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S Apr 16, 2017 at 11:57 history suggested C.F.G CC BY-SA 3.0
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Nov 22, 2010 at 17:43 comment added Johannes Ebert Well, don't there exist forms that are harmonic in the interior of the simplex and coincide with a given form on the boundary (the classical boundary value problem)? The arguments you gave show that these forms are nearly uniquely determined by the boundary values, which to me seems promising rather than discouraging.
Nov 22, 2010 at 15:44 comment added Jeffrey Giansiracusa hi Johannes. I think the extension to simplicial sets is not just a formal matter... An n-simplex is contractible, and rel boundary it looks like an n-sphere. For harmonic forms, Dirichlet boundary conditions give you (up to scalars) the constant function and only a single harmonic form in degree n, and Neumann boundary conditions leave you with only the constant function on each simplex. So it seems unlikely that one represent a middle degree cohomology class by a form that is harmonic on each top-dimensional simplex and satisfies either of these boundary conditions.
Nov 22, 2010 at 10:17 comment added Dmitri Panov You might be interested in the following paper : J. Cheeger. A vanishing theorem for piecewise constant curvature spaces. Curvature and topology of Riemannian manifolds (Katata, 1985), 33–40, Lecture Notes in Math., 1201, Springer, Berlin. (1986).
Nov 22, 2010 at 9:56 comment added Johannes Ebert Hi Jeff, the main step towards this is to understand Hodge theory for manifolds with boundary. If you can do this, then you can treat the standard simplex and then it should be a formal matter to extend the theory to finite simplicial sets. Gilkey's book "Heat Equation, invariance theory and the Atiyah-Singer theorem" contains a section on the de Rham complex on manifolds with boundary. There you probably find the correct boundary condition on the forms and the Hodge decomposition on manifolds with boundary.
Nov 21, 2010 at 16:39 history asked Jeffrey Giansiracusa CC BY-SA 2.5