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.... We can still define the parallel transport, but will it take Lagrangian to Lagrantian? Consider a little two disk in the first fiber, sweep along the parallel transport, get a solid tube. Integrate the $\int d\omega$ on the. The tube side contribution by orthogonal condition is zero, but on the initial. OK, suppose we are just varying symplectic form (within the same cohomology class)
ok, so it is wrong to just take this $S^1$-family of symplectic form and claim you can do stuff. In our actual setup for $T^*M$, we have holomorphic Liouville one-form (which I don't quite like), and we can do $Re(d e^{i\theta} ydx)$ to get the corrections.
In the traditional approach, we have canonical $I, \omega_I, \Omega_I$, and we consider $\omega_I$-Lagrangians (thimble) in the base, and $I$-hol disk.
Let $c_1$ and $c_2$ be two critical point of $f$, with $v_1, v_2$ their values. Let $\gamma$ be the straight segment between $v_1,v_2$. After we made some nice choices of $\omega_I$, we can draw two thimbles over $\gamma$.Unstable manifolds. Over each point in $\gamma$, we have two vanishing cycles above that point and intersects. Let $s \in [0,1]$ parametrize $\gamma$, and we have two intersection points $p(s), q(s)$ above point $s$, and $p(s), q(s) \in V_1(s) \cap V_2(s)$, $V_i(s)$ are vanishing cycles from $c_i$.
OK, for each $s$, we have a bigon, between $V_1, V_2$, easpecially when $s=0,1$, when $V_0$ and $V_1$ are very small sphere, the bigon should be small too... bigon is like saying, we want to correct away mistakes (no, not mistakes, they are relations)
anyway, we have $p_\gamma$ and $q_\gamma$, the two 'instanton' lines between the two critcal points. We can do a mixture, take the hol disk bigon, in the middle fiber, take one side of the bigon, flow it to $c_1$, and the other side, flow to $c_2$. instead, we can do a more uniform disk but with a 'mixed equation', the zeta-instanton equation.
let me stop here.
I see, you want to say the inhomogeneous term comes from the 3rd direction's K-rotation.