conjugation from Cappell, Lee and Miller, and we explain that our result actually is a generalisation of Hu and Portaluri’s Theorem.

Maslov Index and Spectral Flow - a brief recap
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The Maslov Index {#section-Maslov}
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The aim of this section is to briefly recall the definition of the Maslov index, where we follow [@PiccioneBook].\
Let $\operatorname{Sp}(2n,\mathbb{R})$ denote the group of symplectic matrices on $\mathbb{R}^{2n}$, i.e., those $A\in M(2n,\mathbb{R})$ satisfying $A^TJA=J$ or, alternatively, which preserve $\omega_0$. If we identify $\mathbb{R}^{2n}$ with $\mathbb{C}^n$ by $(x_1,\ldots,x_{2n})\mapsto (x_1,\ldots,x_n)+i(x_{n+1},\ldots, x_{2n})$ then the standard hermitian scalar product on $\mathbb{C}^n$ is

$$\langle x,y\rangle_{\mathbb{C}}=\langle x,y\rangle-i\omega_0(x,y).$$ Hence each unitary matrix $U\in U(n)$ preserves $\omega_0$ and so we can regard $U(n)$ as a subset of $\operatorname{Sp}(2n,\mathbb{R})$. Also, the orthogonal matrices $O(n)$ can be seen as a subgroup of $U(n)$ by complexification. Then $O(n)$ consists exactly of those $A\in U(n)$ which leave $\mathbb{R}^n\times\{0\}$ invariant.\
Obviously, $AL\in\Lambda(n)$ if $L\in\Lambda(n)$ and $A\in\operatorname{Sp}(2n,\mathbb{R})$, and it can be shown that the restriction of this action to $U(n)\times \Lambda(n)\rightarrow\Lambda(n)$ is transitive. As the stabiliser subgroup of $\mathbb{R}^n\times\{0\}\in\Lambda(n)$ is $O(n)$, we see that there is a diffeomorphism

$$\begin{aligned}
\label{diffeoU}
U(n)/O(n)\simeq\Lambda(n),\quad A\mapsto A(\mathbb{R}^n\times\{0\}).\end{aligned}$$

Let us now consider the map $d:U(n)\rightarrow S^1$, $d(A)=\det^2(A)$, which descends to the quotient 