Let $M^4$ be a closed Riemannian manifold and $Z:=S\big(\Lambda^2_+(M)\big)$ denote the twistor space of $M,$ i.e., the sphere bundle of the self-dual 2-forms on $M$. Now at a point $(m,J)\in Z$ the tangent space can be thought of as the direct sum of $T_mM$ and the space of 2-forms on $M$ perpendicular to $J$. Hence we have a metric on $TZ,$ using the metric on $TM$ and the inner product on forms. So, we get a splitting $TZ=V\oplus H$. $V$ being the bundle of tangent vectors of the fibers, we call it $\mathcal{O}(2).$ What is the curvature of this complex line bundle w.r.t this metric?
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$\begingroup$ I think that your question is just equivalent to the classification of all 2-sphere bundles over closed riemanniann 4-manifold and this surely include the classification of closed Einstein Riemannian 4-manifold. So I think it too crude to attempt an answer. Nevertheless, to be sure, using EINSTEIN-YANG-MILLS-HIGGS field equations, one can have an answer to the question. The problem (and a great one) is to solve the field equations... $\endgroup$– Eric Arnéo Vespira KengneCommented Jun 1, 2022 at 10:44
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