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--- questions:descent-via-localization [2022/10/26 23:29] – created pzhou
+++ questions:descent-via-localization [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 3: / Line 3: @@
 What does descent mean? I was telling my linear algebra student the following example: suppose you have a linear map $f: V \to W$, and there is a subspace $V' \In V$ such that $f|_{V'}=0$, then $f$ descent to $V/V'$. In general, it means you define something on a 'cover' of an object, and you want to obtain the thing on the object itself.
-** Zariski Descent ** A slightly non-trivial example is the Zariski descent for coherent sheaves. Suppose $X$ is a scheme, and $X = \cup_{i=1}^N U_i$ by some finite open cover $j_i: U_i \to X$, then the restriction
+===== Zariski Descent =====
+A slightly non-trivial example is the Zariski descent for coherent sheaves. Suppose $X$ is a scheme, and $X = \cup_{i=1}^N U_i$ by some finite open cover $j_i: U_i \to X$, then the restriction
 $$ L = \prod_i j_i^*: QCoh(X) \to \prod_i QCoh(U_i) $$
-which is a left-adjoint (hence called L), and its right-adjoint $R = \oplus_i (j_i)_*$ (here I use the finiteness condition to equate  $\prod$ and $\oplus$). Then, we have a comonad
+which is a left-adjoint (hence called L), and its right-adjoint
+$$ R = \oplus_i (j_i)_*: \oplus_i QCoh(U_i) \to QCoh(X) $$
+here I use the finiteness condition to equate  $\prod$ and $\oplus$. Then, we have a comonad
 $$ \Omega := LR, \quad \Omega \to id, \quad \Omega \to \Omega \circ \Omega. $$
+Then Zariski descent says
+$$ QCoh(X) = Comod_{\Omega}( \prod_i QCoh(U_i) ) $$
+===== The mirror side, Viterbo restriction =====
+How do we generalize this? You may ask why? Because by mirror symmetry, we should be able to discover similar phenomenon on the A-side. But you may ask why not directly use mirror symmetry to prove descent? Well, we want to do thing the other way around, namely, prove mirror symmetry locally, then use descent argument to glue them up.
+The kind of question is the following. Let $Y$ be a symplectic manifold, and let $Y = \cup_i U_i$ be covering by certain nice open submanifold, and we want to have 'localization' functor, which is
+$$ L = \prod_i L_i: Fuk(Y) \to \prod_i Fuk(U_i) $$
+where $L_i$ is the Viterbo restriction, which is a categorical localization, a left-adjoint, just like $j_i^*$. Then, we need to define the right-adjoint $R: = \oplus_i R_i \prod_i Fuk(U_i) \to Fuk(Y)$. Under certain enlargement, we can always define this $R$. The question is then, do we have
+$$ Fuk(Y) =  Comod_{\Omega}( \prod_i Fuk(U_i) ) $$
+Here, the thing is a bit funny. The functor, right-adjoint to Viterbo restriction, $R_i: Fuk(U_i) \to Fuk(Y)$ doesn't really lie on $Fuk(Y)^c$, the subcat of compact objects (which contains all geometric ones).
+===== Abstract non-sense =====
+Let $C$ be a dg category, co-complete, compactly generated. Suppose we have a finite collection of localization functors $L_i: C \to C_i$ that are left-adjoint, then we can define the left and the right-adjoint
+$$ L = \oplus L_i C \to D:= \prod_i C_i : R=\oplus R_i $$
+and $\Omega = LR: \prod_i C_i \to \prod_i C_i$.
+** Conjecture ** If $L: C \to \prod_i C_i$ is conservative, i.e., no object got killed, then
+$$ C \cong Comod_\Omega (\prod_i C_i) . $$