An exact raw one-instanton block

Since $B_1=B_2=0$, one has ${\mu }_{\mathbb{C}}=IJ=(\rho ,0)(0,\rho {)}^T=0$ and ${\mu }_{\mathbb{R}}=I{I}^{†}-J^{†}J={\rho }^2-{\rho }^2=0$. The nonzero map $I$ has image $V$, so its word span is $V$ and the datum is stable. The nonzero map $J:V\to W$ is injective, so $\ker J=0$ and the datum is costable. Thus it is regular.

Substitution in the raw homogeneous formulas gives the displayed $\alpha$ and $\beta$. Direct multiplication gives $\beta \alpha =-{\mu }_1{\mu }_2+{\mu }_1{\mu }_2+{\rho }^2{\lambda }_1{\lambda }_2-{\rho }^2{\lambda }_1{\lambda }_2=0$, equivalently the zero-moment polynomial in [F2]. At infinity these maps restrict to $\alpha =(-{\mu }_1,-{\mu }_2,0,0{)}^T$ and $\beta =({\mu }_2,-{\mu }_1,0,0)$, whose middle cohomology is the last two coordinates $W$ with its identity framing. The middle standard Hermitian form is positive.

On the real line over $z$, the incidence equations give $\alpha =a_z{\lambda }_1+b_z{\lambda }_2$ and $\beta =b_z^{†}{\lambda }_1-a_z^{†}{\lambda }_2$, where $a_z=(-z_1,-z_2,0,\rho {)}^T,b_z=({\bar{z}}_2,-{\bar{z}}_1,\rho ,0{)}^T\mathrm{.}$ Multiplication gives $a_z^{†}{a}_z=b_z^{†}{b}_z=f_z$ and $b_z^{†}{a}_z=0$. Thus $a_z,b_z$ are orthogonal isometric copies after rescaling by $f_z^{-1\mathrm{/}2}$. Their common orthogonal complement $E_z=(\mathrm{im}{a}_z\oplus \mathrm{im}{b}_z{)}^{\perp }$ has dimension two. The line complex on the first two copies is the Euler complex $V\mathcal{O}(-1)\to (V\oplus V)\mathcal{O}\to V\mathcal{O}(1)$ and is exact at its ends with zero middle cohomology; the complementary term is $E_z\mathcal{O}$. Hence the line bundle is trivial and its section pairing is the standard positive form on $E_z$. Since $D_z$ has rows $b_z^{†},a_z^{†}$, $E_z=\ker D_z$. Finally direct multiplication gives $D_z{D}_z^{†}=f_z{I}_2$; because $\rho >0$, $f_z>0$ at every $z$.

The matrix $P_z=I_4-D_z^{†}{f}_z^{-1}{D}_z$ is Hermitian and $P_z^2=P_z$, since $D_z{D}_z^{†}=f_z{I}_2$. Also $D_z{P}_z=0$ and $P_z$ acts as the identity on $\ker D_z$, so it is the orthogonal projection onto that rank-two kernel. The connection $\mathrm{\nabla }=Pd$ is unitary. Differentiating $DP=0$ and its adjoint, and using $D{D}^{†}=f{I}_2$, gives $F_{\mathrm{\nabla }}=P(dP\wedge dP)P=Pd{D}^{†}{f}^{-1}\wedge dDP\mathrm{.}$ Here the equality follows by writing $dP$ in the off-diagonal blocks for ${\mathbb{C}}^4=\ker D\oplus \mathrm{im}{D}^{†}$; each off-diagonal block is respectively $-D^{†}{f}^{-1}dDP$ and $-Pd{D}^{†}{f}^{-1}D$, whose product inserts $D{D}^{†}=f{I}_2$.

In the coordinates of the Example, $dD=\left(\begin{array}{cccc}d{z}_2& -d{z}_1& 0& 0\\ -d{\bar{z}}_1& -d{\bar{z}}_2& 0& 0\end{array}\right)\mathrm{.}$ The only nonzero block of $d{D}^{†}\wedge dD$ is its upper-left $2\times 2$ block: $\left(\begin{array}{cc}-2i{\theta }_1& 2({\theta }_2-i{\theta }_3)\\ -2({\theta }_2+i{\theta }_3)& 2i{\theta }_1\end{array}\right),$ where ${\theta }_1=d{x}^0\wedge d{x}^1-d{x}^2\wedge d{x}^3,{\theta }_2=d{x}^0\wedge d{x}^2+d{x}^1\wedge d{x}^3,{\theta }_3=d{x}^0\wedge d{x}^3-d{x}^1\wedge d{x}^2\mathrm{.}$ For the stated positive orientation, direct use of the Hodge-star definition gives ${\ast }_g{\theta }_i=-{\theta }_i$ for each $i$. Every coefficient of $d{D}^{†}\wedge dD$ is therefore anti-self-dual. The scalar $f_z^{-1}$ and the degree-zero matrix factors $P_z$ in step 3.1 only take linear combinations of these coefficient forms, so ${\ast }_g{F}_{\mathrm{\nabla }}=-F_{\mathrm{\nabla }}$ and $F_{\mathrm{\nabla }}^{+}=0$. No sign convention from a differently oriented stereographic chart is used.

The residual $U(1)$ action sends $I\mapsto gI$ and $J\mapsto J{g}^{-1}$; it preserves the raw representative's orbit, and its stabilizer is trivial because $gI=I$ with $I\mathrm{\ne }0$ forces $g=1$. The framed $W$ coordinates at infinity are fixed, so they introduce no extra unitary factor. The finite monad resolution and [F3] give $c_2(F)=H^2$, the unit charge under the positive incidence pullback condition. The example verifies an exact one-instanton block with no singularity because $f_z$ never vanishes.