When migrating the server, I forgot to backup the yiye website, lost a lot of memories, sad. I also dumped a lot of my phd notes to dumpster anyway when I move away from Chicago, it is inevitable to let go.

If I consider coherent sheaf category as the category of B-branes, then how does it depends on the complexified Kahler parameters? One can either do it in the physical way, reading Witten, Hori etc; or one can do it in the math way, thinking about deformation of the Coh category by gerbes.

It turns out Andrei Caldararu's thesis deals exactly with B-field and twisted sheaves. Let's find out the definition.

Roughly speaking, twisted sheaf is a bunch of locally defined sheaves (hmm, aren't they already locally defined?), whose gluing function fails the tricycle condition.

Take an element $\alpha \in H^2(X, \mathcal O^*_X)$. What does this mean? I probably need to resolve this sheaf $\mathcal O_X^*$, here is a short exact sequence $$ 0 \to 2\pi i \Z \to \mathcal O \to \mathcal O^* \to 0 $$ So, if I do $X=\P^1$, then I don't think I have $H^3(\P^1, \Z)$ or $H^2(\P^1, O)$, so I do not have any non-trivial element in $H^2(X, \mathcal O^*_X)$.

It still might make sense, maybe the trivialness of $\alpha$ corresponds to a locally constant deformation. Just like a flat connection gives zero curvature, but still flat connection is useful.

How to take a sheaf of categories and take the global section? Or given a diagram of categories, how to take the limit? Suppose we are trying to get $Coh(\P^1)$ twisted by a complex number $c$ $$ Coh(\P^1 \RM \infty ) \xto{res \otimes O(c)} Coh(\C^*) \gets Coh(\P^1 \RM 0) $$ Now, what is $O(c)$ on $\C^*$? I want to say it is a holomorphic line bundle that does not have any global section (so probably shouldn't be called a coherent sheaf), it is constructed by considering $\C^* \cong \C / \Z$, we consider line bundle given by the quotient $(\C \times \C) / \Z$, where the $\Z$ action is $$(z, \eta) \mapsto (z+ 2\pi i , \eta e^{c z})$$ Indeed, we have $e^z = y$ (hmm, if $c=1$, I am supposed to still get a trivial bundle)

OK, maybe we use analytic topology on $\C^*$.