Take the <torus>
$$
M=T^{2n}=\mathbb R^{2n}/\mathbb Z^{2n}
$$
with the translation-invariant <symplectic form>
$$
\omega=\sum_{j=1}^n dx_j\wedge dy_j.
$$
The coordinate vector fields give a global symplectic frame of $TM$, so $TM$ is symplectically trivial. Passing fiberwise to the <Lagrangian Grassmannian bundle> gives
$$
\mathcal LGr(TM)\cong
T^{2n}\times\operatorname{LGr}(\mathbb R^{2n}),
$$
which is a smooth trivialization over the compact symplectic manifold $T^{2n}$.

Solution

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

{c}
{title2=$\mathcal LGr(E)$}

For a rank-$2n$ symplectic <vector bundle> $E\to B$, the Lagrangian Grassmannian bundle has fiber $\operatorname{LGr}(E_b)$ over $b$. A symplectic trivialization of $E$ induces a trivialization $\mathcal LGr(E)\cong B\times\operatorname{LGr}(\mathbb R^{2n})$.

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