The <unitary group> acts on the <Lagrangian Grassmannian> by $A\cdot L=A(L)$. Every Lagrangian subspace has an orthonormal basis, and adjoining its $J_0$-image gives a unitary basis, so this action is transitive. The stabilizer of the standard real subspace $\mathbb R^n\subset\mathbb C^n$ consists exactly of real unitary matrices, namely $O(n)$. Hence
$$
U(n)/O(n)\longrightarrow\operatorname{LGr}(\mathbb R^{2n}),
\qquad
A,O(n)\longmapsto A\mathbb R^n
$$
is a continuous bijection from a compact space to a Hausdorff space and is therefore a <homeomorphism>.

For $n=1$, every line in $\mathbb R^2$ is Lagrangian. Thus
$$
\operatorname{LGr}(\mathbb R^2)=\mathbb{RP}^1
\cong U(1)/O(1)=S^1/\{\pm1\}\cong S^1.
$$
Concretely, the line making angle $\theta$ with the real axis corresponds to $e^{2i\theta}$.

Solution

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

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The Lagrangian Grassmannian is the space of all <Lagrangian subspaces> of a symplectic vector space. The <unitary group> acts transitively on $\operatorname{LGr}(\mathbb R^{2n})$, and the stabilizer of $\mathbb R^n$ is the <orthogonal group> $O(n)$, giving
$$
\operatorname{LGr}(\mathbb R^{2n})\cong U(n)/O(n).
$$

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