For the usual $4$-dimensional Minkowski space $M$, the standard Dirac operator is given by
$$
D: C^{\infty}(M) \to C^{\infty}(M), ~~~~~ f \mapsto \sum\_{i=1}^4 \gamma\_i\frac{\partial f}{\partial x\_i},
$$
where the $\gamma\_i$ are the usual [gamma matrises](http://en.wikipedia.org/wiki/Gamma_matrices). As we all know, $D$ was originally constructed as a square root of the Lapalcian $\sum\_{i=1}^4 \frac{\partial^2 }{\partial x\_i^2}$. Indeed, routine calculation will verify that $D$ does indeed square to give the Laplacian.


Moreover, there exists another square root of the Laplacian, namely 
$$
d + \ast d \ast: C^{\infty}(M) \to \Omega^1(M),
$$
where $d$ is the usual exterior derivative, and $\ast$ is the Hodge $\ast$-operator. 


How do these two square roots of the Laplacian relate to each other? Are they two separate objects, or just two ways of looking at the same thing?



Dirac's Original Operator and the Hodge--Dirac Operator
As Sebastian and A.J. Tolland point out, the problem is that the two operators act on different bundles. However, the exterior algebra is canonically isomorphic to the Clifford algebra as vector spaces. Using this identification, you find two operators $D$ and $d+d^{\ast}$. Both operators have first order and they have the same principal symbol. Hence their difference is a tensor, i.e. linear over $C^{\infty}$-functions. Finally, both kill translation-invariant forms. Since invariant forms span all forms (over $C^{\infty}$-functions, they agree.


On a general manifold, there might be a problem due to curvature, but if I remeber correctly, they are equal on all Lorentz-manifolds. See Lawson-Michelson.


