Lichnerowicz formula

Remarks

Work at a point with a frame satisfying $\nabla e_i = 0$ there. Then $D^2 = \sum_{i,j} e_i e_j \nabla_{e_i} \nabla_{e_j}$, the terms with $i = j$ give $\nabla^* \nabla$, and the terms with $i \ne j$ give $\frac{1}{2} \sum_{i,j} e_i e_j R^S(e_i, e_j)$, where $R^S$ is the curvature of $S$. The curvature of the spin connection is the Riemann tensor acting through Clifford multiplication, and the first Bianchi identity reduces this sum to $\frac{R}{4}$.

Example: on flat $\R^n$, $R = 0$ and the formula says that $D^2 = -\sum_j \partial_j^2$ on spinor fields, the defining property of the flat Dirac operator. On the round sphere $S^n$ of radius $1$, $R = n(n-1) > 0$, so $D^2 \ge \frac{n(n-1)}{4}$ as an operator, which bounds the eigenvalues of $D$ away from $0$.

Going deeper

For a student

The formula is a special case of the Bochner–Weitzenböck formula for the Dirac operator of a Dirac bundle, like the Weitzenböck identity for forms. Integrating it on a closed manifold gives $\norm{D\psi}^2 = \norm{\nabla\psi}^2 + \frac{1}{4} \int_M R \abs{\psi}^2$, which is the source of the Lichnerowicz vanishing theorem. For a spinor bundle twisted by a bundle $E$ with connection, an extra term appears: Clifford multiplication by the curvature of $E$.