--- title: 'Section 8.6 created: 2023-04-11T23:30 updated: 2023-04-11T23:30 --- [2020 Fall Floer MOC](2020%20Fall%20Floer%20MOC.md) Tags: #geomtop/Floer-theory ## Outline Two Goals: 1. Critical points are discrete and regular points are open/dense. 2. The continuation principle (used elsewhere, see diagram later) - Idea: For $\CC$, a holomorphic function with all derivatives vanishing on a line is identically zero. ## Outline of Statements \begin{center} \includegraphics[width = 0.4\textwidth]{figures/image_2020-05-11-00-55-25.png} \end{center} \begin{center} \includegraphics[width = \textwidth]{figures/image_2020-05-13-18-45-19.png} \end{center} What we'll try to prove: - 8.6.1: Reduction to Cauchy-Riemann equations on $\RR^2$ (short) - 8.6.3 (Partial): $R(v)$ is open. \newpage Statements of "big" theorems for the chapter, in reverse order of implication: - 8.1.5: $(d\mcf)_u$ is a Fredholm operator of index $\mu(x) - \mu(y)$. - 8.1.4: $\Gamma: W^{1, p} \cross C_\eps^\infty \to L^p$ has a continuous right-inverse and is surjective - 8.1.3: $\mcz(x, y, J)$ is a Banach manifold - 8.1.1: For $h\in \mch_{\text{reg}}, H_0 + h$ is nondegenerate and $(d\mcf)_u$ is surjective for every $u\in \mcm(H_0 + h, J)$. - 8.1.2: For $h \in \mch_{\text{reg}}$ and all contractible orbits $x, y$ of $H_0$, $\mcm(x, y, H_0 + h)$ is a manifold of dimension $\mu(x) - \mu(y)$. ## Notation - The Floer equation and its linearization: \[ \mcf(u) &= \dd{u}{s} + J \dd{u}{t} + \grad_u(H) = 0 \\ \qty{d\mcf}_u(Y) &= \dd{Y}{s} + J_0 \dd{Y}{t} + S\cdot Y \\ &Y\in u^* TW,~ S \in C^\infty(\RR\cross S^1; \endo(\RR^{2n})) .\] - $X(t, u): S^1\cross W \to W$ is a time-dependent periodic vector field on $\RR^{2n}$, $J$ an almost-complex structure, both smooth - $u \in C^\infty(\RR \cross S^1; W)$ is a solution to the equation $$\dd{u}{s} + J(t, u) \qty{\dd{u}{t} - X(t, u)} = 0$$ > Note: not sure why we've replaced $\grad_u(H)$ with $-J(t, u) \cdot X(t, u)$ in the Floer equation. - $C(u)$ the set of critical points and $R(u)$ the set of regular points of $u$: \[ (s_0, t_0) \in C(u) \subseteq \RR\cross S^1 &\iff \dd{u}{s}(s_0, t_0) = 0 \\ (s_0, t_0) \in R(u) \subset \RR \cross S^1 &\iff (s_0, t_0) \not\in C(u) ~\& ~s\neq s_0 \implies u(s_0, t_0) \neq u(s, t_0) .\] \newpage ## Goal 1: Discrete Critical Points and Dense Regular Points Goal 1: prove the following theorem Theorem (8.5.4) : \hfill 1. $C(u)$ is discrete and 2. $R(u) \injects \RR \cross S^1$ is open and dense. Outline of the proof: \normalsize - Prove 8.6.1: Reduction to CR - (direct, short) which transforms the Floer(?) equation $$\dd{u}{s} + J(t, u) \qty{\dd{u}{t} - X(t, u)} = 0 \qtext{where} u \in C^\infty(\RR\cross S^1; W)$$ to a Cauchy-Riemann equation on $\RR^2$: $$\dd{v}{s} + J \dd{v}{t} = 0 \qtext{where} v\in C^\infty(\RR^2; W)$$ - Reduce 8.5.4 (Discrete/Open/Dense) to two statements - 8.6.2: $C(v)$ (and thus $C(u)$) is discrete Proved later using 8.6.8: *Similarity Principle*. - 8.6.3 (Injectivity): If $v$ is a smooth periodic solution of CR with $\dd{v}{s}\neq 0$ then $R(v) \subseteq \RR^2$ is open and dense. - Prove 8.6.3 (Injectivity) - Show open (easier) - Show dense (delicate!) - Prove 8.6.8: Similarity Principle - Use similarity principle to prove 8.6.6: Continuation Principle. Yields theorem. \begin{center} \includegraphics[width = 0.8\textwidth]{figures/image_2020-05-13-18-45-19.png} \end{center} \newpage \large ## 8.6.1: Transform to Cauchy-Riemann Proposition (8.6.1, Transform to CR-equation on R2) : If $u$ is a solution to the following equation: \[ \frac{\partial u}{\partial s}+J(t, u)\left(\frac{\partial u}{\partial t}-X(t, u)\right)=0 .\] Then there exists - An almost complex structure $J_1$ - A diffeomorphism $\varphi$ on $W$ ? - A map $v \in C^\infty(\RR^2; W)$ satisfying $$ \dd{v}{s} + J_1(v) \dd{v}{t} = 0 $$ where 1. $v(s, t+1) = \varphi(v(s, t))$ 2. $C(u) = C(v)$, i.e. $u, v$ have the same critical points 3. $R(u) = R(v)$. Proof - Recall the vector field was defined as $X(t, u): S^1\cross W \to W$. - Since $W\cross S^1$ is compact, the flow $\psi_t$ of $X_t$ is defined on all of $W$ - We thus have a map $\psi_t: W\to W$ such that $$\dd{}{t}\psi_t = X_t \circ \psi_t, \quad\quad \psi_0 = \id$$ - Define the (important!) map $$v(s, t) \definedas \qty{\psi_t\inv \circ u}(s, t)$$ \newpage - Since $W\cross S^1$ is compact, the flow $\psi_t$ of $X_t$ is defined on all of $W$ - We thus have a map $\psi_t: W\to W$ such that $$\dd{}{t}\psi_t = X_t \circ \psi_t, \quad\quad \psi_0 = \id$$ - Define the (important!) map $$v(s, t) \definedas \qty{\psi_t\inv \circ u}(s, t)$$ - We can then compute \[ \dd{u}{s} &= \color{blue} \qty{d\psi_t} \qty{\dd{v}{s}} \\ \dd{u}{t} &= \color{orange} \qty{d\psi_t} \qty{\dd{v}{t}} + X_t(u) .\] - Attempt at explanation: rearrange, use chain rule, and known derivative of $\psi_t$: \[ u(s, t) = \qty{\psi_t \circ v}(s, t) &\implies \dd{u}{s}(s, t) = \dd{\psi_t}{s}\qty{v(s, t)} \cdot \dd{v}{s}(s, t) \\ \\ {}_? &\implies \dd{u}{s} = \qty{d \psi_t} \cdot \qty{\dd{v}{s}} \] and \[ \dd{u}{t}(s, t) &= \dd{\psi_t}{t}\qty{v(s, t)} \cdot \dd{v}{t}(s, t) \\ &= \qty{X_t \circ \psi_t}(v(s, t)) \cdot \dd{v}{t}{(s, t)} \\ &= \qty{X_t \circ \color{red}\psi_t\circ v}(s, t) \cdot \dd{v}{t}{(s, t)} \\ &= X_t(u(s, t)) \cdot \dd{v}{t}{(s, t)} \\ &= X_t(u) \qty{\dd{v}{t}} \cdots ???? .\] > Note sure how to relate partial derivatives $\dd{}{\wait} \psi_t$ to differential $d\psi_t$. > Not sure why we're picking up *addition* in the $t$ derivative. \newpage - Given that result, we can compute, \[ 0 &= \frac{\partial u}{\partial s}+J\left(\frac{\partial u}{\partial t}-X_{t}(u)\right) &\text{since $u$ is a solution} \\ &= {\color{blue} \dd{u}{s}} + J {\color{orange} \dd{u}{t}} - J X_t(u) &\text{expanding terms} \\ &= \qty{\color{blue} \qty{d\psi_t} \qty{\dd{v}{s}} } + J \qty{\color{orange}\qty{d\psi_t} \qty{\dd{v}{t}} + X_t(u) } - J X_t(u) &\text{by substitution} \\ &= \left(d \psi_{t}\right)\left(\frac{\partial v}{\partial s}\right)+J(u)\left(d \psi_{t}\right)\left(\frac{\partial v}{\partial t}\right) &\text{cancelling} \\ &= \left(d \psi_{t}\right)\left(\frac{\partial v}{\partial s} + {\color{red} \left( d \psi_{t}\right)^{-1} J(u)\left(d \psi_{t}\right)\left(\frac{\partial v}{\partial t}\right) } \right) &\text{collecting terms}\\ &\definedas \left(d \psi_{t}\right)\left(\frac{\partial v}{\partial s} + {\color{red} \psi_t^* J(v) } \right) &\text{by definition} .\] - Conclude that $v$ is a solution of \[ \frac{\partial v}{\partial s}+\psi_{t}^{\star} J(v) \frac{\partial v}{\partial t}=0 .\] - Set $\varphi \definedas \psi_1$ and $J_1(v) \definedas \psi_1^* J(v)$ to obtain \[ \frac{\partial v}{\partial s}+ J_1(v) \frac{\partial v}{\partial t}=0 \] of which $v$ is still a solution - Property 1, Periodicity: attempt to check directly, using $\psi_{t+1} = \psi_t \circ \psi_1$: \[ v(s, t+1) &\definedas (\psi_t\inv \circ u)\qty{s, t + 1} \\ {}_?&= \qty{ \psi_1 \circ \psi_t\inv \circ u}(s, t) \\ &= \psi_1 \qty{ v(s, t)} \\ &\definedas \varphi(v(s,t)) .\] > Just a guess. If the 1-parameter group is commutative then proving $\varphi(v(s, t-1)) = v(s, t)$ might work. \newpage - Recall definition of $v$: $$v(s, t) \definedas \psi_t\inv \qty{ u(s, t) }$$ - Verifying that $C(v) = C(u)$: not spelled out. Property of flow? - Need to check that $$\dd{u}{s}(s_0, t_0) = 0 \implies \dd{v}{s}(s_0, t_0) = 0$$ where $$\dd{u}{s} = \qty{d\psi_t}\qty{\dd{v}{s}}$$ - How: ? - Verifying that $R(v) = R(u)$ - Need to check that for $(s_0, t_0) \not \in C(u)$ and $s\neq s_0$ we have $$u(s_0,t_0) \neq u(s, t_0) \implies v(s_0, t_0) \neq v(s, t_0)$$ - Follows directly: \[ v(s_0, t_0) \neq v(s, t_0) &\iff \psi_t\inv(u(s_0, t_0)) \neq \psi_t\inv(u(s, t_0)) \quad\text{by definition}\\ &\iff \qty{\psi_t \circ \psi_t\inv} (u(s_0, t_0)) \neq \qty{\psi_t \circ \psi_t\inv} (u(s, t_0)) \quad\text{injectivity of }\psi_t \\ &\iff u(s_0, t_0) \neq u(s, t_0) .\] $\qed$ \newpage ## Splitting the Main Theorem - The main theorem is equivalent to two upcoming statements Proposition (8.6.2: Statement 1, Critical Points are Discrete) : Let $z = s + it$ where $(s, t) \in \RR^1 \cross S^1$ (?). There exists a constant $\delta > 0$ such that \[ 0 < \abs{z} < \delta \implies \qty{dv}_z \neq 0 .\] Proof : Postponed to p.264 because it relies on the 8.6.8 (Similarity Principle). For the second statement, we set up some notation/definitions. - $v\in C^\infty(\RR^2; W)$ is a solution satisfying \[ \dd{v}{s} + J_1(v) \dd{v}{t} &= 0 \\ \\ v(s, t+1) &= \varphi(v(s, t)) \\ \\ C(v)=C(u),&~R(v) = R(u) .\] - The **regular points** are given by \[ R(v) = \theset{(s,t) \in \RR^2 \suchthat \dd{v}{s}(s, t) \neq 0, \quad v(s, t) \neq x^\pm(t),\quad v(s, t) \not\in v(\RR\setminus\theset{s}, t) } .\] > Note: last condition should be equivalent to $s\neq s' \implies v(s, t) \neq v(s', t)$. > For reference, also equivalent to $v(s, t) = v(s', t) \implies s = s'$. \begin{center} \includegraphics[width = 0.5\textwidth]{figures/image_2020-05-13-23-51-21.png} \end{center} \newpage - **Multiple points** are defined as follows: - Set $\bar \RR = \RR \disjoint \theset{\pm \infty}$ - Extend $v: \RR^2 \to W$ to \[ v: \bar \RR \cross \RR &\to W \\ v(\pm \infty, t) &= x^\pm(t) .\] - Define the set of *multiple points* as $$ M(s, t) \definedas \theset{ (s', t) \in \RR^2 \suchthat s\neq s'\in \bar \RR, \quad v(s', t) = v(s, t) } $$ > Note that the same $t$ is used throughout. - Recast definition of $R(v)$ as "points in the complement of $C(v)$ that do not admit multiples". - Potentially incorrect formulation: \[ R(v) = C(v)^c \intersect_{(s, t) \in \bar \RR \cross \RR} M(s, t)^c .\] - Points to remember: - Condition 1, Nonzero $s\dash$derivative. - Condition 2, Proposition (8.6.3: Regular Points Open/Dense, \"Injectivity\") : Let $v$ be a smooth solution of the Cauchy-Riemann equation, then \[ \begin{rcases} v(s, t+1) = \varphi(v(s, t)) \\ \dd{v}{s} \not\equiv 0 \end{rcases} \implies R(v) \subseteq \RR^2 \quad\text{is open and dense} .\] Proof (Long) : Splits into two parts: - Show $R(v)$ is open (easy) - Show $R(v)$ is dense (delicate) ## Regular Points Are Open Proving the first part: $R(v)$ is open. - Want to show $R(v)^c$ is closed. - Toward a contradiction, suppose otherwise: $R(v)^c$ is *open*. - Use limit point definition: $R(v)^c$ is closed $\iff$ it contains all of its limit points - So $R(v)^c$ does *not* contain one of its limit points - Produces a sequence $$R(v)^c \supseteq \theset{(s_n, t_n)}_{n\in \NN} \converges{n\to\infty}\to (s, t) \in R(v)$$ \newpage ## Sequence is Bounded - The first two conditions defining $R(v)$ are open conditions: - The two conditions: \[ \dd{v}{s}(s, t) &\neq 0 &&\text{Condition 1}\\ v(s, t) &\neq x^\pm(t) &&\text{Condition 2} .\] - Thus for $N\gg 1$ we have $$n\geq N \implies \dd{v}{s}(s_n, t_n) &\neq 0, \quad\quad v(s_n, t_n) \neq x^\pm(t)$$ > Note: what is an "open condition"? - But $(s_n, t_n) \not\in R(v)$ for such $n$, so they must fail the last condition: injectivity - Third condition: $$ s \neq s' \implies v(s, t) \neq v(s', t) $$ - Failing this conditions means: $$ \forall n> N,~ \exists s_n' \in \RR \text{ s.t. } s'_n \neq s_n \qtext{and} v(s_n, t_n) = v(s_n', t_n). $$ - Produces a sequence $\theset{s_n'}_{n\in \NN}$, want to show it is bounded. - Toward a contradiction, suppose not, then there is a subsequence $$\theset{s_{n_k}}_{n_k\in \NN} \converges{n_k \to\infty}\to \pm\infty.$$ - This implies \[ v(s, t) &= \lim_{n_k \to \infty} v(s_{n_k}', t_{n_k}') \quad\quad\text{using continuity of } v\\ &= v(\pm \infty, t) \\ &\definedas x^\pm (t) .\] - Why? By definition, precisely because we extended $v$ by setting $v(\pm \infty, t) = x^{\pm}(t)$. - But condition 2 for points in $R(v)$ says $v(s, t) \neq x^\pm(t)$, so this contradicts $(s, t) \in R(v)$. So the sequence is bounded. \newpage ## Reaching a Contradiction - Sequence is bounded, so apply Bolzano-Weierstrass to extract a convergent subsequence converging to some limit: \[ \theset{s_{n_j}'} \converges{n_j\to\infty}\to s' .\] - Use the fact that injectivity failed: \[ \forall n,~ s'_n \neq s_n \qtext{and} v(s_n, t_n) &= v(s'_n, t_n) \\ \implies \lim_{n_k\to \infty} v(s_n, t_n) &= \lim_{n_k\to\infty} v(s'_n, t'_n) \\ \iff v(s, t)&= v(s', t) \hspace{6em}\text{using continuity} .\] - Use the fact that because $(s, t) \in R(v)$ we must have $$s = s'.$$ - *(Minor technical point)* Now have $\theset{s_{n_j}'}_{n_j \in \NN} \to s'$ and $\theset{s_n}_{n\in \NN} \to s$ - Since the latter sequence is convergent, every subsequence converges to the same limit, so $\theset{s_{n_j}}_{n_j \in \NN} \to s$. - Again using failed injectivity, i.e. \[ v(s, t) &= v(s', t) \\ \implies v(s, t) - v(s', t) &= 0 .\] we have \[ s_{n_j}' \neq s_{n_j} \qtext{and} v(s_{n_j}, t_{n_j}) &= v(s_{n_j}', t_{n_j}) \\ .\] - In the last step, we do have equality in the limit, $s= s'$, and $\forall n_j$, \[ v(s_{n_j}, t_{n_j}) - v(s_{n_j}', t_{n_j}) &= 0, \\ s_{n_j} - s_{n_j}' &\neq 0 \] thus \[ \dd{v}{s}(s, t) = \lim_{n_j\to\infty} {v(s_{n_j}, t) - v(s_{n_j}', t) \over s_{n_j} - s_{n_j}' } = 0 .\] - But $(s, t) \in R(v)$ and so this contradicts Condition 1. This proves that $R(v)$ is open. $\qed$ \newpage Lemma 8.6.4 (Failure of Injectivity) : For every $r> 0$ there exists a $\delta >0$ such that \[ \abs{t-t_0},~\abs{s-s_0} < \delta \implies \exists s' \in B_r(s_j) \text{ s.t. } v(s, t) = v(s', t) .\] Proof : Short, include. Lemma (8.6.5: Unique Solutions in a Small Ball) : Let $v_1, v_2$ be two solutions of the CR-equation with $X_t \equiv 0$ on $B_\eps(0)$, $v_1(0, 0) = v_2(0, 0)$. \ Suppose that $\qty{dv_1}_0, \qty{dv_2}_0 \neq 0$. Also suppose \[ \forall \eps ~\exists \delta \text{ s.t. } \forall (s, t) \in B_{\delta}(0), ~\exists s'\in \RR \begin{cases} (s', t) \in B_\eps(0) \\ v_1(s, t) = v_2(s', t) \end{cases} .\] Then \[ z\in B_\eps(0) \implies v_1(s, t) = v_2(s, t) .\] Take perturbed CR equation: \[ \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S \cdot Y=0 .\] Fix $S\in C^\infty(\RR^2; \endo(\RR^{2n}))$ Lemma (Similarity Principle) : Let $Y \in C^\infty(B_\eps; \CC^n)$ be a solution to the perturbed CR equation and let $p>2$. Then there exists $0 < \delta < \eps$ and a map $A\in W^{1, p}(B_\delta, \gl(\RR^{2n}))$ and a holomorphic map $$\sigma: B_\delta \to \CC^n$$ such that \[ \forall (s, t) \in B_\delta \quad Y(s, t) = A(s, t) ~ \sigma(s+it) \qtext{and} J_0 A(s, t) = A(s, t) J_0 .\] Use continuation principle to finish proofs of many old theorems/lemmas. Theorem (8.6.11, Essential property of bar del) : For every $p>1$, the following operator is surjective and Fredholm: \[ \bar\del: W^{1, p}\qty{S^2; \CC^n} \to L^p(\Lambda^{0, 1} T^* S^2 \tensor \CC^n) .\] Lead up to the proof of 8.1.5 in Section 8.7 # Goal 2: Continuation Principle Goal 2: prove a continuation principle: Proposition (8.6.6, Continuation Principle) : On an open $U \subset \RR^2$, let $Y$ be a solution to the perturbed CR equation \[ \dd{Y}{s} + J_0 \dd{Y}{t} + S\cdot Y = 0 \] where $J_0$ is the standard complex structure on $\RR^{2n}$ and $S\in C^\infty(\RR^2, \endo(\RR^{2n}))$. \ Say that $f$ has an *infinite-order zero* at $z_0$ iff \[ \forall k\geq 0,\quad \sup_{\abs{z-z_0}\leq t} {\abs{f(z)} \over r^k} \converges{r\to 0}\to 0 .\] > For $f$ smooth, equivalently $f^{(k)}(z_0) = 0$ for all $k$. Then the set \[ C \definedas \theset{(s, t) \in U \suchthat Y \text{ has an infinite order zero at } (s, t) } \] is clopen. In particular, if $U$ is connected and $Y = 0$ on some nonempty $V \subset U$, then $Y\equiv 0$. Proposition (8.1.4, \"Transversality Property\") : \hfill Define \[ \mathcal{Z}(x, y, J)\definedas \left\{\left(u, H_{0}+h\right) | h \in \mathcal{C}_{\varepsilon}^{\infty}\left(H_{0}\right) \text { and } u \in \mathcal{M}(x, y, J, H)\right\} .\] If $(u, H_0 + h) \in \mathcal{Z}(x, y)$ then the following map admits a continuous right-inverse and is surjective: \[ \Gamma: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \times \mathcal{C}_{\varepsilon}^{\infty}\left(H_{0}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ (Y, h) & \longmapsto\left(d \mathcal{F}^{H_0 + h}\right)_{u}(Y)+\operatorname{grad}_{u} h \] where $\mcf^{H_0 + h}$ is the Floer operator corresponding to $H_+ h$. > Used to show (via the implicit function theorem) that $\mcz(x, y, J)$ is a Banach manifold when $x\neq y$.