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--- blog:2022-12-14 [2022/12/15 06:48] – created pzhou
+++ blog:2022-12-14 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 1: / Line 1: @@
 ====== 2022-12-14 Wed ======
   * Cross my t and dot my i.
 ===== T and I branes =====
+$\gdef\End{\text{End}}$
 Let me be super careful, and state the condition that I need.
@@ Line 26: / Line 29: @@
 Now, the first layer, shall we do $T_i \otimes \Hom(I_i, L)$. Here, I am not just taking $\oplus_{p \in
-I \cap  L} \otimes T_p$. Then,
+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.
+===== Here is a formula =====
+$\gdef\End{\text{End}}$
+$$ T \otimes_{\End(T)} \Hom_{\End(I)}(\Hom(I, T),\Hom(I, L)) $$
+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 $\Hom_{\End(I)}(\Hom(I, T),\Hom(I, T)) = \Hom(T, T)$.