# First chern class of a line bundle and curvature

Let $$L\to X$$ be a complex line bundle ($$U(1)$$-bundle) over a 4-manifold $$X$$, $$A$$ a connection on $$L$$, and $$F_A\in \Omega^2(X;i\Bbb R)$$ its curvature form. I am reading Morgan's book on Seiberg-Witten theory, and in Theorem 5.2.4 it is written that $$c_1(L)^2=\frac{1}{4\pi^2} (||F_A^+||_{L^2}^2 - ||F_A^-||_{L^2}^2)$$ but I can't see why. (Here $$F_A^\pm$$ is the (anti) self-dual part of $$F_A$$.)

Since $$L$$ is a line bundle we have $$c_1(L)=\frac{i}{2\pi}F_A$$ (https://en.wikipedia.org/wiki/Chern_class#Via_the_Chern%E2%80%93Weil_theory), so $$c_1(L)^2=-\frac{1}{4\pi^2} F_A \wedge F_A$$, and $$F_A \wedge F_A = F_A^+\wedge F_A^+ + 2F_A^+\wedge F_A^- + F_A^- \wedge F_A^-$$. If we integrate this over $$X$$, does it gives $$\frac{1}{4\pi^2} (||F_A^+||_{L^2}^2 - ||F_A^-||_{L^2}^2)$$?

The Hodge star isomorphism gives an orthogonal decomposition of the space of $$2$$-form, and so upon integrating, the $$F^+\wedge F^-$$ term vanishes, since they are orthogonal. Integrating the other two terms is, by definition, $$\int_XF^+\wedge F^++\int_XF^-\wedge F^-=\int_XF^+\wedge\star F^+-\int_XF^-\wedge\star F^-=||F^+||^2-||F^-||^2$$