Exact raw ADHM block, real monad and stereographic ASD projection

Matrix multiplication gives $b_z^{†}{a}_z=(-B_2+z_2)(B_1-z_1)+(B_1-z_1)(B_2-z_2)+IJ={\mu }_{\mathbb{C}}$. Taking adjoints gives the other off-diagonal entry of $D{D}^{†}$. Subtracting $a_z^{†}{a}_z$ from $b_z^{†}{b}_z$ cancels every scalar and linear-in-$z$ term and leaves $[B_1,B_1^{†}]+[B_2,B_2^{†}]+I{I}^{†}-J^{†}J={\mu }_{\mathbb{R}}$. This proves both identities with their exact signs.

Failure of surjectivity of $b_z^{†}=(-B_2+z_2,B_1-z_1,I)$ gives a nonzero covector $\mathrm{\ell }$ with $\mathrm{\ell }{B}_i=z_i\mathrm{\ell }$ and $\mathrm{\ell }I=0$, contradicting the stable spanning set. Conversely, if the stable word span $S$ is proper, the induced pair on $V\mathrm{/}S$ commutes since $[B_1,B_2]=-IJ$ has image in $S$. Its dual has a common eigenvector by [F4], and the resulting covector annihilates $I$ and both rows at the corresponding $z$, so surjectivity fails there. A kernel vector of $a_z$ is a common eigenvector of $B_i$ in $\ker J$ and contradicts costability. Conversely, on any nonzero common invariant $S\subset \ker J$, the pair commutes by the complex equation and has a common eigenvector by [F4], producing such a kernel. This proves both equivalences.

Expanding $[A,C]+(I{\lambda }_1-J^{†}{\lambda }_2)(J{\lambda }_1+I^{†}{\lambda }_2)$ gives the stated polynomial with coefficients ${\mu }_{\mathbb{C}},{\mu }_{\mathbb{R}},-{\mu }_{\mathbb{C}}^{†}$. Substitution of $\sigma v$ in $\alpha$, followed by adjoint, gives exactly $(-C,A,I{\lambda }_1-J^{†}{\lambda }_2)=\beta (v)$. On a real line the same substitution gives $\alpha =a_z{\lambda }_1+b_z{\lambda }_2$ and $\beta =b_z^{†}{\lambda }_1-a_z^{†}{\lambda }_2$. The scalar-coordinate matrix $\left(\begin{array}{cc}{\lambda }_1& -{\bar{\lambda }}_2\\ {\lambda }_2& {\bar{\lambda }}_1\end{array}\right)$ is invertible whenever $({\lambda }_1,{\lambda }_2)\mathrm{\ne }0$. Solving the two displayed incidence equations similarly recovers a unique $z$ from each point with $\lambda \mathrm{\ne }0$; their real-linear coefficient matrix has determinant $(\mathrm{\mid }{\lambda }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\lambda }_2{\mathrm{\mid }}^2{)}^2$. The points with $\lambda =0$ form $L_{\mathrm{\infty }}$. Thus these lines cover ${\mathbb{P}}^3$.

At zero real moment, step 1.1 gives $D{D}^{†}=\mathrm{diag}(f,f)$ with $f=a^{†}a=b^{†}b$. Regularity makes $f$ positive definite by injectivity of $a$, and hence makes $D$ onto. Conversely, surjectivity of $D$ makes both diagonal blocks positive definite; $a$ is injective and $b^{†}$ onto, so step 2.1 gives regularity. Direct multiplication shows $P^{†}=P$, $P^2=P$, $DP=0$, and $P$ is the identity on $\ker D$. Inversion of positive matrices is smooth, so [F2] gives the rank-$N$ smooth kernel bundle. Projection of the trivial Hermitian connection is a metric connection because for kernel sections $s,t$, $d⟨s,t⟩=⟨Pds,t⟩+⟨s,Pdt⟩$.

Differentiate $DP=0$ and its adjoint to obtain $dPP=-D^{†}(D{D}^{†}{)}^{-1}dDP$ and $PdP=-Pd{D}^{†}(D{D}^{†}{)}^{-1}D$. Thus [F3] gives $F=PdP\wedge dPP=Pd{D}^{†}\mathrm{diag}(f^{-1},f^{-1})\wedge dDP$. Its nonzero differential blocks are $dD=\left(\begin{array}{cc}d{z}_2& -d{z}_1\\ -d{\bar{z}}_1& -d{\bar{z}}_2\end{array}\right)$. The resulting two-form blocks are multiples of $d{z}_1\wedge d{\bar{z}}_1-d{z}_2\wedge d{\bar{z}}_2$, $2d{z}_1\wedge d{\bar{z}}_2$ and $2d{z}_2\wedge d{\bar{z}}_1$, with coefficient $f^{-1}$; the $W$ blocks vanish. Expanding in real coordinates shows that these forms are complex linear combinations of $d{x}^0\wedge d{x}^1-d{x}^2\wedge d{x}^3$, $d{x}^0\wedge d{x}^2+d{x}^1\wedge d{x}^3$ and $d{x}^0\wedge d{x}^3-d{x}^1\wedge d{x}^2$. Each has Hodge star minus itself for the positive Euclidean orientation. Multiplication by the degree-zero matrices $P$ preserves this property, so $F^{+}=0$.

For regular zero-moment data the maps $a_z,b_z$ have orthogonal images and equal positive Gram matrix $f_z$. Therefore $a_z{\lambda }_1+b_z{\lambda }_2$ is injective for every nonzero $\lambda$, since its squared norm on $v$ is $(\mathrm{\mid }{\lambda }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\lambda }_2{\mathrm{\mid }}^2)⟨v,f_{z}v⟩$. The identity with $\sigma$ makes $\beta$ surjective as well. At $L_{\mathrm{\infty }}$, $\alpha =(-{\mu }_1,-{\mu }_2,0{)}^T$ and $\beta =({\mu }_2,-{\mu }_1,0)$ have the same properties. A nonzero maximal minor locally solves each kernel and image equation by holomorphic matrix inversion, so their kernels, images and middle quotient have holomorphic local frames. Hence the middle cohomology is a holomorphic bundle. On a finite real line, split $K=\mathrm{im}{a}_z\oplus \mathrm{im}{b}_z\oplus \ker D_z$. Using $a_z,b_z$ to identify the first two summands with $V\oplus V$, and multiplying the target by $f_z^{-1}$, the first two summands form $V$ copies of the Euler complex $\mathcal{O}(-1)\stackrel{({\lambda }_1,{\lambda }_2)}{\to }{\mathcal{O}}^2\stackrel{(-{\lambda }_2,{\lambda }_1)}{\to }\mathcal{O}(1)$, whose middle cohomology is zero. The remaining summand is the constant $\ker D_z$, proving the canonical linewise identification. The identical Euler calculation at infinity leaves $W$ with its standard framing.

Under the stated simultaneous torus transformation, $A$ and $C$ acquire factors $t_1,t_2$ and the third component of $\alpha$ acquires $t_1{t}_{2}h$. Dividing the resulting middle map by $t_1{t}_2$ gives $U_K=\mathrm{diag}(t_2^{-1},t_1^{-1},h)$, with source factor $(t_1{t}_2{)}^{-1}$ and target factor $1$; direct substitution verifies both chain squares. At infinity the Euler blocks cancel and the induced framing action is precisely $h$. This projective torus action preserves the real-line incidence and commutes with $\sigma$ projectively. In the chart ${\lambda }_1=1,{\lambda }_2=0$, the holomorphic coordinates are $({\lambda }_2,{\mu }_1,{\mu }_2)$ with $({\mu }_1,{\mu }_2)=(z_1,z_2)$, so vertical complex orientation followed by the positive stereographic orientation is the complex orientation of ${\mathbb{P}}^3$. For $k=0$, $D$ is the unique map to the zero space, $P=1_W$, and all formulas reduce to the framed trivial bundle. No inverse normal-form theorem or global charge calculation was used.