# ASD connection for Line bundle over $4$-manifold

Let $$(M,g)$$ be an oriented closed Riemannian $$4$$ manifold. Let $$L\to M$$ be a complex line bundle.

Q Under what condition, we can find an ASD connection of $$L$$, i.e. a connection $$A$$ such that $$F^+_A=0$$.

PS: Any reference is welcome.

By Chern-Weil theory, $$c_1(L)=\frac{i}{2\pi}[F_A]$$ for any $$U(1)$$-connection $$A$$. For an ASD connection ($$\ast F_A=-F_A$$) we have $$c_1(L)^2=\frac{-1}{4\pi^2}\int_M F_A\wedge F_A=\frac{1}{4\pi^2}\int_M F_A\wedge \ast F_A\equiv-\frac{1}{4\pi^2}||F_A||^2\le0$$ So $$c_1(L)^2>0$$ is an obstruction. Exercise: Find such an $$L$$ on $$M=S^2\times S^2$$.
For an "iff" statement, read sections 1.1.6 and 4.3.3 of Donaldson-Kronheimer's book (The geometry of four-manifolds). Although they talk about ASD connections on $$SU(2)$$ bundles, you can consider reducible $$SU(2)$$ connections which correspond to what we care about on $$U(1)$$ bundles. What we observe is:
2) Main point: Given a metric $$g$$, we have projections $$\pi_\pm:H^2(M;\mathbb R)\to H^2_\pm(M;\mathbb R)$$ and the condition for $$L$$ to admit an ASD connection is that $$\pi_+(c_1(L))=0$$.
3) Remark: If $$b^2_+(M)>0$$ then for generic $$g$$ we have $$\pi_+$$ nonvanishing on the lattice $$H^2(M;\mathbb Z)$$ (which contains $$c_1(L)$$) minus the origin, because the codimension of $$H^2_-(M;\mathbb R)$$ is $$b^2_+(M)$$ and the dimension of the lattice is 0. So we won't have nontrivial line bundles with ASD connections for generic metrics.