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See Floer Reading Group Fall 2020

Summary/Outline

Outline

What we’re trying to prove:

• 8.1.5: $(d{\mathcal{F}})_u$ is a Fredholm operator of index $\mu(x) - \mu(y)$.

What we have so far:

• Define $$\begin{align*} L: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s, t) Y \end{align*}$$ where $$\begin{align*} S: {\mathbb{R}}\times S^1 &\to \operatorname{Mat}(2n; {\mathbb{R}}) \\ S(s, t) &\overset{s\to\pm\infty}\to S^\pm(t) .\end{align*}$$

Outline

• Took $R^\pm: I \to {\mathsf{Sp}}(2n; {\mathbb{R}})$: symplectic paths associated to $S^\pm$

• These paths defined $\mu(x), \mu(y)$

• Section 8.7: $$\begin{align*} R^\pm \in {\mathcal{S}}\coloneqq\left\{{R(t) {~\mathrel{\Big\vert}~}R(0) = \operatorname{id}, ~ \operatorname{det}(R(1) - \operatorname{id})\neq 0}\right\} \implies L \text{ is Fredholm} .\end{align*}$$

• WTS 8.8.1: $$\begin{align*} \operatorname{Ind}(L)\stackrel{\text{Thm?}}{=} \mu(R^-(t)) - \mu(R^+(t)) = \mu(x) - \mu(y) .\end{align*}$$

From Yesterday

• Han proved 8.8.2 and 8.8.4.
  • So we know $\operatorname{Ind}(L) = \operatorname{Ind}(L_1)$
• Today: 8.8.5 and 8.8.3:
  • Computing $\operatorname{Ind}(L_1)$ by computing kernels.

8.8.5: $\dim \ker F, F^*$

Recall

$$\begin{align*} L: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s, t) Y \\ \\ L_{1}: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y \\ \\ L_{1}^{\star}: W^{1, q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Z & \longmapsto-\frac{\partial Z}{\partial s}+J_{0} \frac{\partial Z}{\partial t}+S(s)^t Z \end{align*}$$

Here ${1\over p} + {1\over q} = 1$ are conjugate exponents.

Reductions

$$\begin{align*} L_1^* &= -{\frac{\partial }{\partial s}\,} + J_0 {\frac{\partial }{\partial t}\,} + S(s)^t .\end{align*}$$

• Since $\operatorname{coker}L_1 \cong \ker L_1^*$, it suffices to compute $\ker L_1^*$

• We have

$$\begin{align*} J_0^1 \coloneqq \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \implies J_0 = \begin{bmatrix} J_0^1 & & & \\ & J_0^1 & & \\ & & \ddots & \\ & & & J_0^1 \end{bmatrix} \in \bigoplus_{i=1}^n \operatorname{Mat}(2; {\mathbb{R}}) .\end{align*}$$

• This allows us to reduce to the $n=1$ case.

Setup

$L_1$ used a path of diagonal matrices constant near $\infty$: $$\begin{align*} S(s) \coloneqq\left(\begin{array}{cc} a_{1}(s) & 0 \\ 0 & a_{2}(s) \end{array}\right), \quad \text { with } a_{i}(s)\coloneqq\left\{\begin{array}{ll} a_{i}^{-} & \text {if } s \leq-s_{0} \\ a_{i}^{+} & \text {if } s \geq s_{0} \end{array}\right. .\end{align*}$$

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Statement of Later Lemma (8.8.5)

Let $p>2$ and define $$\begin{align*} F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\end{align*}$$

Note: $F$ is $L_1$ for $n=1$: $$\begin{align*} L_{1}: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) & \longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2 n}\right) \\ Y & \longmapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\end{align*}$$

Statement of Lemma

$$\begin{align*} F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\end{align*}$$

Suppose $a_i^\pm \not \in 2\pi {\mathbb{Z}}$.

• Suppose $a_1(s) = a_2(s)$ and set $a^\pm \coloneqq a_1^\pm = a_2^\pm$. Then

$$\begin{align*} \operatorname{dim} \operatorname{Ker} F &= 2 \cdot {\sharp}\left\{\ell \in \mathbb{Z} {~\mathrel{\Big\vert}~} 2\pi \ell \in (a^-, a+) \subset {\mathbb{R}}\right \} \\ \operatorname{dim} \operatorname{Ker} F^{*} &= 2 \cdot {\sharp}\left\{\ell \in \mathbb{Z} {~\mathrel{\Big\vert}~} 2\pi\ell \in (a^+, a^-) \subset{\mathbb{R}} \right\} .\end{align*}$$

• Suppose $\sup_{s\in {\mathbb{R}}} {\left\lVert {S(s)} \right\rVert} < 1$, then

$$\begin{align*} \operatorname{dim} \operatorname{Ker} F &= {\sharp}\left\{i \in\{1,2\} {~\mathrel{\Big\vert}~}~a_{i}^{-}<0 \text { and } a_{i}^{+}>0\right\}\\ \operatorname{dim} \operatorname{Ker} F^{*} &={\sharp}\left\{i \in\{1,2\} {~\mathrel{\Big\vert}~}~ a_{i}^{+}<0 \text { and } a_{i}^{-}>0\right\} .\end{align*}$$

Statement of Lemma

In words:

• 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 $a^+$.

• Otherwise, if $S$ is uniformly bounded by 1, count the number of entries the flip from positive to negative as $s$ goes from $-\infty\to \infty$.

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Proof of Assertion 1

• Suppose $a_1(s) = a_2(s)$ and set $a^\pm \coloneqq a_1^\pm = a_2^\pm$. Then

$$\begin{align*} \operatorname{dim} \operatorname{Ker} F &= 2 \cdot {\sharp}\left\{\ell \in \mathbb{Z} {~\mathrel{\Big\vert}~} 2\pi \ell \in (a^-, a+) \subset {\mathbb{R}}\right \} \\ \operatorname{dim} \operatorname{Ker} F^{*} &= 2 \cdot {\sharp}\left\{\ell \in \mathbb{Z} {~\mathrel{\Big\vert}~} 2\pi\ell \in (a^+, a^-) \subset{\mathbb{R}} \right\} .\end{align*}$$

Step 1: Transform to Cauchy-Riemann Equations

• Write $a(s) \coloneqq a_1(s) = a_2(s)$.
• Start with equation on ${\mathbb{R}}^2$, $$\begin{align*}\mathbf{Y}(s, t) = \left[ Y_1(s, t), Y_2(s, t) \right].\end{align*}$$
• Replace with equation on ${\mathbb{C}}$: $$\begin{align*}\mathbf{Y}(s, t) = Y_1(s, t) + i Y_2(s, t).\end{align*}$$

Assertion 1, Step 1: Reduce to CR

• Expand definition of the PDE $$\begin{align*} F(\mathbf{Y}) = 0 \leadsto \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu\mathbf{Y} + S \mathbf{Y} = 0 \\ \\ \frac{\partial}{\partial s} \mathbf{Y} +\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right) \frac{\partial}{\partial t} \mathbf{Y} +\left(\begin{array}{cc} a(s) & 0 \\ 0 & a(s) \end{array}\right) \mathbf{Y} =0 .\end{align*}$$

• Change of variables: want to reduce to $\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu\tilde Y = 0$

• Choose $B \in \operatorname{GL}(1, {\mathbb{C}})$ such that $\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5muB + SB = 0$

• Set $Y = B\tilde Y$, which (?) reduces the previous equation to $$\begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu\tilde Y = 0 .\end{align*}$$

Assertion 1, Step 1: Reduce to CR

Can choose (and then solve) $$\begin{align*} B = \begin{bmatrix} b(s) & 0 \\ 0 & b(s) \end{bmatrix} {\quad \operatorname{where} \quad} {\frac{\partial b}{\partial s}\,} = -a(s)b(s) \\ \\ \implies b(s) = \exp{\int_0^s -a(\sigma) ~d\sigma} \coloneqq\exp{-A(s)} .\end{align*}$$

Remarks:

• For some constants $C_i$, we have

$$\begin{align*} A(s) = \begin{cases} C_1 + a^- s, & s \leq -\sigma_0 \\ C_2 + a^+ s, & s \geq \sigma_0 \\ \end{cases} .\end{align*}$$

• 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 $s+it \in {\mathbb{R}}\times S^1$ with $u = e^{2\pi z}$

• Apply Laurent’s theorem to $\tilde Y(u)$ on ${\mathbb{C}}\setminus\left\{{0}\right\}$ to obtain an expansion of $\tilde Y$ in $z$.

• Deduce that the solutions of the system are given by $$\begin{align*} \tilde Y (u) =\sum_{\ell \in \mathbf{Z}} c_{\ell} u^\ell \implies \tilde{Y}(s+i t) =\sum_{\ell \in \mathbf{Z}} c_{\ell} e^{(s+i t) 2 \pi \ell} .\end{align*}$$ where $\left\{{c_\ell}\right\}_{\ell\in{\mathbb{Z}}} \subset {\mathbb{C}}$ 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: $$\begin{align*} \tilde{Y}(s, t)=\sum_{\ell \in \mathbb{Z}} e^{2 \pi s \ell} \begin{bmatrix} \cos(2\pi\ell t) & -\sin(2\pi \ell t) \\ \sin(2\pi\ell t) & \cos(2\pi \ell t) \end{bmatrix} \begin{bmatrix} \alpha_\ell \\ \beta_\ell \end{bmatrix} .\end{align*}$$

Use $$\begin{align*} Y = B\tilde Y = \begin{bmatrix} e^{-A(s)} & 0 \\ 0 & e^{-A(s)} \end{bmatrix} \tilde Y \end{align*}$$

to write $$\begin{align*} Y(s, t)=\sum_{\ell \in \mathbb{Z}} e^{2 \pi s \ell} \begin{bmatrix} e^{-A(s)} & 0 \\ 0 & e^{-A(s)} \end{bmatrix} \begin{bmatrix} \cos(2\pi\ell t) & -\sin(2\pi \ell t) \\ \sin(2\pi\ell t) & \cos(2\pi \ell t) \end{bmatrix} \begin{bmatrix} \alpha_\ell \\ \beta_\ell \end{bmatrix} .\end{align*}$$

For $s\leq s_0$ this yields for some constants $K, K'$: $$\begin{align*} Y(s, t) = \sum_{\ell\in {\mathbb{Z}}} e^{2\pi\ell - a^-} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} .\end{align*}$$

Condition on $L^p$ Solutions

For $s\leq s_0$ we had $$\begin{align*} Y(s, t) = \sum_{\ell\in {\mathbb{Z}}} e^{\qty{2\pi\ell - a^-}s} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} \end{align*}$$

and similarly for $s\geq s_0$, for some constants $C, C'$ we have: $$\begin{align*} Y(s, t) = \sum_{\ell\in {\mathbb{Z}}} e^{\qty{2\pi\ell - a^+}s} \begin{bmatrix} e^C \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{C'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} .\end{align*}$$

Then $$\begin{align*} Y\in L^p \iff \text{exponential terms} \overset{\ell\to\infty}\to 0 .\end{align*}$$

Condition on $L^p$ Solutions: Small Tails

$$\begin{align*} Y(s, t) = \sum_{\ell\in {\mathbb{Z}}} e^{\qty{2\pi\ell - a^-}s} \begin{bmatrix} e^K \qty{\alpha_\ell \cos(2\pi\ell t) - \beta_\ell \sin(2\pi\ell t) } \\ e^{K'} \qty{ \alpha_\ell \sin(2\pi\ell t) + \beta_\ell \cos(2\pi \ell t)} \end{bmatrix} \end{align*}$$

• $\ell \neq 0$: 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

$$\begin{align*} \begin{cases} \alpha_\ell = \beta_\ell = 0 \text{ or } 2\pi\ell \in (a^-, a^+) & \ell\neq 0 \\ \qty{\alpha_0 = 0 {\operatorname{ or }}0 \in (a^-, a^+)} {\operatorname{ and }}\qty{\beta_0 = 0 {\operatorname{ or }}0\in (a^-, a^+)} & \ell = 0 \end{cases} .\end{align*}$$

• Finitely many such $\ell$ that satisfy these conditions
• Sufficient conditions for $Y(s, t) \in W^{1, p}$.

Compute dimension of space of solutions: ` \begin{align*} \operatorname{dim} \operatorname{Ker} F &=2 \cdot {\sharp}\left{{\ell \in \mathbb{Z}^{*} {~\mathrel{\Big\vert}~} 2\pi\ell \in (a^-, a^+) }\right}

• 2\cdot \indic{0 \in (a^-, a^+)} \ &=2 \cdot {\sharp}\left{\ell \in \mathbb{Z} {~\mathrel{\Big\vert}~}2\pi\ell \in (a^-, a^+) \right} .\end{align*} `{=html}

Note: not sure what ${\mathbb{Z}}^*$ is: most likely ${\mathbb{Z}}\setminus\left\{{0}\right\}$.

Counting Solutions

Use this to deduce $\dim \ker F^*$:

• $Y\in \ker F^* \iff Z(s, t) \coloneqq Y(-s, t)$ is in the kernel of the operator $$\begin{align*} \tilde F: W^{1, q}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Z &\mapsto \frac{\partial Z}{\partial s}+J_{0} \frac{\partial Z}{\partial t}+S({\color{red}-s}) Y .\end{align*}$$

• 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 $\sup_{s\in {\mathbb{R}}} {\left\lVert {S(s)} \right\rVert} < 1$, then $$\begin{align*} \operatorname{dim} \operatorname{Ker} F &= {\sharp}\left\{i \in\{1,2\} {~\mathrel{\Big\vert}~}~a_{i}^{-}<0 < a_{i}^{+}\right\}\\ \operatorname{dim} \operatorname{Ker} F^{*} &={\sharp}\left\{i \in\{1,2\} {~\mathrel{\Big\vert}~}~ a_{i}^{+}<0 < a_{i}^{-} \right\} .\end{align*}$$

We use the following:

• Lemma 8.8.7: $$\begin{align*} \sup_{s\in {\mathbb{R}}} {\left\lVert { S(s) } \right\rVert} < 1 \implies \text{the elements in }\ker F,~ \ker F^* \text{ are independent of }t .\end{align*}$$

• Proof: in subsection 10.4.a.

Proof of Assertion 2

$$\begin{align*} F: W^{1, p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) &\longrightarrow L^{p}\left(\mathbb{R} \times S^{1} ; \mathbb{R}^{2}\right) \\ Y &\mapsto \frac{\partial Y}{\partial s}+J_{0} \frac{\partial Y}{\partial t}+S(s) Y .\end{align*}$$

• Given as a fact: ` \begin{align*} \mathbf{Y} \in \ker F \implies {\frac{\partial }{\partial s},}\mathbf{Y} = \mathbf{a}(s)\mathbf{Y} ~~~

  -a_1(s) & 0 \\
  0 & -a_2(s)
  \end{bmatrix}
  \mathbf{Y}
  .\end{align*}
  <span>`{=html}
  

• Therefore we can solve to obtain $$\begin{align*} \mathbf{Y}(s) = \mathbf{c}_0 \exp{-\mathbf{A}(s)}{\quad \operatorname{where} \quad} \mathbf{A}(s) = \int_0^s -\mathbf{a}(\sigma) ~d\sigma .\end{align*}$$

Proof of Assertion 2

• Explicitly in components: $$\begin{align*} \begin{cases} {\frac{\partial Y_1}{\partial s}\,} &= -a_1(s) Y_1 \\ {\frac{\partial Y_s}{\partial s}\,} &= -a_2(s) Y_2 \\ \end{cases} \quad \implies \quad Y_i(s) = c_i e^{-A_i(s)}, \quad A_i(s) &= \int_0^s -a_i(\sigma) ~d\sigma .\end{align*}$$

• As before, for some constants $C_{j, i}$, $$\begin{align*} A_i(s) = \begin{cases} C_{1, i} + a_i^-\cdot s & s \leq -\sigma_0 \\ C_{2, i} + a_i^+\cdot s & s \geq \sigma_0 \\ \end{cases} .\end{align*}$$

• Thus $$\begin{align*} Y_i \in W^{1, p} \iff 0 \in (a_i^-, a_i^+) ,\end{align*}$$

  establishing

$$\begin{align*} \dim \ker F = {\sharp}\left\{i \in\{1,2\} {~\mathrel{\Big\vert}~}0 \in (a_i^-, a_i^+) \right\} .\end{align*}$$

8.8.3: $\operatorname{Ind}(L_1) = k^- - k^+$

Statement and Outline

Statement: let $k^\pm \coloneqq\operatorname{Ind}(R^\pm)$; then $\operatorname{Ind}(L_1) = k^- - k^+$.

• Consider four cases, depending on parity of $k^\pm - n$
• Show all 4 lead to $\operatorname{Ind}(L_1) = k^- - k^+$

• $k^- \equiv k^+ \equiv n \operatorname{mod}2$.
• $k^- \equiv n, k^+ \equiv n-1 \operatorname{mod}2$
• $k^- \equiv n-1, k^+ \equiv n \operatorname{mod}2$.
• $k^- \equiv k^+ \equiv n-1 \operatorname{mod}2$

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Case 1: $k^+ \equiv k^- \equiv n \operatorname{mod}2$

$$\begin{align*} S_{k^-} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -\pi & & & \\ & & & & & -\pi & & \\ & & & & & & (n-1-k^-)\pi & \\ & & & & & & & (n-1-k^-)\pi \\ \end{bmatrix} \\ S_{k^+} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -\pi & & & \\ & & & & & -\pi & & \\ & & & & & & (n-1-{\color{blue}k^+})\pi & \\ & & & & & & & (n-1-{\color{blue}k^+})\pi \\ \end{bmatrix} .\end{align*}$$

Case 1: $k^- \equiv k^+ \equiv n \operatorname{mod}2$

• Take $a_1(s) = a_2(s)$ so $a_1^\pm = a^\pm$
• Apply the proved lemma to obtain

$$\begin{align*} \dim \ker L_1 &= 2\cdot {\sharp}\left\{{\ell \in {\mathbb{Z}}{~\mathrel{\Big\vert}~}2\ell \in (n-1-k^-, n-1-k^+)}\right\} \\ &= \begin{cases} k^- - k^+ & k^- > k^+ \\ 0 & \text{else} \end{cases} \\ \\ \dim \ker L_1^* &= 2\cdot {\sharp}\left\{{ \ell \in {\mathbb{Z}}{~\mathrel{\Big\vert}~}2\ell \in (k^- - n + 1, k^+ - n + 1)}\right\} \\ &= \begin{cases} k^+ - k^- & k^+ > k^- \\ 0 & \text{otherwise} \end{cases} \\ \\ \implies \operatorname{Ind}(L_1) &= \qty{k^- - k^+ \over 2} - \qty{k^+ - k^- \over 2} = k^- - k^+ .\end{align*}$$

Case 2: $k^+ \not\equiv k^- \equiv n \operatorname{mod}2$

$$\begin{align*} S_{k-} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & -{\color{red}{\varepsilon}}\pi & & & \\ & & & & & -{\color{red}{\varepsilon}}\pi & & \\ & & & & & & (n-1-{\color{red}k^-})\pi & \\ & & & & & & & (n-1-{\color{red}k^-})\pi \\ \end{bmatrix} \\ S_{k^+} & = \begin{bmatrix} -\pi & & & & & & & \\ & -\pi & & & & & & \\ & & \ddots & & & & & \\ & & & & {\color{red}{\varepsilon}} & & & \\ & & & & & -{\color{red}{\varepsilon}} & & \\ & & & & & & (n-{\color{red}2}-k^+)\pi & \\ & & & & & & & (n-{\color{red}2}-k^+)\pi \\ \end{bmatrix} .\end{align*}$$

Case 2: $k^+ \not\equiv k^- \equiv n \operatorname{mod}2$

• Take $a_1(s) = a_2(s)$ everywhere except the $n-1$st block, where we can assume $\sup_{s\in {\mathbb{R}}} {\left\lVert {S(s)} \right\rVert} < 1$.
• Assertion 2 applies and we get

$$\begin{align*} \dim \ker L_1 &= 2\cdot {\sharp}\left\{{\ell \in {\mathbb{Z}}{~\mathrel{\Big\vert}~}2\ell \in (n-1-k^-, n-2-k^+)}\right\} + 1 \\ &= \begin{cases} \qty{k^- - k^+ - 1} + 1 & k^- > k^+ \\ 1 & \text{otherwise} \end{cases} \\ \\ \dim \ker L_1^* &= 2\cdot {\sharp}\left\{{\ell \in {\mathbb{Z}}{~\mathrel{\Big\vert}~}2\ell \in (k^- - n + 1, k^+ - n + 2)}\right\} \\ &= \begin{cases} k^+ - k^- + 1, & k^+ > k^- \\ 0 & \text{otherwise} \end{cases} \\ \implies \operatorname{Ind}(L_1) &= \qty{ {k^- - k^+ -1 \over 2} + 1} - \qty{k^+ - k^- + 1 \over 2} = k^- - k^+ .\end{align*}$$

The other 2 cases involve different matrices $S_{k^\pm}$, but proceed similarly.