Is the Cartesian product of Lagrangian submanifolds a Lagrangian submanifold? In my research, I have found a statement where a submanifold of $\mathbb{C}^N$, which is a straight line in each plane $\mathbb{C}$ is a Lagrangian submanifold. Similarly it seems that a submanifold of $\mathbb{C}^N$ which is a circle in each plane $\mathbb{C}$ is also Lagrangian submanifold. 
Another example is a submanifold of $(\mathbb{C}^{\times})^N$ with cylindrical metric which is a straight line parallel to the axis of each cylinder. Why are these Lagrangian submanifolds? Is it such that the Cartesian product of Lagrangian submanifolds is a Lagrangian submanifold?
 A: Yes, a cartesian product of Lagrangians is a Lagrangian (in the cartesian product of the symplectic manifolds).
Let $(M_1, \omega_1), \, \dots, \, (M_n, \omega_n)$ be symplectic manifolds and for $i=1, \dots, n$ let $L_i \subset (M_i, \omega_i)$ be a Lagrangian submanifold. For the cartesian products $M = M_1 \times \dots \times M_n$ and $L = L_1 \times \dots \times L_n$; Of course $L$ is canonically a submanifold of $M$. Denote by $\pi_i : M \to M_i$ the canonical projections. Equip $M$ with the following 'product' symplectic form: $\omega = \sum_{i=1}^n \, \pi_i^{\ast}\omega_i$. That is, for any $m \in M$ and any $V, W \in T_mM$, we have $\omega_m(V, W) = \sum_{i=1}^n [(\pi_i^{\ast}\omega_i)_m(V, W)]$.
$L$ is Lagrangian for the symplectic form $\omega$. Indeed, let $p \in L$ and let $X, Y \in T_pL$. Then, for each $i=1, \dots, n$, we have $(\pi_i^{\ast}\omega_i)_p(X, Y) = (\omega_i)_{\pi_i(p)}((\pi_i)_{\ast}X, (\pi_i)_{\ast}Y) = 0$. This last equality follows from the observation that the vectors $(\pi_i)_{\ast}X$ and $(\pi_i)_{\ast}Y$ belong to $T_{\pi_i(p)}L_i$ (which is a Lagrangian vector space by assumption); This is easily seen once one has chosen a basis for each $T_{\pi_i(p)}L_i$, as these bases lift to form a basis of $T_pL \cong \oplus_{i=1}^n T_{\pi_i(p)}L_i$. Therefore, $\omega_p(X,Y) = 0$. Since this is true for any choice of $p, X, Y$, it follows that $L$ is indeed Lagrangian.
