Suppose $(M,\omega)$ is a $2n$-dimensional symplectic manifold. Consider the $n$-torus $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n.$ Then $\mathbb{T}^n$ is a Lagrangian submanifold of the symplectic $2n$-torus $\mathbb{T}^{2n}$, equipped with the unique symplectic form that pulls back to the canonical symplectic form on $\mathbb{R} ^{2n}$. Now, how does it follow that there exists a Lagrangian submanifold of $M$ that is diffeomorphic to $\mathbb{T}^{n}$?
1 Answer
The fact that $\mathbb T^n$ is a Lagrangian submanifold of $\mathbb T^{2n}$ plays no role. Indeed, by Darboux's theorem, for any symplectic manifold $(M^{2n}, \omega)$ and for any $x\in M$, there is $U_0 \subset \mathbb R^{2n}$ and a local coordinates chart $\phi : U_0 \to U$ so that $\phi^* \omega = \omega_{std}$ and $x\in U$, where $\omega_{std}$ is the standard symplectic structure on $\mathbb R^{2n}$. Since it is easy to find Lagnrangian in $(U_0, \omega_{std})$ diffeomorphic to $\mathbb T^n$: choose $p =(p_1, \cdots, p_n)\in U_0$ and $r_1, \cdots, r_n >0$ small enough, then $$ \mathbb S^1 (p_1) \times \cdots \times \mathbb S^1 (p_n) $$ is such an example.
Of course this also shows the existence of infinitely many Lagrangian tori on any symplectic manifold.