= Lichnerowicz formula for a twisted Dirac operator
{c}
{title2=$D_W^2=\nabla^*\nabla+\frac14\operatorname{Scal}+\sum_{i<j}c(e^i)c(e^j)F^W_{ij}$}
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 $\frac14\sum_{a,b}\langle R(e_i,e_j)e_a,e_b\rangle c(e^a)c(e^b)$ and using the <Clifford algebra> relations yields the scalar term $\operatorname{Scal}/4$: 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.
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