# Existence of second order potential for PDE

There is a statement in the literature (see the paragraph between equations (18) and (19) in http://aip.scitation.org/doi/10.1063/1.523863), which I would like to generalise, but I don't have a nice proof of the original claim. The statement is the following:

Given any symmetric tensor field $T_{ab}$ on a 3-dimensional hyperboloid $H$ with $$D_{[a}T_{b]c}=0, \tag{1}$$ there exists a scalar field $T$ on $H$ satisfying $$T_{ab} = D_a D_b T + T h_{ab}, \tag{2}$$ where $h_{ab}$ is the metric on the hyperboloid and $D$ is the covariant derivative on the hyperboloid. The Riemann tensor on $H$ satisfies $R_{abcd} = h_{ac}h_{bd} - h_{ad} h_{bc}$.

(I am also happy to assume that $T_{ab}$ is traceless, which together with (1) implies that $T_{ab}$ is divergence-less.)

Physically this is saying the the tensor $T_{ab}$ admits a (second order) potential $T$. I would like to generalise it to other dimensions, possibly to other manifolds, etc. But the only proof of this statement which I know, relies on explicit use of spherical harmonics on $H$.

Are there any general methods to determine if given tensor satisfying certain PDE can be written as some differential operator (possibly involving geometric data) acting on some other tensor?

After some search I found Characterizing Hessians among symmetric bilinear tensors, which I can mimic to show that given $T$ then $T_{ab}$ constructed in (2) indeed solves (1). But how can I show that all solutions to (1) are given by (2)?

First of all, you have a sign wrong in your formula for the curvature. The curvature tensor you gave has positive constant sectional curvature +1 while you claim that you want negative sectional curvature (i.e., hyperbolic space), which would flip the sign of $R$. Second, when you speak of spherical harmonics, I believe you must be copying the formula for the unit $n$-sphere, $S^n$, not hyperbolic space $H^n$. This also causes an error in your potential formula (2), which would be correct for the $n$-sphere, but should be $T_{ab} = D_aD_bT - T h_{ab}$ for hyperbolic space.

I'll give the answer for hyperbolic space, since that is what you want, but be aware that you'll need to flip signs to get the same answer for the $n$-sphere.

The result you are seeking follows immediately from the Frobenius theorem, using the techniques mentioned in the MO question you cite at the end. The idea is simply this: Take the tensor $T$ as given. Let $\omega_i$ be any $h$-orthonormal frame field (which can be chosen globally on $H^n$ since it is contractible) and let $\theta_{ij}=-\theta_{ji}$ be the unique $1$-forms satisfying $\mathrm{d}\omega_i = -\theta_{ij}\wedge\omega_j$. (Here and below, I am using the 'Einstein' summation convention.) By the assumption that the sectional curvature is identically $-1$, we have $\mathrm{d}\theta_{ij} = -\theta_{ik}\wedge\theta_{kj} - \omega_i \wedge\omega_j$.

Now, on $X = H^n\times\mathbb{R}\times\mathbb{R}^n$, with projections $u:X\to\mathbb{R}$ and $(u_i):X\to\mathbb{R}^n$ onto the second and third factors, consider the Pfaffian system $\mathcal{I}$ generated by the $(n{+}1)$ linearly independent $1$-forms $$\xi = \mathrm{d}u - u_i\ \omega_i \quad\text{and}\quad \xi_i = \mathrm{d}u_i +\theta_{ij}\ u_j - (T_{ij}+u\,\delta_{ij})\ \omega_j\,.$$ By the hypotheses on $T = T_{ij}\omega_i\omega_j$ and the curvature of $h$, this system is Frobenius, i.e., it satisfies $\mathrm{d}\xi \equiv \mathrm{d}\xi_i\equiv 0 \mod \mathcal{I}$.

Thus, $X$ is foliated by the leaves of $\mathcal{I}$, which are transverse to the fibers of the projection $\pi:X\to H^n$ onto the first factor. Since the system is affine linear in $(u,u_i)$, it follows that each leaf $L\subset X$ of $\mathcal{I}$ becomes a covering space of $H^n$ under the projection $\pi:L\to H^n$. Since $H^n$ is connected and simply-connected, such a projection is a diffeomorphism of the leaf $L$ with $H^n$ and hence has an inverse, which can be written in the form $\sigma:H^n\to L\subset X$ of the form $\sigma(p) = \bigl(p,f(p),f_i(p)\bigr)$. By construction, the function $f = u\circ\sigma:H^n\to\mathbb{R}$ and $(f_i) = (u_i)\circ\sigma:H^n\to\mathbb{R}^n$ must satisfy $$df = f_i\ \omega_i \qquad\text{and}\qquad df_i = -\theta_{ij}\ f_j + (T_{ij}+f\delta_{ij})\ \omega_j\,.$$ The function $f$ is the potential that you seek.

• How do you come up with the second equation in (1)? In particular, does the coefficient in front of $f \delta_{ij}$ term matter? – Yegor Dec 5 '17 at 12:02
• @Yegor: I'm alluding to the fact that the underlying manifold is simply connected and that the affine sheaf of solutions is 'flat'. However, I can rewrite so as to avoid this, so I will. Maybe that will be more useful to you. – Robert Bryant Dec 12 '17 at 7:49
• @Yegor: Yes, these questions can be answered. There are effective tests for when the kernel of a given differential operator can be written as the image of another. This can even be generalized to cover nonlinear operators in some cases. I should say right away that, most of the time, this cannot be done, but when it can, one can usually employ it to good effect. I cannot go into details here, but the most important cases can be described via the machinery of Spencer sequences, resolutions, and cohomology. The Poincaré sequence is the most famous, of course, but there are many others. – Robert Bryant Dec 12 '17 at 9:00
• @Yegor, if you have a linear differential operator $D$ and a linear operator $C$ such that $CD=0$, with the additional property that any other operator $E$ such that $ED=0$ must factor through $C$, $E=E'C$ for some $E'$, then $C$ is a (universal/complete/generating) compatibility operator for $D$. Your problem with $LS=0$ is mapped to the one I described by taking formal adjoints, $D=L^*$, $C=S^*$ and vice versa. As Robert Bryant said, compatibility operators exist under general conditions, and can be analyzed with Spencer cohomologies. ... – Igor Khavkine Dec 12 '17 at 10:50
• ... A comprehensive modern treatment can be found in the book Involution by W. Seiler. Unfortunately, the theory is quite involved. In some cases, if you write out all the operators in coordinates, some computer algebra packages may do the calculations for you (like for instance the Janet package for Maple). – Igor Khavkine Dec 12 '17 at 10:52