1 Lecture 1: Overview (Wednesday, January 13)

1.1 Course Logistics

Note (DZG): Everything in this section comes from Akram!

1.1.1 Description

“I am teaching a topics course about Heegaard Floer homology next semester. Heegaard Floer homology was defined by Peter Ozsváth and Zoltan Szabó around 2000. It is a package of powerful invariants of smooth 3- and 4-manifolds, knots/links and contact structures. Over the last two decades, it has become a central tool in low-dimensional topology. It has been used extensively to study and resolve important questions concerning unknotting number, slice genus, knot concordance and Dehn surgery. It has been employed in critical ways to study taut foliations, contact structures and smooth 4-manifolds. There are also many rich connections between Heegaard Floer homology and other manifold and knot invariants coming from gauge theory as well as representation theory. We will learn the basic construction of Heegaard Floer homology, starting with the definition of the 3-manifold and knot invariants. In the second half of this course, we will turn to computations and applications of the theory to low-dimensional topology and knot theory. In particular, several numerical invariants have been defined using this homological invariants. At the end of the semester, I would expect each one of you to learn the construction of one of these invariants (of course with my help) and present it to the class.”

1.1.2 Expository Papers

1.1.3 Research Papers

1.1.4 Basic Morse Theory, Symplectic Geometry and Floer Homology

1.1.5 Low-dimensional Topology

1.1.6 Suggested Topics for Presentations

1.2 Intro and Motivation

We’ll assume everything is smooth and oriented.

To closed 3-manifolds \(M\) we can assign a graded abelian group \(\widehat{HF}(M)\), which can be computed combinatorially1 . There are several variants:

\(HF^+\) and \(HF^\infty\) can be computed using \(HF^-\). In general, we’ll write \(HF^{-}\) to denote constructions that work with any of the above variants.

Note that \({\mathbb{Z}}_2\) can be replaced with \({\mathbb{Z}}\), but it’s technical and we won’t discuss it here. For the first half of the course, we’ll just discuss \(\widehat{HF}\), and we’ll discuss the latter 3 in the second half.

1.3 Geometric Information

These invariants can be used to compute the Thurston seminorm of a 3-manifold:

A homology class \(\alpha\in H_2(M)\) can be represented as \(\alpha\in [S]\) for \(S\) a closed surface whose fundamental class represents \(\alpha\) where \(S = \bigcup_{i=1}^n S_i\) can be a union of closed embedded surfaces \(S_i\). Then we first compute \begin{align*} \max\left\{{0, - \chi(S_i) }\right\} = \begin{cases} 0 & \text{if } S_i \cong {\mathbb{S}}^2, {\mathbb{T}}^2 \\ \\ - \chi(S_i) = 2g(S_i) - 2 & \text{ else} . \end{cases} .\end{align*} Note that the max checks if \(\chi\) is positive. Then define \begin{align*} {\left\lVert { \alpha } \right\rVert} \mathrel{\vcenter{:}}=\min_S \qty{ \sum_{i=1}^n \max\left\{{0, - \chi(S_i) }\right\} } ,\end{align*} where we sum over the embedded subsurfaces and check which overall surface gives the smallest norm.

Note that this can’t be a norm, since if \({\mathbb{S}}^2, {\mathbb{T}}^2 \in [S] \implies {\left\lVert {\alpha } \right\rVert}= 0\).

\(HF\) detects3 the Thurston seminorm, and there is a splitting as groups/modules \begin{align*} HF^{-}(M) = \bigoplus _{\mathfrak{s} \in {\operatorname{Spin}}^c(M)} HF^{-}(M, S) \end{align*} where \(S \in {\operatorname{Spin}}^c(M)\) is a spin\(^c\) structure: an oriented 2-dimensional vector bundle on \(M\) (up to some equivalence).

The Thurston norm \({\left\lVert {a} \right\rVert}\) can be computed from this data by considering a perturbed version of \(\widehat{HF}\), denoted \(\underline{\widehat{HF}}\), in the following way: taking the first Chern class \(c_1(\mathfrak{s}) \in H^2(M)\) (which can be associated to every 2-dimensional vector bundle), we have \begin{align*} {\left\lVert { \alpha} \right\rVert} = \max_{\underline{\widehat{HF}}(M, \mathfrak{s} ) \neq 0 } {\left\lvert {{\left\langle { c_1(\mathfrak{s}) },~{ \alpha } \right\rangle} } \right\rvert} .\end{align*}

Floer homology groups split over these spin\(^c\) structures and can be used to compute Thurston norms.

Given \(F \subseteq M\) with genus \(g\geq 2\), \(HF\) detects if \(M\) fibers over \(S^1\) with \(F\) as a fiber, i.e. there exists a fiber bundle

This uses the existence of the splitting over spin\(^c\) structures and uses \(HF^+\) in the following way: such a bundles exists if and only if \begin{align*} \bigoplus _{{\left\langle { c_1(\mathfrak{s}) },~{ [F] } \right\rangle} =2g-2} HF^+(M, \mathfrak{s}) = {\mathbb{Z}} .\end{align*}

Equivalently,

The standard contact structure on \({\mathbb{R}}^3\) is given by \begin{align*} \alpha\mathrel{\vcenter{:}}= dz - ydz ,\end{align*} which yields the following 2-plane field \(\xi \mathrel{\vcenter{:}}=\ker \alpha\):

2-Plane Field in {\mathbb{R}}^3

You can see that \(z=0 \implies y=0\), so the \(xy{\hbox{-}}\)plane is in the kernel, yielding the flat planes down the middle:

Flat Planes

To each such \(\xi\) one can associate a contact class \(c(\xi) \in \widehat{HF}(-M)\), where \(-M\) is \(M\) with the reversed orientation.

This gives obstructions for two of the following important properties of contact structures:

We’ll also discuss similar invariants for knots that were created after these invariants for manifolds.

Recall that a knot is an embedding \(S^1 \hookrightarrow M\).

Example: the trefoil knot

Given a knot \(K \subseteq M\) a 3-manifold (e.g. \(M = S^3\)), there is extra algebraic structure on \(\widehat{CF}(M)\): a filtration. These allow defining a new bigraded abelian group \(\widehat{HFK}(M, K)\) (which is also a \({\mathbb{Z}}_2{\hbox{-}}\)vector space) that takes includes the information of \(K\). This yields a decomposition \begin{align*} \widehat{HFK}(M, K) = \bigoplus _{m, a} \widehat{HFK}_m(M, K, a) .\end{align*}

This similarly works for other variants: there is a filtration on \(CF^-(M)\) which yields \(HFK^-(M, K)\), a bigraded \({\mathbb{Z}}_2[u]{\hbox{-}}\)module.

1.3.1 Some properties of Knot Floer Homology

\(\widehat{HFK}(K)\) categorifies the Alexander polynomial \(\Delta_K(t)\) of \(K\), i.e. taking the graded Euler characteristic yields \begin{align*} \Delta_K (t) =\sum_{m, a} (-1)^m\qty{ \dim \widehat{HFK}_m(K, a) } t^a .\end{align*}

\(\widehat{HFK}(K)\) detects the Seifert genus of a knot \(g(K)\), defined as the smallest \(g\) such that there exists an embedded surface5 \(F\) of genus \(g\) in \(S^3\) that bounds \(K\), so \({{\partial}}F = K\).

The unknot bounds a disc, so its genus is zero:

The genus of the unknot

Using the “outside” disc on the trefoil, find 3 bands that show its genus is 1.

The genus of the trefoil

The genus can be computed by setting \(\widehat{HFK}(K, a) \mathrel{\vcenter{:}}=\bigoplus _m \widehat{HFK}_m(K, a)\), which yields \begin{align*} g(k) = \max \left\{{ a {~\mathrel{\Big|}~}\widehat{HFK}(K, a) \neq 0 }\right\} .\end{align*} Note that the \(a\) grading here is referred to as the Alexander grading.

\(\widehat{HFK}\) detects whether or not a knot is fibered, where \(K\) is fibered if and only if it admits an \(S^1\) family \(F_t\) of Seifert surfaces such that \(t\neq s\in S^1 \implies F_t \cap F_s = K\). I.e., there is a fibration on the knot complement where each fiber is a Seifert surface:

The unknot is fibered by \({\mathbb{D}}^2\)s:

The unknot fibered by discs.

This is “detected” in the following sense: \(K\) is fibered if and only if \begin{align*} \widehat{HFK}(k, g(K)) = {\mathbb{Z}}_2 .\end{align*}

Let \(K \subseteq S^3\). We know \(S^3 = {{\partial}}B^4\), so we consider all of the smoothly properly embedded surfaces \(F\) in \(B^4\) such that \({{\partial}}F = K\) and take the smallest genus:

Knot in S^3 bounding a surface in B^4

We thus define the slice genus or 4-ball genus as \begin{align*} g_S(K) \mathrel{\vcenter{:}}= g_4(K) \mathrel{\vcenter{:}}=\min \left\{{ g(F) {~\mathrel{\Big|}~}F\hookrightarrow B^4 \text{ smootherly, properly with } {{\partial}}F = K }\right\} .\end{align*}

Show that \(g_4(K) \leq g(K)\).

Define \(u(K)\) the unknotting number of \(K\) as the minimum number of times that \(K\) must cross itself to become unknotted.

Consider changing the bottom crossing of a trefoil:

Changing one crossing in the trefoil

This in fact produces the unknot:

Unkink to yield the unknot

Thus \(u(K) = 1\), assuming that we know \(K \neq 0\) is not the unknot.

Show that \(g_f(K) \leq u(K)\).

Hint: each crossing change \(K\to K'\) yields some surface that is a cobordism from \(K\) to \(K'\) in \(B^4\), and you can use each step to build your surface.

Surface between K and K'

Define an invariant \(\tau(K) \in {\mathbb{Z}}\) from \(\widehat{HFK}\) such that \({\left\lvert {\tau(K)} \right\rvert} \leq g_4(K) \leq u(K)\).

Recall that we can view \({\mathbb{T}}^2 \mathrel{\vcenter{:}}={\mathbb{R}}^2/{\mathbb{Z}}^2\) where the action is \((x, y) \xrightarrow{(m, n)} (x+m, y+m)\), i.e. we module out by integer translations. Then for $p, q > 0 $ coprime, \(T_{p, q}\) is the image of the line \(y = mx\) in \({\mathbb{T}}^2\) where \(m=p/q\).

The torus knot \(T_{2, 3}\) wraps 3 times around the torus in one direction and twice in the other:

The torus knot T_{2, 3}

\begin{align*} g_4(T_{p, q}) = u(T_{p, q}) = { (p-1)(1-q) \over 2} .\end{align*}

Show that \(u(T_{p. q}) \leq {(p-1)(q-1) \over 2}\), i.e. torus knots can be unknotted with this many crossing changes.

\begin{align*} \tau(T_{p, q}) = {(p-1)(q-1) \over 2} ,\end{align*}

which implies \begin{align*} {(p-1)(q-1) \over 2} \leq g_4(T_{p, q}) \leq u(T_{p, q}) \leq {(p-1)(q-1) \over 2} ,\end{align*} making all of these equal.

There are better lower bounds for \(u(K)\) defined using \(\widehat{HFK}\) which are not lower bounds for the slice genus. There are also other lower bounds for the slice genus with different names (see Jen Hom’s survey), some of which are stronger than \(\tau\).

Another application of having these lower bounds is that we can construct exotic (or fake) \({\mathbb{R}}^4\)s, i.e. 4-manifolds \(X\) homeomorphic to \({\mathbb{R}}^4\) but not diffeomorphic to \({\mathbb{R}}^4\).

All of these invariants work nicely in a \((3+1){\hbox{-}}\)TQFT: we have invariants of 3-manifolds \(M_i\) and knots in them, so we can talk about cobordisms between them: \(W^4\) a compact oriented 4-manifold with \({{\partial}}W^4 = -M_1 {\coprod}M_2\).

A cobordism

Osvath-Szabó define a map \begin{align*} F_{W, t}^{-}: HF^{-}(M_1, { \left.{{t}} \right|_{{M_1}} } ) \to HF^{-}(M_2, { \left.{{t}} \right|_{{M_2}} }) \end{align*} using \(t\) coming from the splitting of spin\(^c\) structure which yields an invariant of closed 4-manifolds referred to as mixed invariants.

Similarly, if we have knots in 3-manifolds we can define a cobordism \((M_1, K_1) \to (M_2, K_2)\) as \((W^4, F)\) where \(W^4\) is a cobordism \(M_1\to M_2\) and \(F\hookrightarrow W\) is a smoothly embedded surface that is a cobordism from \(K_1\to K_2\) with \(F \cap M_i = K_i\) and \({{\partial}}F = -K_1 {\coprod}K_2\).

A cobordism including knots

This similarly yields a map

\begin{align*} F_{W, F t}^{-}: HF^{-}(M_1, K_1, { \left.{{t}} \right|_{{M_1}} } ) \to HF^{-}(M_2, K_2, { \left.{{t}} \right|_{{M_2}} }) \end{align*}

This smoothly embedded surface in the middle can be used to study other smoothly embedded surfaces in 4-manifolds, which has been done recently.

2 Lecture 2 (Tuesday, January 19)

2.1 Constructing Heegard Floer

For Morse Theory, there are some good exercises in Audin’s book – essentially anything other than the existence questions. The first 8 look good on p. 18.

Today:

  1. Overview of the construction of HF, and

  2. A discussion of Morse Theory.

First goal: discuss how the name “Heegard” fits in.

A genus \(g\) handlebody \(H_g\) is a compact oriented 3-manifold with boundary obtained from \(B^3\) by attaching \(g\) solid handles (a neighborhood of an arc).

For \(g=2\) attached to a sphere, we glue \(D^2 \times I\) by its boundary to \(S^2\).

image_2021-01-19-00-35-48

In general, \({{\partial}}H_g = \Sigma_g\) is a genus \(g\) surface, and \(H_g \setminus{\coprod}_{i=1}^g D_i = B^3\). We can keep track of the data by specifying \((\Sigma, \alpha_1, \alpha_2, \cdots, \alpha_g)\) where \({{\partial}}D_i = \alpha_i\).

Attaching a handlebody

A Heegard diagram is \(M = H_1 \cup_{{\partial}}H_2\) where \(H_i\) are genus \(g\) handlebodies and there is a diffeomorphism \({{\partial}}H_1 \to {{\partial}}H_2\).

Every closed 3-manifold has a Heegard decomposition, although it is not unique.

A Heegard diagram is the data \((\Sigma_g, \alpha = \left\{{ \alpha_1, \cdots, \alpha_g}\right\}, \beta = \left\{{ \beta_1, \cdots, \beta_g}\right\})\) where the \(\alpha\) correspond to \(H_1\) and \(\beta\) to \(H_2\) and \(\Sigma_g = {{\partial}}H_1 = {{\partial}}H_2\).

2.2 Lagrangian Floer Homology

This is essentially an infinite-dimensional version of Morse homology.

A symplectic manifold is a pair \((M^{2n}, \omega)\) such that

A Lagrangian submanifold is an \(L^n \subseteq M\) such that \({ \left.{{\omega}} \right|_{{L}} } = 0\).

If \(L_1 \cap L_2\) is finitely many points, case we can define a chain complex \begin{align*} CF(M^{2n}, L_1, L_2) \mathrel{\vcenter{:}}={\mathbb{Z}}_2[L_1 \cap L_2] ,\end{align*} the \({\mathbb{Z}}_2{\hbox{-}}\)vector space generated by the intersection points of the Lagrangian submanifolds. We’ll define a differential by essentially counting discs between intersection points:

Two intersection points

We’ll want to write \({{\partial}}x = c_y y + \cdots\) where \(c_y\) is some coefficient. How do we compute it? In this case, we have half of the boundary on \(L_1\) and half is on \(L_2\)

i

So we can the number of holomorphic discs from \(x\) to \(y\). We’ll get \({\partial}^2 = 0 \iff \operatorname{im}{\partial}\subset \ker {\partial}\), and \(HF\) will be kernels modulo images. In more detail, we’ll have \begin{align*} {{\partial}}x = \sum_y \sum_{\mu(\varphi) = 1} \# \widehat{\mathcal{M}} (\varphi)y && \widehat{\mathcal{M}}(\varphi) = \mathcal{M}(\varphi) / {\mathbb{R}} \end{align*} where \(\widehat{\mathcal{M}}\) will (in good cases) be a 1-dimensional manifold with finitely many points. Note that it’s not necessarily true that \(CF\) has a grading!

Given a 3-manifold \(M^3\), we’ll associate a Heegard diagram \(\Sigma, \alpha, \beta\). Note the \(g{\hbox{-}}\)element symmetric group acts on \(\prod_{i=1}^g \Sigma\) by permuting the \(g\) coordinates, so we can define \(\operatorname{Sym}^g(\Sigma) \mathrel{\vcenter{:}}=\prod_{i=1}^g \Sigma / S_g\).

The space \(\operatorname{Sym}^g(\Sigma)\) is a smooth complex manifold of \({\mathbb{R}}{\hbox{-}}\)dimension \(2g\).

Write \({\mathbb{T}}_{\alpha} \mathrel{\vcenter{:}}=\prod_{i=1}^g \alpha_i \subseteq \prod_{i=1}^g \Sigma\) for a \(g{\hbox{-}}\)dimensional torus; this admits a quotient map to \(\operatorname{Sym}^g(\Sigma)\). We can repeat this to obtain \({\mathbb{T}}_{\beta}\). Then \(HF^{-}(M)\) will be a variation of Lagrangian Floer Homology for \((\operatorname{Sym}^g(\Sigma), {\mathbb{T}}_{\alpha}, {\mathbb{T}}_{\beta} )\).

Consider constructing a genus \(g=1\) Heegard diagram. Recall that \(S^3\) can be constructed by gluing two solid torii.

image_2021-01-19-12-20-16

Here \((T, \alpha, \beta)\) will be a Heegard diagram for \(S^3\).

Show that the following diagram with \(\beta\) defined as some perturbation of \(\alpha\) is a Heegard diagram for \(S^1 \times S^2\).

image_2021-01-19-12-21-56

Consider \(M\) a 3-manifold containing a knot \(K\), we can construct a new 3-manifold by first removing a neighborhood of \(K\) to yield \(M\setminus N(K)\):

image_2021-01-19-12-23-16

Taking a new solid torus \(S \mathrel{\vcenter{:}}={\mathbb{D}}^2 \times S^1\) and a diffeomorphism \(i: {{\partial}}S \to {{\partial}}(M \setminus N(K))\), this yields a new manifold \(M _{\varphi} (K)\), a surgery along \(K\).

image_2021-01-19-12-25-25

Note that the diffeomorphism is entirely determined by the image of the curve \(\alpha\) . The Knot Floer chain complex of \(K\) will allow us to compute any flavor \(HF^{-}M _{\varphi} (K)\) of Floer homology. Why is this important: any closed 3-manifold is surgery on a link in \(S^3\). However there are many more computational tools available here and not in the other theories: combinatorial approaches to compute, exact sequences, bordered Floer homology.


Next time: we’ll talk about “integer surgeries.”

3 Lecture 3: Morse Theory (Thursday, January 19)

3.1 Intro to Morse Theory

Let \(M^n\) be a smooth closed manifold, then the goal is to study the topology of \(M\) by studying smooth functions \(f \in C^ \infty (M, {\mathbb{R}})\). We’ll need \(f\) to be generic in a sense we’ll discuss later.

image_2021-01-19-00-41-55

A point \(p\in M\) is called a critical point if and only if \((df)_p = 0\).

Fixing a critical point \(p\) for \(f\), the second derivative or Hessian of \(f\) at \(p\) is a bilinear form on \(T_pM\) which is defined in the following way: for \(v, w\in T_p M\), extend \(w\) to a vector field \(\tilde w\) in a neighborhood of \(p\) and set \begin{align*} d^2 f_p(v, w) = v\cdot (\tilde w \cdot f)(p) \mathrel{\vcenter{:}}= v \cdot (df)(\tilde w)(p) .\end{align*} where we take the derivative of \(f\) with respect to \(\tilde w\), then take the derivative with respect to \(v\), then evaluate at the point to get a number.

This is only well-defined at critical points (check!). Note that we need \(\tilde w\) so that \(\tilde w \cdot f\) is again a function (and not a number) which can be differentiated again. You can also take e.g. \(\tilde v \cdot (\tilde w \cdot f)\), differentiating with respect to the vector field instead of just the vector \(v\), but we’re plugging in \(p\) in either case.

The second derivative is

  1. Well-defined, and

  2. Symmetric

If you fix a coordinate chart in a neighborhood of \(p\), then the bilinear form is represented by a matrix given by \begin{align*} (d^2 f)_p = H_p = \qty{ {\frac{\partial ^2}{\partial x_j {\partial}x_i}\,}(p)}_{ij} .\end{align*}

We can compute \begin{align*} (d^2 f)_p(v, w) - (d^2 f)_p(w, v) &= v\cdot (\tilde w \cdot f)(p) - w \cdot (\tilde v \cdot f)(p) \\ &\mathrel{\vcenter{:}}= df_p \qty{ [\tilde v, \tilde w]} \\ & = 0 && \text{since $p$ is a critical point and $df_p = 0$} .\end{align*}

This is now easier to prove: we are picking an extension of \(w\) to a vector field, so we need to show that the definition doesn’t depend on that choice. \begin{align*} (d^2 f)(_p(v, w) &= v\cdot (\tilde w \cdot f)(p) && \text{which doesn't depend on }\tilde v\\ &= (d^2 f)_p(w, v) \\ &= w\cdot (\tilde v \cdot f)(p) && \text{which doesn't depend on } \tilde w ,\end{align*} and thus this is independent of both \(\tilde v\) and \(\tilde w\).

Show that the second derivative in local coordinates is given by the matrix \(H_p\) above.

In local coordinates, we can write \(v = \sum_{i=1}^n a_i {\frac{\partial }{\partial x_i}\,}\) and \(w = \sum_{i=1}^n b_i {\frac{\partial }{\partial x_i}\,}\), and thus \begin{align*} (d^2 f)_p(v, w) = \mathbf{b}^t H_p \mathbf{a} = \sum_{1 \leq i,j \leq n} a_i b_j {\frac{\partial ^2 f}{\partial x_i {\partial}x_j}\,}(p) .\end{align*}

A critical point \(p\in M\) is called nondegenerate if the bilinear form \((d^2 f)_p\) is nondegenerate at \(p\), i.e. for all \(v\in T_p M\) there exists a \(w\in T_pM\setminus\left\{{\mathbf{0}}\right\}\) such that \((d^2 f)_p(v, w) \neq 0\). This occurs if and only if \(H_p\) is invertible.

Given a nondegenerate critical point \(p\in M\), define the index \(\operatorname{ind}(p)\) of \(f\) at \(p\) in the following way: since \(H_p\) is symmetric and nondegenerate, its eigenvalues are real and nonzero, so define the index as the number of negative eigenvalues of \(H_p\).

A function \(f\in C^ \infty (M, {\mathbb{R}})\) is called a Morse function if and only if all of its critical points are nondegenerate.

We’ll see that almost every smooth function is Morse, and these are preferable since they have a simple and predictable structure near critical points and don’t do anything interesting elsewhere.

Let \(p\in M\) be a nondegenerate critical point of \(f\) with \(\operatorname{ind}(p) = \lambda\). Then there exists charts \(\varphi:(U, p) \to ({\mathbb{R}}^n, 0)\) such that writing \(f\) in local coordinates yields \begin{align*} (f \circ \varphi ^{-1} )(x) = f(p) - \sum_{i=1}^{\lambda} x_i^2 + \sum_{j= \lambda + 1}^n x_j^2 .\end{align*}

We have \begin{align*} H_p = \begin{bmatrix} -2&&&&&&\\ &\ddots&&&&&\\ &&-2&&&&\\ &&&2&&&\\ &&&&\ddots&&\\ &&&&&2&\\ &&&&&&2 \end{bmatrix} = -2 I_{\lambda} \oplus 2 I_{n- \lambda} .\end{align*}

If \(\lambda=n\)??

??

Consider \(S^2\) with a height function:

Sphere with a height function

Then we have a local minimum at the South pole \(p\) and a local max at the North pole \(q\), where \(\operatorname{ind}(p) = 0\) and \(\operatorname{ind}(q) = 2\). Note that the critical points essentially occur where the tangent space is horizontal

Consider \({\mathbb{T}}^2\) with the height function:

Torus with a height function

This has a similar max/min as the sphere, but also has two critical points in the middle that resemble saddles:

Saddle points

Define \(M_a \mathrel{\vcenter{:}}= f ^{-1} ((- \infty , a])\); we then want to consider how \(M_a\) changes as \(a\) changes:

M_a on the sphere
M_a on the torus

If \(f ^{-1} ([a, b])\) contains no critical points, then \begin{align*} f ^{-1} (a) &\cong f ^{-1} (b) \\ M_a &\cong M_b .\end{align*}

Choose a metric \(g\) on \(M\), then the gradient vector of \(f\) is given by \begin{align*} g(\nabla f, v) = df(v) .\end{align*}

We have \begin{align*} df( \nabla f) = g(\nabla f, \nabla f) = {\left\lVert {\nabla f} \right\rVert}^2 .\end{align*}

We have the following situation:

image_2021-01-21-12-11-16

The gradient vector is always tangent to the level sets, so we can consider the curve \(\gamma\) which satisfies \(\dot\gamma(t) = -\nabla f( \gamma(t))\):

image_2021-01-21-12-12-42

For technical reasons, we want to end up with cohomology instead of homology and will take \(-\nabla f\) instead of \(\nabla f\) everywhere:

image_2021-01-21-12-13-35

So \(\gamma\) will be a trajectory of \(- \nabla f\), and \(f ^{-1} [a, b] \cong f ^{-1} (a) \times[0, 1]\). A problem is that following these trajectories may involve arriving at \(f ^{-1} (a)\) at different times:

image_2021-01-21-12-15-10

We can fix this by normalizing: \begin{align*} V \mathrel{\vcenter{:}}=- \nabla f / {\left\lVert { \nabla f} \right\rVert}^2 \implies (df)(v) = {\left\langle { \nabla f},~{ - \nabla f / {\left\lVert {\nabla f} \right\rVert}^2} \right\rangle} = -1 .\end{align*}

For every \(p \in f ^{-1} (b)\), if \(\gamma(t)\) is the trajectory starting from \(p\), i.e. \(\gamma(0) = p\), then \(\gamma(b-a) \in f ^{-1} (a)\). So define \begin{align*} \Phi: f ^{-1} (b) \times[0, b-a] &\to f ^{-1} ([a, b]) \\ (p, t) &\mapsto \gamma_p (t) ,\end{align*} which will be a diffeomorphism.

Suppose \(f ^{-1} ([a, b])\) contains exactly one critical point \(p\) with \(\operatorname{ind}(p) = \lambda\) and \(f(p) = c\). Then \begin{align*} M_b = M_a \cup_{{{\partial}}} \qty{ D^ \lambda \times D^{n - \lambda} } \end{align*} where \(n \mathrel{\vcenter{:}}=\dim M\).

For \(\lambda= 1, n - \lambda= 2\):

image_2021-01-21-12-32-38
image_2021-01-19-00-53-07

\begin{align*} W_f^u(p) \mathrel{\vcenter{:}}=\left\{{p}\right\} \cup\left\{{ \dot{\gamma(t)} = -\nabla f(\gamma(t)),\, \lim_{t\to -\infty} \gamma(t) = p,\, t\in {\mathbb{R}} }\right\} .\end{align*}

If \(\operatorname{ind}(p) = \lambda\) then \(W_f^u(p) \cong {\mathbb{R}}^ \lambda\).

image_2021-01-19-00-55-24
image_2021-01-19-00-55-41

\begin{align*} W_f^s(p) \mathrel{\vcenter{:}}=\left\{{p}\right\} \cup\left\{{ \dot{\gamma(t)} = -\nabla f(\gamma(t)),\, \lim_{t\to +\infty} \gamma(t) = p,\, t\in {\mathbb{R}} }\right\} .\end{align*}

If \(\operatorname{ind}(p) = \lambda\) then \(W_f^s(p) \cong {\mathbb{R}}^{n- \lambda}\).

\(C^ \infty (M; {\mathbb{R}})\) is defined as smooth function \(M\to |RR\), topologized as:

And a basis for open neighborhoods around \(p\) is given by \begin{align*} N_g(f) = \left\{{ g:M\to {\mathbb{R}}{~\mathrel{\Big|}~} {\left\lvert { {\frac{\partial ^k g}{\partial {\partial}x _{i_1} \cdots {\partial}x _{i_k} }\,}(p) - {\frac{\partial ^k f}{\partial {\partial}x _{i_1} \cdots {\partial}x _{i_k} }\,}(p) } \right\rvert} < \infty\, \forall \alpha,\, \forall p\in h_ \alpha(C_ \alpha) }\right\} .\end{align*}

The set of Morse functions on \(M\) is open and dense in \(C^ \infty (M; {\mathbb{R}})\).

4 Tuesday, January 26

4.1 Attaching Handles

Goal: we want to use Morse functions (smooth, nondegenerate critical points) to study the topology of \(M\). Recall that the torus had 4 critical points,

image_2021-01-26-11-14-32

We defined the index as the number of negative eigenvalues of the Hessian matrix. Here the highest index will be the dimension of the manifold, and by the Morse lemma the two intermediate critical points will be index 1.

We want to use the Morse function to decompose the manifold, so we consider \(M_a \mathrel{\vcenter{:}}= f ^{-1} ((- \infty , a ])\). If \(f ^{-1} [a, b]\) does not contain a critical point, then \(M_a \cong M_b\) and \(f ^{-1} (a) \cong f ^{-1} (b)\). So taking \(M_{1/2}\) and \(M_{3/4}\) here both yield discs:

image_2021-01-26-11-17-46

Passing through critical points does change the manifold, though:

image_2021-01-26-11-19-01

Suppose \(f ^{-1} [a, b]\) contains exactly one critical point of index \(\lambda\) then \begin{align*} M_b \cong M_a \cup_{\varphi} (D_ \lambda \times D_{n - \lambda}) ,\end{align*} where \(\varphi: ({{\partial}}D_ \lambda \times D_{ n - \lambda}) \hookrightarrow{{\partial}}M_a\).

For the case \(\lambda= 1, n = 3\), we have the following situation:

image_2021-01-26-11-24-46

Taking \(\lambda=1, n=2\), we attach \(D^1 \times D^1\) and get the following situation:

image_2021-01-26-11-27-16

Adding on another piece, the new boundary is given by the highlighted region:

image_2021-01-26-11-32-27

And continuing to attach the last pieces yields the following:

image_2021-01-26-11-33-31

There is a deformation retract \(M_b \to M_a \cup C_ \lambda\), where \(C_ \lambda\) is a \(\lambda{\hbox{-}}\)cell given by \(D_ \lambda \times\left\{{0}\right\}\). For example:

image_2021-01-26-11-36-35

4.2 Stable and Unstable Manifolds

Given \(- \nabla f\) for a fixed metric, the unstable manifold for a critical point \(p\) is defined as \begin{align*} W_f^u(p) \mathrel{\vcenter{:}}=\left\{{p}\right\} \cup\left\{{ \gamma(t) {~\mathrel{\Big|}~}\dot \gamma(t) = - \nabla f( \gamma(t) ),\, \gamma(t) \overset{t\to -\infty}\to p }\right\} .\end{align*} Here \(\gamma(t)\) is the trajectory of \(-\nabla(f)\).

The unstable manifold is highlighted in blue here:

image_2021-01-26-11-42-01

The gradient trajectories for other points are given by the yellow lines in the following:

image_2021-01-26-11-44-13

If \(\operatorname{ind}(p) = \lambda\), then the unstable manifold \(W_f^u\) at \(p\) is isomorphic to \({\mathbb{R}}^ \lambda\).

image_2021-01-26-11-46-46

Here the unstable manifold for \(p_2\) will be 2-dimensional, with one flow line ending at \(p_1\) and the rest ending at \(p_0\).

image_2021-01-26-11-47-24

The stable manifold for a critical point \(p\) is defined as \begin{align*} W_f^s(p) \mathrel{\vcenter{:}}=\left\{{p}\right\} \cup\left\{{ \gamma(t) {~\mathrel{\Big|}~}\dot \gamma(t) = - \nabla f( \gamma(t) ),\, \gamma(t) \overset{t\to {\color{red} +} \infty}\to p }\right\} .\end{align*}

The stable manifold for \(p_0\) above is every trajectory ending at \(p_0\). \(W^s(p) = S^2 \setminus W^s(p_1) \cup W_s(p_3)\)? See video?

4.3 Morse Functions

The set of Morse functions is open and dense in \(C^ \infty (M; {\mathbb{R}})\) in a certain topology.6

We’ll use this to define a chain complex \(C_*(f, g)\) where \(g\) is a chosen metric, define a differential, and use this to define a homology theory. For notation, we’ll write \(\operatorname{crit}(f)\) as the set of critical points of \(f\), and given \(p, q\in \operatorname{crit}(f)\) with \(\gamma\) a trajectory running from \(p\) to \(q\), we have \begin{align*} W^u(p) \cap W^s(q) = \left\{{ \gamma(t) {~\mathrel{\Big|}~} \gamma(t) \overset{t\to -\infty }\to p,\, \gamma(t) \overset{t\to +\infty }\to q }\right\} .\end{align*}

Two submanifolds \(X, Y \subseteq M\) intersect transversely if and only if \begin{align*} T_pX + T_p Y \mathrel{\vcenter{:}}=\left\{{v+w{~\mathrel{\Big|}~}v\in T_p X, w\in T_p Y}\right\} = T_p M && \forall p\in X \cap Y .\end{align*} In this case, we write \(X \pitchfork Y\).

An example of a transverse intersection:

image_2021-01-26-12-02-29

An example of an intersection that is not transverse:

image_2021-01-26-12-03-13

A pair \((f, g)\) with \(f\) a Morse function and \(g\) a metric is Morse-Smale if and only if

For a generic metric \(g\), the pair \((f, g)\) is Morse-Smale.

This means that metrics can be perturbed to become Morse-Smale.

The following is not Morse-Smale:

image_2021-01-26-12-06-06

Note that if \(X^a \pitchfork Y^b\), then \(X \cap Y \subseteq M^n\) is a smooth submanifold of dimension \(a+b-n\). In general, we have \(M^s(p) \cong {\mathbb{R}}^{n - \lambda}\) where \(\lambda = \operatorname{ind}(p)\).

If \((f, g)\) is Morse-Smale, then \(M^u(p) \pitchfork M^s(q)\). In this case, \begin{align*} \dim(M^u(p) \cap M^s(q)) = \operatorname{ind}(p) + n - \operatorname{ind}(q) - n = \operatorname{ind}(p) - \operatorname{ind}(q) .\end{align*} Thus if \(\operatorname{ind}(p) = \operatorname{ind}(q)\) then \(\dim M^s(p) \cap M^s(q) = 0\).

There is an \({\mathbb{R}}{\hbox{-}}\)action of \(M^s(p) \cap M^s(q)\): \begin{align*} \qty{ M^s(p) \times M^u(q) } \times{\mathbb{R}}&\to M^s(p) \cap M^u(q) \\ ( \gamma(t), c) &\mapsto \gamma(t+c) .\end{align*} If \(p\neq q\), this action is free and we can thus quotient by it to obtain \begin{align*} \mathcal{M}(p, q) \mathrel{\vcenter{:}}=\qty{ M^s(p) \cap M^u(q)} / {\mathbb{R}} .\end{align*} This identifies all points on the same trajectory, yielding one point for every trajectory, and so this is called the moduli space of trajectories from \(p\) to \(q\).

If \(\operatorname{ind}(p) = \operatorname{ind}(q)\), we have \(\dim M^u(p) \cap M^s (q) = 0\), making \(\dim \mathcal{M}(p, q) = -1\) and thus \(\mathcal{M}(p, q) = \emptyset\) and no gradient trajectories connect \(p\) to \(q\). Referring back to the example, since \(\operatorname{ind}(p_3) = \operatorname{ind}(p_2)\), if \((f, g)\) were Morse-Smale then there would be no trajectory \(p_3 \to p_2\), whereas in this case there is at least one.

If \(\operatorname{ind}(p) - \operatorname{ind}(q) = 1\), then \(\dim \mathcal{M}(p, q) = \operatorname{ind}(p) - \operatorname{ind}(q) - 1 = 0\), making \(\mathcal{M}(p, q)\) a compact 0-dimensional manifold, which is thus finitely many points, meaning there are only finitely many trajectories connecting \(p\to q\) and it becomes possible to define a Morse complex.

Fix \((f, g)\) a Morse-Smale pair, then define \begin{align*} C_i(f, g) \mathrel{\vcenter{:}}={\mathbb{Z}}/2{\mathbb{Z}}\left[\left\{{p {~\mathrel{\Big|}~}\operatorname{ind}p = i}\right\}\right] = \bigoplus_{\operatorname{ind}(p) = i} {\mathbb{Z}}/2{\mathbb{Z}}\left\langle{p}\right\rangle ,\end{align*} with a differential \begin{align*} {{\partial}}: C_i(f, g) &\to C_{i-1}(f, g) \\ p, \operatorname{ind}(p) = i & \mapsto \sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) q ,\end{align*} where we take the count mod 2.

\({{\partial}}^2 = 0\), and thus \(( C(f, g), {{\partial}})\) is a chain complex.

Next time we will work on proving this.

5 Morse Homology and Lagrangian Floer Homology (Thursday, January 28)

5.1 Morse Homology

Last time: defined the Morse complex. Assumed \((f, g)\) was a Morse-Smale pair, where \(f\) is a Morse function and \(g\) is a Riemannian metric, and this guarantees that if \(p, q\in \operatorname{crit}(f)\) with \(\operatorname{ind}(p) - \operatorname{ind}(q) = 1\), then (among other things) there are finitely many gradient trajectories \(p\leadsto q\). We denoted this \(\mathcal{M}(p, q)\). The chain complex was defined by \(C_i(f, g) \mathrel{\vcenter{:}}=\bigoplus_{\operatorname{ind}(p) = i} {\mathbb{Z}}_2 \left\langle{ p }\right\rangle\) with differential \({{\partial}}_i: C_i \to C_{i-1}\) was defined by sending an index \(i\) critical point \(p\) to \(\sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) q \pmod 2\).

\({{\partial}}_{i} \circ {{\partial}}_{i+1} = 0\).

Idea of the proof: we can directly compute \begin{align*} {{\partial}}({{\partial}}p) &= {{\partial}}\qty{ \sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) q } \\ &= \sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) {{\partial}}q \\ &= \sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) \qty{ \sum_{\operatorname{ind}(r) = i-2 \# \mathcal{M}(q, r) r }} \\ &= \sum_{\operatorname{ind}(r) = i-2} \qty{\sum_{\operatorname{ind}(q) = i-1} \# \mathcal{M}(p, q) \# \mathcal{M}(q, r) } r \\ &= \sum_{\operatorname{ind}(r) = i-2} c_{p,q,r} r \\ &= 0 && \text{(claim)} .\end{align*}

This happens if and only if \(c_{p, q, r} = 0 \pmod 2\) for all \(r\) with \(\operatorname{ind}(r) = i-2\). This is multiplication of the number of trajectories:

image_2021-01-28-11-23-19

In other words, this is the total number of trajectories \(p\leadsto r\) that pass through \(q\). These trajectories “break” at \(q\), and so we refer to these as broken trajectories.

Suppose \(\operatorname{ind}(r) = \operatorname{ind}(p) - 2\), then a broken trajectory from \(p\) to \(r\) is a trajectory from \(p\) to \(q\) followed by a trajectory \(q\) to \(r\) where \(\operatorname{ind}(q) = \operatorname{ind}(p)-1 = \operatorname{ind}(r) + 1\).

image_2021-01-28-11-26-25

Why is the number of broken trajectories even?

We can check that \(\dim \mathcal{M}(p, r) = \dim \qty{ W^u(p) \pitchfork W^s(r)}/{\mathbb{R}}= (\operatorname{ind}(p) - \operatorname{ind}(r)) - 1 = 2-1 = 1\). We can compactify \(\mathcal{M}(p, r)\) by adding in all of the broken trajectories to define \begin{align*} \overline{\mathcal{M}(p, r)} \cup\qty{ \bigcup_{\operatorname{ind}(q) = i-1} \mathcal{M}(p, q) \times\mathcal{M}(q, r) } .\end{align*} This is useful here because we can appeal to the classification of smooth compact 1-dimensional manifolds, which are unions of copies of \(S^1\) and \(D_1 = I\). In particular, the number of boundary points \begin{align*} {{\partial}}\overline{\mathcal{M}(p, r)} = \bigcup_{\operatorname{ind}(q) = i-1} \mathcal{M}(p, q) \times\mathcal{M}(q, r) \end{align*} is even:

image_2021-01-28-11-32-34

Suppose you have two critical points of the same index. The Morse-Smale condition implies that there’s no trajectory between them. A counterexample would be \(p_3 \leadsto p_2\) on the torus with the height function:

image_2021-01-28-11-45-16

However, if you perturb this slightly, the trajectories can be made to miss \(p_2\) and end at \(p_1\) instead. All of the trajectories are disjoint, so we end up with a situation like the following after perturbing the metric:

image_2021-01-28-11-48-06

We can cut along a curve on the bottom to better analyze these trajectories:

image_2021-01-28-11-48-57 image_2021-01-28-11-50-27

Now cut this cylinder along the trajectories \(p_1\leadsto p_3 \leadsto p_1\), i.e. the green trajectories here:

image_2021-01-28-11-51-31
image_2021-01-28-11-53-32

Here we can see that as the trajectories approach the corners, they limit to broken trjacetories:

image_2021-01-28-11-54-44

We can compute

Since there are exactly two trajectories \(p_4\) to \(p_2\) or \(p_3\), we get \({{\partial}}_2 = 0\). Similarly \({{\partial}}_1 = 0\), and we get \(HM_i(T) = [{\mathbb{Z}}/2{\mathbb{Z}}, {\mathbb{Z}}/2{\mathbb{Z}}^2, {\mathbb{Z}}/2{\mathbb{Z}}, 0, \cdots]\), which is the same as its singular homology.

\begin{align*} HM_i(f, g) \cong H_i^{{\operatorname{Sing}}}(M; {\mathbb{Z}}/2{\mathbb{Z}}) .\end{align*} In particular, it doesn’t depend on the choice of Morse-Smale pair \((f, g)\). See proof in references, e.g. Audin.

By definition, \(\# \operatorname{crit}_i(f) = \operatorname{rank}C_i(f, g) = \operatorname{rank}HM_i(f, g)\), and in any chain complex the rank of the chain groups are always at least the rank of the homology.

5.2 Lagrangian Floer Homology

Suppose \(L_0^n, L_1^n \subset M^{2n}\) are compact with \(L_0 \pitchfork L_1\), so the intersection is finitely many points.

image_2021-01-28-12-16-27

We can do Morse theory on the space of paths between them: \begin{align*} \mathcal{P}(L_0, L_1) \mathrel{\vcenter{:}}=\left\{{ \gamma: I\to M {~\mathrel{\Big|}~}\gamma(0) \in L_0, \gamma(1) \in L_1}\right\} .\end{align*}

We’ll find analogs of Morse functions on \(P(L_0, L_1)\) such that the critical points are constant paths, i.e. \(L_0 \cap L_1\). The Morse inequalities then gives bounds on the number of intersection points between \(L_0\) and \(L_1\).

A symplectic manifold is a pair \((M^{2n}, \omega)\) with \(\omega\) a 2-form which is

A half-dimensional submanifold \(L^n \subset M^{2n}\) is called Lagrangian if \({ \left.{{ \omega}} \right|_{{L^n}} } = 0\).

The pair \(({\mathbb{R}}^{2n}, \sum_{i=1}^n dx_i \wedge dy_i\) is a symplectic manifold (and also a symplectic vector space). Note that this 2-form is also a bilinear form of the following shape:

\begin{align*} \begin{bmatrix} 0 & \operatorname{id}_n \\ -\operatorname{id}_n & 0 \end{bmatrix} .\end{align*}

This has a Lagrangian submanifold \({\mathbb{R}}^n \mathrel{\vcenter{:}}=\left\{{y_1 = \cdots = y_n = 0}\right\}\).

Note: See Darboux theorem.

The general setup for next time: we’ll have \((M^{2n}, \omega)\) a symplectic manifold, a pair \(L_0, L_1 \subset M\) such that \(L_0 \pitchfork L_1\), and we want to do Morse Homology on \(\mathcal{P}(L_0, L_1)\).

6 Lecture 6 (Tuesday, February 02)

We’re working with a symplectic manifold, i.e. a pair \((M^{2n}, \omega)\) where \(\omega \in \Omega^2\) is closed, i.e. \(d \omega = 0\), and nondegenerate, i.e. \(\bigvee^n \omega \neq 0\). We were also consider \(L^n_0, L^n_1 \subset M\) Lagrangian submanifolds, i.e. \({ \left.{{ \omega }} \right|_{{L_i}} } = 0\). The goal is to do something like Morse homology on \(\mathcal{P}(L_0, L_1)\) where the critical points corresponds to intersection points \(L_0 \cap L_1\), where we’ll assume \(L_0 \pitchfork L_1\).

What is the analog of a Morse function?

The functional \(f\) is defined on the universal cover \(\overline{\mathcal{P}}(L_0, L_1) \to {\mathbb{R}}\). We can get around knowing much about \(f\) because we only ever need derivatives \(df\) and a metric \(g\) on the path space to talk about the gradient \(\nabla_g f\). We’ll define a 1-form \(\alpha: T \mathcal{P}(L_0, L_1) \to {\mathbb{R}}\), where we can define this tangent space as \(T_{\gamma} \mathcal{P}(L_0, L_1)\) where \(\gamma(s): I \to M\). Set \(u(s, t)\) to be a path from \(\gamma\) to \(\gamma'\) where \(u(s, 0) = \gamma\) and \(u(s, 1) = \gamma'\) and \({\frac{\partial u}{\partial t}\,}\Big|_{t=0}\), which is a tangent vector to \(\gamma\) and thus \({\frac{\partial u}{\partial t}\,}(s, 0) \in T_{\gamma(s)}M\).

image_2021-02-02-11-29-54

Upshot: tangent vectors in \(T_{\gamma} \mathcal{P}(L_0, L_1)\) are given by \(\xi(s) \in T_{\gamma(s)}M\) for every \(s \in I\), i.e. a way to push the path off of itself to obtain a new path.

We can thus define \begin{align*} \alpha: T\mathcal{P}(L_0, L_1) &\to {\mathbb{R}}\\ ( \gamma, \xi \in T_{\gamma} \mathcal{P}) &\mapsto \alpha_{\gamma}{\xi} \mathrel{\vcenter{:}}=\int_0^1 \omega( \dot{\gamma}(s), \xi(s)) ds .\end{align*} Does this have the property we want? I.e. is it zero when \(\gamma\) is the constant path?

\begin{align*} \alpha _{\gamma} \equiv 0 \iff \gamma(s) \text{ is constant} \iff \dot{ \gamma(s)} = 0 \text{ for all } \gamma \iff \gamma(s) \in L_0 \cap L_1 .\end{align*}

\begin{align*} \alpha _{\gamma} \equiv 0 \iff \int_0^1 \omega( \dot{\gamma}(s), \xi(s)) ds = 0 \text{ for all } \xi\not\equiv 0 .\end{align*}

If \(\dot{\gamma}(s) \neq 0\) for some \(s\) then this is also true in an open neighborhood by smoothness, so one can find a \(\xi\) such that

We’ll need a few tools:

An almost complex structure is a bundle automorphism \(J: TM\to TM\) such that \(J \circ J = - \one_{TM}\). It is said to be compatible with \(\omega\) if and only if

In this case, there is a Riemannian metric defined by \(g(v, w) = \omega(v, Jw)\). Conversely, given an almost complex structure \(J\) and a metric \(J\), there is a symplectic form defined by \(\omega(v, w) = \omega(Jv, Jw) \mathrel{\vcenter{:}}= g(Jv, w)\).

Check that \(\omega\) is a symplectic form compatible with \(J\) and \(g\) is the corresponding metric.

Given a symplectic form \(\omega\) and a Riemannian metric \(g\) there exists a canonical almost complex structure \(J\) compatible with \(\omega\) such that the previous process sends \((\omega, J)\) to \(g\).

Any symplectic manifold \((M, \omega)\) has a compatible almost complex structures \(J\).

The space of all almost complex structures on \(M\) compatible with \(\omega\) is contractible.

Here we can use that \(\xi(s) = J \dot{ \gamma}(s)\) which implies \(\omega( \dot{\gamma}(s), J \dot{\gamma(s)}) = \omega( \dot{ \gamma(s)}, \xi(s) ) > 0\), which happens if and only if \(\dot{ \gamma}(s) \neq 0\).

So pick an almost complex structure compatible with \(\omega\) and produce a metric \(g\). We’ll define a metric on \(\mathcal{P}(L_0, L_1)\) by the following: for \(\xi, \eta \in T_{\gamma} \mathcal{P}\), recalling that \(\xi = \xi(s), \eta = \eta(s) \in T_{\gamma(s)}M\), set \begin{align*} g_{\gamma}^\mathcal{P}( \xi, \eta) \mathrel{\vcenter{:}}=\int_0^1 g( \xi(s), \eta(s) ) ds = \int_0^1 \omega(\xi, J \eta) ds .\end{align*}

Check that \(g^\mathcal{P}\) is a metric on \(\mathcal{P}(L_0, L_1)\).

We’ll now define a gradient vector field: \begin{align*} g_{\gamma}^\mathcal{P}( -\nabla, {-}) = \alpha({-}) .\end{align*} So here \(\alpha\) will play the role of \(-df\). We can write \begin{align*} \int_0^1 \omega( - \nabla, J \xi) ds = \int_0^1 \omega( \cdot{\gamma}, \xi) ds .\end{align*} Using compatibility, the LHS is equal to \begin{align*} \cdots = \int_0^1 \omega(-J \nabla, J^2 \xi) ds = \int_0^1 \omega(J \nabla, \xi) ds .\end{align*} So the RHS is equal to this for every \(\xi\), which means that \(J \nabla= \dot{\gamma}\). Multiplying both sides by \(J\) yields \(\nabla= -J \dot{\gamma}\) What are the trajectories of \(J \cdot{ \gamma}(s)\in T_{\gamma} \mathcal{P}\)? We can compute \begin{align*} {\frac{\partial u}{\partial t}\,} (s, t) = J {\frac{\partial u}{\partial s}\,}(s, t) .\end{align*} Here \(t\) is the parameter that moves between paths, and \(s\) moves along a given path:

image_2021-02-02-12-34-41

7 Lecture 7 (Thursday February 04)

7.1 Lagrangian Floer Homology

Recall that we had a symplectic manifold \((M^{2n}, \omega)\) with \(L_0, L_1 \subset M\) two Lagrangians. We wanted to do something like Morse theory on \(\mathcal{P}(L_0, L_1)\).

image_2021-02-16-22-21-44

What ingredients do we need?

To define \(\alpha\) we needed to look at \begin{align*} T_{ \gamma} \mathcal{P}= \left\{{ \xi: I\to TM {~\mathrel{\Big|}~}\xi(s) \in T_{\gamma(s)}M }\right\} ,\end{align*} which is like a collection of tangent vectors along \(\gamma\) giving a way to deform the path. Since \(\alpha\in \Omega^1(\mathcal{P})\), for any \(\gamma\) it induces a map \begin{align*} T_{ \gamma} M &\xrightarrow{\alpha} {\mathbb{R}}\\ \xi &\mapsto \alpha_{ \gamma} (\xi) \mathrel{\vcenter{:}}=\int_0^1 \omega( \dot{ \gamma}, \xi)\,ds .\end{align*}

\(\alpha_{ \gamma} = 0 \iff \gamma\) is constant, which happens if and only if \(\gamma\in L_0 \cap L_1\). This corresponds to critical points of the functional yielding intersection points of the Lagrangians.

We wanted to define the gradient, for which we needed a metric on \(\mathcal{P}\). We did this by lifting a metric from \(M\). Pick an almost complex structure \(J\) compatible with \(\omega\), then this yields a Riemannian metric defined by \(g(v, w) = \omega(v, Jw)\). Then we can define \begin{align*} g^\mathcal{P}_{\gamma}(\xi, \eta) \mathrel{\vcenter{:}}=\int_0^1 g( \xi(s), \eta(s) )\,ds .\end{align*} We used this to compute the vector field \(-\operatorname{grad}_{\gamma} J \cdot{\gamma}(s)\). What are its trajectories? These are paths of paths \(u(s, t) \mathrel{\vcenter{:}}= u_t(s)\) such that \({\frac{\partial }{\partial t}\,} u_t(s) = J {\frac{\partial }{\partial s}\,} u_t\). We thus get an equation \begin{align*} {\frac{\partial u}{\partial t}\,}(s, t) = J {\frac{\partial u}{\partial s}\,}(s, t) .\end{align*}

For \(x, y \in L_0 \cap L_1\) trajectories connecting \(x\) to \(y\), we’ll write this as \begin{align*} \mathcal{M}(x, y) \mathrel{\vcenter{:}}=\left\{{ u(s, t) :[0,1] \times{\mathbb{R}}\to M \substack{ u(0, t) \in L_0 \\ u(1, t) \in L_1 \\ u(s, t) \overset{t\to - \infty }\to x \\ u(s, t) \overset{t\to \infty }\to y \\ {\frac{\partial u}{\partial t}\,} = J {\frac{\partial u}{\partial s}\,} } }\right\} .\end{align*}

We can modify this PDE to make things look familiar: multiply both sides with \(J\) to obtain \begin{align*} J {\frac{\partial u}{\partial t}\,} = J^2 {\frac{\partial u}{\partial s}\,} \implies J {\frac{\partial u}{\partial t}\,} = - {\frac{\partial u}{\partial s}\,} \implies {\frac{\partial u}{\partial s}\,} + J {\frac{\partial u}{\partial t}\,} = 0 ,\end{align*} which is the Cauchy-Riemann equation.

Check that this equation can be written as \(J\, du = du \circ i\) where \(i\) is the standard complex structure on \({\mathbb{C}}\supseteq [0, 1] \times{\mathbb{R}}\), so \(du\) commutes with \(i\) and \(J\).

If \(J\, du = du \circ i\), then \(u\) is called a \(J{\hbox{-}}\)holomorphic disc or a pseudoholomorphic disc.

Schematically, the situation is the following:

image_2021-02-16-23-22-40

Using the Riemann mapping theorem, the strip on the left-hand side is biholomorphic to \({\mathbb{D}}\subseteq {\mathbb{C}}\) with \(\pm i\) removed:

image_2021-02-16-23-23-43

Due to the limit conditions at infinity in the strip, we can extend \(u\) to a \(J{\hbox{-}}\)holomorphic map from the entire disc by sending \(i\mapsto y\) and \(-i\mapsto x\).

In Morse homology, we have an \({\mathbb{R}}\) action on the moduli space of trajectories, and that also shows up here. Here \({\mathbb{R}}\curvearrowright\mathcal{M}(x, y)\) by \(u(s, t) \xrightarrow{c} u_c(s, t) \mathrel{\vcenter{:}}= u(s, t+c)\), noting that translating the strip from above still yields a solution.

We define \begin{align*} \widehat{\mathcal{M}}(x, y) \mathrel{\vcenter{:}}=\mathcal{M}(x, y) / {\mathbb{R}} .\end{align*}

We’ll define \begin{align*} CF(L_1, L_2) \mathrel{\vcenter{:}}=\bigoplus_{x\in L_0 \cap L_1} {\mathbb{Z}}/2{\mathbb{Z}}\left\langle{ x }\right\rangle \\ \\ {{\partial}}x \mathrel{\vcenter{:}}=\sum_{y\in L_0 \cap L_1} \# \widehat{\mathcal{M}}(x, y) y .\end{align*}

When is the intersection count \(\# \widehat{\mathcal{M}}(x, y)\) well-defined? In Morse homology, we have two conditions:

  1. \((f, g)\) is Morse-Smale, to ensure that the moduli spaces are smooth manifolds (using Sard’s theorem)

  2. \(\operatorname{ind}(x) - \operatorname{ind}(y) = 1\), ensuring \(\mathcal{M}(x, y)\) is 1-dimensional

  3. Compactness of \(\widehat{\mathcal{M}}(x, y)\) when 1 and 2 hold.

These were enough to guarantee that \(\widehat{ \mathcal{M}} (x, y)\) was a smooth compact 0-dimensional manifold, which allowed for point counts. In Lagrangian Floer homology, we have the following replacements:

For 2 (indices): Recall that the index in Morse homology was the dimension of the negative eigenspace of the Hessian, but we’re in infinite dimensions here. So we won’t have a well-defined index, but we’ll have something that can replace the difference of indices: the Maslov index \(\mu(x, y)\), the expected dimension of \(\mathcal{M}(x, y)\). To actually have this be the dimension will require some conditions, so it’s not always true. This will be the index of some elliptic operator defined using the Cauchy-Riemann equations.

For 1 (transversality): We’ll need some version of transversality, which will imply that for a generic \(J\) that \(\mathcal{M}(x, y)\) is smooth.

For 3 (compactness): We’ll use Gromov compactness and some extra topological assumptions, which will imply that \(\widehat{ \mathcal{M}}(x, y), \mathcal{M}(x, y)\) are both compact.

Taken together, these will make the point-count well-defined.

In order for this to be a chain complex, we’ll need \({{\partial}}^2 = 0\). We’ll look at when \(\mu(x, y) = 2\), and we’ll compactify \(\widehat{ \mathcal{M}}(x, y)\) in order to show this holds. Gromov’s compactness will give us \begin{align*} {{\partial}}\overline{ \mathcal{M}(x, y) } = \bigcup_{\mu(x,z) = \mu(z, y) = 1} \mathcal{M}(x, z) \times\mathcal{M}(z, y) ,\end{align*} much like the broken trajectories from Morse homology. Here we’ll need to add in broken \(J{\hbox{-}}\)holomorphic discs:

image_2021-02-16-23-45-04

Using the same argument as in Morse homology, we can obtain \({{\partial}}^2 = 0\).

Suppose \((M^{2n}, \omega)\) is a compact symplectic manifold with Lagrangians \(L_0, L_1\) such that

  1. \(L_0 \pitchfork L_1\)

  2. \(\pi_2(M) = \pi_2(M, L_0) = \pi_2(M, L_1) = 0\), which are topological conditions on embedded spheres with boundaries mapped to the \(L_i\).

Under these assumptions, \({{\partial}}^2 = 0\) and the homology \begin{align*} HF(L_0, L_1) \mathrel{\vcenter{:}}= H_*( CF(L_0, L_1), {{\partial}}) \mathrel{\vcenter{:}}=\ker {{\partial}}/ \operatorname{im}{{\partial}} \end{align*} is an invariant of \((M, L_0, L_1)\) up to Hamiltonian isotopies of \(L_0, L_1\).

A symplectomorphism is a diffeomorphism \(\psi: M_1 \to M_2\) such \(\psi^* \omega_1 = \omega_2\).

A Hamiltonian vector field is a vector field \(V\) such that \begin{align*} \iota_V \omega\mathrel{\vcenter{:}}=\omega(V, {-}) \in \Omega^1 \end{align*} is exact, and thus equal to \(df\) for some functional \(f\in C^{\infty }(M, {\mathbb{R}})\). Note that if one has a functional \(f\), one can find a symplectic form \(\omega\) such that this holds, so \(V\) is sometimes denoted \(V_f\) to show this dependence.

\({\mathbb{R}}^{2n}\) with the standard symplectic form \(\sum_{i=1}^n dx_i \wedge dy_i\), we have \(V_f = {\frac{\partial f}{\partial y_1}\,}, \cdots, {\frac{\partial f}{\partial y_n}\,}, - {\frac{\partial f}{\partial x_1}\,}, \cdots, -{\frac{\partial f}{\partial x_n}\,}\) for any \(f:{\mathbb{R}}^{2n} \to {\mathbb{R}}\). Note that we can have time-dependent vector fields (i.e. one parameter families) as well.

A Hamiltonian isotopy is a family \(\psi_t\) of diffeomorphisms of \(M\) such that \(\psi_t\) is the flow of a 1-parameter family of Hamiltonian vector fields \(V_t\), so taking the derivative of \(V\) yields this function.

Show that if \(\psi_t\) is a Hamiltonian isotopy, then \(\psi_t^* \omega = \omega\) and is thus a symplectomorphism as well.

Goal: use this as an invariant of closed 3-manifolds in the form of Lagrangian Floer homology, defined by Osvath-Szabo. Note that Floer’s theorem requires topological assumptions which make the homology well-defined, but we don’t have these available in the HF setup. In particular, the assumptions on \(\pi_2\) won’t hold.

8 Lecture 8 (Thursday, February 04)

8.1 Heegard Splittings

Goal: we want to use Lagrangian Floer homology to defined invariants of closed 3-manifolds, where here closed means that \({{\partial}}M^3 = \emptyset\). One example of Lagrangian Floer homology is Heegard Floer homology. We’ll want some symplectic manifold with two Lagrangian submanifolds. Oszvath-Szabo used a 2-dimensional description of closed 3-manifolds called Heegard diagrams. We’ll need Heegard splittings to define these, and handlebodies to define the splittings.

A handlebody of genus \(g\) will mean a compact 3-manifold obtained from \({\mathbb{B}}^3\) by attaching \(g\) solid 1-handles, i.e. \({\mathbb{D}}^1 \times{\mathbb{D}}^2\). These are glued in via two copies of \({{\partial}}{\mathbb{D}}^1 \times{\mathbb{D}}^2\):

image_2021-02-16-19-36-51

Alternatively, these can be defined as a regular neighborhood of \(\bigvee_{i=1}^g S^1 \subset {\mathbb{R}}^3\). We’ll write \(H_g\) for a genus \(g\) handlebody, and \({{\partial}}H_g\) will be a genus \(g\) surface.

A Heegard splitting of genus \(g\) is a decomposition \(M = H_1 {\coprod}_{\varphi} H_2\) where \(\varphi: {{\partial}}H_1 \to {{\partial}}H_2\) is a diffeomorphism.

image_2021-02-16-19-41-17

Explicitly, we have \begin{align*} H_1 {\coprod}_{ \varphi} H_2 \mathrel{\vcenter{:}}={ H_1 {\coprod}H_2 \over \left\langle{ x \sim \varphi(x) {~\mathrel{\Big|}~}\forall x \in {{\partial}}H_1 }\right\rangle } .\end{align*}

We can write \(S^3 = B_3 {\coprod}_{\one} B^3\), where both are just genus \(0\) handlebodies. Note that if you attach a solid 1-handle to \(B^3\), this yields \(S^1 \times{\mathbb{D}}^2\), i.e. a solid torus:

image_2021-02-16-19-42-43

Think of \(S^3\) as the one-point compactification of \({\mathbb{R}}^3\), we can write (and visualize) a decomposition \(S^3 = (S^1 \times{\mathbb{D}}^2) {\coprod}_{\varphi} (S^1 \times{\mathbb{D}}^2)\). The first copy will be a neighborhood of a circle in the plane:

image_2021-02-16-19-44-16

Labeling this circle as \(H^1 \mathrel{\vcenter{:}}=\left\{{ x^2 + y^2 = 1, z = 0}\right\}\), the complement \(H_2 \mathrel{\vcenter{:}}= S^3 \setminus H_1\) will be a regular neighborhood of the \(z{\hbox{-}}\)axis union \(\left\{{\infty }\right\}\):

image_2021-02-16-19-45-36

We can write a Heegard splitting of \(S^1 \times S^2\). Note that \(S^2 = {\mathbb{D}}^2 {\coprod}_{\one} {\mathbb{D}}^2\), so splitting the product over the union yields \((S^1 \times{\mathbb{D}}^2) {\coprod}_{\one} (S^1 \times{\mathbb{D}}^2)\), where the new map is still the identity since it’s just the identity on each factor. This yields two solid torii glued along their boundaries.

Any closed 3-manifold \(M^3\) admits a Heegard splitting.

A fact from Morse theory: there exists a Morse function \(f: M^3\to {\mathbb{R}}\) such that

  1. \(f(p) = i \mathrel{\vcenter{:}}=\operatorname{ind}(p)\) for every \(p\in \operatorname{Crit}(f)\) (i.e. \(f\) is self-indexing), and

  2. \(f\) has exactly one index \(0\) (minimum) and one index \(3\) (maximum) critical point.

We thus have the following situation:

image_2021-02-16-19-51-11

The remaining critical points must occur at 2 and 3:

image_2021-02-16-19-52-00

How can we break this into smaller manifolds? Any time we pass a critical point, we attach a one-handle. Note that we can define a new Morse function \(h \mathrel{\vcenter{:}}= 3-f\) Suppose we have \(g\) critical points of index 1 for \(f\) and \(g'\) critical points of index 1 for \(h\).

Show that \(\operatorname{crit}(f) = \operatorname{crit}(h)\) and if \(p\in \operatorname{crit}(f)\) with \(\operatorname{ind}_f(p) = i\) then \(\operatorname{ind}_h(p) = 3-i\).

Thus \(g'\) is the number of index 2 critical points for \(f\). This means that \({{\partial}}h ^{-1} [0, 3/2] = h ^{-1} (3/2) = f ^{-1} (3/2)\) has genus \(g=g'\), and thus the \(\# \operatorname{crit}(f)_{\operatorname{ind}=1} = \# \operatorname{crit}(h)_{\operatorname{ind}=2} = g\). Even without this, we still have our two handlebodies: \(H_1 \mathrel{\vcenter{:}}= f ^{-1} [0, 3/2]\) and \(H_2 \mathrel{\vcenter{:}}= f ^{-1} [3/2, 3]\) glued over \(\Sigma_g \mathrel{\vcenter{:}}= f ^{-1} (3/2)\), which is a genus \(g\) splitting surface.

We’ll say that two Heegard splittings \(M = H_1 {\coprod}_{\varphi} H_2\) and \(M = H_1' {\coprod}_{\varphi} H_2 '\) are isotopic if and only if there exists an ambient isotopy \(\psi: M \times[0, 1] \to M\) such that \({ \left.{{\psi}} \right|_{{M \times\left\{{ 1}\right\} }} }(H_i) = H_i '\) for each \(i\). Recall that ambient isotopy means

Are any two Heegard splittings isotopic?

No! We can distinguish them by the genus of the splitting surface \(\Sigma\), and we just saw two splittings of \(S^3\), one with genus 0 and one with genus 1.

There are some moves to relate different Heegard splittings.

Given a genus \(g\) Heegard splitting \(M = H_1 {\coprod}_{ \varphi} H_2\), we can produce a genus \(g+1\) splitting \(M = H_1' \cup_{ \varphi} H_2'\) where

\(H_1' = H_1 \cup\overline{ \eta( \gamma) }\), where the new piece is a closed regular neighborhood of an unknotted arc \(\gamma\) in \(H_2\). Here unknotted means that \(\gamma\) is a properly embedded arc in \(H_2 \cup\Sigma\) whose boundary is in \(\Sigma\) which bounds a contractible disc:

image_2021-02-16-21-00-33

Note that adding a regular neighborhood around \(\gamma\) has the effect of adding a 1-handle to \(H_1\). We can then define \(H_2' \mathrel{\vcenter{:}}= H_2 \setminus\eta{\gamma}\). Why is this still a handlebody? We have this situation:

image_2021-02-16-21-04-24

We have the disc below the 1-handle, and if we thicken it to \({\mathbb{D}}^2 \times I\), we have \(B \mathrel{\vcenter{:}}=\eta(\gamma) \cup({\mathbb{D}}^2 \times[0, 1] \cong {\mathbb{B}}^3\):

image_2021-02-16-21-07-07

We then have \(H_2' \mathrel{\vcenter{:}}=(H_2 \setminus B) \cup({\mathbb{D}}^2 \cup[0, 1] )\), and in fact there is something in the intersection of these two terms. The parts that are attached to \(H_2\) are the front and back discs \({\mathbb{D}}^2 \times\left\{{0, 1}\right\}\):

image_2021-02-16-21-07-50

So we can identify this as \(H_2' \mathrel{\vcenter{:}}=(H_2 \setminus B) {\coprod}_{{\mathbb{D}}^2 \times\left\{{ 0, 1 }\right\} } ({\mathbb{D}}^2 \cup[0, 1] )\). Note that \(H_2 \setminus B \cong_{C^\infty} H_2\) are diffeomorphic, and the right-hand side is a 1-handle. To see why this is, consider attaching the middle red part, and then pushing the center part away in order to see the handle:

image_2021-02-16-21-21-12

Show that the isotopy type of \(H_1' \cup H_2 '\) is independent of the choice of \(\gamma\).

Any two Heegard splittings can be made isotopic after sufficiently many stabilizations.

8.2 Heegard Diagrams

2-dimensional pictures of closed 3-manifolds! We have two handlebodies glued along their boundary, so if we can write the handlebodies in terms of 2-dimensional pictures, we can combine them to get a picture of the entire splitting.

Let \(H\) be a genus \(g\) handlebody. A set of attaching curves for \(H\) is a set \(\left\{{ \gamma_1, \cdots, \gamma_g }\right\}\) of pairwise disjoint simple closed curves on \(\Sigma\mathrel{\vcenter{:}}={{\partial}}H\) such that

  1. \(\Sigma\setminus\cup\left\{{\gamma_1, \cdots, \gamma_g}\right\}\) is connected,

  2. All the \(\gamma_i\) bound a disc in \(H\).

For the solid 2-torus, the attaching curves are copies of \(S^1\) that bound discs

image_2021-02-16-21-27-56

Consider \({\mathbb{B}}^3\) with two 1-handles attached, or a solid genus 2 surface:

image_2021-02-16-21-29-36

Note that curves running around each of the two handles also work:

image_2021-02-16-21-30-26

Show that \(\Sigma\setminus\cup\left\{{ \gamma_1, \cdots, \gamma_g }\right\}\) is connected \(\iff\) the classes \([\gamma_1], \cdots, [\gamma_g]\) are linearly independent in \(H_1(\Sigma; {\mathbb{Z}})\).

Given a surface and a set of attaching curves, so the data of \((\Sigma, \left\{{ \gamma_1, \cdots, \gamma_g }\right\} )\) , we can build a handlebody \(H\). Note that we can go the other way: given a genus \(g\) handlebody \(H\), we can take \(\Sigma = {{\partial}}H\) and find \(g\) attaching circles.

The recipe:

  1. Thicken \(\Sigma\) to \(\Sigma \times[0, 1]\) to get a 3-manifold with 2 boundary components, \(\Sigma\times\left\{{ 1 }\right\}\) and \(\Sigma \times\left\{{ 2 }\right\}\).

  2. Attach thickened discs \(\gamma_i \times\left\{{ 0 }\right\}\) for each \(i\), yielding some \(S^2\) boundary components.

  3. Fill the \(S^2\) boundary component with a \({\mathbb{B}}^3\).

This yields a genus \(g\) handlebody \(H\) such that \({{\partial}}H = \Sigma_g \times\left\{{ 1 }\right\}\), where the curves \(\left\{{ \gamma_1 \times\left\{{ 1 }\right\} , \cdots, \gamma_g \times\left\{{ 1 }\right\} }\right\}\).

Note that after attaching the disc on one end of this new cylinder, we have the following:

image_2021-02-16-21-37-08

What’s left on the boundary is the following:

image_2021-02-16-21-37-43

This is a copy of \(S^2\).

Show that for any \(g\) we get a 3-manifold with boundary \(\Sigma \times\left\{{ 1 }\right\} {\coprod}S^2\) after step (2) above.

9 Lecture 9 (Thursday, February 11)

9.1 Heegard Diagrams

Last time we saw that \(M_3 = H_1 {\coprod}_{\varphi} H_2\) as two handlebodies glued along their boundary by a diffeomorphism \(\varphi: {{\partial}}H_1 \to {{\partial}}H_2\). This is referred to as a Heegard splitting for \(M\). We can specify a genus \(g\) handlebody as \(( \Sigma, \left\{{ \gamma_1, \cdots \gamma_g }\right\}\) where \(\Sigma\setminus\left\{{ \gamma_1, \cdots, \gamma_g }\right\}\) is connected and each \(\gamma_i\) bounds a disc in \(H\).

image_2021-02-11-11-15-56

Moreover, we can go backwards: given such data, we can build a handlebody \(H\) by

  1. Thickening \(\Sigma\) to obtain \(\Sigma \times[0, 1]\) This yields \({{\partial}}( \Sigma \times[0, 1] ) = (\Sigma\times\left\{{0}\right\} ) {\coprod}(\Sigma\times\left\{{1}\right\} )\).

  2. Attach thickened discs to \(\gamma_i \times\left\{{0}\right\}\). This makes the boundary \((\Sigma \times\left\{{1}\right\} ) {\coprod}S_2\)

  3. Fill in the \(S^2\) boundary with a \(B^3\).

image_2021-02-11-11-22-43

A Heegard diagram for \(M^3\) compatible with a splitting \(M = H_1 {\coprod}_{ \varphi} H_2\) is a triple \((\Sigma, \alpha, \beta\) where \(\alpha\) and \(\beta\) are attaching circles for \(H_1\) and \(H_2\) respectively.

The following two curves on a torus determine a Heegard splitting for \(S^3\):

image_2021-02-11-11-28-43

Writing \(S^1 \times S^2 = D_2 {\coprod}_{\one_{{{\partial}}D^2}} D^2\), or also \((S^1 \times D^2) {\coprod}_{\one} (S^1 \times D^2)\).

image_2021-02-11-11-30-50

Show that the following diagram is a Heegard diagram for \({\mathbb{RP}}^3\):

image_2021-02-11-11-31-45

Hint: use that \({\mathbb{RP}}^3 \cong L(2, 1)\) and find a Heegard diagram for \(L(p, q)\).

Given a self-indexing Morse function \(f:M \to {\mathbb{R}}\) with exactly one index 0 and one index 3 critical point, pick a generic metric \(g\) so that \((f, g)\) is a Morse-Smale pair (so the stable and unstable submanifolds intersect transversally). Taking \(- \nabla f\), we can obtain a Heegard diagram The stable submanifolds are codimension of their indices, so e.g. for each index critical point there is a 2-dimensional stable submanifold that intersects the next submanifold in a curve:

Stable submanifold

This occurs for (say) the \(g\) critical points of index \(1\) here, and since they are distinct critical points the stable submanifolds are disjoint. So we can obtain a set of attaching circles for the bottom handlebody \(f ^{-1} ([0, 3/2])\): \begin{align*} \left\{{ M^s(p) \cap f ^{-1} (3/2) {~\mathrel{\Big|}~}p \in \operatorname{crit}(f),\, \operatorname{ind}(p) = 1 }\right\} .\end{align*}

So setting these to be the \(\alpha\) curves, repeating with index 2 to get \(\beta\) curves, and setting \(\Sigma\mathrel{\vcenter{:}}= f ^{-1} (3, 2)\) we get a Heegard diagram for \(M\).

Note that given \((\Sigma, \alpha, \beta\) we can construct \(M\) in the following way:

Show that Heegard splittings can be used to compute homology, and \begin{align*} H_1(M; {\mathbb{Z}}) \cong H_1(\Sigma; {\mathbb{Z}}) / \left\langle{ [ \alpha_1] , \cdots, [ \alpha_g], [ \beta_1 ], \cdots, [\beta_g] }\right\rangle .\end{align*}

9.2 Heegard Moves

Given \(M = H_1 \cup H_2 = H_1' \cup H_2'\), we can stabilize to obtain \(M = \tilde H_1 \cup\tilde H_2\). Is there a way to relate the two corresponding Heegard diagrams?

  1. Isotopy. Exchange \(\alpha = \left\{{ \alpha_1, \cdots, \alpha_g }\right\}\) with an ambient isotopy of \(\Sigma\), and similarly \(\beta\), keeping curves of the same type disjoint during the isotopy (where e.g. it’s fine if an \(\alpha\) curve intersects a \(\beta\) curve).
image_2021-02-11-11-49-46
  1. Handleslides (of \(\alpha\) or \(\beta\) curves).
image_2021-02-11-11-51-54

Equivalently, handle sliding \(\alpha_1\) over \(\alpha_2\) replaces \(\alpha_1\) with \(\alpha_1'\) such that the triple \(\alpha_1, \alpha_1', \alpha_2\) bound a pair of pants.

image_2021-02-11-11-53-22
  1. Stabilization. This changes \((\Sigma, \alpha, \beta) \mapsto (\Sigma \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}T^2, \alpha \cup\left\{{ \alpha_{g+1 } , \beta}\right\} \cup\left\{{ \beta_{g+1} }\right\}\), where \(\alpha_{g+1}, \beta_{g+1} \subseteq T^2\) and intersect in exactly on point.
image_2021-02-11-12-10-06

3’. Destabilization. Reversing the stabilization operation.

Show that any two sets of attaching curves for a handlebody \(H\) can be related by a finite sequence of (1) and (2).

Show that stabilization yields a Heegard diagram for the same manifold.

Hint: the new summand is a Heegard diagram for \(S^3\), and connect sums in the diagrams correspond to connect sums of the corresponding manifolds. Moreover, \(M \cong M\mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}S^3\).

Any two Heegard diagrams for \(M\) can be connected by a finite sequence of the above moves.

10 Tuesday, February 16

Note that critical points can be used to compute the Euler characteristic, using the fact the \(\chi(C) = \chi(H_*(C))\), i.e. it can be computed on dimensions of chains or ranks of homology, along with the fact that Morse homology is isomorphic to singular homology. So e.g. for a 3-manifold \(M^3\), we can show \begin{align*} \chi(M^3) &= \sum_{i=0}^3 {\operatorname{rank}}H_i \\ &= \sum_{i=0}^3 {\operatorname{rank}}CM_i \\ &= 1 - \# \operatorname{crit}_1(f) + \# \operatorname{crit}_2(f) - 1 \\ &= 0 ,\end{align*} since the number of index 2 and index 3 critical points will be the same.

10.1 Symmetric Product Spaces

Let \(M^3\) be a closed 3-manifold, then there is a Heegard splitting \begin{align*} (\Sigma_g, \alpha = \left\{{ \alpha_1, \cdots, \alpha_g }\right\}, \beta = \left\{{ \beta_1, \cdots, \beta_g }\right\} =( \Sigma_g, H_ \alpha, H_ \beta) && {{\partial}}(H_ \alpha) = {{\partial}}( H_ \beta) = \Sigma ,\end{align*} where \(M^3 = H_{ \alpha} \coprod_{ \Sigma} H_ \beta\) and \(g\) is the genus of \(HD\). We refer to \(\Sigma\) as a Heegard surface, and this set of data as a Heegard diagram.

We’ll define \(\operatorname{Sym}^g( \Sigma)\) by letting \(S_g \curvearrowright\Sigma^{\times g}\) where if \(\varphi\in S_g\) we set \(\varphi(x_1, \cdots, x_g) = x_{ \varphi(1)}, \cdots, x_{ \varphi(g) }\). Then set \(\Sigma^{\times g} \mathrel{\vcenter{:}}=\Sigma^{\times g} / S_g\). Why does this yield a smooth manifold? Is this action free? The diagonal \(D \subseteq \Sigma^{\times g}\) consists of the points with at least 2 equal coordinates, and it’s easy to see that \(S_g\curvearrowright D\) can not be free. However, this still yields a smooth submanifold!

\(\operatorname{Sym}^g(\Sigma)\) is smooth, and any complex structure \(j\) on \(\Sigma\) will induce a complex structure on the quotient, denoted \(\operatorname{Sym}^g(j)\), which is unique in the sense that the quotient map \(\Sigma^{\times g} \xrightarrow{\pi} \operatorname{Sym}^g(\Sigma)\) is holomorphic.

We’ll check this locally, and then leave it as an exercise to check that it extends globally – this is easy by just considering what happens under transition functions and checking that \(\pi\) is holomorphic. Locally we want to produce a map \begin{align*} \operatorname{Sym}^g({\mathbb{C}}) &\xrightarrow{f} {\mathbb{C}}^g \\ \left\{{ z_1, \cdots, z_g }\right\} &\mapsto \qty{ \prod_{i=1}^g (z-z_i) = z^g +a_1 z^{g-1} + \cdots + a_g \mapsto [a_1, \cdots, a_g] } .\end{align*} This is a bijection, and by the fundamental theorem of algebra, there is an inverse. Equip \(\operatorname{Sym}^g({\mathbb{C}})\) with a complex structure that makes \(f\) biholomorphic, then \(\operatorname{Sym}^g(j)\) is the complex structure locally equal to this one. This structure is obtained by just pulling back the standard complex structure \(i\times i \times\cdots i\) on \({\mathbb{C}}^g\).

\(\operatorname{Sym}^g( \Sigma)\) is a complex manifold of complex dimension \(g\) (or real dimension \(2g\)). We want to find half-dimensional submanifolds to do Lagrangian-Floer homology. Using the Heegard splitting, write \({\mathbb{T}}_ \alpha \mathrel{\vcenter{:}}=\prod_{i=1}^g \alpha_i \subset \Sigma^{\times g}\), which is a \(g{\hbox{-}}\)dimensional torus such that \({\mathbb{T}}_ \alpha \cap D = \emptyset\) since the \(\alpha_i\) are pairwise disjoint. Composing the inclusion above with \(\pi\), we can note that the action of \(S^g\) is free away from the diagonal \(D\), so this composition is an embedding \({\mathbb{T}}_ \alpha \hookrightarrow\operatorname{Sym}^g( \Sigma)\). Similarly, \({\mathbb{T}}_ \beta\mathrel{\vcenter{:}}=\prod_{i=1}^g \beta_i \hookrightarrow\operatorname{Sym}^g( \Sigma)\).

Note that we’re only working with complex structures now, and haven’t upgraded it to a symplectic structure yet. But we don’t really need this to count holomorphic discs. Lagrangians \(L\) were defined as submanifolds where \({ \left.{{\omega}} \right|_{{L}} } = 0\), how do we do this without a symplectic form?

Given a complex manifold \((X, J)\), a submanifold \(L \subseteq X\) is totally real if none of its tangent spaces contains a complex line, i.e. \(T_p L \cap J(T_p L) = \left\{{ p, \mathbf{0} }\right\}\) for all \(p\in L\).

Take a genus \(g\) surface \(\Sigma\):

image_2021-02-16-11-49-52

Here any tangent vector has to get rotated out of the tangent space: if it were an eigenvector for \(J\), then the rank of \(J\) would be too low, contradicting its definition. Note that any 1-dimensional submanifold of \((\Sigma, j )\) is totally real, and so \({\mathbb{T}}_ \alpha, {\mathbb{T}}_ \beta\) are also totally real submanifolds of \(\Sigma^{\times g}\). If you restrict \(\pi\) to \(\Sigma^{\times g}\setminus D \xrightarrow{\pi} \operatorname{Sym}^g(\Sigma) \setminus\pi(D)\), this yields a biholomorphic map.

We’ll write \(\Delta \mathrel{\vcenter{:}}=\pi(D) \subseteq \operatorname{Sym}^g( \Sigma)\). Note that if \(\alpha\pitchfork\beta\), then \({\mathbb{T}}_ \alpha \pitchfork{\mathbb{T}}_ \beta\). Any intersection point \(x \in {\mathbb{T}}_{\alpha} \cap{\mathbb{T}}_{\beta}\) is of the form \(x = \left\{{ x_1, \cdots, x_g}\right\} \subseteq \Sigma\) such that each \(\alpha_i, \beta_j\) contain exactly one of the coordinates of \(x\).

The following is a diagram for \({\mathbb{RP}}^3\):

Heegard diagram for {\mathbb{RP}}^3

Here \(g=1\) and so \(\operatorname{Sym}^1(T^2) = T^2\). We also have \({\mathbb{T}}_{ \alpha} = \alpha, {\mathbb{T}}_{ \beta} = \beta\), and their intersection is \({\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} = \alpha \cap\beta = \left\{{A, B}\right\}\)

Here we have a Poincaré homology sphere \(P^3\), i.e. a 3-manifold with the same homology as \(S^3\), i.e. \(H_*(P^3) = [{\mathbb{Z}}, 0, 0, {\mathbb{Z}}]\) (??)

image_2021-02-16-12-01-57

Compute \(H_*(P^3)\) using this diagram, particularly \(H_1\). Using Poincaré duality here is fine!

The circles with the same color are the “feet” of a handle attachment, or equivalently removing the two circles and identifying their boundary with reversed orientation. The two different colors for circles indicate that this will be genus 2 The arcs between same-colored circles indicate loops that continue through the handle which aren’t shown. Tracing through the lines on the diagram, there are two \(\alpha\) curves and two \(\beta\) curves. Since \(g=2\), we can identify \(\operatorname{Sym}^2( \Sigma) \supseteq \alpha_1 \times\alpha_2 = {\mathbb{T}}_{\alpha}, \beta_1 \times\beta_2 = {\mathbb{T}}_{\beta}\). The two black circles indicate intersection points in \({\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\). However, there are more than just those two!

Show that \({\left\lvert { {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta} } \right\rvert} = 18\).

Computing the intersections:

We’re really working in \(\operatorname{Sym}^g(\Sigma)\), but for computations, we’ll work directly with the Heegard diagram.

For Lagrangian Floer homology, we’ll have a triple \((\operatorname{Sym}^g(\Sigma), {\mathbb{T}}_{ \alpha}, {\mathbb{T}}_{\beta} )\). We’ll define \begin{align*} CF( \Sigma, \alpha, \beta) \mathrel{\vcenter{:}}=\bigoplus_{x\in {\mathbb{T}}_{\alpha} \cap{\mathbb{T}}_{\beta} } {\mathbb{Z}}/2{\mathbb{Z}}\left\langle{ x }\right\rangle \\ \\ {{\partial}}(x) \mathrel{\vcenter{:}}=\sum_{y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta}, \mu = 1} \# \widehat{\mathcal{M}} y .\end{align*}

We’ll first figure out how to count continuous discs up to homotopy classes, since holomorphic discs are much more restrictive. We’ll see that \(\pi_2\) plays a role, and define the topology of \(\operatorname{Sym}^g\).

11 Thursday, February 18

Today: topology of symmetric product spaces \(\operatorname{Sym}^g\). We had an assignment \begin{align*} ( \Sigma_g, \alpha, \beta) &\mapsto ( \operatorname{Sym}^g( \Sigma), {\mathbb{T}}_ \alpha, {\mathbb{T}}_ \beta) ,\end{align*} where if \(\alpha, \beta\) are all transverse then so far \({\mathbb{T}}_ \alpha, {\mathbb{T}}_ \beta\), since e.g. \({\mathbb{T}}_ \alpha = \prod_{i=1}^g \alpha_i\). We wanted to define a chain complex \begin{align*} CF( \sigma, \alpha, \beta) \mathrel{\vcenter{:}}=\bigoplus _{x\in {\mathbb{T}}_ \alpha \cap{\mathbb{T}}_{ \beta } } {\mathbb{Z}}/2{\mathbb{Z}}\left\langle{ x }\right\rangle \\ {{\partial}}x \mathrel{\vcenter{:}}=\sum_{ \substack{ y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta } \\ \mu(x, y) = 1} } \# \mathcal{M}(x, y) y ,\end{align*} where \(\mu\) is the Maslov index and we want to count holomorphic discs. We’ll first talk about continuous (topological) discs.

\begin{align*} \pi_1( \operatorname{Sym}^g( \Sigma ) ) \cong H_1 ( \operatorname{Sym}^g( \Sigma ) ) \cong H_1 (\Sigma) ,\end{align*} so the fundamental group is abelian.

For a proof of the first isomorphism, see Lemma 2.6 in [OSZ04a?]. Idea of proof for the second isomorphism: we’ll define a map \begin{align*} \iota: H_1 ( \Sigma) &\to H_1( \operatorname{Sym}^g( \Sigma) ) \\ x &\mapsto \left\{{ x, z, \cdots, z }\right\} ,\end{align*} for some fixed \(z \in \Sigma\), along with its inverse. Note that we’re identifying an embedding \(\iota( \Sigma ) = \Sigma \times\left\{{ z }\right\}^{\times g-1} \subseteq \operatorname{Sym}^g( \Sigma)\). Now define \(j \mathrel{\vcenter{:}}=\iota_*\) the induced map on homology. \begin{align*} j: H_1( \operatorname{Sym}^g (\Sigma) ) \to H_1 ( \Sigma) \\ .\end{align*} Picking a loop \(\gamma: S^1 \to \operatorname{Sym}^g( \Sigma )\), note that \(\Delta\subset \operatorname{Sym}^g( \Sigma)\) has codimension 2, and so we can perturb \(\gamma\) to be disjoint from \(\Delta\). We can arrange so that \(\gamma\) is the union of \(g\) paths \(\gamma_1, \cdots, \gamma_g\) such that each \(\gamma_i\) connects \(x_i \in \gamma(0)\) to \(x_{ \sigma(i) } \in \gamma(0)\) where \(\gamma_0 = \left\{{ x_1, \cdots, x_g }\right\}\) and \(\sigma\in S_g\) is a permutation.

For example, for \(g=3\):

image_2021-02-18-11-30-51

Then \(\left\{{ \gamma_1(t), \gamma_2(t), \gamma_3(t) }\right\}\) is a loop from \(\gamma(0) \to \gamma(0) \in \operatorname{Sym}^3( \Sigma)\).

This means that \(\bigcup_{i=1}^g \gamma_i\) is a 1-cycle in \(\Sigma\), and thus \([ \cup g_i ] \in H_1( \Sigma)\). So we’ll define this as \(j([ \gamma ]) = [ \cup\gamma_i ]\).

Let \(M \mathrel{\vcenter{:}}=\left\{{ (\mathbf{x}, y) {~\mathrel{\Big|}~}\mathbf{x} \in \operatorname{Sym}^g( \Sigma), y\in \mathbf{x} }\right\}\), then we’ll define a \(g:1\) branched cover away from \(\pi ^{-1} \Delta\) that yields a fiber bundle:

Link to Diagram

This can be restricted to \(M \setminus\pi ^{-1} (\Delta) \xrightarrow{g:1} \operatorname{Sym}^g( \Sigma) \setminus\Delta\). Here \(j([ \gamma ]) = [ \pi_2 \circ \gamma]\) and \(j \circ \iota_* = \one\).

We can use a Heegard diagram and Mayer Vietoris to compute the homology: \begin{align*} H_1( M; {\mathbb{Z}}) = { H_1(\Sigma; {\mathbb{Z}}) \over \left\langle{ [\alpha_1], \cdots, [ \alpha_g], [\beta_1], \cdots, [\beta_g] }\right\rangle } \cong { H_1( \operatorname{Sym}^g( \Sigma ) ) \over \left\langle{ H_1( {\mathbb{T}}_ \alpha ), H_1 ({\mathbb{T}}_ \beta ) }\right\rangle } .\end{align*}

\begin{align*} \pi_2( \operatorname{Sym}^g( \Sigma ) ) \cong {\mathbb{Z}} .\end{align*}

The generator comes from hyperelliptic involution:

image_2021-02-18-11-58-40

Then consider the quotient \(\Sigma / \tau\). To identify this quotient, since the top half is identified with the bottom half, we can first forget about the bottom half, and then forget about half of the arcs along the axis of rotation:

image_2021-02-18-12-01-41

Note that this results in a copy of \(S^2\). We can define a map \begin{align*} \Sigma &\to \Sigma^{\times g} \\ x &\mapsto (x, \tau(x), z, \cdots, z) .\end{align*} This extends to a map to \(\operatorname{Sym}^g( \Sigma)\), since \(\tau(x) \mapsto (\tau(x), x, z, \cdots, z)\) and these will be equal in \(\operatorname{Sym}^g\). So we can factor this through the quotient from above:

Given \(x, y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\), a Whitney disc from \(x\) to \(y\) is a map \begin{align*} \varphi: {\mathbb{D}}^2 \to \operatorname{Sym}^g( \Sigma) \end{align*} such that \begin{align*} \phi(-i) &= x \\ \phi(i) &= y \\ \phi(e_1) &\subseteq {\mathbb{T}}_{ \alpha} \\ \phi(e_2) &\subseteq {\mathbb{T}}_{ \beta } .\end{align*}

image_2021-02-18-12-22-03

We say \(\varphi_1 \sim \varphi_2\) if and only if they are homotopic relative to \({\mathbb{T}}_{ \alpha}, {\mathbb{T}}_{ \beta}\). We’ll write \(\pi_2(x, y)\) for the homotopy class of Whitney discs from \(x\) to \(y\). There is a concatenation operation: \begin{align*} \ast: \pi_2(x, y) \times\pi_2(y, z) \to \pi_2(x, z) .\end{align*}

image_2021-02-18-12-24-03

Note that this is precisely concatenation of paths in the path space \(\mathcal{P}\).

If \(x=y=z\), then this yields an operation on \((\pi_2(x, x), \ast)\) which defines a group.

We can find obstructions to holomorphic discs by just looking at the topology. For \(x, y\in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\), choose two paths connecting them: \begin{align*} a: I &\to {\mathbb{T}}_{ \alpha}\\ b: I &\to {\mathbb{T}}_{ \beta} .\end{align*}

image_2021-02-18-12-27-15

We can consider the homology class \([a-b]\) to investigate \(\pi_1\). This is well-defined as a loop \begin{align*} \varepsilon(x, y) \mathrel{\vcenter{:}}=[a-b] \in { H_1 ( \operatorname{Sym}^g ( \Sigma ) ) \over \left\langle{ H_1( {\mathbb{T}}_{ \alpha } ) \oplus H_1 ( {\mathbb{T}}_{ \beta} ) }\right\rangle } \cong H_1(M) .\end{align*} This turns out to be independent of the choice of \(a, b\), and thus \begin{align*} \varepsilon(x, y) \neq 0 \implies \pi_2(x, y) = \emptyset ,\end{align*} and there are no continuous discs.

12 Tuesday, February 23

12.1 Whitney Discs

For \(x,y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\), recall that we had the following situation:

Whiteney Disc

Then \(\pi_2(x, y)\) was defined to be the homotopy classes of discs connecting \(x\) to \(y\). The obstruction to the existence of such discs was denoted \(\varepsilon(x, y) \in H_1(M)\) for \(M\in {\mathsf{Mfd}}^3\). We’re checking if there exist two paths connecting \(x\) to \(y\), \begin{align*} a: I &\to {\mathbb{T}}_{ \alpha} \\ b: I &\to {\mathbb{T}}_{ \beta} \end{align*} such that \(a-b\) is nullhomotopic. In this case, \(\pi_2(x, y) \neq \emptyset\).

image_2021-02-23-11-17-30

We had a theorem that \(\pi_1(\operatorname{Sym}^g \Sigma) \cong H_1( \operatorname{Sym}^g \Sigma)\), so we can replace nullhomotopic with nullhomologous above. We can also use the fact that \(H_1( \operatorname{Sym}^g \Sigma) \cong H_1 \Sigma\). Note that \([a-b]\) isn’t well-defined, since we can append any loop to \(a\) for example, but the following is well-defined: \begin{align*} \varepsilon(x, y) \mathrel{\vcenter{:}}=[a-b] \in { H_1 \operatorname{Sym}^g \Sigma \over H_1 {\mathbb{T}}_{ \alpha} \oplus H_1 {\mathbb{T}}_{ \beta} } \cong {H_1 \Sigma \over \left\langle{ [ \alpha_1], \cdots, [ \beta_1 ], \cdots }\right\rangle} \cong H_1 M .\end{align*}

How can we compute \(\varepsilon\) using the Heegard diagrams? Recall that a path in \(\operatorname{Sym}^g \Sigma\) was a union of \(g\) paths in \(\Sigma\). So choose arcs \(a_1 \cup\cdots \cup a_g\) on \(\Sigma\) such that \(a_i \subseteq \alpha_i\) is sub-arc and \({{\partial}}( a_1 \cup\cdots \cup a_g ) = y_1 + \cdots + y_g - x_1 - \cdots - x_g\), and similarly choose \(b_1 \cup\cdots \cup b_g\). Note that if \(\varepsilon(x, y) \neq 0\) then \(\pi_2(x, y) = \emptyset\).

The following is a Heegard diagram for \(L(2, 3)\) of minimal genus, where we take \(\alpha\) to be the horizontal line and \(\beta\) will be a line of slope \(2/3\):

image_2021-02-23-11-28-10

Then \({\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} = \left\{{ A, B }\right\}\). Now draw arcs connecting \(A\) and \(B\), e.g. the ones in orange and green here:

image_2021-02-23-11-29-54

Note that we have two generators of homology for the torus, say \(x,y\), and we can write

image_2021-02-23-11-30-37

Then the union of the two arcs is exactly \(x+y\), so we can write \begin{align*} H_1( L(2, 3)) = { {\mathbb{Z}}\left\langle{ x, y }\right\rangle \over \left\langle{ y, 2x + 3y }\right\rangle} .\end{align*} Moreover, \(\varepsilon(A, B) = x + y \neq 0\) in this quotient, so there is not Whitney disc connecting \(A\) to \(B\) and \(\pi_2(A, B) = \emptyset\).

We’ll define \(x\sim y \iff \varepsilon(x, y) = 0\), and this turns out to be an equivalence relation which partitions the set of paths.

Show that \(\varepsilon(x, y) + \varepsilon(y, z) = \varepsilon(x, z)\).

If \(x\sim y\) and \(y\sim z\), so \(\varepsilon(x, y) = \varepsilon(y, z) = 0\), we have \(\varepsilon(x, z) = 0 \implies x \sim z\).

Find the equivalence classes under \(\sim\) for the Poincaré homology sphere using the genus 2 Heegard diagram.

For \(\varphi\in \pi_2(x, y)\), the shadow is the 2-chain \(D( \varphi )\) on \(\Sigma\) defined in the following way: remove the \(\alpha, \beta\) arcs to obtain \begin{align*} \Sigma \setminus(\alpha\cup\sigma) = \displaystyle\coprod_{i=1}^m D_i ,\end{align*}

where \({}^{o}\) denotes that the set is open. Then \(D( \varphi) = \sum_{i=1}^m a_i D_i\).

Given \(z\in \Sigma\setminus(\alpha\cup\beta)\), define a hyperplane \begin{align*} L_z = \left\{{ \mathbf{w} \in \operatorname{Sym}^g( \Sigma) {~\mathrel{\Big|}~}z\in \mathbf{w} }\right\} .\end{align*} Note that this will be codimension 2. Then for a disc \(\varphi\in \pi_2(x, y)\), define \begin{align*} n_z( \varphi ) = \# \qty{ \operatorname{im}( \varphi) \cap L_z } .\end{align*} which is an algebraic (signed) count of how many entries in a tuple contain the point \(z\). We can then define \(a_i \mathrel{\vcenter{:}}= n_{z_i}( \varphi)\) and define \begin{align*} D( \varphi) = \sum_{i=1}^m a_i D_i, && z_i \in {}^{o} D_i .\end{align*}

The following comes from “Introduction to Heegard Floer Homology” (Osvath-Szabo), which we’ve been following relatively closely so far.

Let \(D\) be a domain of a disc connecting \(\left\{{ x_1, x_2 }\right\}\) to \(\left\{{ y_1, y_2 }\right\}\) in the following way:

image_2021-02-23-11-54-09

Attach 1-handles in the following way to obtain \(\beta\) curves:

image_2021-02-23-11-55-13

Use these handles to add curves running through the handles:

image_2021-02-23-11-56-26

What is a Heegard diagram for?

Pick a point in the center of the rectangle and connect it to the 4 vertices, noting that it includes in \(\Sigma\) :

image_2021-02-23-11-58-15

Applying a rotation by \(\pi\) and taking the quotient, we get a 2-fold branched cover of \(S^1\):

image_2021-02-23-11-59-29

Here \(x_1, x_2 \mapsto -i\) and \(y_1, y_2 \mapsto +i\). We can now get a map \(\varphi\) to \(\operatorname{Sym}^2( \Sigma)\):

image_2021-02-23-12-01-02

In the image we get 2 points with multiplicity on \(\Sigma\), and thus an element of \(\operatorname{Sym}^2 \Sigma\). We know \(\varphi(-i) = \left\{{ x_1, x_2 }\right\}\) and \(\varphi(+i) = \left\{{ y_1, y_2 }\right\}\).

Show that \(D'\) is the domain of a disc from \(\left\{{ x_1, x_2 }\right\} \to \left\{{ y_1, y_2 }\right\}\):

image_2021-02-23-12-18-15

We want to make a similar 2-fold cover like in the previous example, so we’ll take the two rectangles bounding the arcs, then taking the rotation by \(\pi\) yields the cover:

image_2021-02-23-12-20-27

As before, we get a map to \(\operatorname{Sym}^2 \Sigma\):

image_2021-02-23-12-22-36

As a result, we again get \(\varphi(-i) = \left\{{ x_1, x_2 }\right\}\) and \(\varphi(+i) = \left\{{ y_1, y_2 }\right\}\).

. Suppose \(x = \left\{{ x_1, \cdots, x_g }\right\}\) and \(y = \left\{{ y_1, \cdots, y_g }\right\}\) such that \(x_i \in \alpha_i \cap\beta_i\) and \(y_i \in \alpha_i \cap\beta_{ \sigma^{-1}(i)}\) for some permutation \(\sigma\in S_g\). Then for any \(\varphi\in \pi_2(x, y)\), show that \begin{align*} {{\partial}}\qty{{{\partial}}D( \varphi) \cap\alpha_i } = y_i - x_i ,\end{align*} where the inner term is a 1-chain in \(\alpha_i\), and \begin{align*} {{\partial}}\qty{ {{\partial}}D( \varphi) \cap\beta_i } = x_i - y_{ \sigma(i) } .\end{align*}

This will characterize the coefficients \(a_i\) for which discs exist. Next time we’ll talk about holomorphic discs.

13 Thursday, February 25

13.1 Whitney Discs

Recall that we discussed the domains of discs: for \(\varphi\in \varphi_2(x, y)\) we defined the 2-chain \(D( \varphi) = \sum_{i=1}^n a_i D_i\) where we’ve written \begin{align*} \Phi \setminus\alpha\cup\beta = {\coprod}_{i=1}^m \overset{\circ}{D_i} \end{align*} and \(a_i\) is the number of points in \(\operatorname{im}( \varphi) \cap L_{z_i}\) for \(z_i \in D_i\).

For \(\varphi\in \pi_2(x, y)\), \({{\partial}}D( \varphi)\) is a 1-chain in \(\alpha \cup\beta\). Then \begin{align*} { \left.{{ {{\partial}}D( \varphi )}} \right|_{{ \alpha}} } = \sum_{i=1}^g y_i - \sum_{i=1}^g x_i { \left.{{ {{\partial}}D( \varphi )}} \right|_{{ \beta}} } = \sum_{i=1}^g x_i - \sum_{i=1}^g y_i \end{align*} where \(x_i, y_i \in \alpha_i\).

For \(\varphi\in \pi_2(x, y)\), consider an intersection point \(w\) which labels 4 nearby regions with coefficients \(a,b,c,d\):

image_2021-02-25-11-28-01

Consider several cases:

  1. \(w\not\in x\) and \(w\not\in y\): Then \({{\partial}}\qty{ {{\partial}}{ \left.{{D( \varphi)}} \right|_{{ \alpha}} } } \not\ni w\). We can expand this out as \begin{align*} D( \varphi) = a D_1 + bD_2 + c D_3 + dD_4 \\ {{\partial}}^2 D( \varphi) = {{\partial}}\qty{ a {{\partial}}D_1 } + {\cdots} .\end{align*} Now restrict this to \(\alpha_i\) to yield \begin{align*} {{\partial}}^2 D( \varphi) = ae_1 + be_2 -ce_2 -de_1 .\end{align*} Checking coefficients of \(w\) contributes \(-aw + bw - cw -d(-w)\), and these should sum to zero. This yields \(a+c = b+d\), and similarly if \(w\cap x \cap y\), this also yields \(a+c = b+d\).

  2. \(w\in x\) and \(w\not \in y\) implies that \(a+c = b +d +1\).

  3. \(w\not\in x\) and \(w\in y\) implies \(a+c+1 = b+d\).

So if you want to check to see if some 2-chain could be the domain of a Whitney disc, this local condition can be checked, i.e. this is an obstruction to existence. It turns out that this is an if and only if condition.

A 2-chain \(A \mathrel{\vcenter{:}}=\sum_{i=1}^m a_i D_i\) connects \(x\) to \(y\) if and only if the following local linear conditions are satisfied: \begin{align*} {{\partial}}^2 { \left.{{A}} \right|_{{ \alpha}} } &= y-x \\ {{\partial}}^2 { \left.{{A}} \right|_{{ \beta}} } &= x-y \\ .\end{align*}

Suppose \(g>1\). If a 2-chain \(A\) connects \(x\) to \(y\) then there exists a Whitney disc \(\varphi\in \pi_2(x, y)\) such that \(D( \varphi) = A\). If \(g>2\), \(\varphi\) is uniquely determined by \(A\).

See proof in Osvath-Szabo paper.

Think of the screen as a plane, and circled letters are handles attached out of the page according to their orientations. Consider the following diagram along with the indicated intersection points:

image_2021-02-25-11-45-11

Set the coefficients of the unlabeled regions to zero, and let \(x \mathrel{\vcenter{:}}=\left\{{ x_1, x_2}\right\}\) and \(y \mathrel{\vcenter{:}}=\left\{{ y_1, y_2 }\right\}\). We can check that if the following yellow region has coefficient 1, it can be the domain of a Whitney disc:

image_2021-02-25-11-47-15

This follows from checking the local conditions (there is a mnemonic involving the diagonal sums for the various cases).

Consider a new diagram, changed by an isotopy (here: a “finger move”):

image_2021-02-25-11-52-12

Is there a Whitney disc connecting \(x \mathrel{\vcenter{:}}=\left\{{ x_1, x_2 }\right\} \xrightarrow{\varphi} y \mathrel{\vcenter{:}}=\left\{{ y_1, y_2 }\right\}\)? Checking the diagonals, all of the local conditions hold, so yes.

Find the 3-manifold that these two diagrams represent.

13.2 Holomorphic Discs

Ultimately these are what we want to define the differential in the chain complex.

image_2021-02-25-12-08-55

We’ll set up a correspondence: \begin{align*} \left\{{\substack{ (\text{Riemann surfaces } F, {\color{red} {{\partial}}_{ \alpha}}F, {\color{blue}{{\partial}}_{\beta}}F) \xrightarrow{\pi_{\Sigma}} ( \Sigma, {\color{red} \alpha}, {\color{blue} \beta }) \\ {\big\Downarrow} \hspace{4em} {\scriptsize \text{$g{\hbox{-}}$fold branched cover $\pi_D$} } \\ (D, e_1, e_2) \\ {{\partial}}F = ({{\partial}}_{ \alpha} F) {\coprod}_{ {{\partial}}} ({{\partial}}_{ \beta} F) \\ \pi_D( {{\partial}}_{ \alpha} ) = e_1 \hspace {2em} \pi_D( {{\partial}}_{ \beta} ) = e_2 }}\right\} &\rightleftharpoons \left\{{\substack{ \text{holomorphic } u: (D^2, e_1, e_2) \to (\operatorname{Sym}^g(\Sigma), {\mathbb{T}}_{ \alpha}, {\mathbb{T}}_{ \beta}) \text{} }}\right\} \end{align*}

To do this, we define \(u(z) = \pi_{\Sigma}( \pi_D ^{-1}(z) ) \in \operatorname{Sym}^g(\Sigma)\). Check that if \(\pi_D, \pi_\Sigma\) are holomorphic, then \(u\) is holomorphic.

Link to Diagram

Then if \(u\) is holomorphic, it can be shown that \(\pi_D, \pi_{\Sigma}\) are also holomorphic. Given \(\varphi\in \pi_2(x, y)\), define \(\mathcal{M}( \varphi)\) to be the moduli space of holomorphic discs connecting \(x\) to \(y\) in the same homotopy class as \(\varphi\) (i.e. such discs represent \(\phi\)). After perturbing the complex structure \(\operatorname{Sym}^g(j)\) to make it generic, \(\mathcal{M}( \varphi)\) will be smooth. We’ll have a notion of dimension, the Maslov index \(\mu( \varphi)\), which is the expected dimension of \(\mathcal{M}( \varphi)\). There will be an \({\mathbb{R}}{\hbox{-}}\)action on \(\mathcal{M}( \varphi)\), where we remember the biholomorphism between the disc and the vertical strip:

image_2021-02-25-12-19-15

We’ll define \(\widehat{ \mathcal{M}}( \varphi) \mathrel{\vcenter{:}}=\mathcal{M}( \varphi) / {\mathbb{R}}\). The chain complex will be defined as \begin{align*} \operatorname{HF}( \Sigma, \alpha, \beta) &\mathrel{\vcenter{:}}=\bigoplus_{x \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} } {\mathbb{Z}}/2 \left\langle{ x }\right\rangle \\ {{\partial}}x &\mathrel{\vcenter{:}}=\sum_{y\in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} } \sum _{ \varphi\in \pi_2(x, y) ?} \# \widehat{ \mathcal{M}}(\varphi) .\end{align*}

We’ll need

This takes a lot of work! Is the homology of this complex interesting? Is this stronger than singular homology?

Let \(M \in \operatorname{ZHS}^3\), so the homology doesn’t distinguish \(M\) from a sphere and \(H_*(M; {\mathbb{Z}}) \cong H_*(S^3; {\mathbb{Z}})\). It turns out that \(H_*( \operatorname{HF}(M^3)) \cong H_*(\operatorname{HF}(S^3))\), so the answer is no!

Osvath-Szabo picked a basepoint \(z\in \Sigma\setminus\qty{ \alpha\cup\beta}\) and work with pointed Heegard diagrams \((\Sigma, \alpha, \beta, z)\). Perturb the differential to obtain \begin{align*} \tilde {{\partial}}x \mathrel{\vcenter{:}}=\sum_{y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} } \sum_{ \substack{ \varphi \in \pi_2(x, y), \\ \mu( \varphi) = 0, \\ n_z( \varphi) = 0 }} \# \widehat{\mathcal{M}}(\varphi) y .\end{align*} where \(n_z\) denotes the coefficient of \(\phi\) at the basepoint \(z\), i.e. the number of intersection points \(\# (\operatorname{im}\varphi \cap L_z)\).

Defining \(\widehat{\operatorname{HF}}\) as the same chain complex with the new differential now gets interesting! We’ll define \(\widehat{\operatorname{HF}}\) as the homology of this new complex.

14 The Heegard-Floer Chain Complex & Maslov Index (Tuesday, March 02)

14.1 Pointed Heegard Diagrams

Last time: to strengthen the homology theory, take a pointed Heegard diagram \((\Sigma, \alpha, \beta, z \in \Sigma\setminus\alpha\cup\beta\) and define a new chain complex \begin{align*} \widehat{\operatorname{HF}}( \Sigma, \alpha, \beta, z) &= \bigoplus_{ x\in {\mathbb{T}}_ \alpha \cap{\mathbb{T}}_ \beta} {\mathbb{Z}}/2 \left\langle{ x }\right\rangle \\ {{\partial}}x &= \sum_{y \in {\mathbb{T}}_ \alpha \cap{\mathbb{T}}_ \beta} \sum_{ \substack{ \varphi\in \pi_2(x, y), \\ \mu( \varphi) = 1, \\ n_z(\varphi) = 0 }} \# \widehat{ \mathcal{M}}(\varphi) y .\end{align*} Note that \(n_z( \varphi) = 0\) means that the coefficient attached to the region containing \(z\) is zero. Recall that we had diagram moves, how do they translate to the pointed setting?

image_2021-03-02-11-18-40

Any two pointed Heegard diagrams for a 3-manifold \(M^3\) can be connected by a sequence of the following moves:

Prove this lemma.

Here is the simplest Heegard diagram from \(S^3\):

image_2021-03-02-11-22-54

Here there is just one one intersection point, so \(\widehat{\operatorname{HF}} = {\mathbb{Z}}/2\left\langle{ x }\right\rangle\) is 1-dimensional, and \({{\partial}}x = 0\). So \(\widehat{\operatorname{HF}} = {\mathbb{Z}}/2\).

We can write \({\mathbb{RP}}^3 = L(2 ,1)\) and produce the following Heegard diagram:

image_2021-03-02-11-38-42

Is there a disc between \(x\) and \(y\)? We can check the obstruction \(\varepsilon(x, y)\) by labeling the generators in homology and tracing the following green path:

image_2021-03-02-11-40-13

We obtain \begin{align*} \varepsilon(x, y) = [B] \in { H_1(T^2) \over \left\langle{ [\alpha] = [A], [\beta] = [A + 2B] }\right\rangle } .\end{align*} In this quotient, \([B] \neq 0\), and this quotient is \({\mathbb{Z}}/2 = \left\langle{ [B] }\right\rangle\) so that \(2B = 0\). So there are no disks in \(\pi_2(x, y)\), making \({{\partial}}x = {{\partial}}y = 0\). So \(\widehat{HF}({\mathbb{RP}}^3) = {\mathbb{Z}}/2 \oplus {\mathbb{Z}}/2\).

Compute \(\widehat{\operatorname{HF}}(L(p, 1))\). Use that \(\varepsilon(x, y) + \varepsilon(y, z) = \varepsilon(x, z)\).

14.2 Maslov Index

Recall that we had a natural concatenation operation on Whitney discs: \begin{align*} \ast: \pi_2(x, y) \times\pi_2(y, z) \to \pi_2(x, z) ,\end{align*} using the identification of these discs with paths in the path space and using concatenation of paths there. Note that the domains of concatenations are given by \(D( \varphi_1 \ast \varphi_2) = D( \varphi_1) + D( \varphi_2)\), since this amounts to adding algebraic intersection numbers.

There is an inverse \begin{align*} \pi_2(x, y) &\to \pi_2(y, x)\\ \varphi &\mapsto \varphi^{-1}(s, t) \mathrel{\vcenter{:}}=\phi(s, -t) ,\end{align*} which reverses the parameterization on \((s, t) \in I \times{\mathbb{R}}\) and runs the path backward. Here \(D( \varphi^{-1}) = -D( \varphi)\).

There is also a sphere addition \begin{align*} \pi_2( \operatorname{Sym}^g( \Sigma), x) \times\pi_2(x, y) &\to \pi_2(x, y) \\ (\Omega, \varphi) &\mapsto \Omega \ast \varphi .\end{align*}

Maps entire boundary to a point, yielding a sphere.

Note that for \(g\geq 2\), the \(\pi_2\) on the left-hand side is isomorphic to \({\mathbb{Z}}\), which came from quotienting by the hyperelliptic involution several lectures ago. Writing the positive generator as \(S\), we have \(\Omega = kS\) for some \(k\in {\mathbb{Z}}\).

Show that \begin{align*} D(S) = \sum_{i=1}^m D_i = [ \Sigma] .\end{align*}

There exists a function \(\mu: \pi_2(x, y) \to {\mathbb{Z}}\) called the Maslov index satisfying:

  1. Additivity: \(\mu( \varphi_1 \ast \varphi_2) = \mu( \varphi_1) + \mu (\varphi_2)\).

  2. Invertibility: \(\mu( \varphi^{-1}) = - \mu( \varphi)\).

  3. Sphere addition: \(\mu( kS \ast \varphi) = \mu( \varphi) + 2k\) where \(k\in {\mathbb{Z}}\) and \(S\in \pi_2( \operatorname{Sym}^g( \Sigma ) )\).

  4. If \(\varphi\in \pi_2(x, x)\) is constant, then \(\mu( \varphi) = 0\).

Note that \(2\implies 4\).

The Maslov index is the “expected” dimension of \begin{align*} \mathcal{M}( \varphi) = \left\{{ u: I \to {\mathbb{R}}\to \operatorname{Sym}^g( \Sigma) {~\mathrel{\Big|}~} [u] = \varphi \,du\circ i = J \circ \,du }\right\} \end{align*} where \(i\) is the standard complex structure on the strip and \(J\) will be a perturbation of the complex structure over the Heegard surface. This will yield an operator \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J: B &\to \mathcal{L} \\ u & \mapsto du \circ i - J \circ du \end{align*} for some appropriate infinite dimensional spaces. The elements of \(\mathcal{M}( \varphi )\) will be in the kernel of this operator. We want 0 to be a regular value (surjective derivative) for \(\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J\), since in finite dimensions the inverse image would be a smooth manifold. In the infinite dimensional setting, we’ll have by the inverse function theorem that \(\mathcal{M} (\phi) = \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu^{-1}_J(0)\) will be a smooth manifold. We’ll want the following derivative to be surjective: \begin{align*} D_u \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J: T_u B \to T_{\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J u} \mathcal{L} \end{align*} for all \(u \in \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu^{-1}_J(0)\), which is referred to as transversality of the operator, and can be made to hold by perturbing the complex structure. Since the dimension of a manifold is the dimension of the tangent spaces, we’ll have \(\mathcal{M}( \varphi)\) smooth of dimension equal to \(\dim \ker D_u \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J\) for any \(u \in \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J^{-1}(0)\). This will be an order 2 elliptic operator (or more generally a Fredholm operator), for which we have a notion of index: \begin{align*} \operatorname{ind}( D \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J) = \dim( \ker D\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J) - \dim (\operatorname{coker}D \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J) .\end{align*} If surjectivity holds, the cokernel will be zero, so it will suffice to compute the dimension of the kernel to get the dimension of the moduli space. The index of this operator will be the Maslov index.

Take a look at Gromov compactness again!

15 Maslov Index Formula (Thursday, March 04)

15.1 Review

Recall that for \(x,y \in {\mathbb{T}}_ \alpha \cap{\mathbb{T}}_ \beta\), there is a map \begin{align*} \mu: \pi_2(x, y) &\to {\mathbb{Z}}\\ &\mu &= \operatorname{ind}(D \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu_J) .\end{align*}

This index is the expected dimension of \(M(\varphi)\). The following theorem can be found in the paper “A cylindrical reformulation of Heegard Floer homology”:

Let \(x = \left\{{ x_1, \cdots, x_g }\right\}\) and \(y = \left\{{ y_1, \cdots, y_g }\right\}\) and \(\varphi\in \pi_2(x, y)\). Then \begin{align*} \mu( \varphi) = e( D( \varphi) ) + n_x( D( \varphi ) ) + n_y( D( \varphi ) ) .\end{align*} where \(e({-})\) is the Euler measure and \(n_x(\cdots), n_y(\cdots)\) is referred to as the point measure. Note that these only depend on the domain of \(\varphi\).

Let \(D( \varphi) = \sum_{i=1}^m n_{z_i} ( \varphi) D_i\), then \begin{align*} e (D (\varphi)) \mathrel{\vcenter{:}}=\sum_{i=1}^m n_{z_i}( \varphi) e(D_i) && e(D_i) \mathrel{\vcenter{:}}=\chi(D_i) + {1\over 4} C_1 - {1\over 4}C_2 .\end{align*} Here we use the fact that all regions are polygons whose corners occur in one of two types:

image_2021-03-04-11-20-25

So we define \(C_1\) to be the number of corners of the first type and \(C_2\) the number of the second type. The point measure is defined as \begin{align*} n_x (D( \varphi ) ) \mathrel{\vcenter{:}}=\sum_{i=1}^g n_{x_i}( D( \varphi ) ) = {n_1 + n_2 + n_3 + n_4 \over 4} ,\end{align*}

where the \(n_i\) are the surrounding regions’ coefficients:

image_2021-03-04-11-24-27

Let \(x = \left\{{ x_1, x_2 }\right\}, y = \left\{{ y_1, y_2 }\right\}\) and compute \(\mu( \varphi)\) where \(D( \varphi)\) is one of the following domains:

  1. The first type:
image_2021-03-04-11-26-34
image_2021-03-04-11-30-34
  1. A second type:
image_2021-03-04-11-32-08
  1. A third type:
image_2021-03-04-11-36-09

Another example calculation:

image_2021-03-04-11-47-57

Does this domain have a holomorphic representative?

15.2 Positivity Principle

For \(\varphi\in \pi_2(x, y)\), if \(\mathcal{M}( \varphi) \neq \emptyset\) then \(D( \varphi) \geq 0\), i.e. \(D( \varphi ) = \sum n_i D_i\) where \(n_i \geq 0\). This happens if and only if \(n_w( \varphi) \geq 0\) for all \(w \in \Sigma \setminus\alpha\cup\beta\).

If \(u \in M( \varphi)\) then \(u:D \to \operatorname{Sym}^g( \Sigma )\) is holomorphic and \(\operatorname{im}(u)\) is a complex submanifold. If \(w \in \Sigma \setminus\alpha\cup\beta\) then \(\mathcal{L}_w\) is holomorphic.

Show that transverse complex submanifolds intersect non-negatively, i.e.  \begin{align*} n_w( \varphi) \mathrel{\vcenter{:}}=\# \qty{ \operatorname{im}(u) \cap\mathcal{L}_w } \geq 0 .\end{align*}

Consider \(S^1 \times S^2\) with the following Heegard diagram:

image_2021-03-04-12-18-25

We have \(\widehat{\operatorname{HF}} (\Sigma, \alpha, \beta, z) = {\mathbb{Z}}/2 \left\langle{ x,y }\right\rangle\). Then for \(\varphi\in \pi_2(x, y)\) with \(\mu( \varphi) = 1\) and \(n_z( \varphi) = 0\), we can write \(D( \varphi) = a D_1 + b D_2\). Now checking the diagonals:

image_2021-03-04-12-20-40

Since the sum of multiplicities NW \(\to\) SE should be 1 more than the sum NE \(\to\) SW, we have \(a+b=1\) and by the positivity principle, \(D( \varphi) \geq 0\) implies \(a, b \geq 0\). We then obtain \begin{align*} \begin{cases} a = 0, b = 1 & \implies D( \varphi) = D_2 \ni \phi_2 \\ a = 1, b = 0 & \implies D( \varphi) = D_1 \ni \phi_1 \end{cases} .\end{align*}

For example, if \(\mu( \varphi_1) = \mu( \varphi_2) = 1\), we’re looking for holomorphic maps

image_2021-03-04-12-23-52

For any \(p\) on the \(\alpha\) circle from \(y\) to \(x\), there exists a unique holomorphic map with \(\mu(1) = p\) by the Riemann mapping theorem. After taking the quotient \(\widehat{\mathcal{M}} ( \varphi_1) = \mathcal{M}(\varphi_1) / {\mathbb{R}}\), we obtain \(\# \widehat{\mathcal{M}} ( \varphi_1 ) = 1 = \# \mathcal{M}( \varphi_2 )\). Then note that \begin{align*} {{\partial}}x = \qty{ \# \widehat{\mathcal{M}} ( \varphi_1) + \# \widehat{ \mathcal{M} } ( \varphi_2) } y = 0 ,\end{align*} since we are taking coefficients mod 2. Then \(\varphi\in \mu(x, y)\) implies that \(a+b=-1\), so there is no non-negative disk and \({{\partial}}y = 0\).

Show that there is no non-negative disc in \(\pi_2(x, x)\) and \(\pi_2(y, y)\) by looking at local coefficients.

So \({{\partial}}=0\) which implies that \(\widehat{\operatorname{HF}}( \Sigma, \alpha, \beta, z) = ({\mathbb{Z}}/2)^{\oplus 2}\).

What if we used an isotopic diagram?

image_2021-03-04-12-30-18

The only difference between this and the first is an isotopy of \(\beta\), and we’ll see that there’s an invariance and a condition called admissibility to help decide which to use.

Do another isotopy to create 4 intersection points and show that the ranks of homology are unchanged.

16 Tuesday, March 09

Recall that we were working with a diagram for \(S^1 \times S^2\):

image_2021-03-09-11-14-10

Here we have \({{\partial}}x = 2y = 0\) since we’re working mod 2, and \({{\partial}}y = 0\), so we have \begin{align*} \widehat{\operatorname{HF}}(H_1) = {\ker {{\partial}}\over \operatorname{im}{{\partial}}} = { \left\langle{ x, y }\right\rangle\over 1} = ({\mathbb{Z}}/2)^{\oplus 2} .\end{align*}

However, with a different diagram, we get a different result:

image_2021-03-09-11-15-46

Here \(\widehat{\operatorname{HF}}(H_2) = 0\). To prevent this, we’ll have some class of admissible diagrams.

A 2-chain \(P = \sum_{i=1}^m a_i D_i\) is called a periodic domain if and only if

  1. The local multiplicity of \(P\) at \(z\) is zero, i.e. \(n_z(P) = 0\), and
  2. \({{\partial}}P\) is a linear combination of \(\alpha, \beta\).

Note that for (2), the boundary could involve 1-chains, so this condition avoids corners on \({{\partial}}P\). The local picture is the following:

image_2021-03-09-11-19-12

In this picture, \(P = nD_1\) will be a periodic domain for any \(n\);

image_2021-03-09-11-20-54

Labeling the first picture, we have

image_2021-03-09-11-21-32

We should have \(n_1 + n_2 = 0\), so any \(P = n(D_1 - D_2)\) will be a periodic domain. Checking the boundary yields \({{\partial}}P = n \alpha \pm n \beta\). In fact there is single “generator” for the periodic domains here:

image_2021-03-09-11-23-43

A Heegaard diagram \(H = ( \Sigma, \alpha, \beta, z)\) is called weakly admissible if any periodic domain \(P\) has both positive and negative coefficients.

\(H_1\) from above is weakly admissible, but \(H_2\) is not.

For any Whitney disc \(\varphi\in \pi_2(x, x)\) with \(n_z( \varphi) = 0\), \(D( \varphi)\) is a periodic domain. For any periodic domain \(P\), we can associate a homology class \(H(P) \in H_2(M)\). Writing \begin{align*} {{\partial}}P = \sum_{i=1}^g a_i \alpha_i + \sum_{i=1}^g b_i \beta_i \xrightarrow{H} H(P) \mathrel{\vcenter{:}}=[ P + \sum_{i=1}^g a_i A_i + \sum_{i=1}^g b_i B_i] .\end{align*} using that each \(\alpha_i\) is the boundary of some disc \(A_i\) in one handlebody, and \(\beta_i = {{\partial}}B_i\) similarly. Noting that \(P\) is a boundary, this amounts to adding a number of discs to get a closed nontrivial cycle.

Show that if \(H(P) =0\) the \(P=0\), and that \(H\) is a bijection.

Let \(P = \sum_{i=1}^m n_i D_i\) be a 2-chain that satisfies condition 2, so \({{\partial}}P = \sum_{i=1}^m a_i \alpha_i + \sum_{i=1}^m b_i \beta_i\). Then we can obtain a periodic domain: \begin{align*} P_0 \mathrel{\vcenter{:}}= P - n_z(P) \qty{ \sum_{i=1}^m D_i } \mathrel{\vcenter{:}}= P - n_z(P) [ \Sigma ] .\end{align*}

Show that if \(g>2\), then \begin{align*} \pi_2(x, x) &\xrightarrow{\sim} {\mathbb{Z}}\oplus H_2(M)\\ P = P_0 + n_z(P)[ \Sigma ] &\mapsto (n_z(P), H(P_0)) .\end{align*}

Alternatively, given \(\varphi\in n_z( \varphi) S\) where \(S\) is the positive generator of \(\pi_2( \operatorname{Sym}^g( \Sigma ) ) \ast\varphi_0\) (i.e. the hyperelliptic involution) where \(D(\phi_0)\) is a periodic domain.

Use that for \(g\geq 2\) there is a bijection between Whitney discs and domains, and domains of Whitney discs are domains satisfying condition (2) above.

Show that for a closed 3-manifold \(M\in \operatorname{QHS}^3\), \(H_2(M; {\mathbb{Z}}) = 0\).

If \(H_2(M) =0\) (e.g. if \(M \in \operatorname{QHS}^3\)) then any Heegard diagram is weakly admissible.

This is because \(H_2(M) = 0\) means there are no periodic domains.

If \(H\) is weakly admissible, then for any \(x, y \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\) there are finitely many Whitney discs \(\varphi\in \pi_2(x, y)\) with \(D( \varphi) \geq 0\).

Any Heegard diagram can be made admissible using finitely many isotopies.

For \(g=1\), we have \(\operatorname{Sym}^1( \Sigma) = \Sigma\). We’ll use this in what follows.

For any \(x, y\in \alpha \cap\beta\), the 0-dimensional moduli space of holomorphic disks connecting \(x\) to \(y\) correspond to orientation-preserving immersions of the following form which satisfy:

image_2021-03-09-12-06-12
  1. \(u(e_1) \subseteq \beta, u(e_2) \subseteq \alpha, u(-i) = x, u(i) = y\).
  2. There are \(\pi/2\) radian corners at \(x, y\), but these are smooth immersions at other boundary points.

Prove this lemma using the Riemann mapping theorem.

Consider the following example:

image_2021-03-09-12-09-05

List all of the bigons in this picture that will contribute to the differential.

17 Thursday, March 11

Recall the example from last time: we are trying to show that changing a diagram by isotopy doesn’t change the homology.

image_2021-03-11-11-16-15

Here we have \(g=1\) and so \(\operatorname{Sym}^1(T^2) = T^2\), and \(\alpha \cap\beta = \left\{{ a,b,c,d,e }\right\}\). So \(\widehat{\operatorname{HF}}( \Sigma, \alpha, \beta, z) = {\mathbb{Z}}/2 \left\langle{ a,b,c,d,e }\right\rangle\).

First mark the component that contains the base point \(z\) and give it a coefficient of zero:

image_2021-03-11-11-18-34

We can make this part bigger, and find that there are only two bigons involving \(a\).

image_2021-03-11-11-23-41

This is because starting at a point and following the orientation should yield red first and then blue, matching up with the orientation on the disc.

image_2021-03-11-11-35-44

So \({{\partial}}a = {\color{yellow} b} + {\color{purple} d}\), since we require 90 degree corners. Similarly,

We can simplify this information with a graph with arrows pointing toward boundaries:

image_2021-03-11-11-29-41

Then any linear combination with the same image will have zero boundary, so we have \begin{align*} \ker {{\partial}}&= \left\langle{ a + e, b + d, c }\right\rangle \\ \operatorname{im}{{\partial}}&= \left\langle{ b+d, c }\right\rangle ,\end{align*} and thus \(\widehat{\operatorname{HF}}(\Sigma, \alpha, \beta, z) = {\mathbb{Z}}/2\).

image_2021-03-11-11-37-28

Drawing this on a surface yields the following:

image_2021-03-11-12-01-09

One useful trick here is labeling the points along one curve with letters and the other with numbers. Another is making a table like the following:

image_2021-03-11-12-05-41

From this it’s easy to read off the 4 possible generators \(\left\{{ ae, ce, bf, bd }\right\}\). The regions the contain \(z\) can be seen in the latter picture:

image_2021-03-11-12-07-48

Translating this to the original picture yields these regions:

image_2021-03-11-12-09-26

Note that the half-bigons in the diagram actually pair to a bigon on the surface, so consider this simplified drawing of the surface:

image_2021-03-11-12-11-38

Setting \(D_1 = D( \varphi)\) for \(\varphi\in \pi_2( ae, bf )\), we have \(\mu( \varphi) = 1\) since we showed that rectangular regions have Maslov index 1. Are there any holomorphic representatives? The claim is that \(\# \widehat{\mathcal{M}}( \varphi)\). Checking boundaries yields the following:

image_2021-03-11-12-19-43

Then \begin{align*} \ker {{\partial}}&= \left\langle{ ce, bf }\right\rangle \\ \operatorname{im}{{\partial}}&= \left\langle{ ce + bf }\right\rangle \\ &\implies \widehat{HF}(\Sigma, \alpha, \beta, z) \cong {\mathbb{Z}} .\end{align*} This is good, since some valid moves will make this into a standard diagram for \(S^3\) (?).

Recall that given a rectangle, there is a 2-to-1 branched cover:

image_2021-03-11-12-25-02

Such branched coverings bijectively correspond to biholomorphic involutions \begin{align*} a &\rightleftharpoons e \\ b &\rightleftharpoons f .\end{align*}

This is because there is a unique involution exchanging them by the Schwarz lemma, since any pole of the involution must lie along the line connecting points it exchanges, and exchanging each pair of corners in the rectangle forces to pole to be precisely the point in the center of the rectangle. So these correspond got biholomorphic involutions of \({\mathbb{D}}\) using complex analysis.

Next week: more about the Maslov index and \(\mathrm{Spin}^{\mathbb{C} }\) structures, then invariance under diagram moves.

18 Maslov Grading and \(\mathrm{Spin}^{\mathbb{C} }\) Structures (Tuesday, March 16)

Let \(M\in {\mathsf{Mfd}}^3({\mathbb{R}})\) be a closed oriented 3-manifold and \(\mathcal{H} = (\Sigma, \alpha, \beta, z)\) a Heegaard diagram for \(M\). Letting \(b_i\) be the Betti numbers, note that \(b_1 = 0 \iff M \in \operatorname{QHS}^3\) is a rational homology 3-sphere, i.e. \(H_i(M; {\mathbb{Q}}) \cong H_i(S^3; {\mathbb{Q}})\) for all \(i\). This also implies that \(H_2(M; {\mathbb{Z}}) = 0\). Under this condition, we can define a relative \({\mathbb{Z}}{\hbox{-}}\)grading (i.e. we have a difference of grading between any two elements) on \(\widehat{CF}\) in the following way: for \(x, y\) two generators, we set \begin{align*} {\mathsf{gr}\,}(x) - {\mathsf{gr}\,}(y) \mathrel{\vcenter{:}}=\mu( \varphi) -2n_z( \varphi) && \text{for some } \varphi\in \pi_2(x,y) .\end{align*}

Recall that \(\mu({-})\) denotes the Maslov index, \(n_z( {-})\) is the local multiplicity of a Whitney disc at \(z\), and \(x, y\) denote tuples of points.

This involves a choice of disc, so why is it well-defined? We’ll also see why we need \(M\in \operatorname{QHS}^3\).

Let \(\varphi, \varphi' \in \pi_2(x, y)\). We have \begin{align*} \varphi \ast (-\varphi') \in \pi_2(x, x) = {\mathbb{Z}}\oplus H_2(M) = {\mathbb{Z}}\oplus 0 ,\end{align*} so this is some multiple \(kS\) where \(S\) is the positive generator of \(\pi_2\operatorname{Sym}^g \Sigma\). So \begin{align*} \mu( \varphi \ast (- \varphi') ) = \mu( \varphi) - \mu( \varphi') = k \mu(S) = 2k .\end{align*} Similarly, \begin{align*} n_z( \varphi \ast (- \varphi')) = n_z( \varphi) - n_z( \varphi') = k n_z(S) = k ,\end{align*} where we’ve used \(\mu(S) = 2, n_z(S) = 1\). Then \begin{align*} \mu( \varphi) - \mu ( \varphi') &= 2( n_z( \varphi) - n_z( \varphi') ) \\ \implies \mu( \varphi) - 2n_z( \varphi) &= \mu( \varphi') - 2 n_z( \varphi') .\end{align*}

Note that the relative grading is only defined if \(\pi_2(x, y) \neq \emptyset \iff \varepsilon(x, y) = 0 \in H_1(M; {\mathbb{Z}})\). This generated an equivalence relation of elements in \({\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\) by \(x\sim y \iff \varepsilon(x, y) = 0\), so we have a decomposition \begin{align*} \widehat{CF}( \mathcal{H} ) = \bigoplus _{?} \widehat{CF}( \mathcal{H}, ?) .\end{align*} which is preserved by \({{\partial}}\), so \(\widehat{HF}(\mathcal{H})\) will split similarly as \begin{align*} \widehat{HF}( \mathcal{H} ) = \bigoplus _{?} \widehat{CF}( \mathcal{H}, ?) .\end{align*}

It turns out that the right thing to replace the “?” with will be \(\mathrm{Spin}^{\mathbb{C} }\) structures.

18.1 \(\mathrm{Spin}^{\mathbb{C} }\) Structures

We’ll discuss Turaev’s (?) reformulation of \(\mathrm{Spin}^{\mathbb{C} }\) structure for \({\mathsf{Mfd}}^3\). Note that \(\chi(M) = 0\), so there exists nowhere vanishing vector fields on \(M\) by Poincaré-Hopf.

Let \(v_1, v_2\) be nowhere vanishing vector fields on \(M\). We say \begin{align*} v_1 \sim v_2 \iff { \left.{{v_1}} \right|_{{M\setminus B}} } \simeq{ \left.{{v_2}} \right|_{{M\setminus B}} } ,\end{align*} i.e. their restrictions to \(M\setminus B\) are homotopic, and here \(B\) is a 3-ball in \(M\). Equivalently, \(v_1\sim v_2 \iff v_1, v_2\) are homotopic in the complement of finitely many 3-balls in \(M\).

\begin{align*} \mathrm{Spin}^{\mathbb{C} }(M) \mathrel{\vcenter{:}}=\left\{{ \text{Nowhere vanishing vector fields on } M }\right\}_{/\sim} .\end{align*}

Let \(\mathcal{H} = ( \Sigma, \alpha, \beta, z)\) be a Heegard diagram for \(M\), then define a map \begin{align*} S_z: {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta} \to \mathrm{Spin}^{\mathbb{C} }(M) .\end{align*}

Step 1: Choose a self-indexing Morse function \(f\) with \(\# \operatorname{Crit}^0(f) = \# \operatorname{Crit}^3(f) = 1\) such that its corresponding Heegaard diagram is \(\mathcal{H}\):

image_2021-03-16-11-47-19

Note that we have a surface in \(f ^{-1} (3/2)\) and there are exactly \(q\) critical points along each of \(f ^{-1} (1), f ^{-1} (2)\). For each \(x = \left\{{ { {x}_1, {x}_2, \cdots, {x}_{g}} }\right\} \cap{\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta}\), we have \(x_i \in \alpha_i \cap\beta_{\sigma(i)}\) for some permutation \(\sigma\in S_g\) Then \(\alpha\mapsto p_i\) and \(\beta_{\sigma(i)} \mapsto q_{\sigma(i)}\):

image_2021-03-16-11-49-42

Trajectories of \(-\nabla f\) that pass through \({ {x}_1, {x}_2, \cdots, {x}_{g}}\) are \(g\) pairwise disjoint arcs connecting \({ {q}_1, {q}_2, \cdots, {q}_{g}}\) to \({ {p}_1, {p}_2, \cdots, {p}_{g}}\), so there is a one-to-one correspondence between these intersection points.

Now taking tubular neighborhoods of the \(g+1\) disjoint arcs yields \(g+1\) pairwise disjoint 3-balls in \(M\), so write this as \(B\mathrel{\vcenter{:}}= B_1 \coprod \cdots \coprod B_{g+1}\).

image_2021-03-16-11-53-04

Note that

Show this!

Hint: the trajectories of \(-\nabla f\) in each ball connect critical points of different parities, and so each \({ \left.{{-\nabla f }} \right|_{{{{\partial}}B_i }} }\) has index zero.

Define \(S_z(x) \in \mathrm{Spin}^{\mathbb{C} }(M)\) to be the equivalence class represented by this vector field. This is well-defined since outside of the finitely many balls, this vector field is just equal to \(-\nabla f\).

Show that this does not depend on which Morse function is chosen.

There is a one-to-one correspondence \begin{align*} \mathrm{Spin}^{\mathbb{C} }(M) \rightleftharpoons H^2(M; {\mathbb{Z}}) .\end{align*} Picking a trivialization \(\tau: TM \to M \times{\mathbb{R}}^3\) and a Riemannian metric on \(M\), then \begin{align*} \left\{{\substack{ \text{Nowhere vanishing vector fields} \\ \text{on } M }}\right\} &\rightleftharpoons \left\{{\substack{ \text{functions } f: M\to S^2 }}\right\} \\ v:M\to {\mathbb{R}}^3\setminus\left\{{0}\right\}&\mapsto x \xrightarrow{f_v} \widehat{\mathbf{v}_x} \end{align*}

Let \(\alpha\in H^2(S^2; {\mathbb{Z}})\) be the positive generator, then define \begin{align*} \delta^{ \tau} (v) \mathrel{\vcenter{:}}= f_v^*( \alpha) \in H^2(M; {\mathbb{Z}}) .\end{align*} Note that if \(v_1 \sim v_2\), we have \(\delta^{ \tau}(v_1) = \delta^{ \tau}(v_2)\) since they are homotopic on the complement of a ball: \begin{align*} M\setminus B \overset{i}\hookrightarrow M \xrightarrow{f_v} S^2 .\end{align*} Conclude that \begin{align*} (f_{v_1} \circ i)^* ( \alpha) = (f_{v_2} \circ i)^* (\alpha) ,\end{align*} \(i^*\) is an isomorphism, so \(f_{v_1}^*( \alpha) = f_{v_2}^* ( \alpha)\), yielding the identification.

  1. Show that \(\delta^{ \tau}\) is a bijection.

  2. \(\delta(v_1, v_2) \mathrel{\vcenter{:}}=\delta^{ \tau}(v_1) - \delta^{ \tau}(v_2) \in H^2(M; {\mathbb{Z}})\) is well-defined and independent of the choice of \(\tau\), and satisfies \begin{align*} \delta(v_1, v_2) + \delta(v_2, v_3) = \delta(v_1, v_3) .\end{align*}

Thus we also have a relative map \begin{align*} \mathrm{Spin}^{\mathbb{C} }(M) &\rightleftharpoons H^2(M; {\mathbb{Z}})\\ s_1, s_2 \mathrel{\vcenter{:}}=[v_1] - [v_2] &\mapsto s_1 -s_2 \mathrel{\vcenter{:}}=\delta(v_1, v_2) .\end{align*}

19 Thursday, March 18

19.1 \(\mathrm{Spin}^{\mathbb{C} }\) Structures and Invariance

Recall that given a Heegard diagram \(( \Sigma, \alpha, \beta, z )\) gives an equivalence relation \begin{align*} x \sim y \iff \varepsilon(x, y) = 0 \in H_1(M) \overset{\mathrm{PD}}{=} H^2(M) .\end{align*} This yields a decomposition of \(\widehat{\operatorname{HF}}\) into a direct sum over equivalence classes of subcomplexes defined by \(\mathrm{Spin}^{\mathbb{C} }\) structures. Note that the differential will preserve each direct summand. We defined \(\mathrm{Spin}^{\mathbb{C} }(M)\) as the set of nowhere vanishing vector fields on \(M\) modulo being homotopic outside finitely many 3-balls in \(M\). We had a map \begin{align*} {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta} \xrightarrow{s_z} \mathrm{Spin}^{\mathbb{C} }(M) ,\end{align*} recalling that the left-hand side are the generators of \(\widehat{\operatorname{HF}}\). We took a self-indexing Morse function on \(M\), took the inverse image of \(3/2\) to get the Heegard surface, and each intersection point \(x_i\) gave a flow line from an index 2 critical point to an index 1 critical point passing through \(x_i\):

Trajectories of negative gradient flow

We proceeded by cancelling adjacent flow lines (at the level of vector fields), and then modifying \(\gamma_z\) (the flow line passing through the basepoint \(z\) connecting the index 0 to the index 3) to get a nowhere vanishing vector field. We then took a trivialization \(\tau: TM \to M \times{\mathbb{R}}^3\) defined a map \begin{align*} \mathrm{Spin}^{\mathbb{C} }(M) &\xrightarrow{\gamma^ \tau} H^2(M) \\ s = [v] &\mapsto f_v^*( \alpha) .\end{align*} where \(\alpha\) is the volume form of \(S^2\) and \begin{align*} f_v: M &\to S^2 \\ x &\mapsto \widehat{v_x} \mathrel{\vcenter{:}}={ v_x \over {\left\lVert {v_x} \right\rVert} } .\end{align*} Note that \(\delta^\tau\) a priori depends on \(\tau\), but \begin{align*} \delta(s_1, s_2) = \delta^{ \tau}(s_1) - \delta^{ \tau}(s_2) \in H^2(M) ,\end{align*} and the difference is independent of \(\tau\).

For \(x, y\in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}\), defining \(s_1 - s_2 = \delta(s_1, s_2) \in H^2(M)\), we have \begin{align*} s_z(y) - s_z(x) = \mathrm{PD}[\varepsilon(x, y) ] .\end{align*}

As corollaries,

  1. If \(x\sim y\) then \(s_z(y) = s_z(x)\), and
  2. If \(x\not\sim y\) then the above equation holds.

Prove this!

Hint, take the Poincaré dual of the link below to get the formula: \begin{align*} s_z(y) - s_z(x) = \mathrm{PD}[ \gamma_y \cup(- \gamma_x)] .\end{align*} This implies that the two vector fields are equal everywhere outside of a tubular neighborhood of the link. Then show that \([ \gamma_x \cup(-\gamma_x) = [ \varepsilon(x, y) ]\).

We thus have \begin{align*} \widehat{\operatorname{HF}}( \Sigma, \alpha, \beta, z) = \bigoplus _{{\mathfrak{s}}\in \mathrm{Spin}^{\mathbb{C} }(M)} \widehat{\operatorname{HF}}( \Sigma, \alpha, \beta, z, {\mathfrak{s}}) .\end{align*}

We have several properties of \(\mathrm{Spin}^{\mathbb{C} }\) structures. There is a map \begin{align*} J: \mathrm{Spin}^{\mathbb{C} }(M) &\to \mathrm{Spin}^{\mathbb{C} }(M) \\ s = [v] &\mapsto {\overline{{s}}} \mathrel{\vcenter{:}}=[-v] .\end{align*} There is also a first Chern class \begin{align*} c_1: \mathrm{Spin}^{\mathbb{C} }(M) &\to H^2(M) \\ s &\mapsto s - {\overline{{s}}} ,\end{align*} i.e. \(c_1(s) = \delta(s, {\overline{{s}}})\).

The association \begin{align*} ( \Sigma, \alpha, \beta, z), J \leadsto \widehat{\operatorname{HF}} ( \Sigma, \alpha, \beta, z) \end{align*} does not depend on the choice of Heegard diagram or the almost complex structure \(J\), so this yields a well-defined invariant of \(M\) which we’ll denote \(\widehat{\operatorname{HF}}(M)\) for \(M\in {\mathsf{Mfd}}^3({\mathbb{R}})\).

There are few things to discuss:

  1. The almost complex structure \(J\):

  2. Isotopies

  3. Handle slides

  4. Stabilization

Remarks on these:

  1. This involves a standard argument from Lagrangian Floer homology.
  2. There are two cases:
    • If the isotopy doesn’t create a new intersection, we have a 1-to-1 correspondence between generators for any two choices, and changing \(J\) to \(J'\) will give a correspondence between the differentials. This just involves picking a diffeomorphism that maps \(\alpha\) circles to \(\alpha'\) circles, and so on. So this reduces to showing 1.
    • If is does create new intersection points, there are again standard arguments in Lagrangian Floer homology for this.
  3. This involves the following situation, which induces a map
image_2021-03-18-11-51-38

For an appropriate choice of \(J\) on \(\Sigma \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}T^2\), the map \(f\) above will induce a chain homotopy equivalence \begin{align*} \tilde f: \widehat{\operatorname{HF}} (\Sigma, \alpha, \beta, z) \xrightarrow{\sim} \widehat{\operatorname{HF}} (\Sigma \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}T^2, \alpha', \beta', z) .\end{align*}

  1. What’s the picture?
image_2021-03-18-11-56-59

This will yield a map \begin{align*} (\Sigma, \alpha, {\color{blue} \beta}, z) \leadsto (\Sigma, \alpha, {\color{green} \gamma}, z) .\end{align*} For \(i = 1, \cdots, g-1\), we’ll have \(\gamma_i\) isotopic to \(\beta_i\), and for \(i=g\), \(\gamma_g\) is obtained by sliding \(\beta_g\) over \(\beta_{g-1}\). We’ll combine these into the same diagram with different colors to compare them, yielding a Heegard triple: \begin{align*} (\Sigma, {\color{red} \alpha}, {\color{blue} \beta}, { \color{green} \gamma}, z) .\end{align*} We can think of this as three separate diagrams: \begin{align*} (\Sigma, {\color{red} \alpha}, {\color{blue} \beta}, z) &\leadsto M\\ (\Sigma, {\color{blue} \beta}, { \color{green} \gamma}, z) &\leadsto ? \\ (\Sigma, {\color{red} \alpha}, { \color{green} \gamma}, z) &\leadsto M .\end{align*} What does the middle one represent?

Heegard diagram

Here this is a diagram for \((S^1 \times S^2)^{\mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}2}\).

Note that we draw \(\gamma_i\) such that it intersects \(\beta_i\) in two transverse intersection points to make sure the diagram is admissible.

Give a Heegard triple \(( \Sigma, \alpha, \beta, \gamma, z\), pick three intersection points \begin{align*} x &\in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta} \\ y &\in {\mathbb{T}}_{ \beta} \cap{\mathbb{T}}_{ \gamma} \\ w &\in {\mathbb{T}}_{ \gamma} \cap{\mathbb{T}}_{ \alpha} .\end{align*} We can use Whitney triangles to connect \(x,y,w\):

image_2021-03-18-12-23-07

We then define \(\pi_2(x,y,z)\) to be the homotopy class of Whitney triangles connecting \(x,y,w\). We can similarly define \({\mathcal{M}}( \psi )\) to be the moduli space of \(J{\hbox{-}}\)holomorphic representatives of \(\psi\in \pi_2(x,y,w)\), along with a chain map \begin{align*} f_{\alpha \beta \gamma}: \widehat{\operatorname{HF}}(\Sigma, \alpha, \beta, z) \otimes \widehat{\operatorname{HF}}(\Sigma, \beta, \gamma z) &\to \widehat{\operatorname{HF}}(\Sigma, \alpha, \gamma, z) \\ x\otimes y &\mapsto \sum_{ w \in {\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{ \beta}} \sum_{\substack{ \psi\in \pi_2(x,y,w) \\ \mu( \psi) = 0 \\ n_z( \psi) = 0 }} \# {\mathcal{M}}( \psi) \cdot w .\end{align*}

\(f_{ \alpha \beta \gamma}\) is a chain map.

Next time: we’ll show how to get a chain homotopy equivalence from the first tensor term above to the codomain. We’ll also see surgery exact triangles.

20 Thursday, March 25

Recall that we have several variants: namely \(\widehat{\operatorname{HF}}, \operatorname{HF}^-, \operatorname{HF}^+, \operatorname{HF}^{\infty }\). Let \(M\in {\mathsf{Mfd}}^3\) and take a Heegard diagram \((\Sigma, \alpha, \beta, z)\). Note that \(\operatorname{HF}^-( \Sigma, \alpha, \beta, z)\) is the free \({\mathbb{Z}}/2[u]{\hbox{-}}\)module generated by \({\mathbb{T}}_{ \alpha} \cap{\mathbb{T}}_{\beta}\) with differential given by ?.

A Heegaard diagram is called nice if every connected component of \(\Sigma\setminus\alpha\cup\beta\) that does not contain \(z\) is either a bigon or a rectangle.

For nice Heegaard diagrams, the Maslov index 1 holomorphic discs with \(n_z( \varphi) = 0\) are embedded bigons and rectangles.

Any 3-manifold has a nice Heegaard diagram, so computing \(\widehat{\operatorname{HF}}\) is combinatorial.

Some properties:

  1. \(\mathrm{Spin}^{\mathbb{C} }\) structures: we have a decomposition \begin{align*} \operatorname{HF}^-(M) = \bigoplus _{s\in \mathrm{Spin}^{\mathbb{C} }(M) } \operatorname{HF}^-(M, s) \end{align*} which induces \begin{align*} \operatorname{HF}^\star(M) = \bigoplus_{s\in \mathrm{Spin}^{\mathbb{C} }(M)} \operatorname{HF}^\star(M, s) \end{align*} where \(\star = +, -, \infty\).

  2. Maslov grading: For a \(\operatorname{QHS}^3\), \(\operatorname{HF}^-(M)\) is relatively \({\mathbb{Z}}{\hbox{-}}\)graded. The degree of \(u\) is -2, and this grading can be lifted to an absolute \({\mathbb{Q}}{\hbox{-}}\)grading.

  3. There is a SES \begin{align*} 0 \to \operatorname{HF}^-(M, s) \xrightarrow{\cdot u} \operatorname{HF}^-(M, s) \to \widehat{\operatorname{HF}}(M, s) \to 0 .\end{align*}

This yields an exact triangle

Link to Diagram

  1. There is a short exact sequence

\begin{align*} 0 \to \operatorname{HF}^-(M) \to \operatorname{HF}^{\infty }(M) \to \operatorname{HF}^+(M) \to 0 .\end{align*}

yielding an exact triangle

Link to Diagram

  1. \({\mathbb{Z}}/2[u]\) is a PID, so by the structure theorem, any module over it will decompose and we have \begin{align*} \operatorname{HF}^-(M, s) = \bigoplus_{i} {\mathbb{Z}}/2[u] \oplus \bigoplus_j {{\mathbb{Z}}/2[u] \over \left\langle{u^{n_j}}\right\rangle } .\end{align*}

Supposing that \(M \in \operatorname{QHS}^3\), then by Osvath-Szabo, for any \(s\in \mathrm{Spin}^{\mathbb{C} }(M)\) there is exactly one free summand. Let \(d\) be the Maslov grading of the free generator, and \(c_j\) be the grading of the torsion part. We write the \(u{\hbox{-}}\)torsion part as \(\operatorname{HF}_{ \text{red} }(M, s)\).

The Maslov grading of the free summand \(d = d(M, s)\) is referred to as the \(d{\hbox{-}}\)invariant or correction term, and \begin{align*} d(M, s) = \max \left\{{ {\mathsf{gr}\,}( \alpha) {~\mathrel{\Big|}~}\alpha\in \operatorname{HF}^-(M, s),\, u^n \alpha \neq 0 \forall n }\right\} .\end{align*}

The rational homology cobordism group is denoted \begin{align*} \qty{\Theta_{\mathbb{Q}}^3 \mathrel{\vcenter{:}}=\left\{{ M \in \operatorname{QHS}^3 }\right\} /\sim, \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}} \end{align*} where \(M_1\sim M_2\) if and only if they are rationally homology cobordant, i.e. 

  1. There exists an \(W \in {\mathsf{Mfd}}^4\) (connected, oriented) such that \({{\partial}}W = -M_1 {\coprod}M_2\), i.e. \(W\) is a cobordism from \(M_1\) to \(M_2\).

  2. \(H_i(W; {\mathbb{Q}}) = 0\) for \(i=1, 2\), so \(W\) is a rational homology cylinder.

Note that this is only a monoid without the equivalence relation, but this equivalence creates inverses.

Define the \(d{\hbox{-}}\)invariant of \(M\) as \begin{align*} d(M) = \sum_{s\in \mathrm{Spin}^{\mathbb{C} }(M) } d(M, s) .\end{align*}

This induces a homomorphism \begin{align*} d: \Theta_{\mathbb{Q}}^3 \to {\mathbb{Q}} .\end{align*}

21 Tuesday, March 30

21.1 \(L{\hbox{-}}\)spaces

Today: \(L{\hbox{-}}\)spaces and the surgery exact triangle. We’ve been loosely following [OS-1?], references for upcoming topics include [OS-2?] and Jen Hom’s survey [H?].

Recall that we were discussing \(\operatorname{HF}^-(M, s)\) for \(M\in {\mathsf{Mfd}}^3({\mathbb{R}})\) and \(s\) a \(\mathrm{Spin}^{\mathbb{C} }\) structure, and if \(M\in \operatorname{QHS}\) this decomposes as \({\mathbb{Z}}/2[u] \oplus \qty{\bigoplus_i {{\mathbb{Z}}/2[u] \over \left\langle{u^{}n_i}\right\rangle }} \mathrel{\vcenter{:}}={\mathbb{Z}}/2[u] \oplus \operatorname{HF}_{ \text{red} }(M, s)\). The Maslov grading of \(1\) in the first summand is the \(d{\hbox{-}}\)invariant, \(d(M, s)\). If one defines \(d(M) \mathrel{\vcenter{:}}=\sum_{s\in \mathrm{Spin}^{\mathbb{C} }(M)} d(M, s)\), then \(d: \Theta^3_{\mathbb{Q}}\to {\mathbb{Q}}\) is a group homomorphism. We want to talk about \(X\in {\mathsf{Mfd}}^3\) which have the “simplest” Floer theory, in the sense that the torsion summand above vanishes.

A manifold \(M\in\operatorname{QHS}^3\) is an \(L{\hbox{-}}\)space if \(\operatorname{HF}_{ \text{red} }(M, s) = 0\), which happens if and only if \(\operatorname{HF}^-(M, s) = {\mathbb{Z}}/2[u]\).

Recall that there is an exact triangle

Link to Diagram

Since multiplication by \(u\) is injective, we obtain \begin{align*} \widehat{\operatorname{HF}} = \operatorname{im}p \cong {\operatorname{HF}^-(M, s) \over \ker p} \cong {{\mathbb{Z}}/2[u] \over u {\mathbb{Z}}/2[u] } \cong {\mathbb{Z}}/2 .\end{align*}

So \(\operatorname{HF}^-(M, s) \cong {\mathbb{Z}}/2[u] \implies \widehat{\operatorname{HF}}(M, s) \cong {\mathbb{Z}}/2\). Show the converse is also true.

\(M\) is an \(L{\hbox{-}}\)space if and only if \(\widehat{\operatorname{HF}}(M, s) \cong {\mathbb{Z}}/2\) for all \(s\in \mathrm{Spin}^{\mathbb{C} }(M)\). This happens if and only if \({\operatorname{rank}}_{{\mathbb{Z}}[u]} \widehat{\operatorname{HF}}(M, s) = 1\), if and only if \(\# \mathrm{Spin}^{\mathbb{C} }(M) = \# H^2(M) \overset{\mathrm{PD}}{=} \# H^1(M)\), which is finite for \(\operatorname{QHS}\).

Any \(M\in \operatorname{QHS}^3\) is an \(L{\hbox{-}}\)space if and only if \({\operatorname{rank}}_{{\mathbb{Z}}[u]} \widehat{\operatorname{HF}}(M) = \# H^1(M)\).

Note that we’ve proved the forward implication but not the reverse. This is sometimes used as a definition in talks!

Sketch of the proof (\(\impliedby\)): A computation will show that \(\chi \widehat{\operatorname{HF}}(M, s) = \pm 1\) for all \(s\), since the grading can be shifted. This is proved in [OS-2?], and is the main ingredient in this proof. This implies that \({\operatorname{rank}}\widehat{\operatorname{HF}}(M, s) \geq 1\), and so adding all summands yields \({\operatorname{rank}}\widehat{\operatorname{HF}}(M) \geq \# H_1(M)\). This implies that \(\widehat{\operatorname{HF}}(M, s) \cong {\mathbb{Z}}/2\) for all \(s\), making \(M\) an \(L{\hbox{-}}\)space.

Here note that \(C_*\) is \({\mathbb{Z}}/2{\hbox{-}}\)graded, as is \((\widehat{\operatorname{HF}}(M), {{\partial}})\), so we define \(\chi(C_*) = {\operatorname{rank}}C_0 - {\operatorname{rank}}C_1\). Since we have a relative \({\mathbb{Z}}{\hbox{-}}\)grading given by \(\mu\), we get a relative \({\mathbb{Z}}/2{\hbox{-}}\)grading given by \({\mathsf{gr}\,}_{{\mathbb{Z}}/2}(x, y) = {\mathsf{gr}\,}_{\mathbb{Z}}(x, y)\), which gives us \(\chi \widehat{\operatorname{HF}}(M)\) up to sign.

We have seen lens spaces, here’s an example of \(L(2, 3)\):

image_2021-03-30-11-46-39

Here \(\widehat{\operatorname{HF}}L(2, 3) \cong ({\mathbb{Z}}/2)^{\oplus 2}\) and \({\operatorname{rank}}\widehat{\operatorname{HF}}L(2, 3) = 2 = \# H_1(L(2, 3); {\mathbb{Z}}/2)\).

In general, every \(L(p, q)\) is a lens space, hence the name! Note that \(H_1(L(p, q)) \cong {\mathbb{Z}}/p\) is not a \(\operatorname{ZHS}^3\).

A Poincaré homology sphere \(\pm P^3\) (with either the standard orientation or its reverse) will be an \(L{\hbox{-}}\)space.

Poincaré-type conjecture in Heegard Floer homology: the only irreducible \(\operatorname{ZHS}^3\) \(L{\hbox{-}}\)spaces are \(S^3\) and \(\pm P^3\). Still open!

So \(\widehat{\operatorname{HF}}\) can detect these two among all integral homology spheres using \(\widehat{\operatorname{HF}}\).

21.2 Surgery

Let \(K \subseteq M \in {\mathsf{Mfd}}^3\) be a knot, i.e. the image of an embedding \(S^1 \hookrightarrow M\). Remove a tubular neighborhood of \(K\), and set \(X = M\setminus\nu(K)\). Fill in the torus boundary with a solid torus \(S^1 \times{\mathbb{D}}^2\) using a diffeomorphism \begin{align*} \phi: {{\partial}}(S^1 \times{\mathbb{D}}^2) \xrightarrow{\text{diffeo}} {{\partial}}X .\end{align*}

Any surgery will be determined by the image of the red circle \(\gamma \mathrel{\vcenter{:}}={\operatorname{pt}}\times{{\partial}}{\mathbb{D}}^2\) in the following:

image_2021-03-30-12-01-24

So \(\phi\) is determined by \(\phi( \gamma)\), and in fact only depends on its class in homology since \(\pi_1T^2 = H_1T^2\). If \(\phi ( \gamma) = \lambda\), we denote the resulting manifold as \(M_{\lambda}(K)\), the Dehn surgery on \(K\).

A meridian \(\mu\) of \(K\) will be a simple closed curve on \({{\partial}}X\) that bounds a disk in the tubular neighborhood \(\nu(K)\):

image_2021-03-30-12-05-02

Here we orient \(\mu\) on the boundary of this disk.

A longitude will be a nullhomotopic simple closed curve such that \(\#( \mu \cap\lambda) = -1\). For example:

image_2021-03-30-12-07-30?

A knot \(K\) along with a choice of longitude \(\lambda\) is called a framed knot.

This allows us to specify Dehn surgeries by a rational number.

Let \(K\) be a framed knot, then \(M_{p\over q}(K) = M_{\gamma}(K)\) where \([\gamma] = p [\mu] + q[ \lambda]\). We’ll use the notation \(\gamma = p\mu + q\lambda\).

If \(K\) is nullhomologous, for example when \(M = S^3\) since \(H^1S^3 = 0\), there is a canonical choice for \(\lambda\) by assuming that

This longitude is called the Seifert framing.

Show this equivalence, and find the Seifert framing for the trefoil in \(S^3\).

IF \(K\neq U\) and \({p\over q} \neq \infty\), then \(S_{p\over q}^3(K) \neq S^3\).

Can we get everything 3-manifold this way, as surgery on a knot? The answer is no, but yes if you allow links!

There is a bijection \begin{align*} \left\{{\substack{ M\in {\mathsf{Mfd}}^3 {~\mathrel{\Big|}~}\text{closed, oriented, connected} }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Integer $\pm 1$ surgeries on links in $S^3$} }}\right\} \end{align*}

22 Tuesday, April 06

22.1 Surgery Exact Triangle

Recall that for \(K \subseteq M^3\) a knot, surgery on \(K\) involves the following: take a tubular neighborhood of \(K\), \(\operatorname{nd}(K)\), and set \(X \mathrel{\vcenter{:}}= M \setminus\operatorname{nd}(K)\), whose boundary is a solid torus \(S^1 \times{\mathbb{D}}^2\):

image_2021-04-06-11-18-17

Take a basis for its homology:

image_2021-04-06-11-19-05

We can then write any simple closed curve \(\gamma\) as \([\gamma] = p[\mu] + q[\lambda]\), where for shorthand we’ll just write \(\gamma= p \mu + q \lambda\). We’ll refer to the pair \((K, \lambda)\) as a framed knot and the corresponding surgery as \(M_{p\over q}(K)\), which is surgery on \(K\) such that the following curve is \(p\mu + q\lambda\):

image_2021-04-06-11-22-32

We’ll write \(M_{ \lambda}(L) = M_0(K)\) for the surgery that sends \(\lambda\) to this curve instead, since this corresponds to \(p=0, q=1\).

Suppose \(\gamma_0, \gamma_1, \gamma_{\infty }\) are oriented simple closed curves on \({{\partial}}X = M - \operatorname{nd}(K)\) such that \begin{align*} -1 = \#( \gamma_0 \cap\gamma_1) = \#( \gamma_1 \cap\gamma_{\infty }) = \#( \gamma_{\infty } \cap\gamma_0) ,\end{align*} and we have a cyclic ordering

Link to Diagram

Then writing \(M_{i } \mathrel{\vcenter{:}}= M_{ \gamma_{i }}(K)\), the triple \((M_{\infty }, M_0, M_1)\) is a tried of 3-manifold.

Let \(\gamma_{\infty } = \mu, \gamma_0 = \lambda, \gamma_1 = a\mu - b\lambda\), and the punch line is that the third is determined by the other two. What are \(a\) and \(b\)? We have \begin{align*} \#(\gamma_0, \cap\gamma_1) = \#( \lambda, a\mu + b \lambda) = -a \#(\mu \cap\lambda) = (-1) a \implies a = -1 \\ \\ \#( \gamma_1 \cap\gamma_{\infty }) = \#( a \mu + b\lambda, \mu) = b \#(\lambda \cap\mu) = b(1) = b \implies b = -1 .\end{align*} We thus obtain the following picture, which has the curves arrange in a clockwise fashion:

image_2021-04-06-11-32-35

Here we get the triad \((M_{\infty }(K) = M, M_0(K), M_1(K)\).

Show that \((M, M_{-1}(K), M_0(K))\) is also a triad.

Let \(\lambda_{\infty } = p\mu + q\lambda\) and \(\lambda_0 = r\mu + s\lambda\), then \begin{align*} -1 &= \#( \gamma_{\infty } \cap\gamma_0) \\ &= \#( p\mu + q \lambda, r \mu + s \lambda) \\ &= ps \#( \mu \cap\lambda) + qr \# (\lambda \cap\mu) \\ &= -ps + qr \\ &\implies qr - ps = 1 .\end{align*}

Similarly, \begin{align*} -1 = \#( \gamma_0 \cap\gamma_1) &= sa -rb = -1 \#( \gamma_1 \cap\gamma_{\infty }) &= bp - aq = -1 .\end{align*}

Pick a framed knot \(K\) (or really just a fixed longitude), then pick \(\gamma_{\infty } = \mu, \gamma_0 = p\mu + \lambda\). Then \(\gamma_1 = (p+1) \mu + \lambda\), and we get the triad \(M_{\mu}(K) = M, M_{\gamma_0}(K) = M_p(K), M_{p+1}(K)\).

Suppose \((M, M_0, M_1)\) is a triad, then there exist exact triangles:

Link to Diagram

Here exactness means that e.g. \(\ker(\widehat{F}_1 = \operatorname{im}( \widehat{F}_0)\). There is a similar triangle for \(\operatorname{HF}^+\), as well as \(\operatorname{HF}^{\infty }\) and \(\operatorname{HF}^-\), although these are more complicated. However, \(\operatorname{HF}^-\) becomes easier to work with when one is looking at knot invariants instead.

Let \(K\) be the unknot in \(S^3\), and take as before \((S^3, S_0^3(K) = S^1 \times S^2, S_{+1}^3(K) = S^3\). We get the exact triangle

Link to Diagram

Suppose the following is an exact triangle of vector spaces for some cyclic ordering:

Link to Diagram

Then \begin{align*} V_{\infty } = V_0 \oplus V_1 \iff {\operatorname{rank}}V_{\infty } = {\operatorname{rank}}(V_0) + {\operatorname{rank}}(V_1) ,\end{align*} and if \(f_0 = 0\) then \(f_1\) is injective and \(f_{\infty }\) is surjective.

\begin{align*} {\operatorname{rank}}V_{\infty } &\cong {\operatorname{rank}}\ker f_{\infty } \bigoplus {\operatorname{rank}}\operatorname{im}f_{\infty } \\ &= {\operatorname{rank}}\operatorname{im}f_1 \oplus \operatorname{im}f_{\infty } \\ &\leq {\operatorname{rank}}V_1 + {\operatorname{rank}}V_0 .\end{align*} Equality holds if and only if \({\operatorname{rank}}V_1 = {\operatorname{rank}}\operatorname{im}f_1\), which implies \(f_1\) is injective, and similarly \({\operatorname{rank}}V_0 = {\operatorname{rank}}\operatorname{im}f_{\infty } \implies f_{\infty }\) is surjective. These together would imply that \(f_0 = 0\).

For \(K\) the unknot in \(S^3\), take the triad \((S^3, S^3_p(K) = L(p, 1), L(p+1, 1)\). This yields the exact triangle

Link to Diagram

Note that this gives a way to produce \(L{\hbox{-}}\)spaces.

Any \(M\in \operatorname{QHS}^3\) is called an \(L{\hbox{-}}\)space if \begin{align*} {\operatorname{rank}}\widehat{\operatorname{HF}}(M) = {\left\lvert { H_1(M; {\mathbb{Z}}) } \right\rvert} .\end{align*}

If \(p, q\) are coprime then \(\widehat{\operatorname{HF}}L(p, q) = ({\mathbb{Z}}/2)^{\oplus p}\) since there is one \(\mathrm{Spin}^{\mathbb{C} }\) class for each element of \(H^1\). So \({\operatorname{rank}}\widehat{\operatorname{HF}}= p\), and on the other hand, \({\left\lvert {H_1} \right\rvert} = {\left\lvert {{\mathbb{Z}}/p} \right\rvert} = p\).

For any triad \((M_{\infty }, M_0, M_1)\) there exists a cyclic reordering such that \begin{align*} {\left\lvert { H_1(M_{\infty })} \right\rvert} = {\left\lvert { H_1(M_{0})} \right\rvert} = {\left\lvert { H_1(M_{1 })} \right\rvert} ,\end{align*} where we define \begin{align*} {\left\lvert { H_1(M) } \right\rvert} = \begin{cases} \# H_1(M) & \text{ if this is a finite group} \\ 0 & \text{otherwise}. \end{cases} \end{align*}

For the triad \((S^3, L(p, 1), L(p+1, 1))\) we have \begin{align*} p+1 {\left\lvert { H_1(L(p+1, 1) ) } \right\rvert} = {\left\lvert { H_1(S^3) } \right\rvert} + {\left\lvert { H_1( L(p, 1)) } \right\rvert} = 1 + p .\end{align*}

This exercise is useful because it can be used to prove the following:

Suppose \((M, M_0, M_1)\) is a triad with an ordering fixed such that \begin{align*} {\left\lvert { H_1 M } \right\rvert} = {\left\lvert { H_1 M_0} \right\rvert} + {\left\lvert { H_1 M_1 } \right\rvert} .\end{align*} If \(M_0, M_1\) are \(L{\hbox{-}}\)spaces, then \(M\) is also an \(L{\hbox{-}}\)space.

We have the exact triangle

Link to Diagram

Thus \begin{align*} {\operatorname{rank}}\widehat{\operatorname{HF}}M \leq {\operatorname{rank}}\widehat{\operatorname{HF}}M_0 + {\operatorname{rank}}\widehat{\operatorname{HF}}M_1 \\ &\leq {\left\lvert { H_1 M_0} \right\rvert} + {\left\lvert { H_1 M_1 } \right\rvert} \\ = {\left\lvert { H_1 M } \right\rvert} .\end{align*}

In general, \({\left\lvert { H_1 M} \right\rvert} \leq {\operatorname{rank}}\widehat{\operatorname{HF}}M\), so we get an equality \({\operatorname{rank}}\widehat{\operatorname{HF}}M = {\left\lvert { H_1 M } \right\rvert}\).

Let \(K \subseteq S^3\) be a knot, and take the triad \((S^3, S_p^3(K), S_{p+1}^3(K)\). So if \(S_p^3(K)\) is an \(L{\hbox{-}}\)space, so is \(S_{p+1}^3(K)\). Inductively this shows that \(S^3_n(K)\) is an \(L{\hbox{-}}\)space for all \(n\geq p\).

For \(K = T_{p, q}\), the surgery \(S^3_{pq-1}(T_{p, q})\) is a lens space. Thus \(S_n^3(T_{p, q})\) is an \(L{\hbox{-}}\)space for all \(n\geq pq-1\).

23 Surgery Exact Triangle and Knot Diagrams (Thursday, April 15)

Recall: let \((M, M_0, M_1)\) be a triple of 3-manifolds corresponding to a knot \(K \subseteq M\), where \(M_0\) is 0-surgery, \(M_1\) is 1-surgery, and \(M_{\infty}\) is \(\infty{\hbox{-}}\)surgery. Here \(M\) can be chosen such that M

Then there exists an exact triangle:

Link to Diagram

Our goal is to define \(f: \widehat{\operatorname{HF}}(M) \to \widehat{\operatorname{HF}}(M_0)\).

Note that \(M\) admits a Heegard diagram \begin{align*} ( \Sigma_g, \vec{ \alpha} = [\alpha_1, \cdots, \alpha_g], \vec{ \beta} = [\alpha_1, \cdots, \alpha_g] ) \end{align*} such that \((\Sigma_g, \vec \alpha, [\beta_1, \cdots, \beta_{g-1}]\) is a “diagram” for \(M - \operatorname{nd}(K)\). Recall the notion of handlebodies, where each handle bounds a disc:

image_2021-04-15-11-20-29

We can generalize this to a compression body:

image_2021-04-15-11-22-13

This yields a cobordism from \(\Sigma'_{g'} \times\left\{{ 0 }\right\}\) to \(\Sigma_{g' + k}\). So we can write \({{\partial}}C = \Sigma' \times\left\{{ 0 }\right\} {\coprod}\Sigma\). Label the curves bounding the embedded discs as \(\gamma_i\):

image_2021-04-15-11-24-22

Then we can form a diagram \((\Sigma_g, \left\{{ \gamma_1, \cdots, \gamma_k }\right\}\) where \(k\leq g\) will specify the compression body. If these are pairwise disjoint simple closed curves that are linearly independent in \(H_1( \Sigma )\), this will be a compression body from a surface with genus \(g-k\) to \(\Sigma_g\).

image_2021-04-15-11-31-50

In this case, \((\Sigma, \vec \alpha, \left\{{ \beta_1, \cdots, \beta_{g-1} }\right\}\) will be a diagram for \(M\setminus\operatorname{nd}(K)\).

The cobordism from \Sigma to the compressionbody

Consider \(S^3 \setminus\operatorname{nd}(T)\) for \(T\) the trefoil. Behold the beautiful trefoil:

image_2021-04-15-11-37-56

After thickening, we obtain the following:

image_2021-04-15-11-39-41

We can push the top down:

image_2021-04-15-11-40-24

And wrap part of it around:

image_2021-04-15-11-43-42

We can keep moving this to undo the crossing:

image_2021-04-15-11-45-58
image_2021-04-15-11-47-36

So the blue curve gets complicated, but the neighborhood of \(T\) is a genus 2 surface, since the outer two circles bound discs. So in summary, we have the following process:

image_2021-04-15-11-50-12

We can represent this with a planar picture:

image_2021-04-15-11-56-01

Following the longitude, we obtain:

image_2021-04-15-12-15-41

Here \(\lambda\) has been wrapped twice, and to do \(n{\hbox{-}}\)surgery, we wrap \(n\) times.

image_2021-04-15-12-16-46

Draw a diagram for \(S_n^3\) (the figure eight).

24 Bibliography


  1. See Sarkour-Wang↩︎

  2. This is the strongest variant.↩︎

  3. What does “detect” mean? This is slightly technical.↩︎

  4. Note that wedging to a nontrivial top form is equivalent to being nowhere integrable here.↩︎

  5. These are referred to as Seifert surfaces.↩︎

  6. See Akram’s notes for details.↩︎