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--- blog:2024-12-04 [2024/12/05 06:54] – created pzhou
+++ blog:2024-12-04 [2024/12/05 09:51] (current) – [Twisted Sheaves] pzhou
@@ Line 5: / Line 5: @@
 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.
+===== Twisted Sheaves =====
+It turns out Andrei Caldararu's [[https://people.math.wisc.edu/~caldararu/publications/ThesisSingleSpaced.pdf|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 $\C^*$.