--- title: 'Floer Talk 4 created: 2023-04-11T23:30 updated: 2023-04-11T23:30 --- Tags: #projects/my-talks #projects/reading-groups #projects/floer-paper See [2020 Fall Floer MOC](2020%20Fall%20Floer%20MOC.md) # Summary/Outline [@AT17] ## Outline What we're trying to prove: - 8.1.5: $(d\mcf)_u$ is a Fredholm operator of index $\mu(x) - \mu(y)$. What we have so far: - Define \[ L: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s, t) Y \] where \[ S: \RR\cross S^1 &\to \mat(2n; \RR) \\ S(s, t) &\converges{s\to\pm\infty}\to S^\pm(t) .\] ## Outline - Took $R^\pm: I \to \Sp(2n; \RR)$: symplectic paths associated to $S^\pm$ - These paths defined $\mu(x), \mu(y)$ - Section 8.7: \[ R^\pm \in \mcs \definedas \theset{R(t) \suchthat R(0) = \id, ~ \det(R(1) - \id)\neq 0} \implies L \text{ is Fredholm} .\] - WTS 8.8.1: \[ \ind(L)\stackrel{\text{Thm?}}{=} \mu(R^-(t)) - \mu(R^+(t)) = \mu(x) - \mu(y) .\] ## From Yesterday - Han proved 8.8.2 and 8.8.4. - So we know $\ind(L) = \ind(L_1)$ - Today: 8.8.5 and 8.8.3: - Computing $\ind(L_1)$ by computing kernels. # 8.8.5: $\dim \ker F, F^*$ ## Recall \[ L: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s, t) Y \\ \\ L_{1}: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y \\ \\ L_{1}^{\star}: W^{1, q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Z & \longmapsto-\frac{\partial Z}{\partial s}+J_{0} \frac{\partial Z}{\partial t}+S(s)^t Z \] Here ${1\over p} + {1\over q} = 1$ are conjugate exponents. ## Reductions \[ L_1^* &= -\dd{}{s} + J_0 \dd{}{t} + S(s)^t .\] - Since $\coker L_1 \cong \ker L_1^*$, it suffices to compute $\ker L_1^*$ - We have \[ J_0^1 \definedas \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \implies J_0 = \begin{bmatrix} J_0^1 & & & \\ & J_0^1 & & \\ & & \ddots & \\ & & & J_0^1 \end{bmatrix} \in \bigoplus_{i=1}^n \mat(2; \RR) .\] - This allows us to reduce to the $n=1$ case. ## Setup $L_1$ used a path of diagonal matrices constant near $\infty$: \[ S(s) \definedas \left(\begin{array}{cc} a_{1}(s) & 0 \\ 0 & a_{2}(s) \end{array}\right), \quad \text { with } a_{i}(s)\definedas \left\{\begin{array}{ll} a_{i}^{-} & \text {if } s \leq-s_{0} \\ a_{i}^{+} & \text {if } s \geq s_{0} \end{array}\right. .\] \begin{center} \includegraphics[width = \textwidth]{figures/image_2020-05-27-20-10-07.png} \end{center} ## Statement of Later Lemma (8.8.5) Let $p>2$ and define \[ F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\] > Note: $F$ is $L_1$ for $n=1$: \[ L_{1}: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\] ## Statement of Lemma \[ F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\] Suppose $a_i^\pm \not \in 2\pi \ZZ$. 1. Suppose $a_1(s) = a_2(s)$ and set $a^\pm \definedas a_1^\pm = a_2^\pm$. Then \[ \operatorname{dim} \operatorname{Ker} F &= 2 \cdot \size \left\{\ell \in \mathbb{Z} \suchthat 2\pi \ell \in (a^-, a+) \subset \RR \right \} \\ \operatorname{dim} \operatorname{Ker} F^{*} &= 2 \cdot \size \left\{\ell \in \mathbb{Z} \suchthat 2\pi\ell \in (a^+, a^-) \subset\RR \right\} .\] 2. Suppose $\sup_{s\in \RR} \norm{S(s)} < 1$, then \[ \operatorname{dim} \operatorname{Ker} F &= \size \left\{i \in\{1,2\} \suchthat ~a_{i}^{-}<0 \text { and } a_{i}^{+}>0\right\}\\ \operatorname{dim} \operatorname{Ker} F^{*} &=\size \left\{i \in\{1,2\} \suchthat ~ a_{i}^{+}<0 \text { and } a_{i}^{-}>0\right\} .\] ## Statement of Lemma In words: 1. If $S(s)$ is a scalar matrix, set $a^\pm = a_1^\pm = a_2^\pm$ to the limiting scalars and count the integer multiples of $2\pi$ between $a^-$ and $a^+$. 2. Otherwise, if $S$ is uniformly bounded by 1, count the number of entries the flip from positive to negative as $s$ goes from $-\infty\to \infty$. \begin{center} \includegraphics[width = \textwidth]{figures/image_2020-05-27-20-10-07.png} \end{center} ## Proof of Assertion 1 1. Suppose $a_1(s) = a_2(s)$ and set $a^\pm \definedas a_1^\pm = a_2^\pm$. Then \[ \operatorname{dim} \operatorname{Ker} F &= 2 \cdot \size \left\{\ell \in \mathbb{Z} \suchthat 2\pi \ell \in (a^-, a+) \subset \RR \right \} \\ \operatorname{dim} \operatorname{Ker} F^{*} &= 2 \cdot \size \left\{\ell \in \mathbb{Z} \suchthat 2\pi\ell \in (a^+, a^-) \subset\RR \right\} .\] Step 1: Transform to Cauchy-Riemann Equations - Write $a(s) \definedas a_1(s) = a_2(s)$. - Start with equation on $\RR^2$, $$\vector Y(s, t) = \left[ Y_1(s, t), Y_2(s, t) \right].$$ - Replace with equation on $\CC$: $$\vector Y(s, t) = Y_1(s, t) + i Y_2(s, t).$$ ## Assertion 1, Step 1: Reduce to CR - Expand definition of the PDE \[ F(\vector Y) = 0 \leadsto \bar \del\vector Y + S \vector Y = 0 \\ \\ \frac{\partial}{\partial s} \vector Y +\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right) \frac{\partial}{\partial t} \vector Y +\left(\begin{array}{cc} a(s) & 0 \\ 0 & a(s) \end{array}\right) \vector Y =0 .\] - Change of variables: want to reduce to $\bar \del \tilde Y = 0$ - Choose $B \in \GL(1, \CC)$ such that $\bar\del B + SB = 0$ - Set $Y = B\tilde Y$, which (?) reduces the previous equation to \[ \bar\del \tilde Y = 0 .\] ## Assertion 1, Step 1: Reduce to CR Can choose (and then solve) \[ B = \begin{bmatrix} b(s) & 0 \\ 0 & b(s) \end{bmatrix} \qtext{where} \dd{b}{s} = -a(s)b(s) \\ \\ \implies b(s) = \exp{\int_0^s -a(\sigma) ~d\sigma} \definedas \exp{-A(s)} .\] Remarks: - For some constants $C_i$, we have \[ A(s) = \begin{cases} C_1 + a^- s, & s \leq -\sigma_0 \\ C_2 + a^+ s, & s \geq \sigma_0 \\ \end{cases} .\] - The new $\tilde Y$ satisfies CR, is continuous and $L^1_{\text{loc}}$, so elliptic regularity $\implies C^\infty$. - The real/imaginary parts of $\tilde Y$ are $C^\infty$ and harmonic. ## Assertion 1, Step 2: Solve CR - Identify $s+it \in \RR\cross S^1$ with $u = e^{2\pi z}$ - Apply Laurent's theorem to $\tilde Y(u)$ on $\CC\smz$ to obtain an expansion of $\tilde Y$ in $z$. - Deduce that the solutions of the system are given by \[ \tilde Y (u) =\sum_{\ell \in \mathbf{Z}} c_{\ell} u^\ell \implies \tilde{Y}(s+i t) =\sum_{\ell \in \mathbf{Z}} c_{\ell} e^{(s+i t) 2 \pi \ell} .\] where $\theset{c_\ell}_{\ell\in\ZZ} \subset \CC$ converges for all $s, t$. ## Assertion 1, Step 2: Solve CR Use $e^{s+it} = e^s\qty{\cos(t) + i\sin (t)}$ to write in real coordinates: \[ \tilde{Y}(s, t)=\sum_{\ell \in \mathbb{Z}} e^{2 \pi s \ell} \begin{bmatrix} \cos(2\pi\ell t) & -\sin(2\pi \ell t) \\ \sin(2\pi\ell t) & \cos(2\pi \ell t) \end{bmatrix} \begin{bmatrix} \alpha_\ell \\ \beta_\ell \end{bmatrix} .\] Use \[ Y = B\tilde Y = \begin{bmatrix} e^{-A(s)} & 0 \\ 0 & e^{-A(s)} \end{bmatrix} \tilde Y \] to write \[ Y(s, t)=\sum_{\ell \in \mathbb{Z}} e^{2 \pi s \ell} \begin{bmatrix} e^{-A(s)} & 0 \\ 0 & e^{-A(s)} \end{bmatrix} \begin{bmatrix} \cos(2\pi\ell t) & -\sin(2\pi \ell t) \\ \sin(2\pi\ell t) & \cos(2\pi \ell t) \end{bmatrix} \begin{bmatrix} \alpha_\ell \\ \beta_\ell \end{bmatrix} .\] For $s\leq s_0$ this yields for some constants $K, K'$: \[ Y(s, t) = \sum_{\ell\in \ZZ} e^{2\pi\ell - a^-} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} .\] ## Condition on $L^p$ Solutions For $s\leq s_0$ we had \[ Y(s, t) = \sum_{\ell\in \ZZ} e^{\qty{2\pi\ell - a^-}s} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} \] and similarly for $s\geq s_0$, for some constants $C, C'$ we have: \[ Y(s, t) = \sum_{\ell\in \ZZ} e^{\qty{2\pi\ell - a^+}s} \begin{bmatrix} e^C \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{C'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} .\] Then \[ Y\in L^p \iff \text{exponential terms} \converges{\ell\to\infty}\to 0 .\] ## Condition on $L^p$ Solutions: Small Tails \[ Y(s, t) = \sum_{\ell\in \ZZ} e^{\qty{2\pi\ell - a^-}s} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} \] - $\ell \neq 0$: Need $\alpha_\ell = \beta_\ell = 0$ **or** $2\pi \ell < a^+$ - $\ell = 0$: Need both - $\alpha_0 = 0$ or $a^+ > 0$ and - $\beta_0 = 0$ or $a^+ > 0$. ## Counting Solutions \[ \begin{cases} \alpha_\ell = \beta_\ell = 0 \text{ or } 2\pi\ell \in (a^-, a^+) & \ell\neq 0 \\ \qty{\alpha_0 = 0 \txor 0 \in (a^-, a^+)} \txand \qty{\beta_0 = 0 \txor 0\in (a^-, a^+)} & \ell = 0 \end{cases} .\] - Finitely many such $\ell$ that satisfy these conditions - Sufficient conditions for $Y(s, t) \in W^{1, p}$. Compute dimension of space of solutions: \[ \operatorname{dim} \operatorname{Ker} F &=2 \cdot \size \theset{\ell \in \mathbb{Z}^{*} \suchthat 2\pi\ell \in (a^-, a^+) } + 2\cdot \indic{0 \in (a^-, a^+)} \\ &=2 \cdot \size \left\{\ell \in \mathbb{Z} \suchthat 2\pi\ell \in (a^-, a^+) \right\} .\] > Note: not sure what $\ZZ^*$ is: most likely $\ZZ\smz$. ## Counting Solutions Use this to deduce $\dim \ker F^*$: - $Y\in \ker F^* \iff Z(s, t) \definedas Y(-s, t)$ is in the kernel of the operator \[ \tilde F: W^{1, q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Z &\mapsto \frac{\partial Z}{\partial s}+J_{0} \frac{\partial Z}{\partial t}+S({\color{red}-s}) Y .\] - Obtain $\ker F^* \cong \ker \tilde F$. - Formula for $\dim \ker \tilde F$ almost identical to previous formula, just swapping $a^-$ and $a^+$. ## Assertion 2 **Assertion 2**: Suppose $\sup_{s\in \RR} \norm{S(s)} < 1$, then \[ \operatorname{dim} \operatorname{Ker} F &= \size \left\{i \in\{1,2\} \suchthat ~a_{i}^{-}<0 < a_{i}^{+}\right\}\\ \operatorname{dim} \operatorname{Ker} F^{*} &=\size \left\{i \in\{1,2\} \suchthat ~ a_{i}^{+}<0 < a_{i}^{-} \right\} .\] We use the following: - Lemma 8.8.7: \[ \sup_{s\in \RR} \norm{ S(s) } < 1 \implies \text{the elements in }\ker F,~ \ker F^* \text{ are independent of }t .\] - Proof: in subsection 10.4.a. ## Proof of Assertion 2 \[ F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\] - Given as a fact: \[ \vector Y \in \ker F \implies \dd{}{s}\vector Y = \vector a(s)\vector Y ~~~ ~~~\definedas \begin{bmatrix} -a_1(s) & 0 \\ 0 & -a_2(s) \end{bmatrix} \vector Y .\] - Therefore we can solve to obtain \[ \vector Y(s) = \vector c_0 \exp{-\vector A(s)}\qtext{where} \vector A(s) = \int_0^s -\vector a(\sigma) ~d\sigma .\] ## Proof of Assertion 2 - Explicitly in components: \[ \begin{cases} \dd{Y_1}{s} &= -a_1(s) Y_1 \\ \dd{Y_s}{s} &= -a_2(s) Y_2 \\ \end{cases} \quad \implies \quad Y_i(s) = c_i e^{-A_i(s)}, \quad A_i(s) &= \int_0^s -a_i(\sigma) ~d\sigma .\] - As before, for some constants $C_{j, i}$, \[ A_i(s) = \begin{cases} C_{1, i} + a_i^-\cdot s & s \leq -\sigma_0 \\ C_{2, i} + a_i^+\cdot s & s \geq \sigma_0 \\ \end{cases} .\] - Thus \[ Y_i \in W^{1, p} \iff 0 \in (a_i^-, a_i^+) ,\] establishing \[ \dim \ker F = \size \left\{i \in\{1,2\} \suchthat 0 \in (a_i^-, a_i^+) \right\} .\] # 8.8.3: $\ind(L_1) = k^- - k^+$ ## Statement and Outline Statement: let $k^\pm \definedas \ind(R^\pm)$; then $\ind(L_1) = k^- - k^+$. - Consider four cases, depending on parity of $k^\pm - n$ - Show all 4 lead to $\ind(L_1) = k^- - k^+$ 1. $k^- \equiv k^+ \equiv n \mod 2$. 2. $k^- \equiv n, k^+ \equiv n-1 \mod 2$ 3. $k^- \equiv n-1, k^+ \equiv n \mod 2$. 4. $k^- \equiv k^+ \equiv n-1 \mod 2$ \begin{center} \includegraphics[width = 0.3\textwidth]{figures/image_2020-05-27-22-54-44.png} \end{center} ## Case 1: $k^+ \equiv k^- \equiv n \mod 2$ \[ S_{k^-} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -\pi & & & \\ & & & & & -\pi & & \\ & & & & & & (n-1-k^-)\pi & \\ & & & & & & & (n-1-k^-)\pi \\ \end{bmatrix} \\ S_{k^+} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -\pi & & & \\ & & & & & -\pi & & \\ & & & & & & (n-1-{\color{blue}k^+})\pi & \\ & & & & & & & (n-1-{\color{blue}k^+})\pi \\ \end{bmatrix} .\] ## Case 1: $k^- \equiv k^+ \equiv n \mod 2$ - Take $a_1(s) = a_2(s)$ so $a_1^\pm = a^\pm$ - Apply the proved lemma to obtain \[ \dim \ker L_1 &= 2\cdot \size \theset{\ell \in \ZZ \suchthat 2\ell \in (n-1-k^-, n-1-k^+)} \\ &= \begin{cases} k^- - k^+ & k^- > k^+ \\ 0 & \text{else} \end{cases} \\ \\ \dim \ker L_1^* &= 2\cdot \size \theset{ \ell \in \ZZ \suchthat 2\ell \in (k^- - n + 1, k^+ - n + 1)} \\ &= \begin{cases} k^+ - k^- & k^+ > k^- \\ 0 & \text{otherwise} \end{cases} \\ \\ \implies \ind(L_1) &= \qty{k^- - k^+ \over 2} - \qty{k^+ - k^- \over 2} = k^- - k^+ .\] ## Case 2: $k^+ \not\equiv k^- \equiv n \mod 2$ \[ S_{k-} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -{\color{red}\eps}\pi & & & \\ & & & & & -{\color{red}\eps}\pi & & \\ & & & & & & (n-1-{\color{red}k^-})\pi & \\ & & & & & & & (n-1-{\color{red}k^-})\pi \\ \end{bmatrix} \\ S_{k^+} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & {\color{red}\eps} & & & \\ & & & & & -{\color{red}\eps} & & \\ & & & & & & (n-{\color{red}2}-k^+)\pi & \\ & & & & & & & (n-{\color{red}2}-k^+)\pi \\ \end{bmatrix} .\] ## Case 2: $k^+ \not\equiv k^- \equiv n \mod 2$ - Take $a_1(s) = a_2(s)$ everywhere except the $n-1$st block, where we can assume $\sup_{s\in \RR} \norm{S(s)} < 1$. - Assertion 2 applies and we get \[ \dim \ker L_1 &= 2\cdot \size \theset{\ell \in \ZZ \suchthat 2\ell \in (n-1-k^-, n-2-k^+)} + 1 \\ &= \begin{cases} \qty{k^- - k^+ - 1} + 1 & k^- > k^+ \\ 1 & \text{otherwise} \end{cases} \\ \\ \dim \ker L_1^* &= 2\cdot \size \theset{\ell \in \ZZ \suchthat 2\ell \in (k^- - n + 1, k^+ - n + 2)} \\ &= \begin{cases} k^+ - k^- + 1, & k^+ > k^- \\ 0 & \text{otherwise} \end{cases} \\ \implies \ind(L_1) &= \qty{ {k^- - k^+ -1 \over 2} + 1} - \qty{k^+ - k^- + 1 \over 2} = k^- - k^+ .\] > The other 2 cases involve different matrices $S_{k^\pm}$, but proceed similarly.