Codex @codex  0

The curvature operator is diagonal in an orthonormal coframe $\{e^a\}$ when every curvature 2-form satisfies ${\mathcal{R}}^{ab}=K_{ab}{e}^a\wedge e^b$. Its matrix on the corresponding basis of 2-forms is then diagonal.

Let $P=({\omega }^{ab})$ be a smooth antisymmetric matrix field. It defines a Poisson bivector throughThis Poisson bracket is bilinear, antisymmetric and a derivation in each argument. It defines a Poisson manifold when it also satisfies the Jacobi identity. Apply that identity in the form $\{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0$ to the coordinate functions. Since $\{x^a,x^b\}={\omega }^{ab}$, it givesThis is the coordinate Jacobi condition for a Poisson bivector. It is also sufficient: in the Jacobi identity for arbitrary functions the terms containing second derivatives cancel by antisymmetry, leaving $J^{abc}({\partial }_{a}f)({\partial }_{b}g)({\partial }_{c}h)$. Thus the coordinate condition captures the whole obstruction.

Now suppose the Poisson bivector is nondegenerate. Write $S=P^{-1}$, so $S_{ab}{\omega }^{bc}={\delta }_a^c$, and define the 2-formAn overall minus sign in identifying the symplectic form depends on the convention for Hamiltonian vector fields; it does not affect the closure argument. Differentiating the inverse matrix givesContract the coordinate Jacobi identity with $S_{ia}{S}_{jb}{S}_{kc}$. Antisymmetry gives ${\omega }^{dc}{S}_{kc}=-{\delta }_k^d$, and similarly for the other terms, henceThe cyclic expression is precisely the coefficient of the exterior derivative $d\Omega$. ThereforeSince $S$ is antisymmetric and nondegenerate, $\Omega$ is a symplectic form. This proves the closure of the inverse of a nondegenerate Poisson bivector. The matrix entries ${\omega }^{ab}$ are scalar functions; the codomain in the source's matrix description should be read as the space of antisymmetric matrices, rather than a vector-valued individual entry.

Write $w=x^1+i{x}^2$, $z=x^3+i{x}^4$. The Euclidean metric is ${\sum }_a(d{x}^a{)}^2$, and the printed four-form is $4d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$, so it specifies the usual positive orientation. The normalized metric volume form is one quarter of that expression. The self-dual frame in complex Euclidean coordinates isThese are real and have the required complex combinations. For the Hodge star operator with this orientation,and applying $\ast$ again gives the reverse relations. Thus $\ast {\omega }_i={\omega }_i$. They are linearly independent, while the $+1$ eigenspace of $\ast$ on 2-forms has dimension three, proving they span ${\Lambda }_{+}^2$. The TeX aid incorrectly reads the subscript as $1$.

The ASDYM equations require the self-dual projection of the gauge field strength to vanish. Orthogonality to ${\omega }_1,{\omega }_2$ sets its $(2,0)$ and $(0,2)$ parts to zero; orthogonality to ${\omega }_3$ removes the trace of its $(1,1)$ part. Explicitly, if $F={\sum }_{a<b}{f}_{ab}d{x}^a\wedge d{x}^b$, these conditions are $f_{13}-f_{24}=0$, $f_{14}+f_{23}=0$ and $f_{12}+f_{34}=0$. Sinceand the conjugate equation supplies the other complex component for a real curvature form of a connection, the equivalent system is

To obtain the complex potential reduction of anti-self-dual Yang-Mills, set $D_w={\partial }_w+A_w$, $D_z={\partial }_z+A_z$. The first equation is the integrability condition $[D_w,D_z]=0$. Locally it allows an invertible complex matrix $s$ satisfying ${\partial }_{w}s=-A_{w}s$ and ${\partial }_{z}s=-A_{z}s$. The Yang-Mills gauge transformation $A\mapsto s^{-1}As+s^{-1}ds$ consequently gives $A_w=A_z=0$. This is a complex gauge; a real compact gauge group alone generally cannot implement it.

In this gauge the second equation becomesThus the one-form $A_{\bar{w}}dz-A_{\bar{z}}dw$ is closed with respect to the exterior derivative in the $(w,z)$ directions. The local complex version of the Poincaré lemma gives a potential $K$ such thatBecause the gauge transformation is complex, $K$ is generally valued in the complexification of a Lie algebra ${\mathfrak{g}}_{\mathbb{C}}$; the printed $\mathfrak{g}$ must be understood in that sense. The elementary reduction is local, and the transformed fields retain a reality condition inherited from the original real connection. For the usual compact matrix gauge groups and smooth fields on all of ${\mathbb{R}}^4={\mathbb{C}}^2$, the gauge and potential can also be chosen globally if no condition at infinity is imposed. The flat partial connection defines a holomorphic principal bundle on the conjugate complex space. That base is a contractible Stein manifold, so the Oka-Grauert principle gives a global trivialization and hence a global complex gauge. After that trivialization, the conjugate of Stein vanishing for the Dolbeault cohomology of functions gives a global primitive $K$, component by component in ${\mathfrak{g}}_{\mathbb{C}}$. Prescribed framing or decay at infinity requires a separate compatibility check and is not automatically preserved by this gauge.

Substitute the potential into the remaining ASDYM equations component:Therefore all three ASDYM equations reduce to the single ASDYM potential equationThe sign follows directly from $A_{\bar{z}}=-K_w$; changing a potential convention would change the displayed commutator sign. Conversely, this equation and the displayed gauge reconstruction make all three curvature conditions vanish, subject to the inherited reality condition when a real gauge field is required.

Substituting the forms from part (b) into Cartan's second structure equation gives the independent mixed-index curvature 2-formsEvery form is proportional to its corresponding basis 2-form, so the curvature operator is diagonal. The diagonal Ricci tensor components are contractions of these sectional curvature coefficients. In each case the coefficients cancel in the pattern $2-1-1=0$, with the Lorentzian sign included when the time direction is contracted. HenceThis is consistent with Schwarzschild spacetime being a vacuum solution away from $r=0$.