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S Dec 30, 2015 at 10:43 history suggested John B CC BY-SA 3.0
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S Dec 30, 2015 at 10:43
Feb 26, 2011 at 20:40 vote accept Pavol S.
Feb 22, 2011 at 14:12 comment added Pavol S. If $V_F=V$ for all $F$'s we still get de Rham cohomology tensored with $V$, and de Rham with compact support in the interior tensored with $V^*$, so the duality holds. The motivation for the question was from symplectic form on the moduli space of flat connections on a surface, with various boundary conditions, and Lagrangian subspaces coming from cobordisms.
Feb 20, 2011 at 0:14 answer added Johannes Nordström timeline score: 6
Feb 18, 2011 at 3:44 comment added John Klein I guess I don't really have a feel for your question. What is the relation between $V$ and the geometry of the stratification of the manifold? What if $V_F = V$ for all faces $F$? How did this question arise?
Feb 14, 2011 at 21:58 comment added Pavol S. @John Klein: Then $V$ is not important. One cohomology is de Rham's tensored with $V^*$ and the other one is (isomorphic to) de Rham's with compact support in the interior of $M$ tensored with $V$. We can safely put $V=\mathbb R$ in this case. More-dimensional $V$ is only needed for those more complex boundary conditions. For $V=\mathbb{R}$ we can only have $V_F=0$ or $V_F=\mathbb{R}$. In that case, if say $\partial M$ is divided by a hypersurface to two faces, one with $V_F=0$ and the other with $V_F=\mathbb{R}$, then we do get perfect pairing (at least I hope :)
Feb 14, 2011 at 21:16 comment added John Klein You write: "If $V_F = 0$ for all $F's$ we get the standard Poincare duality for manifolds with boundary." But what is the role of $V$ in that statement?
Feb 14, 2011 at 15:14 history edited Pavol S. CC BY-SA 2.5
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Feb 13, 2011 at 20:52 history edited Pavol S. CC BY-SA 2.5
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Feb 13, 2011 at 16:02 history edited Pavol S. CC BY-SA 2.5
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Feb 12, 2011 at 23:55 history edited John Klein
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Feb 12, 2011 at 15:10 history edited Pavol S. CC BY-SA 2.5
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Feb 11, 2011 at 21:04 history asked Pavol S. CC BY-SA 2.5