I am interested in whether each component of a divergenceless vector can itself be written as a divergence. My motivation for this question is the characterization of so-called trivial conservation laws in physics. I suspect this representation is general, but I have not made much progress in a proof.


Consider the following differential equation for a vector $M$ depending on a set of independent variables $x=(x_1,x_2,...,x_n)$ and an arbitrary smooth vector field $u=(u^1,u^2,...,u^p)$ and its derivatives $\partial u=(u^1_1,u^1_2,...,u^p_n)$: \begin{align}\tag{1} \sum_{i=1}^nD_iM^i(x,u,\partial u(x),...,\partial^m u(x))=0. \end{align} Here, $D_i=\partial_i+u^j_i\partial_{u^j}+...$ denotes a "total derivative", and $\partial^ku$ denotes the set of $k$-th order derivatives of $u$.


The general solution to (1) for $M$ is as follows: \begin{align}\tag{2} M^i=\sum_{j\neq i}D_jN^{ij}(x,u,\partial u,...,\partial^q u),\quad 1\le i\le n, \end{align} where each $N^{ij}=-N^{ji}$ is some function in an anti-symmetric $n\times n$ matrix, and $q\ge m-1$ is some integer.

By "general solution", I mean that if (1) has a solution $M$, then there exist functions $N$ such that (2) holds.


It is clear that (2) is a particular solution to (2). That it is the general solution is straightforward to derive when $m=1$. We can do this by expanding (1) using the chain rule and equating coefficients of higher order derivatives to zero. However, this problem quickly becomes intractable as $m$ is increased. I am wondering if there is a more elegant and general way to approach this problem.

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    $\begingroup$ Just a complement to Robert Bryant's answer. Once you convert (using Hodge duality) your $M^i$ and $N^{ij}$ tensor to differential forms and divergence operators to exterior derivatives, you are essentially asking whether there exists a "Poincaré lemma" for the de Rham complex on $u$-dependent differential forms. The theory of the variational bicomplex, expounded in I. Anderson's notes, was created to precisely answer that question. $\endgroup$ – Igor Khavkine Sep 14 '15 at 9:18
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    $\begingroup$ An impatient physicist might also appreciate explicit and accessible formulas that go along with a more pedagogical introduction. Such formulas can be found, for instance, in the appendices of the following papers: hep-th/9601124, hep-th/0111246. $\endgroup$ – Igor Khavkine Sep 14 '15 at 9:20

The answer to your question is 'yes', that is the general solution. This is one of the basic results in the theory of the variational bicomplex. It is a statement of the vanishing of a certain cohomology group in the variational bicomplex. A good place to look for the proof would be in say, these lectures by Ian Anderson on the variational bicomplex.

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