Peng Zhou

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
====== 2025-02-16 ======

===== A new approach on Fukaya-Seidel category =====
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.