Timeline for Dual space of vector fields with null divergence

Current License: CC BY-SA 3.0

14 events

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Jun 3, 2016 at 9:32	history	edited	Asaf Karagila♦		edited tags
Jun 3, 2016 at 8:55	history	edited	Bazin	CC BY-SA 3.0	added 2 characters in body
Jun 1, 2016 at 21:14	history	edited	Bazin	CC BY-SA 3.0	A brief point on Navier-Stokes weak solutions.
Jun 1, 2016 at 21:01	comment	added	Bazin		@Jean Duchon I added more on this topic in my question, since I haven't enough room inside a comment.
Jun 1, 2016 at 13:42	comment	added	Jean Duchon		Why should it be bothering that a solution is obtained modulo a gradient? Gradients with null divergence are special enough that there is none in $L^2$, right?
Jun 1, 2016 at 13:07	comment	added	Igor Khavkine		The answer to the updated question 2 is also Yes. As pointed out by Jochen, Poincaré duality for currents gives $\ker \mathrm{div} = \mathrm{im } \mathrm{grad }$ for $\mathscr{D}'$. It only remains to check that the dual space of $\widetilde{\mathscr{D}} \subset \mathscr{D}$ is given by the quotient $\mathscr{D}'/(\widetilde{\mathscr{D}})^\perp$. This is true for any closed subspace of a locally convex topological vector space, as a consequence of the Hahn-Banach theorem. Note that $\widetilde{\mathscr{D}} = \ker \mathrm{div}$ is closed because $\mathrm{div}$ is continuous.
Jun 1, 2016 at 10:08	history	edited	Bazin	CC BY-SA 3.0	Thanks to the answers, I propose a more realistic second question.
Jun 1, 2016 at 9:49	comment	added	Bazin		Thanks for your answers. I have modified the formulation of the questions.
May 31, 2016 at 11:50	history	edited	Jochen Wengenroth		Tag fa added
May 31, 2016 at 7:07	comment	added	Jochen Wengenroth		The second equality in question 2 seems right by the Poincare lemma for currents.
May 30, 2016 at 14:22	comment	added	Jochen Wengenroth		I think, you have to be more precise concerning the second question. The dual of a subspace is (canonically isomorphic to) a quotient of the dual. In which sense do you want to identify $(\tilde{\mathscr D})'$ with a subspace of $\mathscr D'(\mathbb R^3,\mathbb R^3)$?
May 30, 2016 at 13:39	comment	added	Jochen Wengenroth		The compactly supported de Rham cohomology is, according, e.g., to the book of Bott and Tu, page 19, $H^2_c(\mathbb R^3) =0$. Thus, the answer to your first question is: Yes.
May 30, 2016 at 13:34	comment	added	Thomas Rot		For Q1): isn't $\tilde{\mathcal {D}}/\mathrm{curl}\mathcal{D}$ the second de Rham cohomology group with compact support, suitably dualized? If so, yes this is true.
May 30, 2016 at 12:52	history	asked	Bazin	CC BY-SA 3.0