Peng Zhou

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====== Matrix Factorization ======
$\gdef\ccal{\mathcal C}$
===== [N] =====

==== section 2.1   ====

By a dg category $\ccal$, we mean a stable differential $\Z$-graded category. The hom complex is a $\Z$-graded cochain complex. We have a shift functor $[n]$ for any $n \in \Z$. 

By a 2-periodic dg category $\ccal$, we mean a $\Z$-graded category that the functor $[2]$ is equivalent to identity. Note that the hom complex is $\Z$-graded. 

By a $\Z/2$-dg category, we mean a stable differential $\Z/2$-graded complex. 

To any dg category, we can associate to it the $\Z/2$ folding, denoted as $\ccal_{\Z/2}$, which takes the same object, and only remembering the $\Z/2$-grading on the complex. 

To any $\Z/2$-graded category, we can associate to it a 2-periodic $\Z$-graded category, by unfurling. (It is like $p^*p_!$ where $p: \Z \to \Z/2$ )

==== section 2.2  ====
$\gdef\Spec{\text{Spec}}$

Suppose $A = \C[x_1, \cdots, x_n]$, $W \in A$ is a polynomial with only one singular value $0$. And $B = A/(W)$. $X = \Spec B$. 

We want to say $Coh(X) / Perf(X)$ is a 2-periodic dg category. 

Why? What is perf? What is perfect complex? bounded complex of finite projective B-modules. But, what is projective? 

++++ Example: $A = \C[x,y], W = xy, B = \C[x,y]/(xy). $ |

Q1: is $M = B / (x)$ a projective $B$-module? A sufficent condition is that $\Hom(M, -)$ should have no higher cohomology, but, look at the free resolution of $M$, we get
$$ \cdots B \xto{x} B \xto{y} B \xto{x} B \to M$$
Then, if we $\Hom(M, M)$, we get
$$ M \xto{x} M \xto{y} M \cdots $$
taking cohomology, we get
$$ M \xto{0} 0 \xto{0} \C\xto{0} 0 \xto{0} \C \cdots $$
So, $M$ is not projective! 

I guess, the only projective guy is $B$ itself. We know that, over any commutative ring, fin gen module is projective, iff it is locally free. 

Consider $M = (x+1)B$. It contains $y(x+1)B = y B$. Is $M$ locally free? Consider the prime ideal $(x,y)$. If we invert this prime, meaning, anything not in this prime ideal can be placed in the denominator. Then $M_p = B_p$ so it is locally free. Suppose $p = (x+1)$, then $M_p = (x+1) B_p$ also free. 

So, is $(x+1)\C[x]$ a free module? I think yes! And, in fact, if $(x+1)$ is not a zero-divisor, then $(x+1) B$ is a free module. 


So, many objects in $Coh(X)$ is not in $Perf(X)$. For example, $M_1 = B/(x), M_2 = B/(y), M_0 = B/(x,y)$. 
The $y$-axis, $x$-axis, and the origin. 

I think, the tail of the infinite resolution of $M_1$ and $M_2$ are the same, we have $[M_1] = [M_2] [1]$.

On the other hand, what about $M_0$? We have resolution
$$ B^2 \xto{y,x} B^2 \xto{x,y} B \to M_0 $$
So, it is like $[M_0] = [M_1] \oplus [M_2]$. That's pretty fun! 

So, what's the A-model picture? I think $Coh(X)$ is the Fuk of 

++++

===== MF example =====
consider the simplest matrix factorization example on toric 3-folds. They are both gluing two copies of $\C^3$ together. 

The first one is given by the fan with ray generator $(0,0,1), (1,0, 1), (0,1,1), (1,1,1)$. yes, a square in $x,y$ direction on level $z=1$. It is the total space of two line bundles on $\P^1$, but which two? Let's compute the face conormal, we have
  * $a=(-1, 0, 0), b=(0, -1, 0) $ and $c=(1,0, -1), d=(0, 1, -1)$. 
one can see that $a+c =b+d$. That means the coordinate ring of the affine space is given by $AC = BD$, wher $A,B,C,D\in \C^4$. Now, we can do blow-up, I guess we can do two patches with coordinate $A, B, u = C/B=D/A$ and the other with $C, D, v = B/C = A/D$. OK, great! Now, how does the fiber coordinate tranform? We have $u B = C, uA = D$, so it is like, $Tot[O(-1) \oplus O(-1)]$. 

There is another way to resolve. Let's not go there. 

The second one is given by the fan on generators $(0, 0, 1), (1, 0, 1), (-1, 0, 1), (0, 1, 1)$. with dual cone generator being $a=(0, -1, 0), b=(1, 1, -1), c=(-1,1,-1)$. The problem with this set of generator is that, $b,c$ span a unsaturated sublattice, we have $d=(b+c)/2 = (0, 1, -1)$, which should be added. so the relation is $2d = b+c$. Then the coordinates are $A,B,C,D$ with $D^2 = BC$, and we do $\P^1$ coordinate like $u=B/D = D/C$ and $v = 1/u = D/B = C/D$. So $v^2 = C/B$, the local charts are
  * $(A, B, v)$ and $(A, C, u)$, with change of coordinates like $u = 1/v$ and $C = v^2 B$. We have $Tot[O(0) \oplus O(-2)]$

Now, for the superpotential, we know we want the coordinate $(0,0,-1)$. 
  * In the first case, we have function $ABu = AB(C/B) = AC = AB(D/A)=BD$, similarly, we have $CDv=CD(B/C)=BD..$, so $u = c-b=(1,1,-1)$ and $a+b+u= a+c=(0,0,-1)$ hmm, it works. 
  * For the other case, we have $a+d = (0,0,-1)$, so the function is $AD = ABv = ACu$.
 
OK, great. Now, how do we compute its matrix factorization? We are no longer on an affine scheme, although, the function $W$ still is a well-defined function on the affinization. 




===== References =====
  * [N] [[https://arxiv.org/abs/1604.00114 | Wrapped microlocal sheaves on pairs of pants ]] by [[https://math.berkeley.edu/~nadler/ | David Nadler]]. Section 2 on LG B-model
  * [L] [[https://arxiv.org/abs/1608.04473 | Homological mirror symmetry for open Riemann surfaces from pair-of-pants decompositions]] by [[https://heather-math.github.io | Heather Lee]]. See also the [[ https://people.math.harvard.edu/~auroux/schms2015-2020/Miami2017/H.%20Lee%20-%2028-01-17%20-%20Homological%20mirror%20symmetry%20for%20open%20Riemann%20surfaces%20from%20pair-of-pants%20decompositions.pdf | slide ]]