# Helmholtz-Decomposition on compact Riemannian manifolds

For smooth domains $\Omega$ in $\mathbb{R}^n$ it is known that one can decompose vector fields in $L^p(\Omega)^n$, $1 < p <\infty$ into a "gradient"- and a "divergence-free"-part such that

$L^p(\Omega)^n=G^p(\Omega) \oplus D^p(\Omega)$,

where $G^p(\Omega)= \{ w\in L^p(\Omega)^n; w= \nabla p$ for some $p\in W^{1,p}(\Omega)\}$, and $D^p(\Omega)$ is the completion of $\{ u\in \mathcal{C}^\infty_0(\Omega)^n; \nabla \cdot u=0 \}$ in $L^p$.

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 $q$)?

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For closed manifolds this is established in jstor.org/stable/2154923 –  Willie Wong Sep 2 '11 at 12:01
It seems to me that the formula for L^p(Ω)^n should have direct sum instead of tensor product. –  Dmitri Pavlov Sep 3 '11 at 19:48
Thank you - this is what I meant. –  Sören Sep 5 '11 at 7:16
Concerning the comment of Willie Wong: This paper deals only with manifolds without boundary. A later paper including manifolds with boundary would be springerlink.com/content/c0737x2130116676. However, I have some doubts about the results on orthogonal complements (see the table after Theorem 5.7), since the proof seems to rely on a L^p-decomposition of a function in L^q (without having the appropriate embedding, see at the end of page 67). –  Sören Sep 5 '11 at 12:15