Dirac bundle

Remarks

Clause 2 is equivalent to the skew-adjointness of Clifford multiplication by each tangent vector. Clauses 2 to 4 make $D$ formally self-adjoint, and its square satisfies $D^2 = \nabla^* \nabla + \mathfrak{R}$ for an explicit curvature endomorphism $\mathfrak{R}$. The spinor bundle of a spin manifold, with the spin connection, is a Dirac bundle, and its Dirac operator is the Atiyah–Singer operator.

Example: let $S = \Lambda^* T^*M$ with the metric and connection induced by $g$, and let $v \cdot \omega = v^\flat \wedge \omega - \iota_v \omega$. Then $v \cdot v \cdot \omega = -(v^\flat \wedge \iota_v \omega + \iota_v(v^\flat \wedge \omega)) = -g(v, v) \, \omega$, so clause 1 holds, and clauses 2 to 4 hold because the Levi-Civita connection is metric and commutes with wedge and contraction. Its Dirac operator is $\sum_i (e_i^\flat \wedge \nabla_{e_i} - \iota_{e_i} \nabla_{e_i}) = d + \delta$, whose square is the Hodge Laplacian.

Non-example: on $M = \R^n$ with $n \ge 2$, take the same bundle and Clifford multiplication, but the connection $\nabla'_X = \nabla_X + dx_1(X) \, A$, where $A$ is Clifford multiplication by $e_1 = \partial_1$. Since $A$ is skew-adjoint, $\nabla'$ still preserves the metric, so clause 3 holds. Clause 4 fails: for $X = e_1$ and $\varphi = e_2$, it would require $A(e_2 \cdot \psi) = e_2 \cdot A\psi$, but $e_1 e_2 = -e_2 e_1 \ne e_2 e_1$.