Section \[section-spectralflow\] and the fact that $\ker(\mathcal{A}_\lambda)=\gamma_1(\lambda)\cap\gamma_2(\lambda)$. Also, (ii’) follows from (iii) in Section \[section-spectralflow\]. Finally, (ii’) is an immediate consequence of the homotopy invariance (i) of the spectral flow and Proposition \[prop-continuity\].\
Hence it remains to show that $\mu(\gamma_{nor},L_1)=1$ and $\mu(L_0,\gamma'_{nor})=-1$, which will be a direct consequence of the following lemma.

The spectra of the operators $\mathcal{A}_\lambda$ in are

-   for $(\gamma_1,\gamma_2)=(\gamma_{nor},L_1)$ $$\sigma(\mathcal{A}_\lambda)=\left\{\pi\lambda-\frac{\pi}{2}+\pi k:\,k\in\mathbb{Z}\right\}\cup\left\{\frac{\pi}{2}+k\pi:\,k\in\mathbb{Z} \right\},$$

-   for $(\gamma_1,\gamma_2)=(L_0,\gamma'_{nor})$

    $$\sigma(\mathcal{A}_\lambda)=\left\{-\pi\lambda+\frac{\pi}{2}+\pi k:\,k\in\mathbb{Z}\right\}\cup\left\{\frac{\pi}{2}+k\pi:\,k\in\mathbb{Z}\right\}.$$

We consider $Ju'=\mu u$ and note that the solutions of this equation are

$$u(t)=\exp(-\mu tJ)v,\quad t\in I, v\in\mathbb{R}^{2n}.$$ Let us first consider the path in (i). Then $u$ belongs to the domain of $\mathcal{A}_\lambda$ if and only if

$$\begin{aligned}
\label{equ}
u(0)=v\in\gamma_{nor}(\lambda),\quad u(1)=\exp(-\mu J)v\in L_1.\end{aligned}$$

As $\exp(-\mu J)v\in L_1$ if and only if $v\in\exp(\mu J)L_1$, and $\exp(\mu J)=\cos(\mu)I_{2n}+\sin(\mu)J$, we see that is equivalent to

$$(\cos(\mu)I_{2n}+\sin(\mu)J)(\{0\}\times\mathbb{R}^n)\cap\left(\mathbb{R}(\cos(\pi\lambda)e_1+\sin(\pi\lambda)e_{n+1})+\sum^{n}_{j=2}{\mathbb{R}e_j}\right)\neq\{0\}.$$ There are two different cases where these spaces intersect non-trivially. Firstly, if $\cos(\mu)=0$, i.e. $\mu=\frac{\pi}{2}+k\pi$ for $k\in\mathbb{Z}$. Secondly, if there is an $\alpha\neq 0$ such that $\sin(\pi\lambda)e_{n+1}=\alpha\cos(\mu)e_{n+1}$ and $\cos(\pi\lambda)e_1=-\alpha\sin(\mu)e_1$, where we use that $Je_{n+1}=-e_1$. Of course, the latter equations are equivalent to $\sin(\pi\lambda)=\alpha\cos(\mu)$ and $\cos(\pi\lambda)=-\alpha\sin(\mu)$, which can be rewritten as

$$\begin{aligned}
e^{i\pi\lambda}&=\cos(\pi\lambda)+i\sin(\pi\lambda)=\alpha (-\sin(\mu)+i\cos(\mu))=\alpha i e^{i\mu}=\alpha e^{i(\mu+\frac{\pi}{2})}.\end{aligned}$$

Hence $|\alpha|=1$, and this equation holds if and only 