--- aspectratio: 169 author: D. Zack Garza colortheme: default date: April 2020 fontfamily: noto-sans fontsize: 10pt header-includes: - | ```{=tex} \usepackage{cmbright} ``` subtitle: Section 8.4 Follow-Up theme: Berkeley title: Linearization Continued created: 2023-04-11T23:30 updated: 2023-04-11T23:31 --- ```{=tex} \usepackage{cmbright} ``` [Floer%20Reading%20Group%20Fall%202020.html](2020%20Fall%20Floer%20MOC.md) # Review ## Definitions - The Floer equation is given by ` \begin{align*} \frac{\partial u}{\partial s}+J(u) \frac{\partial u}{\partial t}+\operatorname{grad} H_{t}(u)=0 .\end{align*} `{=html} - We fixed a solution and lifted it to a sphere: ` \begin{align*} u \in C^\infty(S^1\times{\mathbb{R}}; W) \quad\mapsto\quad \tilde u \in C^\infty(S^2; W) \end{align*} `{=html} - We use the assumption: > For every $w\in C^\infty(S^2, W)$ there exists a symplectic trivialization of the fiber bundle $w^* TW$, i.e. ${\left\langle {c_1(TW)},~{\pi_2(W)} \right\rangle} = 0$ where $c_1$ denotes the first Chern class of the bundle $TW$. - We use this trivialize the pullback $\tilde u ^* TW$ to obtain an orthonormal unitary frame ` \begin{align*}\left\{{Z_i}\right\}_{i=1}^{2n} \subset T_{u(s, t)} W\end{align*} `{=html} ## The Frame - We used the chosen frame $\left\{{Z_i}\right\}$ to define a chart centered at $u$ of ${\mathcal{P}}^{1, p}(x, y)$ given by ` \begin{align*} \iota: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow \mathcal{P}^{1, p}(x, y) \\ Y = \left(y_{1}, \dots, y_{2 n}\right) & \longmapsto \exp _{u}\left(\sum y_{i} Z_{i}\right) .\end{align*} `{=html} - We regard $Y(s, t)$ as a tangent vector to $W$ in some Euclidean embedding. ## Charts - We seek to compute the composite map in charts: ```{=tex} \begin{center} \includegraphics[height= 0.5\textheight]{figures/image_2020-04-16-01-11-02.png} \end{center} ``` ## Add a Tangent ```{=tex} \begin{center} \begin{tikzcd}[ampersand replacement=\&, matrix scale=0.002] {\mathcal{F}}(u) = \& \frac{\partial u}{\partial s} \& + J(u) \frac{\partial u}{\partial t} \& - J(u)X_{t}(u) \\ \mathcal{F}(u+Y) = \& \frac{\partial(u+Y)}{\partial s} \&+ J(u+Y) \frac{\partial(u+Y)}{\partial t} \&- J(u+Y) X_{t}(u+Y) \end{tikzcd} \end{center} ``` Extract the part that is linear in $Y$ and collect terms: ` \begin{align*} &(d \mathcal{F})_{u}(Y) \\ & = {\color{red}\frac{\partial Y}{\partial s}} +(d J)_{u}(Y) \frac{\partial u}{\partial t} + {\color{red} J(u) \frac{\partial Y}{\partial t}} -(d J)_{u}(Y) X_{t}-J(u)\left(d X_{t}\right)_{u}(Y) \\ & = \qty{ {\color{red}\frac{\partial Y}{\partial s}} + {\color{red} J(u) \frac{\partial Y}{\partial t}}} \\ & \hspace{2em} + \qty{ (d J)_{u}(Y) \frac{\partial u}{\partial t} - (d J)_{u}(Y) X_{t}-J(u)\left(d X_{t}\right)_{u}(Y)} \\ .\end{align*} `{=html} # 8.4.1: Leibniz Rule ## Leibniz Rule Recall the Leibniz rule ` \begin{align*}(dJ)(Y) \cdot v = d(Jv)(Y) - J dv(Y)\end{align*} `{=html} ` \begin{align*} (d \mathcal{F})_{u}(Y) & = \qty{ {\color{red}\frac{\partial Y}{\partial s}} + {\color{red} J(u) \frac{\partial Y}{\partial t}}} \\ & \quad + \qty{ (d J)_{u}(Y) \frac{\partial u}{\partial t} - (d J)_{u}(Y) X_{t}-J(u)\left(d X_{t}\right)_{u}(Y)} \\ &= \sum_{i=1}^{2n} \qty{ {\frac{\partial y_i}{\partial s}\,} Z_i + {\frac{\partial y_i}{\partial t}\,} J(u) Z_i} \\ & + \sum_{i=1}^{2n} y_i \Bigg( {\frac{\partial Z_i}{\partial s}\,} + J(u) {\frac{\partial Z_i}{\partial t}\,} + (dJ)_u (Z_i) {\frac{\partial u}{\partial t}\,} \\ &\hspace{5em}- J(u) (dX_t)_u Z_i - (dJ)_u (Z_i) X_t \Bigg) .\end{align*} `{=html} Use the fact that this is $O_1 + O_0$ in $Y$. ## Order 1 Study $O_1$ first, which (claim) reduces to ` \begin{align*} O_1 = \sum_{i=1}^{2n} \qty{{\frac{\partial y_i}{\partial s}\,} + J_0 {\frac{\partial y_i}{\partial t}\,} }Z_i = \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu(y_1, \cdots, y_{2n}) .\end{align*} `{=html} where $J_0$ is the standard complex structure on ${\mathbb{R}}^{2n} = {\mathbb{C}}^n$ Use this to write ` \begin{align*} (d{\mathcal{F}})_u &= {\color{red} \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5muY} + SY \end{align*} `{=html} where $S\in C^\infty({\mathbb{R}}\times S^1; \endo({\mathbb{R}}^n))$ is a linear operator of order 0. # Order 0 Part is Symmetric in the Limit ## 8.4.4: Order 0 Part is Symmetric in the Limit Theorem (8.4.4, CR + Symmetric in the Limit) : If $u$ solves Floer's equation, then ` \begin{align*} (d{\mathcal{F}})_u = \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu+ S( s, t) \end{align*} `{=html} where 1. $S$ is linear 2. $S$ tends to a symmetric operator as $s\to \pm \infty$, and 3. We have the limiting behavior ` \begin{align*}{\frac{\partial S}{\partial s}\,}(s, t) \overset{s\to\pm\infty}\to 0 {\quad \operatorname{uniformly in $t$} \quad}\end{align*} `{=html} ## Proof Collect terms in the order zero part: ` \begin{align*} O_0 = S(y_1, \cdots , y_{2n}) &= \sum_{i=1}^{2n} y_i \Bigg[ {\color{orange} {\frac{\partial Z_i}{\partial s}\,}} + J(u) {\frac{\partial Z_i}{\partial t}\,} + {\color{blue} (dJ)_u (Z_i) {\frac{\partial u}{\partial t}\,}} \\ &\hspace{5em} - J(u) (dX_t)_u Z_i - {\color{blue} (dJ)_u (Z_i) X_t} \Bigg] \\ &= \sum_{i=1}^{2n} y_i \Bigg[ {\color{orange} {\frac{\partial Z_i}{\partial s}\,} } + { \color{blue} (dJ)_u (Z_i) \qty{ {\frac{\partial u}{\partial t}\,} - (Z_i) X_t} } \\ &\hspace{5em} + {\color{red} J(u) {\frac{\partial Z_i}{\partial t}\,} - J(u) (dX_t)_u Z_i} \Bigg] .\end{align*} `{=html} - Claim: the terms in blue and orange vanish in the limit $s \to \pm \infty$, so it suffices to prove that the red term limits to a symmetric operator. ## Proof: Blue Term Vanishes ` \begin{align*} { \color{blue} (dJ)_u (Z_i) \qty{ {\frac{\partial u}{\partial t}\,} - (Z_i) X_t} } \to 0 \end{align*} `{=html} The term in blue vanishes: since $u$ is a solution and ` \begin{align*}{\frac{\partial u}{\partial s}\,} \overset{s\to \pm \infty}\to 0 {\quad \operatorname{uniformly} \quad}\end{align*} `{=html} as do its derivatives, we have ` \begin{align*} \qty{ {\frac{\partial u}{\partial t}\,} - X_t(u)} \overset{s\to \pm\infty}\to 0 \end{align*} `{=html} > This seems to be the full argument for the blue term. ## Proof: Orange Term Vanishes (1 and 3) ` \begin{align*} {\color{orange} {\frac{\partial Z_i}{\partial s}\,} } \overset{s\to \pm \infty}\to 0 \end{align*} `{=html} Follows since the frame $Z_i$ was chosen such that ` \begin{align*}{\frac{\partial }{\partial s}\,}, \quad \frac{{\partial}^2}{{\partial}s^2}, \quad \frac{{\partial}^2}{{\partial}s~{\partial}t} \quad \curvearrowright Z_i \overset{s\to \pm\infty}\to 0\quad \text{for each } i\end{align*} `{=html} This also implies ` \begin{align*} {\frac{\partial S}{\partial s}\,} \overset{s\to \pm\infty}\to 0.\end{align*} `{=html} > This shows parts (1) and (3) of the theorem: linearity and limits to zero uniformly in $t$? ## Proof: Symmetry Write the remaining red term as ` \begin{align*} A \coloneqq A\left(y_{1}, \ldots, y_{2 n}\right)=\sum y_{i}\left( {\color{red} J(u) \frac{\partial Z_{i}}{\partial t}-J(u)\left(d X_{t}\right)_{u}\left(Z_{i}\right) } \right) .\end{align*} `{=html} Extract the $j$th component: ` \begin{align*} A_j = \sum y_{i}\left\langle J(u) \frac{\partial Z_{i}}{\partial t}-J(u)\left(d X_{t}\right)_{u}\left(Z_{i}\right), \quad Z_{j}\right\rangle .\end{align*} `{=html} We'll show that ` \begin{align*} \lim _{s \rightarrow \pm \infty} &\left\langle J(u) \frac{\partial Z_{\color{red} i}}{\partial t}-J(u)\left(d X_{t}\right)_{u}\left(Z_{\color{red} i}\right), Z_{\color{red} j}\right\rangle \\ - &\left\langle J(u) \frac{\partial Z_{\color{red} j}}{\partial t}-J(u)\left(d X_{t}\right)_{u} Z_{\color{red}j}, ~~~Z_{\color{red} i}\right\rangle=0 .\end{align*} `{=html} ## Proof: Symmetry Use the fact that the frame $\left\{{Z_i}\right\}$ is unitary: ` \begin{align*} 0 &=\frac{\partial}{\partial t}\left\langle J(u) Z_{i}, Z_{j}\right\rangle \\ &=\left\langle(d J)_{u}\left(\frac{\partial u}{\partial t}\right) Z_{i}, Z_{j}\right\rangle+\left\langle J(u) \frac{\partial Z_{i}}{\partial t}, Z_{j}\right\rangle+\left\langle {\color{red} J(u) } Z_{i}, \frac{\partial Z_{j}}{\partial t}\right\rangle \\ &= \left\langle(d J)_{u}\left(\frac{\partial u}{\partial t}\right) Z_{i}, Z_{j}\right\rangle+\left\langle J(u) \frac{\partial Z_{i}}{\partial t}, Z_{j}\right\rangle-\left\langle Z_{i}, {\color{red} J(u)} \frac{\partial Z_{j}}{\partial t}\right\rangle .\end{align*} `{=html} ## Proof: Symmetry Therefore it suffices to show ` \begin{align*} -&\left\langle J(u)\left(d X_{t}\right)_{u}\left(Z_{i}\right), \quad Z_{j}\right\rangle \\ +&\left\langle J(u)\left(d X_{t}\right)_{u}\left(Z_{j}\right), \quad Z_{i}\right\rangle \\ -&\left\langle(d J)_{u}\left(\frac{\partial u}{\partial t}\right) Z_{i}, \quad Z_{j}\right\rangle \\ \\ &\overset{s\to \pm\infty}\to 0 .\end{align*} `{=html} ## Proof: Symmetry Using the fact that ` \begin{align*} \qty{ {\frac{\partial u}{\partial t}\,} - X_t(u)} \overset{s\to \pm\infty}\to 0 \end{align*} `{=html} we can equivalently show ` \begin{align*} &- \left\langle J(u)\left(d X_{t}\right)_{u}\left(Z_{i}\right), \quad Z_{j}\right\rangle \\ &+ \left\langle J(u)\left(d X_{t}\right)_{u} \left(Z_{j}\right), \quad Z_{i}\right\rangle \\ &- \left\langle(d J)_{u}\left(X_{t}\right) Z_{i}, \hspace{2.5em} Z_{j}\right\rangle \\ \\ &\overset{s\to \pm\infty}\to 0 \end{align*} `{=html} > For a fixed $(s, t)$, this expression only depends on $Z_i$ at the point $u(s, t)$. # Lemma That Concludes the Proof ## Lemma **Lemma**: For $p\in W$, $\left\{{Z_i}\right\}$ a unitary basis of $T_p W$, ` \begin{align*} &- \left\langle J(p)\left(d X_{t}\right)_{p}\left(Z_{i}\right), \quad Z_{j}\right\rangle \\ &+ \left\langle J(p)\left(d X_{t}\right)_{p}\left(Z_{j}\right), \quad Z_{i}\right\rangle \\ &- \left\langle(d J)_{p}\left(X_{t}\right) Z_{i}, \hspace{2.5em} Z_{j}\right\rangle \\ &= 0 .\end{align*} `{=html} **Claim**: This lemma immediately concludes the previous proof? ## Proof of Lemma Extend $\left\{{Z_i}\right\}$ to a chart containing $p$ and use the Leibniz rule to rewrite ```{=tex} \small ``` ` \begin{align*} - \left\langle J(p)\left(d X_{t}\right)_{p}\left(Z_{i}\right), Z_{j}\right\rangle + \left\langle J(p)\left(d X_{t}\right)_{p}\left(Z_{j}\right), Z_{i}\right\rangle - \left\langle(d J)_{p}\left(X_{t}\right) Z_{i}, Z_{j}\right\rangle = 0 \end{align*} `{=html} `\normalsize`{=tex} as `\footnotesize`{=tex} ` \begin{align*} &- \left\langle J\left(d X_{t}\right)\left(Z_{i}\right),Z_{j}\right\rangle + \left\langle J\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle + \left\langle J\left(d Z_{i}\right)\left(X_{t}\right), Z_{j}\right\rangle - {\color{red} \left\langle d\left(J Z_{i}\right)\left(X_{t}\right), Z_{j}\right\rangle } \\ \\ &=\left\langle J\left[X_{t}, Z_{i}\right], Z_{j}\right\rangle + \left\langle J\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle - {\color{red} \left\langle d\left(J Z_{i}\right)\left(X_{t}\right), Z_{j}\right\rangle } .\end{align*} `{=html} `\normalsize`{=tex} where we'll rewrite the red terms. ## Proof of Lemma ```{=tex} \scriptsize ``` Now use ` \begin{align*} X_t {\left\langle {JZ_i},~{Z_j} \right\rangle} = 0 \implies \left\langle d\left(J Z_{i}\right)\left(X_{t}\right), Z_{j}\right\rangle+\left\langle J Z_{i},\left(d Z_{j}\right)\left(X_{t}\right)\right\rangle=0 .\end{align*} `{=html} We now rewrite the RHS from before: ` \begin{align*} \left\langle J\left[X_{t}, Z_{i}\right], Z_{j}\right\rangle &+\left\langle J\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle+\left\langle J Z_{i},\left(d Z_{j}\right)\left(X_{t}\right)\right\rangle \\ \\ &=\left\langle J\left[X_{t}, Z_{i}\right], Z_{j}\right\rangle+\left\langle J\left(d X_{t}\right)\left(Z_{j}\right)-J\left(d Z_{j}\right)\left(X_{t}\right), Z_{i}\right\rangle \\ &=\left\langle J\left[X_{t}, Z_{i}\right], Z_{j}\right\rangle-\left\langle J\left[X_{t}, Z_{j}\right], Z_{i}\right\rangle \\ &=\omega\left(\left[X_{t}, Z_{i}\right], Z_{j}\right)-\omega\left(\left[X_{t}, Z_{j}\right], Z_{i}\right) .\end{align*} `{=html} The symmetry follows from $\omega$ being closed and ` \begin{align*} 0 &= d \omega\left(X_{t}, Z_{i}, Z_{j}\right) \\ &= X_{t} \cdot \omega\left(Z_{i}, Z_{j}\right)-Z_{i} \cdot \omega\left(X_{t}, Z_{j}\right)+Z_{j} \cdot \omega\left(X_{t}, Z_{i}\right) \\ &\quad -\omega\left(\left[X_{t}, Z_{i}\right], Z_{j}\right)+\omega\left(\left[X_{t}, Z_{j}\right], Z_{i}\right)-\omega\left(\left[Z_{i}, Z_{j}\right], X_{t}\right) \\ \\ &= -X_{t} \cdot\left\langle Z_{i}, J Z_{j}\right\rangle+Z_{i} \cdot\left(d H_{t}\right)\left(Z_{j}\right)-Z_{j} \cdot\left(d H_{t}\right)\left(Z_{i}\right) \\ &\quad -\left(d H_{t}\right)\left(\left[Z_{i}, Z_{j}\right]\right)-\omega\left(\left[X_{t}, Z_{i}\right], Z_{j}\right)+\omega\left(\left[X_{t}, Z_{j}\right], Z_{i}\right) \\ \\ &= d\left(d H_{t}\right)\left(Z_{i}, Z_{j}\right)-\omega\left(\left[X_{t}, Z_{i}\right], Z_{j}\right)+\omega\left(\left[X_{t}, Z_{j}\right], Z_{i}\right) \\ &= -\omega\left(\left[X_{t}, Z_{i}\right], Z_{j}\right)+\omega\left(\left[X_{t}, Z_{j}\right], Z_{i}\right). \hspace{9em} \hfill\blacksquare \end{align*} `{=html} # 8.4.6: Linearization of Hamilton's Equation ## Linearization of Hamilton's Equation Recall ` \begin{align*} \qty{d{\mathcal{F}}}_u = \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5muY + SY = \qty{\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu+ S}Y \end{align*} `{=html} Now think of $S$ as a map $Y \mapsto S\cdot Y$, so $S\in C^\infty({\mathbb{R}}\times S^1; \endo({\mathbb{R}}^{2n}))$ and define the symmetric operators ` \begin{align*} S^{\pm} \coloneqq\lim_{s\to \pm\infty} S(s, {-}) {\quad \operatorname{respectively} \quad} \end{align*} `{=html} Theorem : The equation ` \begin{align*} {\partial}_t Y = J_0 S^{\pm} Y \end{align*} `{=html} is a linearization of Hamilton's equation ` \begin{align*} {\frac{\partial z}{\partial t}\,} = X_t(z){\quad \operatorname{at} \quad} \begin{cases} x = \lim_{s\to -\infty} u & \text{for } S^- \\ y = \lim_{s\to \infty} u & \text{for } S^+ \end{cases}\quad\text{respectively} .\end{align*} `{=html} ## Proof We first linearize Hamilton's equation at $x$: ` \begin{align*} {\frac{\partial z}{\partial t}\,} = X_t(z) \quad \stackrel{\text{linearized}}\implies \quad {\frac{\partial Y}{\partial t}\,} = (dX_t)_x Y .\end{align*} `{=html} So write $Y = \sum y_i Z_i$ to obtain ` \begin{align*} \sum_{i} \frac{\partial y_{i}}{\partial t} Z_{i} &=\sum_{i} y_{i}\left(-\frac{\partial Z_{i}}{\partial t}+\left(d X_{t}\right)\left(Z_{i}\right)\right) \\ &=\sum_{i} \sum_{j} y_{i}\left\langle-\frac{\partial Z_{i}}{\partial t}+\left(d X_{t}\right)\left(Z_{i}\right), Z_{j}\right\rangle Z_{j} \\ &=\sum_{i} \sum_{j} y_{j}\left\langle-\frac{\partial Z_{i}}{\partial t}+\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle Z_{i} \\ \\ \implies \frac{\partial y_{i}}{\partial t} &=\sum_{j}\left\langle-\frac{\partial Z_{j}}{\partial t}+\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle y_{j} .\end{align*} `{=html} ## Proof Thus we can rewrite the linearized equation as `\scriptsize`{=tex} ` \begin{align*} {\frac{\partial Y}{\partial t}\,} = (dX_t)_x Y = B^- \cdot Y, \quad b_{i j}=\left\langle-\frac{\partial Z_{j}}{\partial t}+\left(d X_{t}\right)_{x}\left(Z_{j}\right), Z_{i}\right\rangle .\end{align*} `{=html} `\normalsize `{=tex} Recall `\scriptsize`{=tex} ` \begin{align*} A \coloneqq A\left(y_{1}, \ldots, y_{2 n}\right)=\sum y_{i}\left( {\color{red} J(u) \frac{\partial Z_{i}}{\partial t}-J(u)\left(d X_{t}\right)_{u}\left(Z_{i}\right) } \right) .\end{align*} `{=html} `\normalsize`{=tex} Now take $s\to -\infty$ and look at the order zero part of $(d{\mathcal{F}})_u$. By the proof of 8.4.4, we have `\scriptsize`{=tex} ` \begin{align*} A\left(\sum y_{i} Z_{i}\right) &=\sum_{i}\left(J(x) \frac{\partial Z_{i}}{\partial t}-J(x)\left(d X_{t}\right)_{x}\left(Z_{i}\right)\right) \\ &=\sum_{i} \sum_{j} y_{i}\left\langle J \frac{\partial Z_{i}}{\partial t}-J\left(d X_{t}\right)\left(Z_{i}\right), Z_{j}\right\rangle Z_{j} \\ &=\sum_{i} \sum_{j} y_{j}\left\langle J \frac{\partial Z_{j}}{\partial t}-J\left(d X_{t}\right)\left(Z_{j}\right), Z_{i}\right\rangle Z_{i} \\ &=\sum_{i} \sum_{j}\left\langle-\frac{\partial Z_{j}}{\partial t}+\left(d X_{t}\right)\left(Z_{j}\right), J Z_{i}\right\rangle {y_{j}} Z_{i} .\end{align*} `{=html} `\normalsize`{=tex} ## Proof Deduce that ` \begin{align*} S = (s_{ij}),\quad & s_{ij} = \left\langle-\frac{\partial Z_{j}}{\partial t}+\left(d X_{t}\right)_{x}\left(Z_{j}\right), ~~J(x) Z_{i}\right\rangle \\ \\ \\ JZ_i = \begin{cases} Z_{i+n} & i \leq n \\ -Z_{i-n} & i \geq n+1 \end{cases} &\implies s_{ij} = \begin{cases} b_{i+n, j} & i \leq n \\ -b_{i-n, j} & i \geq n+1 \end{cases} \\ \\ &\iff S^- = - J_0 B^- .\end{align*} `{=html}