During a talk I heard it was claimed (without proof) that the canonical connection $\nabla^{S^-}$ on the bundle $S^-\to X^4$ of negative chirality spinors over a spin, Ricci-flat, anti-self-dual (ASD) 4-manifold $(X^4,g)$ is flat. Why is this true, i.e. why does the ASD Weyl curvature $W^-$ not contribute to the curvature of $\nabla^{S^-}$?
I think it should depend on $W^-$: by Ricci-flatness and anti-self-duality of $g$, the curvature operator $R:\Lambda^2\to \Lambda^2$ of $g$ is equal to $W^-$. Thus for any $v,w\in \Gamma(TX)$ and $\psi\in \Gamma(S^-)$,
\begin{align} \mathcal{R}^{S^-}_{v,w}\cdot\psi &=\frac{1}{2}\sum_{i<j}\langle R_{v,w} e_i,e_j\rangle e_i\cdot e_j \cdot \psi \\ &=\frac{1}{2}\sum_{i<j}\langle W^-_{v,w}e_i,e_j\rangle e_i\cdot e_j \cdot \psi, \end{align} which (as far as I can tell) is nonzero provided $W^-\neq 0$ and $v\wedge w \in \Lambda^{2,-}$ is a an ASD 2-form. Where is my mistake?
Thanks for your help.