Multiplicative inverse of Weyl symbol and invertibility of operator

If the Weyl symbol $$A_W$$ of an operator $$\hat{A}$$ has a multiplicative inverse at every point of the phase-space, can I conclude that $$\hat{A}$$ is invertible?

• Weyl symbols compose through the $\star$-product. What makes you suspect plain multiplication suffices for $\star$-invertibility, which is tantamount to operator invertibility? Are you looking for counterexamples? – Cosmas Zachos Feb 15 at 15:08
• Thank you for your reply, @CosmasZachos. I know about the Moyal product composition, but I was just wondering if the answer to my question was positive. In case the idea does not hold, a counter-example would suffice. – ilp Feb 15 at 16:03
• I frankly don't know the answer, but it would be hard to ascertain. I might look for a counterexample, AB=I but A⋆B=0, but it too might be tricky to cook up.., – Cosmas Zachos Feb 15 at 16:10
• I will keep thinking about it. Perhaps something on the line you suggested could lead to a counter-example. Thanks again. – ilp Feb 15 at 16:13
• You might mean that if $A_W(x,p)$ has no zeros, then it has no zero star-gen-values, $A_W\star f(x,p)=0$...? – Cosmas Zachos Feb 15 at 20:25