by

$$\overline{d}:U(n)/O(n)\rightarrow S^1,\quad A\cdot O(n)\mapsto \det{^2}(A).$$ Note that $$\ker(d)/O(n)\hookrightarrow U(n)/O(n)\xrightarrow{\overline{d}} S^1$$ is a fibre bundle, and it is not difficult to see that $\ker(d)/O(n)\simeq SU(n)/SO(n)$, where the latter space is simply connected. It follows from the long exact sequence of a fibre bundle that the induced map

$$\overline{d}_\ast:\pi_1(U(n)/O(n))\rightarrow\pi_1(S^1)\cong\mathbb{Z}$$ is an isomorphism. Consequently, we obtain from an isomorphism

$$\mu_{Mas}:\pi_1(\Lambda(n))\rightarrow\mathbb{Z},$$ which is the *Maslov index* for closed paths in $\Lambda(n)$. Roughly speaking, given an arbitrary $L_0\in\Lambda(n)$, the Maslov index counts the total number of intersections of a loop in $\Lambda(n)$ with $L_0$. This is independent of the particular choice of $L_0$, which however is no longer the case if we extend the definition to non closed paths in $\Lambda(n)$ as follows.\
We fix $L_0\in\Lambda(n)$ and note at first that $L_0$ yields a stratification

$$\Lambda(n)=\bigcup^n_{k=0}\Lambda_k(L_0),$$ where

$$\Lambda_k(L_0)=\{L\in\Lambda(n):\, \dim(L\cap L_0)=k\}.$$ From the fact that $\Lambda_0(L_0)$ is contractible (see e.g. [@PiccioneBook Rem. 2.5.3]) and the long exact sequence of homology, we see that the inclusion induces an isomorphism

$$H_1(\Lambda(n))\rightarrow H_1(\Lambda(n),\Lambda_0(L_0)).$$ Also, as $\pi_1(\Lambda(n))$ is abelian, $H_1(\Lambda(n))$ is isomorphic to $\pi_1(\Lambda(n))$ and so we obtain a sequence of isomorphisms

$$\begin{aligned}
\label{Maslovsequence}
H_1(\Lambda(n),\Lambda_0(L_0))\rightarrow H_1(\Lambda(n))\rightarrow \pi_1(\Lambda(n))\rightarrow\pi_1(U(n)/O(n))\rightarrow\mathbb{Z}.\end{aligned}$$

Finally, every path in $\Lambda(n)$ having endpoints in $\Lambda_0(L_0)$ canonically yields an 