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Mar 13, 2019 at 0:22 comment added Will Sawin I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)!
Mar 13, 2019 at 0:15 comment added yue he @WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much!
Mar 12, 2019 at 6:55 history became hot network question
Mar 12, 2019 at 6:22 answer added Denis Nardin timeline score: 16
Mar 12, 2019 at 5:20 comment added Will Sawin I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R \Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful.
Mar 12, 2019 at 4:29 comment added user74900 maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{\infty}$-ring spectrum level
Mar 12, 2019 at 3:50 review First posts
Mar 12, 2019 at 6:58
Mar 12, 2019 at 3:49 history asked yue he CC BY-SA 4.0