--- aspectratio: 169 author: D. Zack Garza colortheme: default date: April 2020 fontfamily: noto-sans fontsize: 10pt header-includes: - | ```{=tex} \usepackage{cmbright} ``` subtitle: Sections 8.3 and 8.4 theme: Berkeley title: Linearization and Transversality 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 ## Recalling Notation - 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} - Critical points of the action functional ${\mathcal{A}}_H$ are given by orbits, i.e. contractible loops $x, y \in {\mathcal{L}}W$ - In general, $x, y$ are two periodic orbits of $H$ of period 1. - Solutions are functions $u \in C^\infty({\mathbb{R}}\times S^1; W) = C^\infty({\mathbb{R}}; {\mathcal{L}}W)$ - ${\mathcal{M}}(x, y)$ is the moduli space of solutions of the Floer equation connecting orbits $x$ and $y$. - $W^{1, p}(x, y)$ and ${\mathcal{P}}^{1, p}(x, y)$ were completions of $C^\infty(?)$ with respect to certain norms. ## The "Program" for Chapter 8 - Show that ${\mathcal{M}}(x, y)$ is a manifold of dimension $\mu(x) - \mu(y)$ - Define ${\mathcal{M}}(x, y)$ as the inverse image of a regular value of some map - Perturb $H$ to apply the Sard-Smale theorem - Show the tangent maps are Fredholm operators and compute their index. # Section 8.3: The Space of Perturbations of H ## Goal **Goal**: Given a fixed Hamiltonian $H\in C^\infty(W\times S^1; {\mathbb{R}})$, perturb it (without modifying the periodic orbits) so that ${\mathcal{M}}(x, y)$ are manifolds of the expected dimension. ## Goal `\tiny `{=tex}Start by trying to construct a subspace ${\mathcal{C}}_{\varepsilon}^{\infty}(H) \subset {\mathcal{C}}^\infty(W\times S^1; {\mathbb{R}})$, the space of perturbations of $H$ depending on a certain sequence ${\varepsilon}= \left\{{{\varepsilon}_k}\right\}$, and show it is a dense subspace. ```{=tex} \begin{center} \includegraphics[width = 0.8\textwidth]{figures/image_2020-04-14-23-36-58.png} \end{center} ``` ## Define an Absolute Value Idea: similar to how you build $L^2({\mathbb{R}})$, define a norm ${\left\lVert {{-}} \right\rVert}_{\varepsilon}$ on $C_{\varepsilon}^{\infty}(H)$ and take the subspace of finite-norm elements. - Let $h(\mathbf{x}, t) \in C_{\varepsilon}^\infty(H)$ denote a perturbation of $H$. - Fix ${\varepsilon}= \left\{{{\varepsilon}_k \mathrel{\Big|}k\in {\mathbb{Z}}^{\geq 0}}\right\} \subset {\mathbb{R}}^{>0}$ a sequence of real numbers, which we will choose carefully later. - For a fixed $\mathbf{x} \in W, t\in {\mathbb{R}}$ and $k \in {\mathbb{Z}}^{\geq 0}$, define ` \begin{align*} {\left\lvert {d^k h(\mathbf{x}, t)} \right\rvert} = \max \left\{{d^\alpha h(\mathbf{x}, t) {~\mathrel{\Big\vert}~}{\left\lvert {\alpha} \right\rvert} = k}\right\} ,\end{align*} `{=html} the maximum over all sets of multi-indices $\alpha$ of length $k$. > Note: I interpret this as ` > > \begin{align*} > d^{\alpha_1, \alpha_2, \cdots, \alpha_k}h = \frac{{\partial}^k h}{{\partial}x_{\alpha_1} ~{\partial}x_{\alpha_2} \cdots {\partial}x_{\alpha_k}} > ,\end{align*} > `{=html} the partial derivatives wrt the corresponding variables. ## Define a Norm - Define a norm on $C^\infty(W\times S^1; {\mathbb{R}})$: ` \begin{align*} |h|_{\mathbf{{\varepsilon}}} &= \sum_{k \geq 0} {\varepsilon}_{k} \sup _{(x, t) \in W \times S^{1}}\left|d^{k} h(x, t)\right| .\end{align*} `{=html} - Since $W\times S^1$ is assumed compact (?), fix a finite covering $\left\{{B_i}\right\}$ of $W\times S^1$ such that ` \begin{align*} \cup_i B_i^\circ = W \times S^1 .\end{align*} `{=html} - Choose them in such a way we obtain charts ` \begin{align*} \Psi_i: B_i &\to \mkern 1.5mu\overline{\mkern-1.5muB(0, 1)\mkern-1.5mu}\mkern 1.5mu \subset {\mathbb{R}}^{2n + 1} ~(?) .\end{align*} `{=html} - Obtain the computable form ` \begin{align*} {\left\lVert {h} \right\rVert}_{\mathbf{{\varepsilon}}} &= \sum_{k \geq 0} {\varepsilon}_{k} \sup _{(x, t) \in W \times S^{1}} \sup_{i, z\in B(0, 1)} {\left\lvert { d^k (h\circ \Psi_i^{-1})(z)} \right\rvert} .\end{align*} `{=html} ## Define a Banach Space - Define ` \begin{align*} C_{\varepsilon}^\infty = \left\{{h\in C^\infty(W\times S^1; {\mathbb{R}}) {~\mathrel{\Big\vert}~}{\left\lVert {h} \right\rVert}_{\varepsilon}< \infty}\right\} \subset C^{\infty}(W\times S^1; {\mathbb{R}}) ,\end{align*} `{=html} which is a Banach space (normed and complete). - Show that the sequence $\left\{{{\varepsilon}_k}\right\}$ can be chosen so that $C_{\varepsilon}^\infty$ is a *dense* subspace for the $C^\infty$ topology, and in particular for the $C^1$ topology. Theorem : Such a sequence $\left\{{{\varepsilon}_k}\right\}$ can be chosen. Lemma : $C^\infty(W \times S^1; {\mathbb{R}})$ with the $C^1$ topology is separable as a topological space (contains a countable dense subset). ## Sketch Proof of Theorem - By the lemma, produce a sequence $\left\{{f_n}\right\} \subset C^\infty(W\times S^1; {\mathbb{R}})$ dense for the $C^1$ topology. - Using the norm on $C^n(W\times S^1; {\mathbb{R}})$ for the $f_n$, define ` \begin{align*} \frac 1 {{\varepsilon}_n} = 2^n \displaystyle\max \left\{{{\left\lVert {f_k} \right\rVert} {~\mathrel{\Big\vert}~}k\leq n}\right\} \implies {\varepsilon}_n \sup {\left\lvert {d^n f_k(x, t)} \right\rvert} \leq 2^{-n} \end{align*} `{=html} which is summable. $\hfill\blacksquare$ > Why does this imply density? I don't know. ## Modified Theorem The next proposition establishes a version of this theorem with compact support: Theorem : For any $(\mathbf{x}, t) \subset U \in W \times S^1)$ there exists a $V\subset U$ such that every $h\in C^{\infty}(W\times S^1; {\mathbb{R}})$ can be approximated in the $C^1$ topology by functions in $C_{\varepsilon}^\infty$ supported in $U$. Then fix a time-dependent Hamiltonian $H_0$ with nondegenerate periodic orbits and consider ` \begin{align*} \left\{{ h\in C_{\varepsilon}^\infty(H_0) {~\mathrel{\Big\vert}~}h(x, t) = 0 \text{ in some $U\supseteq$ the 1-periodic orbits of $H_0$}}\right\} \end{align*} `{=html} Then $\mathop{\mathrm{supp}}(h)$ is "far" from $\mathop{\mathrm{per}}(H_0)$, so ` \begin{align*} {\left\lVert {h} \right\rVert}_{\varepsilon}\ll 1 \implies \mathop{\mathrm{per}}(H_0 + h) = \mathop{\mathrm{per}}(H_0) \end{align*} `{=html} and are both nondegenerate. # Section 8.4: Linearizing the Floer Equation: The Differential of F ## Goal Choose $m> n = \dim (W)$ and embed $TW \hookrightarrow{\mathbb{R}}^m$ to identify tangent vectors (such as $Z_i$, tangents to $W$ along $u$ or in a neighborhood $B$ of $u$) with actual vectors in ${\mathbb{R}}^m$. > Why? Bypasses differentiating vector fields and the Levi-Cevita connection. We can then identify ` \begin{align*} \operatorname{im}{\mathcal{F}}= C^\infty ({\mathbb{R}}\times S^1; {\mathbb{R}}^m) {\quad \operatorname{or} \quad} L^p({\mathbb{R}}\times S^1; W) ,\end{align*} `{=html} and we seek to compute its differential $d {\mathcal{F}}$. > We've just replaced the codomain here. ## Definitions Recall that - $x, y$ are contractible loops in $W$ that are nondegenerate critical points of the action functional ${\mathcal{A}}_H$, - $u \in {\mathcal{M}}(x, y) \subset C_{\mathsf{loc}}^\infty$ denotes a fixed solution to the Floer equation, - $C_{\searrow}(x ,y) \subset \left\{{u \in C^\infty(R\times S^1; W)}\right\}$ is the set of smooth solutions $u: {\mathbb{R}}\times S^1 \to W$ satisfying some conditions: ` \begin{align*} &\lim_{s\to -\infty}u(s, t) = x(t),~ \lim_{s\to\infty}u(s, t) = y(t) \\ \\ &\text{and } {\left\lvert {{\frac{\partial u}{\partial t}\,}(s, t)} \right\rvert}, ~~ {\left\lvert {{\frac{\partial u}{\partial t}\,}(s, t) - X_H(u)} \right\rvert} \sim \exp({\left\lvert {s} \right\rvert}) \end{align*} `{=html} ## Compactify to Sphere Fix a solution ` \begin{align*} u\in {\mathcal{M}}(x, y) \subset C_{{\mathsf{loc}}}^\infty({\mathbb{R}}\times S^1; W) .\end{align*} `{=html} We lift each solution to a map ` \begin{align*} \tilde u: S^2 \to W \end{align*} `{=html} in the following way: The loops $x, y$ are contractible, so they bound discs. So we extend by pushing these discs out slightly: ## Lift to 2-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} ```{=tex} \begin{center} \includegraphics[width = 1.02\textwidth]{figures/image_2020-04-15-18-10-40.png} \end{center} ``` ## Trivialize the Pullback From earlier in the book, we have **Assumption (6.22)**: 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 $TW$. > `\tiny `{=tex}Note: I don't know what this pairing is. The top Chern class is the Euler class (obstructs nowhere zero sections) and are defined inductively: ` > > \begin{align*} > c_1(TW) = e(\Lambda^n(TW)) \in H^2(W; {\mathbb{Z}}) > \end{align*} > `{=html} Assumption is satisfied when all maps $S^2 \to W$ lift to $B^3$ $\iff \pi_2(W) = 0$. We have a pullback that is a symplectic fiber bundle: ```{=tex} \begin{center} \begin{tikzcd}[ampersand replacement=\&] \tilde u^* TW \ar[r, "d\tilde u"] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{\color{black}$\lrcorner$}" , very near start, color=black] \& TW\ar[d] \\ S^2 \ar[r, "\tilde u"] \& W \end{tikzcd} \end{center} ``` ## Choose a Frame - Using the assumption, 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} where - The frame depends smoothly on $(s, t) \in S^2$, - $\lim_{s\to \infty} Z_i$ exists for each $i$. - ` \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} Claim: such trivializations exist, "using cylinders near the spherical caps in the figure". ## Define "Banach Manifold Charts" Recall we had $W^{1, p}(x, y)$ a completion of $C^\infty$ ` \begin{align*} {\mathcal{M}}(x, y) \subset C_{\searrow}^\infty(x, y) \subset {\mathcal{P}}^{1, p}(x , y) \underset{\text{defn}}\subset \left\{{ (s, t) \xrightarrow{\varphi} \exp_{w(s, t)} Y(s, t)}\right\} .\end{align*} `{=html} where we restrict to - $Y \in W^{1, p}(w^* TW)$, - $w\in C_{\searrow}^\infty(x, y)$ Use 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) \\ \mathbf{y} = \left(y_{1}, \dots, y_{2 n}\right) & \longmapsto \exp _{u}\left(\sum y_{i} Z_{i}\right) .\end{align*} `{=html} - Note that the derivative at zero is $\sum_{i=1}^{2n} y_i Z_i$. ## Define the Floer Map in Charts Define and compute the differential of the composite map $\tilde{\mathcal{F}}$ defined as follows: ```{=tex} \begin{center} \begin{tikzcd}[ampersand replacement=\&] \mathcal{P}^{1, p}(x, y) \ar[r, "\mathcal{F}"] \ar[rr, dotted, "\tilde {\mathcal{F}}", bend left] \& L^{p}\left(\mathbb{R} \times S^{1} ; T W\right) \ar[r] \& L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{m}\right) \\ u \ar[rr, "\tilde {\mathcal{F}}"] \& \&\frac{\partial u}{\partial s}+J(u)\left(\frac{\partial u}{\partial t}-X_{t}(u)\right) \end{tikzcd} \end{center} ``` - From now on, let ${\mathcal{F}}$ denote $\tilde {\mathcal{F}}$. ## Add a Tangent - Take the vector ` \begin{align*}Y(s, t) \coloneqq(y_1(s, t), \cdots) \in {\mathbb{R}}^{2n} \subset {\mathbb{R}}^m\end{align*} `{=html} - View $Y$ as a vector in ${\mathbb{R}}^m$ tangent to $W$, given by $Y = \sum_{i=1}^{2n} y_i Z_i$. - Plug $u + Y$ into the equation for ${\mathcal{F}}$, directly yielding ## 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 Linear Part 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}}} \\ & \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)} \\ .\end{align*} `{=html} - This is a sum of two differential operators: - One of order 1, one of order 0 (Perspective 1) - The Cauchy-Riemann operator, and one of order zero (Perspective 2, not immediate from this form) ## Leibniz Rule - Now compute in charts. Need a lemma: Lemma (Leibniz Rule) : For any source space $X$ and any maps ` \begin{align*} J: X &\to \endo({\mathbb{R}}^m) \\ Y, v: X &\to {\mathbb{R}}^m \end{align*} `{=html} we have ` \begin{align*} (dJ)(Y) \cdot v = d(Jv)(Y) - J dv(Y) .\end{align*} `{=html} ## Sketch: Proof of Leibniz Rule Differentiate the map ` \begin{align*} J\cdot v: X &\to {\mathbb{R}}^m \\ x &\mapsto J(x)\cdot v(x) \end{align*} `{=html} to obtain ` \begin{align*} &J(x + Y) v(x + y) \\ &= \qty{J(x) + (dJ)_{x} (Y) } ~\cdot~ \qty{v(x) + (dv)_x(Y)} + \cdots \\ &= J(x) \cdot v(x) + {\color{blue} J(x) \cdot (dv)_x(Y) + (dJ)_x(Y) \cdot v(x)} \\ &\quad + (dJ)_x(Y)\cdot (dv)_x(Y) + \cdots \\ \\ &\implies d(J\cdot v)_x(Y) = (dJ)_x(Y) \cdot v(x) + J(x) \cdot (dv)_x(Y) .\end{align*} `{=html} $\hfill\blacksquare$ ## Decompose by Order Using the chart $\iota$ defined by $\left\{{Z_i}\right\}$ to write $Y = \sum_{i=1}^{2n} y_i Z_i$ and thus ` \begin{align*} (d{\mathcal{F}})_u(Y) = O_0 + O_1 \end{align*} `{=html} where $O_0$ are order 0 terms ("they do not differentiate the $y_i$") and the $O_1$ are order 1 terms: ` \begin{align*} O_1 &= \sum_{i=1}^{2n} \qty{ {\frac{\partial y_i}{\partial s}\,} Z_i + {\frac{\partial y_i}{\partial t}\,} J(u) Z_i} \\ \\ O_0 &= \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}\,} \\ &\quad\quad\quad\quad - J(u) (dX_t)_u Z_i - (dJ)_u (Z_i) X_t \Bigg) .\end{align*} `{=html} ## Order One - 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 ${\mathbb{R}}^{2n} = {\mathbb{C}}^n$ - The second equality follows from the assumption that the $Z_i$ are symplectic and orthonormal. - Note that this writes $(d{\mathcal{F}})_u(Y) = O_0 + O_{CR}$, a sum of an order zero and a Cauchy-Riemann operator. ## Recap Note that since we've computed in charts, we have actually computed the differential of ${\mathcal{F}}_u$ in the following diagram ![image_2020-04-16-01-11-02](Archive/Floer%20Reading%20Group/Talk1/figures/image_2020-04-16-01-11-02.png) So we've technically computed $(dF_\mu)_0$. ## Order 0 Term is Linear ` \begin{align*} (d{\mathcal{F}})_u &= \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)} \\ \\ &\coloneqq{\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 Symmetry 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 - $S$ is linear - $S$ tends to a symmetric operator as $s\to \pm \infty$, and - ` \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 Omitted -- $S$ is exactly $O_0$ from before: ` \begin{align*} O_0 &= \sum_{i=1}^{2n} y_i \Bigg( {\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}\,}} \\ &\quad\quad\quad\quad - J(u) (dX_t)_u Z_i - {\color{blue} (dJ)_u (Z_i) X_t} \Bigg) \\ &= \sum_{i=1}^{2n} y_i \Bigg() {\frac{\partial Z_i}{\partial s}\,} + { \color{blue} (dJ)_u (Z_i) \qty{ {\frac{\partial u}{\partial t}\,} - (Z_i) X_t} } \\ &\quad\quad\quad\quad + J(u) {\frac{\partial Z_i}{\partial t}\,} - J(u) (dX_t)_u Z_i \Bigg) .\end{align*} `{=html} - The term in blue vanishes as $s\to \pm \infty$ - Using the fact that $u$ is a solution - Uses ${\frac{\partial u}{\partial s}\,} \to 0$ uniformly (as do its derivatives?) - Suffices to show the remaining part is symmetric in the limit ## Proof - Write the remaining part as ` \begin{align*} A(y_1, \cdots, y_{2n}) = \cdots \implies A_{ij} = A_{ji} \end{align*} `{=html} using inner product calculations - Uses the fact the $Z_i$ needed to be chosen to be unitary and symplectic. $\hfill\blacksquare$ ## asdas Write $O_0$ 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} linearizes 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} ## Image Reminder the $x, y$ were the top/bottom pieces of the original cylinder/sphere: ```{=tex} \begin{center} \includegraphics[height= 0.8\textheight]{figures/image_2020-04-15-21-53-36.png} \end{center} ``` ## Proof Sketch - Use the fact that ${\frac{\partial Y}{\partial t}\,} = (dX_t)_x Y$ - Expand $\sum {\frac{\partial y_i}{\partial t}\,} Z_i$ in the $Z_i$ basis (roughly) to write ${\frac{\partial y_i}{\partial t}\,} = \sum b_{ij} y_j$ for some coefficients $b_{ij}$. - Collect terms into a matrix/operator $B^\mp$ for $x,y$ respectively to write ` \begin{align*}{\frac{\partial Y}{\partial t}\,} = B^- \cdot Y\end{align*} `{=html} - Write $(d{\mathcal{F}})_u = \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu+ S$ where $S$ is zero order and symmetric in the limit ## Proof Sketch - Get the corresponding operator $A$ in coordinates - Expand in a basis (roughly) as $A(\sum y_i Z_i) = \sum s_{ij} y_j Z_i$ - Check that $s_{ij} = \pm b_{i\pm n, j}$ - This implies ` \begin{align*}S^- = - J_0 B^- \quad S^+ = -J_0 B^+ \implies {\frac{\partial Y}{\partial t}\,} = J_0 S^\pm Y\end{align*} `{=html} ## Final Remarks - Given a solution $u$, we have a right ${\mathbb{R}}{\hbox{-}}$action, so for $s\in {\mathbb{R}}$, ` \begin{align*} u \cdot s \in C^\infty({\mathbb{R}}\times S^1; W) \\ (\sigma, t) \mapsto u(\sigma + s, t) \end{align*} `{=html} is also a solution, so ${\mathcal{F}}(u \cdot s) = 0$ for all $s$. > In other words: we can flow solutions? ## Final Remarks **Punchline**:**${\frac{\partial u}{\partial s}\,}$ is a solution of the linearized equation**, since ` \begin{align*} 0 = {\frac{\partial }{\partial s}\,} {\mathcal{F}}(u\cdot s) = (d{\mathcal{F}})_u \qty{{\frac{\partial u}{\partial s}\,}} .\end{align*} `{=html} - Along any nonconstant solution connecting $x$ and $y$, $\dim \ker (d{\mathcal{F}})_u \geq 1$.