Lichnerowicz formula for a twisted Dirac operator ($D_W^2={\nabla }^{\ast }\nabla +\frac{1}{4}\mathrm{Scal}+{\sum }_{i<j}c(e^i)c(e^j)F_{ij}^W$)

At the centre of geodesic normal coordinates, expand the square of the Twisted Dirac operator. The diagonal terms give the rough Laplacian; the off-diagonal terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. The tensor-product vector-bundle curvature is the sum of spin curvature and $F^W$. Substituting spin curvature $\frac{1}{4}{\sum }_{a,b}⟨R(e_i,e_j)e_a,e_b⟩c(e^a)c(e^b)$ and using the Clifford algebra relations yields the scalar term $\mathrm{Scal}/4$: the four-generator contribution vanishes by the algebraic Bianchi identity and the contracted terms are the scalar curvature. This proves the displayed identity at every point.