On the quotient from part (b), the <Maslov map>
$$
\mu:U(n)/O(n)\longrightarrow S^1,
\qquad
[A]\longmapsto\det(A)^2
$$
is well defined because $\det(Q)^2=1$ for $Q\in O(n)$. Consider the loop of <Lagrangian subspaces>
$$
\ell(t)=e^{\pi it}\mathbb R\oplus\mathbb R^{n-1},
\qquad 0\leq t\leq1.
$$
Its endpoints agree as unoriented subspaces, and $\mu\circ\ell(t)=e^{2\pi it}$ has degree one. If $\eta$ is the generator of $H^1(S^1;\mathbb Z)$, then
$$
\langle\mu^*\eta,[\ell]\rangle=1.
$$
Consequently $\mu^*\eta\ne0$, proving
$$
H^1(\operatorname{LGr}(\mathbb R^{2n});\mathbb Z)\ne0.
$$
The same integer is the <Maslov index> of $\ell$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 1 / c / Solution

{c}
{title2=$\det^2:U(n)/O(n)\to S^1$}

The Maslov map is well defined because every matrix in $O(n)$ has determinant $\pm1$. The loop
$$
t\longmapsto e^{\pi it}\mathbb R\oplus\mathbb R^{n-1}
$$
maps to $e^{2\pi it}$ and therefore has Maslov index one.


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