Lagrangian manifolds are totally real.
What if I take a complex manifold $M$ and then take its cotangent bundle $T^*M$, which is also a complex manifold with complex structure induced from the base $M$. $T^*M$ is a symplectic manifold and $M$, viewed as the zero-section, is Lagrangian in $T^*M$, and thus totally real. But $M$ is also a complex submanifold of $T^*M$.
Q: Isn't this a contradiction?
I'm thinking that this might mean that the complex structure on $T^*M$ (as defined above) is not compatible with the symplectic structure. I've tried to show this, but am stuck:
If $(z_1, \dots, z_n)= (x_1 +i y_1, \dots, x_n+iy_n)$ are complex coordinates on $M$ and $a_j$, $b_j$ are the coordinates for $dx_j$ and $dy_j$, respectively, then the symplectic form $\omega_0$ can be expressed as $$ \omega_0 =\sum_j dx_j \wedge da_j +dy_j \wedge d b_j. $$
If $\zeta_j = c_j +i d_j$ is the (complex) coordinate on $dz_j$, then the complex structure $J$ is given by $$ J(\frac{\partial}{\partial x_j}) = \frac{\partial}{\partial y_j}, \quad J(\frac{\partial}{\partial y_j}) = -\frac{\partial}{\partial x_j},$$ and $$ J(\frac{\partial}{\partial c_j}) = \frac{\partial}{\partial d_j}, \quad J(\frac{\partial}{\partial d_j}) = -\frac{\partial}{\partial c_j}. $$
Next, I want check if $\omega_0 (\cdot,J \cdot) $ is riemannian. However, I am stuck here because I don't see a relationship between the $a_j$ and $b_j$ and the $c_j$ and $d_j$. I think if $a_j = c_j$ and $b_j = d_j$ that would give compatibility, but it looks like that is impossible in general because: $$ dz_j = dx_j +i dy_j $$ and given any $a_j$ and $b_j$ we cannot always write $$ a_j dx_j + i b_j dy_j = \zeta_j dz_j,$$ for some $\zeta_j$.