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--- blog:2023-03-30 [2023/03/30 07:31] – created pzhou
+++ blog:2023-03-30 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 20: / Line 20: @@
 Now, $Y_0 \cong T^*S^2$, we know its Fukaya category, which is $Loc(S^2)$, somehow is $\C[x]$-mod, where $|x|=-1$ (who know they can be of negative degree??). To test if my theory is correct, we compute the LG-model, we first compute what is
-$$Fuk( (\C^*)^4, \eta( x^2+y^2+z^2 - 1) ) \cong \Coh((\C / \Z_2)^3 \times \C) $$
+$$Fuk( (\C^*)^4, \eta( x^2+y^2+z^2 - 1) ) \cong Coh((\C / \Z_2)^3 \times \C) $$
-Then, you partially compactify on the A-side, and turn on superpotentialon the B-side. The divisor $x=0$, corresponds having $$W_B = z_1^2 + z_2^2 + z_3^2 + z_1 z_2 z_3 z_4$$
+Then, you partially compactify on the A-side, and turn on superpotentialon the B-side.
+$$W_B = z_1^2 + z_2^2 + z_3^2 + z_1 z_2 z_3 z_4$$
+If we considering $W_A = \eta(x+y+z-1)$, so $m_1, \cdots, m_4$. They are basis of the $M_A$. we need to have $n_1, n_2, n_3, n_4$ for compactification, so that we get
+==== Simplest ====
+What's the Fukaya category of $(\C, z^2)$? Well, you can try to view it as $(\C^*, z^2)$ deformation. The former has mirror $[\C / \Z_2]$. Then, you put in a divisor, and you turn on a superpotential. What superpotential do you want to turn on? I guess it is like $[ (\C, z^2) / \Z_2]$. The reason is $MF(\C, z^2)$, I know its critical loci is one point, so Vect (but I am not sure why quotient by $\Z_2$ doesn't give it more structure.
+No, I think it should be $MF([\C / \Z_2], w)$, where $w$ is the coordinate after quotient.
+Consider the Fukaya category of $(\C, z^n)$, then it is the representation of $A_{n-1}$ quiver (with n node and n-1 arrows). However, if we view it as $(\C^*, z^n)$ deformation, then we are making it affine, namely $n$-node with with $n$ arrows making it a circle quiver.
+==== Another one ====
+What's the Fukaya category of $(\C^2, y x^2)$? Not a nice Lefschetz fibration. Don't even know the definition. But, the stop should be a Legendrian curve.
+How about, just $Fuk(\{x^2=0\})$?