TITLE: Moyal $\star$-product inverse?
QUESTION [8 upvotes]: On a 2n-dimensional phase-space with coordinates $x$ and $p$, the Moyal product can be written explicitly as
$$g(x,p) \star h(x,p) = g(x,p) e^{\frac{i}{2}\left( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overrightarrow{\partial_x} \cdot \overleftarrow{\partial_p}\right)} h(x,p) \, .$$
Assume now that we have something like $f(x,p)=g(x,p)\star h(x,p),$ where we have to solve for $g(x,p)$. I suppose it's not always possible, but how would one go about solving that when it is? And what would the invertibility conditions even be? I can't seem to find the answer anywhere.
Maybe even a simpler problem, $f(x,p)=g(x,p)\star g(x,p) $ with $f(x,p)$ known. How would one take the $\star$-square root and when would it be possible?

REPLY [4 votes]: The inversion is conveniently described in terms of the Fourier transform
$$g(x,p)=\int dy\,e^{-iyp}G(x+y/2,x-y/2).$$ Then the composition $f(x,p)=g(x,p)\star h(x,p)$ is a matrix multiplication [1],
$$F(x,y)=\int dz\, G(x,z)H(z,y).\qquad(\ast)$$
So to find $h$ if $f$ and $g$ are given one would first calculate the Fourier transforms $F$ and $G$, 
$$G(x+y/2,x-y/2)=\frac{1}{2\pi}\int dp\,e^{iyp}g(x,p),$$
then solve the integral equation $(\ast)$ for $G$, and finally transform back to $g$. Whether this is doable will of course entirely depend on the details of the particular problem. But this is the general recipe.
[1] Map of Witten's $\star$ to Moyal's $\star$, Itzhak Bars, 2001.