$\gdef\ccal{\mathcal C}$

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 $p: \Z \to \Z/2$ )

$\gdef\Spec{\text{Spec}}$

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?

$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.

• [N] Wrapped microlocal sheaves on pairs of pants by David Nadler. Section 2 on LG B-model
• [L] Homological mirror symmetry for open Riemann surfaces from pair-of-pants decompositions by Heather Lee. See also the slide