--- title: 'Floer Talk: 8.6 - 8.8, Setup for Computing the Index' --- Tags: #geomtop/symplectic-topology #geomtop/Floer-theory #projects/notes/reading #projects/my-talks Outline - Sketch proof of 8.6.3 - Statement of Somewhere Injectivity - Statement of Continuation Principle - Statement of Similarity Principle - 8.7 - 8.8 What we're trying to prove 8.1.5: $(d\mcf)_u$ is a Fredholm operator of index $\mu(x) - \mu(y)$. # 8.8 - Define \[ L: W^{1, p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s, t) Y \] - By the end of 8.8: replace the Fredholm operator $L$ by an operator $L_1$ with the same *index* (not the same kernel/cokernel) - Compute the index of $?$ because we can explicitly describe its kernel and cokernel - Use the fact $S: \RR \cross S^1 \to \mat(2n;\RR)$ and $$ S(s, t) \converges{s\to\pm\infty}\to S^{\pm}(t) $$ which are symmetric. - Take corresponding symplectic paths $R^\pm: I \to \Sp(2n; \RR)$. - If $$ R^\pm \in \mcs \definedas \theset{R: I \to \Sp(2n; \RR) \suchthat R(0) = \id, \det(R(1) - \id)\neq 0} ,$$ then $L$ is a Fredholm operator - Theorem 8.8.1: $\ind(L) = \mu(R^-(t)) - \mu(R^+(t)) = \mu(x) - \mu(y)$. - Prop 8.8.2: Define an operator \[ L_{1}: W^{1, p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y \] where $S: \RR \to\mat(2n; \RR)$ is a path of diagonal matrices depending on $\ind(R^\pm(t))$; then $\ind(L_1) = \ind(L)$. - Then $\ind(L_1) = \ind(R^-(t)) - \ind (R^+(t))$. - Idea of proof: take a homotopy of operators \[ L_{\lambda}: W^{1, p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S_{\lambda}(s, t) Y \] which are all Fredholm and all have the same index, then take time 1. - Use the fact that $\coker L_1 \cong \ker L_1^*$, and we can explicitly write the adjoint of $L_1$. - Get a formula that resembles the Morse case: counting the number of eigenvalues that change sign. - Summary: - Replace $L$ by $L_0$, which is modified in a neighborhood of zero in the $s$ variable. Use invariance of index under small perturbations. - Homotope $L_0$ to $L_1$, where $S$ is replaced by a diagonal matrix $S(s)$ that is a constant matrix outside the neighborhood of zero in $s$. Use invariance of index under homotopy. # 8.7 - Goal: Toward 8.1.5, show that $L \definedas \bar \del + S(s, t): W^{1, p} \to L^p$ is a Fredholm operator, i.e. the index makes sense (finite-dimensional kernel and cokernel). - Statement: if $\det(\id - R^\pm_1) \neq 0$ then the operator \[ L: W^{1, p}(\RR\cross S^1; \RR^{2n}) \to L^p(\RR\cross S^1; \RR^{2n})\\ .\] given by $L = \bar\del + S(s, t)$ is Fredholm for every $p>1$. > Most of the work goes into showing that $\dim(\ker L) < \infty$ and $\im(L)$ is closed. - $\dim \ker L < \infty$: - Use elliptic regularity and consequence of Calderón-Zygmund inequality. - Elliptic regularity: If $Y \in L^p$ is a weak solution of the linearized Floer equation $LY = 0$, then $Y\in W^{1, p}$ and is smooth. - Consequence: If $Y\in W^{1, p}$ and $p>1$ then $\norm{Y}_{W^{1, p}} = O(\norm{LY}_{L^p} + \norm{Y}_{L^p} )$. - Case 1: $S(s, t) = S(t)$ doesn't depend on $s$. - Then $L$ is a bijective for every $p>1$ - Invertibility allows replacing the weak solution $Y \in L^p(\RR \cross S^1; \cdot)$ with $Y\in L^p([-M, M] \cross S^1, \cdot)$ for $M\gg 0$ - Restriction $\bar L: W^{1, p}(\RR\cross S^1) \to L^p([-M, M] \cross S^1)$ is a compact operator, it is "semi-Fredholm" - Apply a theorem: if the inequality is satisfies, the kernel is finite-dimensional and the image is closed. - $\dim \coker L < \infty$: - Take \[ L: W^{1, p}(\RR\cross S^1; \RR^{2n}) \to L^p(\RR\cross S^1; \RR^{2n})\\ \] and consider the adjoint operator \[ L^{\star}: W^{1, q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \longrightarrow L^{q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \] where $p\inv + q\inv = 1$, which appeared in 8.5.1. - Use the fact that $\coker L = \ker L^\star$ - Show $\dim \ker L^\star < \infty$ since it satisfies the conditions of 8.7.4. # 8.6 - Start with $u \in C^\infty(\RR \cross S^1; W)$ a solution to the equation $$\dd{u}{s} + J_1(t, u) \qty{\dd{u}{t} - X(t, u)} = 0$$ - "Unwrap the cylinder" to a map $v\in C^\infty(\RR^2; W)$ which is a solution to $\qty{\dd{}{s} + J\dd{}{t}}u = 0$ - No longer periodic; instead $v(s, t+1) = \phi(v(s, t))$. - Built by precomposing $u$ with the inverse flow $\psi_t$ of $X_t$ - Conjugate original $J$ with $\psi$ to get $J_1$ - Define regular points $R(v)$ as $\dd{}{s}v \neq 0$ with injectivity condition: $s \neq s' \implies v(s, t) \neq v(s', t)$. - Prove "injectivity result": $R(v) \subseteq \RR^2$ is dense and open ## 8.6.3, Part 1: R(v) is Open - Prove $R(v)$ is open: contradict zero derivative - Proof uses sequential characterization of being a closed set - Construct a sequence in the complement converging to a regular point - Since first two conditions of $R(v)$ are open, extract a sequence of points failing injectivity - Show it is bounded - Apply Bolzano-Weierstrass to extract a convergent subsequence - Use quotient definition of derivative, show it is zero, contradiction. ## 8.6.3, Part 2: R(v) is Dense in R2 (p.258) ## Step 1: Exclude critical points that are also multiple points - Definition: *Multiple points* are where injectivity fails in $s$. - Characterize $R(v)$ as those in $C(v)^c$ that are not multiples. - Suffices to show $R(v)$ is dense in $C(v)^c$ -- Every $(s, t) \in C(v)^c \subseteq \RR^2$ is the limit of $(s_n, t) \in C(v)^c$ where $v(s_n, t) \neq x^\pm(t)$. - Why? $v(s + {1\over n}, t) = x^+(t) \implies \dd{v}{s} = 0 \implies (s+{1\over n}, t)\in C(v)$. - Then suffices to show every $(s_0, t_0) \in \RR \cross I$ with (importantly) \[ \dd{v}{s}{(s_0, t_0)} \neq 0 \qtext{and} v(s_0, t_0) \neq x^{\pm}(t_0) .\] {#eq:eq1} is the limit of a sequence of points in $R(v)$. - Proceed by assuming this is not the case, toward a contradiction. ### A Small Ball Avoids Critical Points in the Image - Surround every point $(s_0, t_0)$ by a ball $B_\eps(s_0, t_0)$ missing $R(v)$ - We can choose $\eps$ small enough and $M\gg 1$ big enough (defining $\mathbf{M} = [-M, M] \subset \RR$) such that 1. Translate to far enough to get a point outside the image of the ball: \[ (s, t) \in \mathbf{M}^c &\cross [t_0-\eps, t_0+\eps]\subset \RR \cross I \implies \\ v(s, t) \intersect v(B_\eps(s_0, t_0) = \emptyset &\qtext{and} x^\pm(t) \not \in v(B_\eps(s_0, t_0)) .\] - Idea: If not, can cook up sequences that force $v(s_0, t_0) = x^\pm(t_0)$, a contradiction to @eq:eq1. 2. For $t \in B_\eps(t_0)$, $B_\eps(s_0) \injects W$ is an injective immersion - Combine 1 and 2 to show that - $v$ is locally constant - $(s_0 ,t_0) \in C(v)$ - Every point in $B_\eps(s_0, t_0)$ satisfies [@eq:eq1] 3. - Show $$\abs{\mathbf{M}_C}\definedas \abs{\qty{\mathbf{M} \cross I} \intersect C(v)} < \infty$$ since it's the intersection of a compact and a discrete set - Perturb $(s_0, t_0)$ so that $(s, t) \in \mathbf{M}_C \implies v(s_0, t_0) \neq v(s, t)$. - Possible since $(s_0, t_0) \not\in C(v) \implies v$ is non-constant in a neighborhood of $(s_0, t_0)$. - Decrease $\eps$ so that $$v(B_\eps(s_0, t_0)) \intersect v(\mathbf{M}_C) = \emptyset.$$ - This means that in the thick strip containing $(s_0, t_0)$, no critical points land in its image. - Conclude that we only have to consider injectivity, not critical points that are also multiple points ## Step 2: Failure of Injectivity in Small Boxes - Define $$\mathbf{S}_M(t_0) = \theset{s_1, \cdots, s_N} =\theset{s_i \in [-M, M] \suchthat v(s_i, t_0) = v(s_0, t_0)},$$ the set of multiple points over $s_0$. - This is finite, since infinite $\implies$ has a limit point $\implies \dd{v}{s} = 0$, contradiction. - Lemma: For every $\eps >0$ there exists a $\delta > 0$ such that defining $$\Delta_0 = [s_0-\delta, s_0 + \delta] \cross [t_0-\delta, t_0 + \delta]$$ then $$(s, t)\in \Delta \implies \exists s' \in B_\eps(s_j) \text{ s.t. } v(s, t) = v(s', t)$$ for some $1 \leq j \leq N$. - In English: for every epsilon, we can find a delta box $\Delta_0 \ni (s_0, t_0)$ such that every point in $\Delta_0$ is a multiple point over *some* point in an epsilon ball around *some* point in $\mathbf{S}_M(t_0)$. - Proof idea: otherwise, construct a sequence $(\sigma_n, t_n) \to (s_0, t_0)$, use properties 1 and 2 from earlier, extract a limit point $\sigma$ *not* in $\mathbf{S}_M(t_0)$ which satisfies $v(\sigma, t_0) = v(s_0, t_0)$, a contradiction (since that set contained exactly the multiple finitely many points). - Fix $\eps' <\eps/2$ form Step 1 and apply the lemma to $\eps'$ to produce a $\delta$ and a box $\Delta_0$. - Apply the lemma: shrink the delta box $\Delta_0$ to a closed delta ball $\bar B_\delta(s_0, t_0)$. - Every point in the delta box is a multiple point over some $s_j$. - So partition the ball up: define $\Sigma_j$ to be all of the multiple points over $s_j \in \mathbf{S}_(t_0)$. - Take a smaller $\rho\dash$ball around some point $(s_\star, t_\star) \in \Sigma_1^\circ$, making sure to choose $\eps'$ small enough such that $$ B_\rho(s_\star, t_\star) \intersect \qty{ [s_1-\eps', s_1 + \eps'] \cross [t_0-\delta, t_0 + \delta] } = \emptyset .$$ In other words, the shaded region is disjoint from the $\rho\dash$ball. - Then every point in the $\rho\dash$ball is a multiple point over some point in the box around $(s_1, t_0)$. Pick on such point $(s_\star', t_\star)$ on the $t_\star$ line. ## Step 3: Final Contradiction - Construct $v_1, v_2$ which - Satisfy the same Cauchy-Riemann equations - Are equal at the origin - Have nonzero derivative at the origin. - We want to show they are equal on $\RR^2$ - Constructing them: use points from step 2 to translate - Obtain $(s_\star, t_\star)$ and $(s_\star', t_\star)$ from previous step. - Define \[ \begin{array}{ll} v_{1}(s, t)=v\left(s+s_{\star}, t+t_{\star}\right) &\implies v_1(z) = v(z + w_1)\\ v_{2}(s, t)=v\left(s+s_{\star}^{\prime}, t+t_{\star}\right) &\implies v_2(z) = v(z + w_2) \end{array} \] - Satisfy the same CR equations - By construction, they coincide at $(0, 0)$ since $v(s_\star, t_\star) = v(s_\star', t_\star)$. - Derivatives at the origin are nonzero, coming from the fact that $\dd{v}{s}(s_\star, t_\star) \neq 0$. - Now work at zero: for every $(s, t) \in B_\rho(0, 0)$ there exists a multiple point $s' \in B_{2\eps'}(0)$ over $s$. - Use the following extension lemma, consequence of **Continuation Principle**: in this situation, with $X_t \equiv 0$ on $B_\eps(\vector 0)$, then $$z\in B_\eps(\vector 0) \implies v_1(z) = v_2(z).$$ - Define \[ \mathcal{F}: \mathrm{C}^{\infty}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) & \longrightarrow \mathrm{C}^{\infty}\left(\mathbf{R} \times S^{1} ; \mathbf{R}^{2 n}\right) \\ w &\longmapsto \qty{\dd{}{s} + J\cdot \dd{}{t} }w \] - Since $v_1, v_2$ satisfy the same CR equation, $\mcf(v_1) = \mcf(v_2)$ - Linearize $\mcf$ as we did for the Floer operator to obtain \[ \qty{d\mcf}_{\cdots}(Y) = \qty{ \dd{}{s} + J_0 \cdot \dd{}{t} + \tilde S } Y .\] where $\tilde S: I \cross \RR^2 \to \endo(\RR^{2n})$ - Set $Y = v_1 - v_2$, then $$ S = \int_{[0, 1]} \tilde S \implies S\qty{ \dd{}{s} + J_0 \cdot \dd{}{t} + S } Y = 0 $$ - From above, $Y \equiv 0$ in $B_\eps(\vector 0)$, apply **Continuation Principle** to obtain $v_1 = v_2$ on $\RR^2$ - Inductive argument to show $$ \forall k\in\ZZ,\quad v(s, t) = v(k(s_\star' - s_\star), t) \converges{k\to \infty}\longrightarrow x^\pm(t), $$ which is the desired contradiction. $\qed$ BREAK ## The Continuation Principle - Take the perturbed CR equation \[ \qty{\dd{}{s} + J_0 \cdot \dd{}{t} + S }Y &= 0 \qtext{where} S: \RR^2 \longrightarrow \endo(\RR^{2n}) \] where $J_0$ is the standard complex structure on $\RR^{2n}$. - Define an *infinite-order zero* $z$ of an arbitrary function $f$ as \[ z_0 \in Z_\infty \iff \sup_{B_r(z_0)} {\abs{f(z)} \over r^k} \converges{r\to 0}\to 0 \quad \forall k\in \ZZ^{\geq 0} ,\] or for a smooth function, \[ z_0\in Z_\infty \iff f^{(k)}(z_0)= 0 \quad \forall k \in \ZZ^{\geq 0} .\] - Statement: If $Y$ solves CR on $U \subset \RR^2$ then the set \[ C \definedas \theset{(s, t)\in U \suchthat ~Y \text{ is an infinite-order zero at } (s, t)} .\] - Explanation: for $f$ smooth, $Z_\infty$ is closed. For $f$ holomorphic, it is clopen. - From complex analysis: for a connected domain $\Omega$, the only clopen subsets are $\emptyset,\Omega$, so nonempty and $f =g$ on a connected subset implies $f = g$ on $\Omega$. - In particular, $Y = 0$ on $U' \subseteq U$ implies $Y = 0$ on $U$. - Prove is a consequence of the **Similarity Principle** ## Similarity Principle - Statement: - Recall definition of perturbed CR equation: \[ \qty{\dd{}{s} + J_0 \cdot \dd{}{t} + S }Y &= 0 \qtext{where} S: \RR^2 \longrightarrow \endo(\RR^{2n}) \] - Let - $Y \in C^\infty(B_\eps; \CC^n)$, or more generally $W^{1, p}( B_\eps; \CC^n)$, be a solution - $S \in C^\infty(\RR^2, \endo(\RR^2) )$ or more generally $L^p(B_\eps; \endo_\RR(\RR^{2n}))$ - $p > 2$ Then there exist \[ \delta &\in (0, \eps), \\ \sigma &\in C^\infty(B_\delta, \CC^n) \\ A &\in W^{1, p}(B_\delta, \GL(\RR^{2n})) .\] such that \[ \forall(s, t) \in B_{\delta}, ~~\quad Y(s, t)=A(s, t) \cdot \sigma(s+i t) \qtext { and } J_0 A(s, t) = A(s, t)J_0 .\] - Used to prove: - $C(v)$ is discrete - "Extension" lemma used to prove $R(v)$ is dense - Some ideas from proof: - Matrix $A$ will look like the fundamental matrix of solutions to the equation - Compactify to get $B_\delta \subset S^2$, if $Y: S^2 \to \CC^n$ then we can consider the section \begin{center} \begin{tikzcd} \left(A^{0,1} T^{\star} S^{2}\right)^{n}=\Lambda^{0,1} T^{\star} S^{2} \otimes \mathbb{C}^{n} \ar[d]\\ X \ar[u, "\bar \del Y"', bend right=50, dotted] \end{tikzcd} \end{center} - $\bar Y = 0$ makes $Y$ a holomorphic sphere in $\CC^n$. ## Odds and Ends - Theorem: the following is a surjective Fredholm operator for every $p> 1$: \[ \bar{\partial}: W^{1, p}\left(S^{2}, \mathbb{C}^{n}\right) \longrightarrow L^{p}\left(\Lambda^{0,1} T^{\star} S^{2} \otimes \mathbb{C}^{n}\right) .\] - Computation will show that $\dim \ker \bar \del = \dim \coker \bar \del = 2n$, so $\ind \bar \del = 0$.