# normal form of currents?

(this question did not get any answers on math.SE, so I am reposting it here)

Let $M$ be an $n$-dimensional manifold. Then the space of currents $\mathcal D^k(M)$ of degree $k$ on $M$ is the space of continuous linear functionals on the space of test $n-k$-forms. Two typical examples of currents are

where $\Gamma$ is a $n-k$-chain, and

where $\eta$ is a $k$-form.

One can extend the action of Lie derivative from forms to currenst simply by

So the algebra of differential operators on $M$ acts on the space of $k$-currents.

There are also currents that are obtained by contraction of a $k$-vector with the test form, but I am not sure whether they cannot be obtained by applying differential operators to currents of the two kinds mentioned above.

Is it true that the examples of currents above generate the space of currents as a module over the ring of differential operators, i.e. any current can be obtained from a current like $\mathcal{F}$ or $\mathcal{G}$ by successive applications of Lie derivatives? If no, can the statement be repaired by adding more closure operations/generators? Or is it a completely wrong attitude, and a general current is a significantly more “singular” object?

It may be helpful to look at a simple example. $\newcommand{\bR}{\mathbb{R}}$ Assume for simplicity that $M=\bR^n$. Fix a Radon measure $\mu$ on $\bR^n$. It defines a $0$-dimensional current on $\bR^n$. (I define the dimension of a current to be $n-\deg$.) If $\mu$ is not absolutely continuous with respect to the Lebesgue measure, then it cannot have the form $\mathcal{G}$. Also, if the support of $\mu$ is very complicated, this $0$-current is not given by the integration along a $0$-chain.
Also, it is not clear to me why there would exist a locally integrable function $f$ on $\bR^n$ and a differential operator $L$ such that $L f=\mu$ in the sense of distributions.