Dirac operator on a spin manifold

Remarks

The right side does not depend on the frame, because it is the contraction of $\nabla^S \psi \in \Gamma(T^*M \otimes S)$ with Clifford multiplication, after identifying $T^*M$ with $TM$ by the metric. The principal symbol of $D$ at $\xi$ is Clifford multiplication by the vector dual to $\xi$, up to a factor $i$ in some conventions, and its square is $-\abs{\xi}^2$. So $D$ is an elliptic operator and a Dirac operator in the general sense. It is formally self-adjoint for the Hermitian metric on $S$. In even dimension $S = S^+ \oplus S^-$, and $D$ maps $\Gamma(S^\pm)$ to $\Gamma(S^\mp)$.

Example: on $\R^n$ with the flat metric and the trivial spin structure, $S = \R^n \times \C^N$, the spin connection is the ordinary derivative, and $D = \sum_j \gamma_j \partial_j$, where $\gamma_j$ are complex $N \times N$ matrices with $\gamma_j \gamma_k + \gamma_k \gamma_j = -2 \delta_{jk} I$. For $n = 1$ one can take $N = 1$ and $\gamma_1 = i$, which gives $D = i \, d/dx$ and $D^2 = -d^2/dx^2$.

Non-example: the complex projective plane $\C P^2$ with the Fubini–Study metric has no spin structure, since its second Stiefel–Whitney class is nonzero. So it has no spinor bundle, and the definition does not apply; one uses a spin$^c$ structure instead.