Clifford algebras and spinors part 3: Bochner identity
This post has been migrated from my old blog, the math-physics learning seminar.
Let \(M\) be a Riemannian manifold, and let \(Cl(M)\) be its Clifford bundle. Let \(E \to M\) be any vector bundle with connection, and assume that \(C^\infty(M, E)\) is a \(Cl(M)\)-module. We can define a Dirac operator \(\mathcal{D}\) acting on sections of \(E\) via the formula
\[ \mathcal{D} \sigma = \sum_{i=1}^n e_i \cdot \nabla_i \sigma \]
for any orthonormal frame \({e_1, \dots, e_n}\) on \(M\), and where \(\cdot\) denotes the Clifford module action. We demand that the connection on \(E\) is compatible with Clifford multiplication in the following sense:
\[ \nabla_j (e_i \cdot \sigma) = (\nabla_j e_i) \cdot \sigma + e_i \cdot \nabla_j \sigma. \]
Let \(R\) denote the curvature of \(E\), i.e. we have
\[ [\nabla_i, \nabla_j] \sigma = R(e_i, e_j) \sigma+ \nabla_{[e_i, e_j]} \sigma \]
We can define an endomorphism \(\mathcal{R}\) on \(E\) by
\[ \mathcal{R} = \frac{1}{2} \sum_{ij} R(e_i, e_j). \]
Theorem. We have the identity \(\mathcal{D}^2 = -\Delta + \mathcal{R}\).
Proof. We compute
\[ \begin{aligned} \mathcal{D}^2 \sigma &= \sum_{ij} e_i \nabla_i \left( e_j \nabla_j \sigma \right) \cr &= \sum_{ij} e_i e_j \nabla_i \nabla_j \sigma + e_i ( \nabla_i e_j ) \nabla_j \sigma \cr &= -\Delta \sigma + \frac{1}{2}\sum_{ij}e_i e_j [\nabla_i, \nabla_j] \sigma + \sum_{ij} e_i ( \nabla_i e_j) \nabla_j \sigma \cr &= -\Delta \sigma + \frac{1}{2}\sum_{ij}e_i e_j R(e_i, e_j) \sigma + \frac{1}{2}\sum_{ij} e_i e_j \nabla_{[e_i, e_j]} \sigma+ \sum_{ij} e_i ( \nabla_i e_j) \nabla_j \sigma \cr &= (-\Delta + \mathcal{R})\sigma + \frac{1}{2} \sum_{ij} \left( e_i e_j \nabla_{[e_i, e_j]}\sigma + e_i (\nabla_i e_j) \nabla_j + e_j (\nabla_j e_i) \nabla_i \right)\sigma \end{aligned} \]
We will be done provided we can show that the last term vanishes. Notice that it is fully tensorial, since it can be expressed as \(\mathcal{D}^2 + \Delta - \mathcal{R}\). On the other hand, the terms \([e_i, e_j]\) and \(\nabla_j e_i\) are (by definition!) proportional to Christoffel symbols. Since we can always choose a frame so that these vanish at a point, these terms must vanish identically. Hence we have \(0 = \mathcal{D}^2 + \Delta - \mathcal{R}\), as desired.