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Does it make sense to do Helmholtz decomposition of a vector field defined on a surface or on a manifold? I am mostly interested in the surface case. I was trying to find a reference for this and found only a handful of them mostly from electromagnetics literature. e.g.

Scharstein, Robert W. "Helmholtz decomposition of surface electric current in electromagnetic scattering problems." System Theory, 1991. Proceedings., Twenty-Third Southeastern Symposium on. IEEE.

On the related note, is there a well defined curl operator for a (tangential) vector field defined on a surface?

AnandJ
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    For general manifolds, Helmholtz decomposition is sort of generalized to Hodge Decomposition of differential forms. i.e. ${\omega ^k} = \operatorname{d} {\alpha ^{k - 1}} + \delta {\beta ^{k + 1}} + {\gamma ^k}$ where d is the exterior derivative, delta is the "coderivative", and gamma is harmonic. –  Apr 28 '16 at 23:33

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The presence of a curl term in the Helmholtz decomposition is very much a three-dimensional phenomenon. While it is reasonable to define the curl of a vector field tangent to a surface, what you get is either a field normal to the sphere or a scalar field, so it's not going to help us decompose tangent vector fields.

The Hodge decomposition tells us that any vector field on a surface can be decomposed as $$ X = \nabla f + J \nabla g + \sum_{i=1}^m c_iY_i$$ where $f,g$ are functions, $J$ is rotation by $\pi /2$ in the tangent plane, $c_i$ are some real coefficients determined by $X$ and the $Y_i$ are some finite collection of curl-free, divergence-free vector fields (independent of $X$) which describe the relevant features of the topology of the surface.

Very roughly, there is one field $Y_i$ going around each independent "loop" in the surface - so a plane or a sphere would have none, an infinite cylinder would have one and a torus would have two. If you're interested in some search terms to learn the details: the $Y_i$ are vector fields dual to a basis of the space of harmonic one-forms, which provide representatives of every cohomology class.

In three-dimensional manifolds you get the more familiar

$$ X = \nabla f + \operatorname{curl} V + \sum_{i=1}^m c_iY_i. $$

In higher dimensions, scalar fields and vector fields no longer suffice to describe the whole de Rham complex, so it's necessary to switch to talking about differential forms.

Anthony Carapetis
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  • Thank you for the detailed answer. What is your opinion on the paper cited in my question where they have defined curl of a function multiplied by surface normal? – AnandJ Apr 29 '16 at 01:45
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    @AnandJ: it's equivalent to the $J \nabla g$ term I've written above - if you extend $\Omega \hat n$ to a neighbourhood of the surface so that it is parallel in the normal direction, then the curl of this extension will be the tangent vector field $J \nabla \Omega$. Thus the "surface curl" sends tangent fields to normal ones and vice versa. The downside is that this interpretation as a curl requires the embedding in 3-space, whereas the "twisted gradient" $J \nabla$ (a special case of the codifferential) is expressed purely in terms of the two-dimensional geometry. – Anthony Carapetis Apr 29 '16 at 01:59