Construction of Hodge decomposition 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.
 A: 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.
