# Normal form of currents

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 $$\mathcal{F}(\omega) = \int_\Gamma \omega$$ where $\Gamma$ is a $n-k$-chain, and $$\mathcal{G}(\omega) = \int_M \eta \wedge \omega$$ where $\eta$ is a $k$-form.

One can extend the action of Lie derivative from forms to currenst simply by $$L_X \mathcal{H}(\omega) = \mathcal{H}(L_X \omega)$$ So the algebra of differential operators on $M$ acts on the space of $k$-currents.

Is it true that the two 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.