• Cross my t and dot my i.

$\gdef\End{\text{End}}$

Let me be super careful, and state the condition that I need.

Let $X$ be a Weinstein manifold. Let $T_p$ be a collection of cocores, one for each critical point of the psh function $\varphi$. Let $I_p$ be the collection of cores. We have $\Hom(I_p, T_q) = \delta_{p,q}$, and similarly $\Hom(T_p, I_q) = \delta_{pq}[-n]$.

We assume that our skeleton is a union of smooth Lagrangians. This is true in the case of cotangent bundle, or in the case of plumbing of cotangent bundles.

We can ask, if every object in the wrapped Fukaya category can be expressed as a finite twisted complex of the cocores. CDGG says yes.

If you give me a Lagrangian $L$, then I will intersect it with the core. I assume the intersection is transverse.

If I am brave enough, I will take the core as a Lagrangian, and I allow disk to end on it, as if my core is a smooth Lagrangian.

How to remember the given Lagrangian $L$. I first, remember the intersections between $I$ and $L$. But, that's not good enough. There are disks bounding the core and $L$.

Consider the example, where $I = S^n$, and $L =S^n$ perturbed. The intersection $I \cap L$ are two points. There is a morphism $\gamma \in \End^{-n+1}(T) = C_{n-1}(\Omega S^n)$. The fundamental class of $S^{n-1}$ corresponds to $\gamma$. Now, consider endomorphism of that complex, we have $$\End( T[-n] + T) \supset \Hom(T, T[-n]) = \End(T)[-n] \ni \gamma[-n]$$ Here $\gamma[-n]$ is a degree $1$ map.

There is no rigid holomorhpic disks, that bounds $L$ and $I$, and create a differential for the bottom degree guy, to the top one. In the case of $S^1$, we have two disks, worth $S^0$. In the case of $S^2$, we have $S^1$-family of disks from bottom intersection to the top one. But, that information is not remembered by the usual rigid disks.

What are we doing? The Lagrangian $L$ evaluate out a collection of 'path on the core'. Given a collection of $T$ branes with bounding cochains, we can also get a collection of paths on the core. The requirement is that, $T$ and $L$ intersects $I$ at the same points.

Now, the first layer, shall we do $T_i \otimes \Hom(I_i, L)$. Here, I am not just taking $\oplus_{p \in I \cap L} T_p[d_p]$. Wait, I think it is OK, we can do that. Pretend, we are computing the $A_\infty$ structure of $\Hom(I, L)$, we will first count the intersection points, and build the chain complex, but here, for an intersection point $p$, instead of putting just we also include the factor $T_p$ tensor with $\C[-d_p]$.

Are we just trying to get the $\End(I)$ module structure of $L$? Suppose $I$ is a single smooth compact Lagrangian. It is a funny adjoint, we want to probe with $I$, but we want to output with $T$.

So, we are looking at structure of $$T \otimes_{\End(T)} \Hom(T, I) \otimes_{\End(I)} \Hom(I, L)$$ whatever the output is. Or, we can do $$T \otimes_{\End(T)} \Hom_{\End(I)}(\Hom(I, T),\Hom(I, L))$$

Now, let's test if it works on $I = S^2, L = S^2$. Well, $\Hom(I, L)$, cohomologically, is just the free $\End(I)$ module. We just use a holomorphic disk. Not quite enough.

One need higher dimensional moduli space of disks.

Well, a family of disks is a family of disks, there is no way out. We are no longer doing just 1 or 0 dimensional disks counting. Good bye. We could introduce more strata in the middle, If we have Maslov indices of the two points, differ by $k$, then the moduli of disks going between them should be of dimension $k$ as well, $\R \times N^{k-1}$, something like that.

What's the story of a torus? We look at the Morse gradient flowline. Do we have higher degree morphisms between $T$ branes? This time, no.

$\gdef\End{\text{End}}$ $$T \otimes_{\End(T)} \Hom_{\End(I)}(\Hom(I, T),\Hom(I, L))$$

In principle, this should work, since we are doing nothing more than Koszul duality. For example, in the cotangent bundle case, if we do $L=T$, then we should get back $\Hom_{\End(I)}(\Hom(I, T),\Hom(I, T)) = \Hom(T, T)$.