The positive-FI and zero-FI rank-one chambers

Since stability says $I\mathrm{\ne }0$, $IJ=0$ forces $J=0$. The real equation is then $\mathrm{\mid }I{\mathrm{\mid }}^2=\zeta$, so $I$ lies on the circle of radius $\sqrt{\zeta }$. The $U(1)$ gauge action multiplies $I$ by its phase and leaves scalar $B_1,B_2$ unchanged. It is transitive on that circle with trivial stabilizer; hence the quotient is exactly the unconstrained pair $(B_1,B_2)\in {\mathbb{C}}^2$. By [F2], this is the positive-FI stable quotient.

At zero real level, $J=0$ still follows from $I\mathrm{\ne }0$ and $IJ=0$, while the real equation becomes $\mathrm{\mid }I{\mathrm{\mid }}^2=0$. This contradicts $I\mathrm{\ne }0$, so the stable zero-level locus is empty. In the positive chamber, $J=0$ makes all of $V$ a nonzero invariant subspace in $\ker J$; the data are not costable. By [F3], their framed sheaves are not locally free.

Let $P\to S^4$ be any smooth principal $U(1)$ bundle. Smooth local connection forms glue by a smooth partition of unity [F5]. Parallel transport from the center of each closed hemisphere along radial segments gives a smooth trivialization: each transported frame solves a linear smooth ODE, and [F5] supplies its smooth dependence on the endpoint. Their transition is a map $g:S^3\to U(1)$. Pulling back the universal cover $\mathbb{R}\to U(1)$ gives a covering of $S^3$; its connected component through a chosen point is a connected cover, hence is one-sheeted by [F4]. Thus $g$ lifts to $\theta :S^3\to \mathbb{R}$. Since the covering is a local diffeomorphism and $g$ is smooth, the lift is smooth. The homotopy ${\theta }_s=(1-s)\theta +s\theta (x_0)$ makes $g$ null-homotopic. Reparameterize this homotopy to be constant near its final parameter and use it on a collar of one hemisphere; this changes that trivialization so the transition is constant and the local frames glue to a global frame. Hence $P$ is trivial. In that frame write the real connection form as $a$ and its curvature as $f=da$, so $df=0$. Then $f\wedge f=d(a\wedge f)$, and Stokes [F6] gives ${\int }_{S^4}f\wedge f=0$. If it is ASD, $f=-\ast f$, so $f\wedge f=-f\wedge \ast f=-\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mathrm{v}\mathrm{o}\mathrm{l}$. The zero integral forces $f=0$, and the charge in the Statement is zero. Thus no smooth charge-one $U(1)$ ASD connection exists, and that empty ordinary sector cannot equal the positive-FI quotient ${\mathbb{C}}^2$.