Let $X$ be a smooth affine complex manifold, and let $f: X \to \C$ be a holomorphic function. From this data, we should get a category (up to equivalences).

At this moment, we did not specify the Kahler form (which is not important), just as Riemannian metric is not important for homology of a manifold. It does not matter which car you use to drive from A to B.

Conceptually, one way is to consider $M = T^*X$, and two Lagrangians, $L_0$ the zero section $X$, and $L_1$ the graph Lagrangian. Both of them are holomorphic Lagrangian for $\Omega$.

What do I do with holomorphic Lagrangians?

• We can take the real part of the complex $\Omega$. $\omega_\theta = Re(e^{i\theta} \Omega)$, what does it mean? You have a smooth manifold, you put on some complex structure $J$, automorphism of the tangent bundle. We complexify the tangent bundle, so that $J$ has eigenspaces. We can decompose the complexified tangent bundle into direct sum of two bundles, then we can decompose the tensor products.
• What does it mean by the 'real part' of something? Well, we can do $\R^2 \otimes_\R \C$, say $J(e_1) = e_2, J(e_2) = -e_1$. Then this complex vector space is two dimensional, it can either have a basis of $e_1, e_2$; or it can have $J$ eigenbasis like $e_1 + i e_2$ and $e_1 - i e_2$, both of them are good complex vector spaces. If one just have a complex vector space, it does not make sense to say the 'real part' of a vector. The best way to say this, is to have a 'real structure', meaning saying that this complex vector space as a real vector space as a distinguished real subspace. Then, $J$ gives a direct sum decomposition, and there is a real linear map (not complex linear map), that takes the projection.
• We have $S^1$-family of real symplectic form $\omega_\theta$, and holomorphic Lagrangian. If we do $Re(\lambda dx \wedge dy)$ on $\C^3$, and map it down to $\C^*_\lambda$, do we get a symplectic fibration?

What is the 'paralell' transport? Wait, but, it is not a closed 2-form on the total space....