Is the square root of curl^2-1/2 a natural (Dirac-)operator?

In current computations on a particular $3$-dimensional Riemannian manifold, a first order differential operator $D:\Gamma^\infty(TM,M)\to \Gamma^\infty(TM,M)$ acting on vector fiels shows up, with the property $$D^2=\mathrm{curl}^2 - \frac{1}{2}I,$$ where $I$ is the identity and $\mathrm{curl}:\Gamma^\infty(TM,M)\to \Gamma^\infty(TM,M)$ is the familiar Riemannian curl operator. Considered instead as an operator acting on $1$-forms, curl corresponds to $\ast\circ d$, where $\ast$ is the Hodge star operator and $d$ the exterior derivative. Thus, my question can be formulated as:

Suppose that, on a $3$-dimensional compact connected Riemannian manifold, there exist the two equivalent operators $$D:\sqrt{\mathrm{curl}^2-\frac{1}{2}I}: \Gamma^\infty(TM,M)\to \Gamma^\infty(TM,M),\qquad \tilde D=\sqrt{\ast\circ d-\frac{1}{2}I}:\Omega^1(M)\to\Omega^1(M).$$ Can one express them in terms of well-known geometrical/topological operators such as $\ast$ and $d$, or does one rather have to consider them as 'basic' operators of Dirac-type? In any case, what can we learn about the de Rham cohomology of $M$ from the spectrum of $D$ (or $\tilde D$)?