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Dec 14 at 13:12 answer added Shuo Zhang timeline score: 2
Dec 4, 2020 at 18:10 comment added Merlin Christ A functor assigns objects to objects and morphisms to morphisms. Two functors are pointwise equivalent if they assing to each object equivalent objects.
Dec 3, 2020 at 17:04 comment added Filip What does the word "pointwise" in "pointwise both twist functors are equivalent" mean?
Nov 19, 2020 at 14:29 comment added Zack Consider the Lefschetz fibration $W_L$ with fiber $M$ and single vanishing cycle $L$. Then $\mathcal{FS}(W_L)$ is the ground ring $\mathbb k$, and the cup/Orlov functor is the coYoneda map $\hom(L, \cdot)$. It follows that the dual cotwist of the cup functor is the algebraic twist along $L$, while the monodromy is Dehn twist along $L$ (with a shift).
Nov 19, 2020 at 7:04 comment added Merlin Christ I don't really understand, could you be more specific why Theorem 1.2 is relevant for my question?
Nov 19, 2020 at 4:21 comment added Zack I doubt this is the first proof, but in the case where $M$ is Weinstein I think this follows from Theorem 1.2 of arxiv.org/abs/1908.02317. In the non-Weinstein case something similar should be possible by identifying the semiorthogonal gluing as a full subcategory of $\mathcal{FS}(W_L^2)$, where $W_L^2$ is the square of the Lefschetz fibration with vanishing cycle $L$, but I haven't worked it out.
Nov 18, 2020 at 18:07 history edited Merlin Christ CC BY-SA 4.0
added 6 characters in body; edited title
Nov 18, 2020 at 18:05 review First posts
Nov 18, 2020 at 19:12
Nov 18, 2020 at 18:02 history asked Merlin Christ CC BY-SA 4.0