First chern class of a line bundle and curvature

Question

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 $\frac{1}{4\pi^2} (||F_A^+||_{L^2}^2 - ||F_A^-||_{L^2}^2)$?

Answer 1

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, $$\int_XF^+\wedge F^++\int_XF^-\wedge F^-=\int_XF^+\wedge\star F^+-\int_XF^-\wedge\star F^-=||F^+||^2-||F^-||^2$$