Raw monad full rank and ADHM regularity

The adjoint of $b_z^{†}$ is the column $b_z=(-B_2^{†}+{\bar{z}}_2,B_1^{†}-{\bar{z}}_1,I^{†}{)}^T$. A covector $v^{†}$ annihilates the image of $b_z^{†}$ exactly when $B_i^{†}v={\bar{z}}_{i}v$ and $I^{†}v=0$, equivalently $v^{†}{B}_i=z_i{v}^{†}$ and $v^{†}I=0$. Such a nonzero covector annihilates every word vector $B_{i_1}\cdots B_{i_m}I(w)$, so stability implies surjectivity for every $z$. Conversely, if the word span $Q$ is proper, it is invariant under both $B_i$ and contains $I(W)$. The complex moment equation makes the induced pair on $V\mathrm{/}Q$ commute, since $IJ(V)\subseteq I(W)\subseteq Q$. By [F2] its dual has a common eigenvector; pulling that covector back to $V^{\ast }$ gives $v^{†}I=0$ and $v^{†}{B}_i=z_i{v}^{†}$ for some $z$, so $b_z^{†}$ is not surjective.

A nonzero vector in $\ker a_z$ is a common eigenvector of $B_1,B_2$ contained in $\ker J$, so costability implies injectivity for every $z$. Conversely, if a nonzero common invariant subspace $Q\subseteq \ker J$ exists, then $IJ(Q)=0$ and the complex moment equation gives $[B_1,B_2]{\mathrm{\mid }}_Q=0$. The restrictions commute and have a common eigenvector by [F2]; this vector lies in $\ker a_z$ for its eigenvalues $z_1,z_2$. Thus the two affine rank criteria are equivalent to stability and costability.

At the point $[{\lambda }_1:{\lambda }_2:{\mu }_1:{\mu }_2]=[1:0:z_1:z_2]$, the displayed homogeneous maps reduce to $\alpha =a_z$ and $\beta =b_z^{†}$. If the homogeneous maps have their monad ranks at every projective point, step 1.1 gives stability and step 1.2 gives costability.

Now suppose both moments vanish and the datum is regular. Step 1.1–1.2 give surjectivity of $b_z^{†}$ and injectivity of $a_z$ for every finite $z$. Direct multiplication gives $b_z^{†}{a}_z={\mu }_{\mathbb{C}}=0,b_z^{†}{b}_z-a_z^{†}{a}_z={\mu }_{\mathbb{R}}=0.$ Hence $f_z=a_z^{†}{a}_z=b_z^{†}{b}_z$ is positive definite. On the real line over $z$, $\alpha =a_z{\lambda }_1+b_z{\lambda }_2$ and $\beta =b_z^{†}{\lambda }_1-a_z^{†}{\lambda }_2$; the cross Gram terms vanish by $b_z^{†}{a}_z=0$, so ${\alpha }^{†}\alpha =(\mathrm{\mid }{\lambda }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\lambda }_2{\mathrm{\mid }}^2)f_z$ and $\beta {\beta }^{†}=(\mathrm{\mid }{\lambda }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\lambda }_2{\mathrm{\mid }}^2)f_z$. Therefore both maps have full rank whenever $({\lambda }_1,{\lambda }_2)\mathrm{\ne }(0,0)$.

Every projective point with $({\lambda }_1,{\lambda }_2)\mathrm{\ne }(0,0)$ lies on a unique real twistor line: after conjugating the second incidence equation, the system for $(z_1,{\bar{z}}_2)$ has matrix $\left(\begin{array}{cc}{\lambda }_1& -{\lambda }_2\\ {\bar{\lambda }}_2& {\bar{\lambda }}_1\end{array}\right)$ and determinant $\mathrm{\mid }{\lambda }_1{\mathrm{\mid }}^2+\mathrm{\mid }{\lambda }_2{\mathrm{\mid }}^2>0$. On the line at infinity $({\lambda }_1={\lambda }_2=0)$ the maps are $\alpha =(-{\mu }_1,-{\mu }_2,0{)}^T$ and $\beta =({\mu }_2,-{\mu }_1,0)$, which are injective and surjective respectively whenever $({\mu }_1,{\mu }_2)\mathrm{\ne }(0,0)$. Direct expansion gives

Thus the maps have the required ranks everywhere and compose to zero, so they form a bundle monad.

If $k=0$, then $V=0$, every endomorphism and map to or from $V$ is the unique zero map, and the conditions of stability and costability both hold vacuously. The affine map $a_z:0\to W$ is injective and $b_z^{†}:W\to 0$ is surjective; the homogeneous source and target are zero, so their required ranks hold at every projective point. The moments vanish and the monad is $0\to W\otimes \mathcal{O}\to 0$. This verifies the zero-dimensional boundary case without using an eigenvector argument on the zero space.