the curves, so \begin{align*} R_i \in \pi_0 \Sigma \setminus\left\{{\vec \alpha, \vec \beta}\right\} .\end{align*} A formal sum of regions is a 2-chain.

Given 2 generators \(\mathbf{x}, \mathbf{y}\) define \(\pi_2(\mathbf{x}, \mathbf{y})\) to be the set of all 2-chains \(\phi\) satisfying \({{\partial}}^2 \phi = \mathbf{y} - \mathbf{x}\). Call such 2-chains domains.

Define \(n_p(\phi)\) to be the coefficient of the region \(R_i \ni p\), then \(\phi\) is a positive domain if \(n_p(\geq 0)\) for all \(p \in \Sigma\setminus\left\{{ \vec \alpha, \vec \beta }\right\}\).

Define \(\pi_2^0(\mathbf{x}, \mathbf{y})\) to be all domains \(\phi\) such that \(n_{\vec w} = 0\). A Heegaard diagram \(\mathcal{H}\) is admissible if for every generator \(\mathbf{x} \in \operatorname{CF}\), every positive domain \(\varphi\in \pi_2^0(\mathbf{x}, \mathbf{x})\) is trivial.

A region is good if it is an \(n{{\hbox{-}}\mathrm{gon}}\) with \(n\leq 4\), and bad if \(n\geq 5\).

For a disc region \(D\), define the badness \begin{align*} b(D) \coloneqq\max\left\{{n-2, 0}\right\} .\end{align*}

Note

  • \(D\in 2{{\hbox{-}}\mathrm{gon}}\implies b(D) = 0\)
  • \(D\in 4{{\hbox{-}}\mathrm{gon}}\implies b(D) = 0\)
  • \(D\in 6{{\hbox{-}}\mathrm{gon}}\implies b(D) = 1\)
  • \(D\in 8{{\hbox{-}}\mathrm{gon}}\implies b(D) = 2\)

Goal: do moves where

  • \(d(\mathcal{H}' ) \leq d( \mathcal{H} )\), so total distance doesn’t increase.
  • \(b( \mathcal{H}' ) \leq b( \mathcal{H} ) -1\), so badness decreases.

Algorithm Overview

  • Start from an admissible pointed Heegaard diagram, end up with an admissible nice pointed Heegaard diagram using isotopies and handleslides on the \(\beta\) curves.

  • Overview of strategy:

    • Isotope all regions to disks
    • Define a complexity for the diagram
    • Show it’s minimized iff all regions not containing basepoints are good
    • Do an isotopy or handleslide that strictly decreases the complexity.
  • Recipe:

    • Kill non-disk regions (easier)
    • Make all but one region bigons or squares (harder)

Step 1: Killing Non-disk Regions

Short procedure: ensure every \(\alpha\) curve intersects some \(\beta\) curve and vice-versa

  • Fix \(j\), what is in between \(\alpha_j\) and the nearest \(\beta\) curve?

    figures/image_2021-05-07-15-39-22.png

  • Case 1: \(\alpha, \alpha, \beta\).

    • Connect \(p\) to \(q\), do a finger move of \(\beta\) curve to introduce 2 intersections with chosen \(\alpha\)

    figures/image_2021-05-07-15-40-21.png

    • Yields 4 new intersections.
  • Case 2: \(\alpha, \beta, \beta\).

    • Finger move on \(\beta\) curve, pulling all other \(\beta\) curves with it.

    figures/image_2021-05-07-15-41-23.png

Now every \(\alpha_j\) intersects some \(\beta_k\) All non-disk regions have \(\geq 2\) boundary components, so we’ll try to reduce the number of boundary components to one. Problem: poly-annuli regions

figures/image_2021-05-07-15-47-09.png

How to fix:

figures/image_2021-05-07-15-47-29.png

  • Creates no new non-disks
  • Decreases number of boundary components by 1.
  • Repeat until equal to 1.

Step 2: Converting Regions to Bigons or Squares

Now all regions are discs.

\(\alpha\) curves should be red, \(\beta\) should be blue!

For a Heegaard diagram \(\mathcal{H}\) define the distance of \(D\) as \begin{align*} d(D) \coloneqq\min \left\{{ {\sharp}(\gamma\cap\vec\beta) {~\mathrel{\Big\vert}~}w'\in D, w \xrightarrow{\gamma} w', \gamma\in \vec\alpha^c }\right\} \in {\mathbb{Z}}_{\geq 0} .\end{align*}

This is the smallest number of intersection points on any arc connecting the basepoint \(w\) to \(w'\in D\). Define the total distance \begin{align*} d( \mathcal{H} ) \coloneqq\max\left\{{ d(D)}\right\} \in {\mathbb{Z}}_{\geq 0} .\end{align*}

Computing the distance via a path.

figures/image_2021-05-07-15-49-13.png

Can only cross \(\beta\) curves, need to stay in \(\vec \alpha^c\).

For a fixed distance \(d\) define the distance \(d\) complexity as \begin{align*} c_d \coloneqq{\left[ { \sum b(D_i), -b(D_1), -b(D_2), \cdots} \right]} && b(D_1) \geq b(D_2) \geq \cdots \end{align*} For a fixed \(d\), order various \(c_d( \mathcal{H}' )\) lexicographically.

Main theorem

figures/image_2021-05-07-15-51-57.png

Setup: fix \(D_0\) to be the region containing the basepoint, \(D_m\) to be the least bad region. Find an adjacent region \(D_*\) with 1 smaller distance:

figures/image_2021-05-07-15-53-06.png

Idea: we will finger move \(b_*\) through \(D_m\) to reduce the badness of \(D_m\):

figures/image_2021-05-07-15-53-42.png

Look at what happens locally: we introduce some new regions, usually less bad:

figures/image_2021-05-07-15-54-37.png

We can push through \(4{{\hbox{-}}\mathrm{gon}}\) regions:

figures/image_2021-05-07-15-54-57.png

So continue, then do some casework:

figures/image_2021-05-07-15-55-19.png

Case 1: Reach a Bigon

figures/image_2021-05-07-15-57-00.png

figures/image_2021-05-07-15-57-19.png

figures/image_2021-05-07-15-57-30.png

Case 2: Smaller Distance

figures/image_2021-05-07-15-57-53.png

figures/image_2021-05-07-15-58-10.png

figures/image_2021-05-07-15-58-22.png

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