# Hodge decomposition in Minkowski space

This question is motivated by the physical description of magnetic monopoles. I will give the motivation, but you can also jump to the last section.

Let us recall Maxwell’s equations: Given a semi-riemannian 4-manifold and a 3-form $j$. We describe the field-strength differential form $F$ as a solution of the equations

$\mathrm{d}F=0$

$\mathrm{d}\star F=j$ (where $\star$ denotes the Hodge star).

If the second de-Rham-cohomology vanishes (for example in Minkowski space), $F$ is exact and we can write it as $F=\mathrm{d}A$, where $A$ denotes a 1-form.

Now let us consider monopoles: We use two 3-forms $j_m$ (magnetic current) and $j_e$ (electric current) and consider the equations

$\mathrm{d}F=j_m$

$\mathrm{d}\star F=j_e$.

Essentially, it is described in this paper, but the author Frédéric Moulin (a physicist) uses coordinates. Now he assumes that (in Minkowski space) $F$ can be decomposed using two potentials — into an exact (in the image of the derivative) and a coexact (in the image of the coderivative) form: $F=\mathrm{d}A-\star\mathrm{d}C$. Is there a mathematical justification for this assumption (maybe it is just very pragmatic)?

### The actual question

Given a 2-form $F$ on 4-dimensional Minkowski space (more generally: semi-riemannian manifolds)—are there any known conditions such that $F$ decomposes into an exact and a coexact form: $F=\mathrm{d}A+\star\mathrm{d}C$)?

For compact riemannian manifolds there is the well-known Hodge decomposition: There is always a decomposition into an exact, a coexact and a harmonic form. In the non-compact case you might be able to get rid of the harmonic form by only considering “rapidly decaying” forms (Wikipedia suggests that, but I do not have a good reference, in euclidean space there is the Helmholtz decomposition, and non-trivial (smooth) harmonic 1-forms do not vanish at infinity).

That is why I also ask: Are there “rapidly decaying” harmonic 2-forms in Minkowski space? Any references where I could see what is known about harmonic forms and Hodge theory in the semi-riemannian case are also welcome.

• I understand this is a math question on a math site, but it might be worth mentioning that the question has little to do with how physicists actually understand the possible existence of magnetic monopoles. The simplest construction of well defined magnetic monopoles involves embedding the Standard Model group $H=SU(3) \times SU(2) \times U(1)$ into a simple, compact Lie group $G$ like $SU(5)$ or $SO(10)$ and in that situation the symmetry breaking of $G$ to $H$ by the Higgs mechanism leads to a classification of magnetic monopoles by $\pi_2(G/H)$ and the equations of motion are not Maxwell's. – Jeff Harvey Jun 6 '13 at 14:12
• @Jeff, as you probably know, magnetic monopoles can exist in pure electromagnetism as well, as first demonstrated by Dirac using his "string of dipoles" construction. So I think the question's physical motivation can stand on its own. – Igor Khavkine Jun 7 '13 at 10:12
• I think it depends on what you mean by "exist." I do understand Dirac's argument and construction, but monopoles in pure electromagnetism are singular objects and there is no way to compute their mass or study any other properties they might have. For example scattering of charged particles off of magnetic monopoles is sensitive to the short distance description of the monopole. So to embed monopoles into a well defined, predictive physics framework you need to do something like what I described (or variants thereof). – Jeff Harvey Jun 7 '13 at 14:10

1. There are no "rapid decaying harmonic 2-forms" in Minkowski space.

Consider the expression $$0 = \mathrm{d} \star \mathrm{d} F = \partial^i \partial_{[i}F_{jk]} = \frac13 (\Box F_{jk} + \partial_j \partial^i F_{ki} + \partial_k \partial^i F_{ij})$$ The second and third terms in the parentheses vanish by $\mathrm{d}\star F = 0$, so we see that over Minkowski space, Maxwell's equation implies that the components of the Faraday tensor must solve a linear wave equation.

(In the case you have source, you also see that the equations are equivalent to $\Box F_{jk} = J_{jk}$ where $J_{jk}$ is built out of the magnetic and electric currents in the appropriate way.)

Now, the linear decay estimates for the wave equation (that $|F| \sim \frac{1}{1+|t|}$) is sharp. In fact, by the finite speed of propagation + strong Huygens principle, if $F$ decays too rapidly in time, conservation of energy immediately implies that $F$ must vanish identically!

(If you don't want to go through the wave equation, you can also argue through the Maxwell equation by considering the energy integrals.)

2. Conservation of energy also implies that over Minkowski space $\mathbb{R}^{1,3}$ there cannot exist a non-vanishing "harmonic" form with finite space-time $L^2$ norm. So harmonic forms cannot exist, even in a class that is not necessary "rapidly" decaying.

3. Now, under the influence of external sources $j_e$ and $j_m$, you can pose the ansatz (by linearity of the equations) $F = G + H$, where \begin{align} \mathrm{d} G & = j_e & \mathrm{d} H & = 0 \newline \mathrm{d} \star G & = 0 & \mathrm{d} \star H &= j_m \end{align} Then the cohomology result you mentioned implies that $H = \mathrm{d}A$ and $\star G = \mathrm{d}C$ for $A,C$. The difference of the true solution with your ansatz is $\tilde{F} - F$ a harmonic two form, which can be completely absorbed into the electric potential $A$ if you want...

So the claimed decomposition is available whenever it is possible to solve the equations (possibly nonuniquely) for $G$ and $H$. For compatible source terms (you need $\mathrm{d} j_e =0 = \mathrm{d} j_m$ as a necessary condition) which are $L^1$ in space-time, the existence of a solution follows from the well-posedness of the corresponding Cauchy problem, for example.

4. Much of "Hodge theory" is false for "compact semi-Riemannian manifolds" in general. For example, there is no good connection between the cohomology and the number of harmonic forms. (Simplest example: on any compact Riemannian manifold the only harmonic functions are constants. But take, for example, the torus $\mathbb{T}^2 = \mathbb{S}^1\times\mathbb{S}^1$ with the Lorentzian metric $\mathrm{d}t^2 - \mathrm{d}x^2$. Then for any $k$ the function $\sin (kx) \sin(kt)$ is "harmonic", and they are linearly independent over $L^2$. )

5. Number 4 above also implies that the Hodge decomposition is not true for general compact semi-Riemannian manifolds. Consider the exact same $\mathbb{T}^2$ as above. Let $\alpha$ be the one form $\sin(t) \cos(x) \mathrm{d}t$. We have that $$\mathrm{d}\alpha = \sin(t) \sin(x) \mathrm{d}t\wedge \mathrm{d}x \qquad \mathrm{d}\star \alpha = \cos(t) \cos(x) \mathrm{d}t \wedge \mathrm{d}x$$ To allow a Hodge decomposition requires that there exists $A$ and $C$, scalars, such that $$\mathrm{d}\star\mathrm{d}C = \mathrm{d}\alpha \implies \Box C = \pm \sin(t)\sin(x)$$ and $$\mathrm{d}\star \mathrm{d}A = \mathrm{d}\star\alpha \implies \Box A = \pm \cos(t) \cos(x)$$ where $\Box$ is the wave operator $\partial_t^2 - \partial_x^2$.

Now taking Fourier transform (since we would desire $C$ and $A$ be in $L^2$ at least), we see that we have a problem. If $f\in L^2(\mathbb{T}^2)$ we have $$\widehat{\Box f} = -(\tau^2 - \xi^2)\widehat{f} = 0 \qquad\text{ whenever } |\tau| = |\xi|$$ But the Fourier support of $\sin(t)\sin(x)$ and $\cos(t)\cos(x)$ are precisely when $\tau = \xi = 1$. So there in fact does not exist a Hodge-like decomposition for the $C^\infty$ form $\alpha$.

The above just goes to say that

1. You are not going to find anything in the literature concerning Hodge decomposition for general compact semi-Riemannian manifolds, as such as theory does not exist
2. You are also unlikely to find literature that studies the Hodge decomposition in the non-compact case in general: on non-compact manifolds causality violations (closed time-like curves, for example) can, in principle, also cause problems with solving the wave equation (both homogeneous and inhomogeneous), which will prevent an analogue for Hodge decomposition to hold.
3. Your best bet, in so far as two-forms are concerned, are precisely the mathematical-physics literature concerning Maxwell equations over globally hyperbolic Lorentzian manifolds. For these manifolds, at least in principle, some version of the Hodge decomposition can be had by solving the appropriate initial value problems (something like $\Box C = j_e, \Box A = j_m, \mathrm{d}F_0 = 0 = \mathrm{d}\star F_0$ where $F_0$ is the harmonic part, with initial data prescribed so that the data for $C$ and $A$ vanishes on the initial surface and $F_0$ is equal to $F$ on the initial surface).
• I notice that your counter-example 3 only works because the lengths of the space-like and time-like geodesic on your torus are commensurate. Out of curiosity, might there be Hodge decomposition for generic Lorentzian manifolds? (For example, $\mathbb{R}^2/\mathbb{Z}^2$ where the metric on $\mathbb{R}^2$ is chosen generically.) – David E Speyer Nov 11 '14 at 17:16
• @DavidSpeyer: I don't know. On $\mathbb{T}^2$ with irrational speed of light, the only "harmonic" forms are constants; this can be seen immediately using the method of characteristics. Or you can see it on the Fourier side where the lines $\tau \pm c\xi = 0$ for $c\not\in \mathbb{Q}$ only intersects $\mathbb{Z}^2$ at the origin. – Willie Wong Nov 12 '14 at 8:47
• On the other hand, the div-curl equation $\mathrm{d} \alpha = 0$, $\mathrm{d}\star \alpha = f$ can be reduced to solving in characteristic coordinates $\partial_u \alpha_v = \partial_v \alpha_u = f / 2$. So whether the Hodge decomposition exists now rapidly becomes a problem in ergodic theory, I think. – Willie Wong Nov 12 '14 at 8:55

Willie's answer is of course correct. But the part that I think is most interesting in the physical context is only briefly mentioned in point 3. of the last paragraph. Let me expand on that.

The relevant condition that make everything work nicely is that the background Lorentzian spacetime be globally hyperbolic. There are two important spaces of forms, $\Omega^p_0$ and $\Omega^p_{SC}$, $p$-forms with compact and spacelike compact supports, respectively. A set $X$ is spacelike compact if it is contained in the causal influence set of a compact set $Y$, $X\subseteq J(Y)$.

On Lorentzian manifolds, the D'Alembertian operator $\square = d\delta + \delta d$ is the analog of the Laplacian operator $\Delta$ in Riemannian signature. Note that I'm using the notation $\delta = (-1)^p{\star d \star}$. It is not elliptic, but it is hyperbolic with very nice properties. Instead of the nice analytical properties of the elliptic Laplacian, we have the following exact sequence (the image of each map is equal to the kernel of the next one): $$0 \to \Omega^p_0 \stackrel{\square}{\to} \Omega^p_0 \stackrel{G}{\to} \Omega^p_{SC} \stackrel{\square}{\to} \Omega^p_{SC} \to 0 ,$$ where $G=G^+-G^-$ is the causal Green function, defined as the difference of the retarded ($G^+$) and advanced ($G^-$) Green functions. The retarded and advanced Green functions of the D'Alembertian always exist on Globally hyperbolic manifolds (see, for instance, here or here).

A version of the result that you want is this: if $F\in \Omega^2_{SC}$, $d F = 0$ and $\delta F = 0$, then there exist two forms $A\in \Omega^1_{SC}$ and $B\in \Omega^3_{SC}$, with $\delta A = 0$ and $d B = 0$, such that $F = d A + \delta B$. Moreover, there exist $\alpha \in \Omega^1_0$ and $\beta \in \Omega^3_0$ such that $A=G\alpha$ and $B=G\beta$. This result can be found, for instance, as Prop.2.2 here. I think you can straight forwardly adapt the proof to the case with non-vanishing currents $j_e$ or $j_m$.

• How to read the Green function as an operator $\Omega^p_0\to\Omega^p_{SC}$? – The User Jun 6 '13 at 12:11
• Just a remark, in the case with non-vanishing currents you no longer have the the $A = G\alpha, B = G\beta$ results; the "moreover" part does not adapt to the non-vanishing currents case. (BTW, that seems to be the only part that uses the exact sequence you mentioned; to get the actual "Hodge-like" decomposition for non-vanishing currents, Prop 2.1 in your last link is more relevant than Prop 2.2) – Willie Wong Jun 6 '13 at 12:38
• @The User, a Green function $G$ can be represented as an integral kernel $G(x,y)$. Then you need only use the usual formula $(G\alpha)(x) = \int G(x,y) \alpha(y) \, dy$. For example, the integral kernel of the retarded Green function for the scalar wave equation in Minkowski space is $G^+(x,y) = \Theta(x_0-y_0)\delta((x-y)^2)$. – Igor Khavkine Jun 6 '13 at 13:08