We know Hodge decomposition splits any $k$-form into three $L^2$ components. And I see some proofs, none of them provide an explicit constructive method. Is there any general method to construct one? Or it is just a theorem of existence.

  • I believe it is just a theorem of existence. – user122283 Mar 26 '14 at 2:59

Let $\partial$ be the vector differentiation operator on some region $M$ that is $n$-dimensional. In the calculus of clifford algebra, this is a single operator that incorporates both the exterior derivative $d$ and the interior derivative $\delta$. That is, given a $k$-vector field (or equivalently, a $k$-form) $F$, we can arrive at a $k+1$-vector field $j$ and a $k-1$-vector field $\rho$ from the following:

$$\partial F = \rho + j \implies \partial \cdot F = \delta F = \rho, \quad \partial \wedge F = dF = j$$

The reason to consider $\partial$ as an object in itself is that it usually has a Green's function. Suppose $G$ is that Green's function. Then one merely needs to invoke the fundamental theorem of calculus:

$$\oint_{\partial M} G(r-r') dS' F(r') = \int_M G(r-r') dV' \partial' F(r') + \int_M \dot G(r-r') \, dV' \dot \partial' F(r')$$

Some notes: $dS'$ should be interpreted as a tangent $n-1$-vector. It is absolutely not an $n-1$-form. Similarly, $dV'$ is an $n$-vector. The overdots on the last integral denote that $G$ is being differentiated, not $F$. This is a consequence of the product rule.

Denote the LHS by $i\gamma(r)$, where $i$ is the unit $n$-vector. Observe that $\partial \gamma = 0$, as only $G$ will be differentiated, but $\partial G = \delta_d$, and since the surface integral will never contain the point $r= r'$, the Dirac delta will kill this term.

Moving $\partial$ through $dV'$ at the cost of $(-1)^{n-1}$ allows us to get to $\dot G \dot \partial$, which is also equal to $\delta_d$; this simplifies the last integral, and we can write

$$i\gamma(r) = \int_M G(r-r') \, dV' [\rho (r') + j(r')] + (-1)^{n-1} i F(r)$$

again, $i$ is the unit $n$-vector.

Finally, we can separate the $\rho$ and $j$ terms, calling their respective integrals $i\alpha(r)$ and $i\beta(r)$. Note that $\partial \alpha = \partial \cdot \alpha = \rho$ and $\partial \beta = \partial \wedge \beta = j$. By the equality of mixed partial derivatives, we know that $\partial \cdot (\partial \cdot \alpha) = 0$ and $\partial \wedge (\partial \wedge \beta) = 0$.

This completes the construction. $F = \alpha + \beta + \gamma$, such that $\partial \cdot (\partial \cdot \alpha) = 0$, $\partial \wedge (\partial \wedge \beta) = 0$, and $\partial \gamma = 0$.

For reference, the explicit forms of $\alpha, \beta, \gamma$ are below:

$$\begin{align*} \gamma(r) &= \oint_{\partial M}\langle G(r-r') \hat n' F(r') \rangle_k |dS'|\\ \alpha(r) &= \int_M G(r-r') \wedge [ \partial' \cdot F(r') ]|dV'| \\ \beta(r) &= \int_M G(r-r') \cdot [\partial' \wedge F(r')] |dV'| \end{align*}$$

Some more notes:

1) In the expression for $\gamma$, I use the grade projection $\langle \rangle_k$, which is just a fancy way of saying "multiply this using clifford products, then take the $k$-vector part". An expression using more elementary products would be hideous.

2) There's actually considerably more content to the equation I gave than just the Hodge decomposition. In particular, consider $k+2$ and $k-2$-vector terms. These give various integral theorems that, alas, would be somewhat tedious to list.

3) The Hodge decomposition is usually presented as $F = d\alpha + \delta \beta + \gamma$. I've presented it slightly differently--I took $\alpha$ and $\beta$ as multivector fields of the same grade as $F$. This was for convenience; since $\alpha$ and $\beta$ obey the constraints I gave earlier, it's not hard to find potentials for them.

I've done this calculation dozens of times, though I never remain particularly sure on the signs of the terms. It's actually a lot easier (in my mind) to prove this than it is to prove the Helmholtz decomposition.

If you find this approach using geometric calculus powerful or interesting as a supplement to traditional differential forms, I encourage you to look into it further. The literature should be quite useful even to someone who's only been schooled in differential forms.

  • This is a great answer. Thanks for putting this together. – Mike Miller Mar 26 '14 at 3:44
  • Wow, it's worthwhile for me to compare this with mine. I followed the proof of Helmholtz's theorem and wrote a proof generalized to $n$ dimensional vector fields decomposition, same with geometric algebra. But I don't see the harmonic term, also I don't know know the required condition for topology on which vector field define. – Shuchang Mar 26 '14 at 8:23
  • Helmholtz is a little different in that they use the Green's function for $\nabla^2$ instead of $\nabla$, so it's somewhat hard to directly compare. – Muphrid Mar 26 '14 at 14:42

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