I am trying to prove the following special case of the Hodge decomposition theorem in differential geometry for an $n$ component vector field $V_i$ in $\mathbb{R}^n$. I have very little knowledge of differential geometry.

any vector can be written as the following combination $V_i = −\partial_i \phi + \epsilon^{ii_2 i_3···i_n}\partial_{i_2}F_{i_3i_4···i_n}$

where $F$ is a rank $n-2$ anti-symmetric tensor.

I have read the proof of the Helmholtz theorem, but I don't know how I can generalize it.

More specifically the Helmholtz theorem uses the identity $$\bar{\nabla} \times (\bar{\nabla} \times \bar{V})=\bar{\nabla}(\bar{\nabla} \cdot \bar{V} )-\nabla^2 \bar{V}$$

What is the generalization of this identity to $n$ dimensions?

Do I have to assume some Ansatz? What would that be? It would be great if somebody could show a stepwise method showing tensor calculations, and stating the necessary theorems?

Lastly, how do I prove the uniqueness that given the curl and divergence ($\partial_i V_i=s$ and $\partial_iV_j-\partial_jV_i=c_{ij}$), and the normal component at the boundary, the vector is uniquely specified.

  • $\begingroup$ The identity will be $\Delta = \mathrm{d}\mathrm{d}^* + \mathrm{d}^*\mathrm{d}$. $\endgroup$
    – Shuhao Cao
    Jun 4, 2013 at 3:36

2 Answers 2


Shuhao Cao has provided a strong answer in terms of the usual language of differential forms and tensor calculus. I present here an alternative in terms of geometric calculus which I hope will be enlightening.

As I related in my answer to your other question, geometric calculus can look somewhat different from differential forms, but most of the basic concepts are the same. For instance, there is an exterior derivative:

$$\mathrm d\omega \equiv \nabla \wedge \omega$$

And a coderivative, or interior derivative,

$$\delta \omega \equiv \nabla \cdot \omega$$

These are simply how these concepts are denoted in geometric calculus. The preference of using them over $d, \delta$ is that the geometric product makes meaningful the statement

$$\nabla \omega = \nabla \wedge \omega + \nabla \cdot \omega$$

which, written in terms of $\mathrm d\omega, \delta \omega$ might otherwise come off as patent gibberish. But the use of multivectors in geometric algebra and calculus makes this a perfectly valid mathematical statement. It allows us to separate what would otherwise be some tensor $\partial_i \omega_j$ into two distinct, meaningful parts. If $\omega$ is a $k$-vector, then $\nabla \cdot \omega$ is a $k-1$-vector and $\nabla \wedge \omega$ is a $k+1$-vector, and both pieces can be kept track of in the equation using the clifford algebra framework.

Again, treating $\nabla$ as an operator in and of itself is useful in geometric calculus because it has a Green's function:

$$\nabla G = \delta$$

where here, $\delta$ should be understood as the Dirac delta function. The form of the Green's function in 3d is $G(r) = \frac{r}{4\pi |r|^3}$, and you can ultimately relate it to having the same form in higher dimensions, just dependent on the hyperarea of the unit sphere and so on. Having a Green's function makes $\nabla$ invertible through integration in a way that the exterior and interior derivatives by themselves are not.

With that in mind, the Helmholtz decomposition can be expressed using geometric calculus quite simply. First, you ask about an identity that is used in the derivation. This is actually a general result:

$$\nabla(\nabla \omega) = \nabla^2 \omega = \nabla \wedge (\nabla \cdot\omega) + \nabla \cdot (\nabla \wedge \omega)$$

This is the geometric calculus analogue of your identity, and it unifies the Laplace-Beltrami and Laplace-de Rham operators that Shuhao Cao refers to into a single, scalar operator $\nabla^2$.

Now we can go about proving the Helmholtz decomposition. Let $E$ be a vector field. Let $M$ be some region with boundary $\partial M$. Let us presume that there exist a scalar function $\phi$ and a bivector field $A$ such that

$$E = \nabla \wedge \phi + \nabla \cdot A = \nabla \psi$$

such that $\nabla \wedge F = 0$. There is nothing particularly lacking in generality with these assumptions, as I will show. Consider the following integral:

$$\oint_{\partial M} G(r-r') dS' \, \psi(r') = (-1)^{n} \int_{M} \delta(r-r') \, dV' \psi(r') + \int_{M} G(r-r') \, dV' \, E(r')$$

This is where keeping track of grades, as they're used in clifford algebra, is important: even if $\nabla \wedge A$ were nonzero, we could separate it out of this equation by considering that 3-vector term alone. So I have done nothing that violates generality (yes, I've assumed some things about integrability and smoothness and such; we'll go with all those necessary things being applicable, for otherwise you wouldn't even be able to do the problem).

What this use of the fundamental theorem does is give us a formula for $\psi = \phi + A$, and since separate grades must always independently obey such an equation, we can write down the equation as

$$\begin{align*} \phi(r) &= \int_M G(r-r') \cdot E(r') \, |dV'|- \oint_{\partial M} |dS'| \left[ G(r-r') \cdot \hat n' \phi(r') - [G(r-r') \wedge \hat n'] \cdot A(r') \right] \\ A(r) &= \int_M G(r-r') \wedge E(r') \, |dV'| - \oint_{\partial M} |dS'| \left[G(r-r') \wedge \hat n' \phi(r') + G(r-r') \cdot \hat n' \, A(r')\right] \end{align*}$$

We're not actually going to use these formulas; I'm merely arguing that, given these potentials defined on some surface, the functions themselves can be constructed as well.

What we really want to do is look at the Green's function for $\nabla^2$ instead. There does exist such a function (which I will call $H$ such that $\nabla H = G$). In 3d, $H(r) = (4\pi |r|)^{-1}$, so this is exactly what you're used to seeing in expressions of the Helmholtz decomposition.

Again, our preferred hammer for deriving integral theorems is the fundamental theorem. Use $H$ and $E$ and observe that

$$\nabla \oint_{\partial M} H(r-r') \, dS' \, E(r') = (-1)^n (i E) + \nabla \int_M H(r-r') \, dV' \, \nabla' E(r')$$

(If you can't see how this is so, just pull the $\nabla$s into the integrals and remember $\nabla H = G$.)

But remember that $E = \nabla \psi$. The invertibility of $\nabla$ means that we can, through some integrals, essentially cancel $\nabla$ to get

$$(-1)^n \phi(r) = \oint_{\partial M} H(r-r') \, \hat n' \cdot E(r') |dS'| - \int_{M} H(r-r') \nabla' \cdot E(r') \, |dV'|$$

and the bivector potential $A$ follows by replacing dots with wedges. This is the Helmholtz decomposition. You need to know the Green's function for the Laplacian. You need to know the dimension of the space. But most importantly, since you seem hung up on this point, you need to know the tangential components of the vector field, not just the normal component.

  • $\begingroup$ +1 Nice to know Laplace-Beltrami and Laplace-de Rham operators are unified like that. Btw, when dimension is 2 or 3, $$\nabla \times u = f,\\ \nabla \cdot u = 0,\\ u\cdot n|_{\partial \Omega} =0$$ yields a unique solution too. For 2 is trivial, for 3 $f$ has to satisfy certain boundary constraint. Not for higher dimension though. $\endgroup$
    – Shuhao Cao
    Jun 13, 2013 at 15:23
  • $\begingroup$ You wrote $∇∧ϕ$ for a scalar field $ϕ$. What does the curl of a scalar field mean? $\endgroup$
    – Luke Burns
    Feb 18, 2015 at 18:48
  • $\begingroup$ In geometric calculus, we use $\nabla \wedge$ to denote the exterior derivative. Hence, $\nabla \wedge \phi = d\phi$ is the gradient (well, depending on whether you consider gradient to be a vector or 1-form, but in clifford algebra, the distinction is typically not observed anyway). This can also be denoted $\nabla \phi$ because $\nabla \phi \equiv \nabla \wedge \phi + \nabla \cdot \phi$. The last term is the codifferential and is automatically zero for a scalar field. $\endgroup$
    – Muphrid
    Feb 18, 2015 at 18:53
  • $\begingroup$ Ah. Thanks! The general result is: for a vector $a$ and r-vector $B_r$, we have $a \cdot B_r = \frac{1}{2}(aB_r - (-1)^r B_r a)$. Hence, $a \cdot B_0 = \frac{1}{2}(aB_0 - (-1)^0B_0a) = \frac{1}{2}(aB_0 - B_0a) = 0$. $\endgroup$
    – Luke Burns
    Feb 18, 2015 at 21:12

Partial answer: In general, if we ignore music isomorphism at the moment (i.e., we are not that into identifying $k$-forms with $k$-vectors). In $\Omega\subset \mathbb{R}^n$, the Hodge-Laplacian (Laplace-de Rham operator) for $k$-form is define as: $$ \Delta := \mathrm{d} \delta + \delta \mathrm{d},\tag{1} $$ and Laplace-Beltrami operator is defined as: $\Delta_l = \delta \mathrm{d}$ (see Marsden's book Manifolds, Tensor Analysis, and Applications Page 443), where $$ \mathrm{d}^k : \Lambda^k(\Omega) \to \Lambda^{k+1}(\Omega) $$ is the exterior derivative of $k$-forms, and $$\delta_k := (-1)^{nk+n+1} \star^{n-k+1}\mathrm{d}^{n-k}\star^k, $$ which is $\delta_{k}: \Lambda^k\to\Lambda^{k-1}$ (again ignoring musical isomorphism at this moment). It can be viewed as the adjoint of $\mathrm{d}^{k-1}: \Lambda^{k-1}(\Omega)\to\Lambda^{k}(\Omega)$ with respect to the inner product through Hodge star operator $\star$. Then to be precise (1) is actually: $$ \Delta =\mathrm{d}^{k-1}\delta_k + \delta_{k+1} \mathrm{d}^k : \Lambda^k(\Omega) \to \Lambda^{k}(\Omega) .\tag{2} $$ This is the formula in $n$ dimension.

Example: For 3 dimensional case, $\omega$ is $1$-form, (2) reads: $$ \Delta \omega = (\mathrm{d}^0 \delta_1 + \delta_2 \mathrm{d}^1)\omega = - \nabla (\nabla \cdot \omega) + \nabla \times (\nabla \times \omega) . $$ Above is from the cochain complex $$ \Lambda^0(\Omega)\ \stackrel{\mathrm{d}^0 = \nabla}{\longrightarrow}\ \Lambda^1(\Omega) \ \stackrel{\mathrm{d}^1=\nabla \times}{\longrightarrow}\ \Lambda^2(\Omega)\ \stackrel{\mathrm{d}^2=\nabla \cdot}{\longrightarrow}\ \Lambda^3(\Omega), $$ and its dual complex $$ \Lambda^3(\Omega)\ \stackrel{\delta_3 = -\nabla}{\longrightarrow}\ \Lambda^2(\Omega) \ \stackrel{\delta_2 = \nabla\times}{\longrightarrow}\ \Lambda^1(\Omega)\ \stackrel{\delta_1 = -\nabla \cdot}{\longrightarrow}\ \Lambda^0(\Omega). $$

Remark: if using musical isomorphism, curl is $$\big( \star\mathrm{d}(\omega^{\flat})\big)^{\sharp} = \partial_i v_j − \partial_j v_i $$ which is an $n(n-1)/2$-vector for all $1\leq i<j \leq n$.

For Helmholtz decomposition in arbitrary dimension, please see this paper page 2 (1.1), it specifically deals with the Sobolev spaces, and $L^p$-theory for Maxwell systems. Without ambiguity, it is $$ \omega = \mathrm{d}^{k-1} \alpha + \delta_{k+1} \beta + \gamma.\tag{3} $$

In (3), $\delta_{k-1}\alpha =0$, $\mathrm{d}^{k+1} \beta = 0$, and $\mathrm{d}^{k}\gamma = \delta_k \gamma = 0$.

For the last question, Mitrea's book Layer Potentials, the Hodge Laplacian and Global Boundary Problems in Nonsmooth Riemannian Manifolds has a presentation. Also this paper, section 3 has a very nice and readable presentation in addressing your question in 3 dimension. For the $n$-dimensional problem, the paper refered above has discussed a related problem(see Proposition 4.4-4.6), in which the main tool used is Helmholtz decomposition.

The other way of viewing Helmholtz decomposition is through weak formulation of PDEs to construct the decomposition, for the following problem: $$\left\{ \begin{aligned} \mathrm{d}^k\omega &= f \quad \text{in }\Omega, \\ \delta_k \omega &= 0\quad \text{in }\Omega, \\ \omega\wedge \nu &= 0\quad \text{on }\partial \Omega, \end{aligned} \right.$$ having a solution actually relies on the coercivity of the corresponding bilinear form $$ B(\cdot,\cdot) := \langle\mathrm{d}(\cdot),\mathrm{d} (\cdot)\rangle + \langle\delta(\cdot),\delta(\cdot)\rangle, \tag{4} $$ which in 3 dimensional setting is just $$B(u,v) = \langle\nabla \times u,\nabla \times v\rangle + \langle\nabla \cdot u,\nabla \cdot v\rangle.$$

(4)'s coercivity relies on a Poincaré type inequality in the corresponding Sobolev spaces, namely, $$ \|\omega\|_{L^2} \leq \|\mathrm{d}^k\omega\|_{L^2} +\|\delta_k\omega\|_{L^2}, $$ which is true when $\omega\wedge \nu = 0$. More generally, whenever we get tensor field when taking exterior derivative, the Poincaré type inequality should be $$ \|\omega\|_{L^2} \leq \|\text{symmetric part of } \nabla \omega \|_{L^2} +\|\text{anti-symmetric part of } \nabla \omega\|_{L^2}, $$ of which this paper has a dicussion.

Lastly, Marsden's book chapter 6 through 9 is a good ride.

  • $\begingroup$ @user79365 Thanks for the correction. $\endgroup$
    – Shuhao Cao
    Jun 12, 2013 at 20:48

You must log in to answer this question.