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Dec 3, 2019 at 17:29 vote accept asv
Dec 3, 2019 at 1:58 answer added Zach Teitler timeline score: 10
Dec 2, 2019 at 21:31 history edited YCor CC BY-SA 4.0
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Dec 2, 2019 at 19:50 comment added Zach Teitler $A$ is finite dimensional if and only if the radical of the ideal $I=(f_1,\dotsc,f_l)$ is the maximal graded ideal $(x_1,\dotsc,x_k)$. Assume $A$ is finite dimensional. Then $A$ has Poincaré duality if and only if $I$ is Gorenstein, equivalently there exists a homogeneous polynomial $F$ such that the ideal of $g=g(x_1,\dotsc,x_k)$ satisfying $g(\partial/\partial x_1,\dotsc,\partial/\partial x_l)(F) = 0$, is precisely $I$. There is a book by Meyer and Smith, Poincaré duality algebras, Macaulay's dual systems, and Steenrod operations.
Dec 2, 2019 at 19:39 comment added asv @JohannesHahn: Thanks, added.
Dec 2, 2019 at 19:38 history edited asv CC BY-SA 4.0
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Dec 2, 2019 at 19:31 comment added Johannes Hahn One way of making this question more concrete would be to describe more explicitly what you mean by "satisfies Poincaré duality" ? Do you just want any isomorphism $A_i \cong A_{n-i}$, i.e. do you only want equality of dimensions? Or do you want something more, like the isomorphism being a (member of a) specific (class of) map(s)? What about the top-dimension $n$ ? Do you want $n$ to be a pre-determined value (like the dimension of the manifold) or just whatever the last highest non-zero degree happens to be?
Dec 2, 2019 at 19:18 history edited asv CC BY-SA 4.0
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Dec 2, 2019 at 19:12 history asked asv CC BY-SA 4.0