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.


1 Answer 1


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 $M=S^2\times S^2$.

On the other hand, the trivial connection on the trivial bundle is ASD.

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:
1) We really care about cohomology thanks to Hodge-deRham theory, noting that ASD connections have harmonic curvature.
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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.