What is the relation between totally real submanifold and Lagrangian submanifold?

Question

By definition, for a complex manifold M, totally real submanifold X of M is satisfying

1) $2 dim X$ = $dim M$ and

2) $T_pM \cap J T_pM =\{0\}$ for $\forall p \in X $with integrable complex structure $J$.

I saw the question Lagrangian submanifold totally real

The Question is : For a Kahler manifold M (Since totally real submanifold is defined on complex manifold and lagrangian submanifold is defined on symplex manifold, consider Kahler manifold)

(1) lagrangian submanifold of M is totally real submanifold. How can prove it?

(2) The converse is false? I think, for lagrangian submanifold X of M, I could find complex coordinate chart ($z_1=x_1 + \sqrt{-1} y_1, \cdots, z_n=x_n+\sqrt{-1}y_n$) s.t for $\zeta \in X$, $y_1(\zeta)= \cdots = y_n(\zeta)=0$ hold. the converse if false?

Answer 1

1) Suppose that $u\in T_pM\cap J(T_pM)$, $J(u)\in T_pM\cap J(T_pM)$ since $J^2=-Id_{TM}$ $\langle u,u\rangle=\omega(u,J(u))=0$ since $\omega$ is Lagrangian and its restriction to $T_M$ vanishes, we deduce that $u=0$, since $\omega(.,J(.))$ is a scalar product.

Consider on $\mathbb{R}^4$ the symplectic form $e_1\wedge e_2+e_3\wedge e_4$ and $J(e_1)=-e_2, J(e_3)=-e_4$. Let $M=Vect(e_1,e_2+e_3)$, $J(M)=Vect(e_2,e_1-e_4)$ and $\omega(e_1,e_2+e_3)\neq 0$.