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=$\mu$}
{wiki}

The Maslov index assigns an <integer> to a loop of <Lagrangian subspaces>. Under $\operatorname{LGr}(\mathbb R^{2n})\cong U(n)/O(n)$, it is the degree of the map induced by $A\mapsto\det(A)^2$.

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 Incoming links: Maslov index