Helmholtz-Decomposition on compact Riemannian manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:49:35Z http://mathoverflow.net/feeds/question/74334 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74334/helmholtz-decomposition-on-compact-riemannian-manifolds Helmholtz-Decomposition on compact Riemannian manifolds Sören 2011-09-02T08:49:46Z 2012-09-01T11:20:07Z <p>For smooth domains $\Omega$ in $\mathbb{R}^n$ it is known that one can decompose vector fields in $L^p(\Omega)^n$, $1 &lt; p &lt;\infty$ into a "gradient"- and a "divergence-free"-part such that</p> <p>$L^p(\Omega)^n=G^p(\Omega) \oplus D^p(\Omega)$,</p> <p>where <code>$G^p(\Omega)= \{ w\in L^p(\Omega)^n; w= \nabla p$</code> for some <code>$p\in W^{1,p}(\Omega)\}$</code>, and $D^p(\Omega)$ is the completion of <code>$\{ u\in \mathcal{C}^\infty_0(\Omega)^n; \nabla \cdot u=0 \}$</code> in $L^p$.</p> <p>Is such a decomposition also available on a compact Riemannian manifold (with boundary) $M$ with respect to the gradient- and divergence-operator induced by the Riemannian metric? Does one additionally have a "annihilator"-property in the spirit of $D^p(\Omega)^\perp = G^q(\Omega)$ (with dual exponent $q$)?</p> http://mathoverflow.net/questions/74334/helmholtz-decomposition-on-compact-riemannian-manifolds/87887#87887 Answer by Martin Gisser for Helmholtz-Decomposition on compact Riemannian manifolds Martin Gisser 2012-02-08T12:56:04Z 2012-02-08T12:56:04Z <p>Günter Schwarz, <em>Hodge Decomposition - A Method for Solving Boundary Value Problems</em>, Lecture Notes in Maths <strong>1607</strong> (1995)</p>