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For a connected base and a chosen frame at one point, a flat connection gives a representation of the fundamental group by parallel transport. Conversely, $(\tilde{X}\times G)/{\pi }_1(X)$ with monodromy representation has a flat product connection. Based gauge classes correspond to representations, and unbased classes to conjugacy classes. On a fixed underlying bundle only representations whose associated flat bundle has that bundle type are allowed.

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The holonomy of a flat connection is invariant under based homotopy of loops, giving a group homomorphism from the fundamental group into the structure group after choosing a frame at the base point. Changing that frame conjugates the representation. Its associated bundle is constructed from the universal cover and this monodromy. It differs from the natural representation of a Riemannian holonomy group on a tangent fiber.

Write the flat connection as $d+A_{x}dx+A_{y}dy$ in a global unitary frame. Solve the matrix ordinary differential equationSmooth dependence on parameters gives a smooth solution on the entire open square. Since $A_x$ is skew-Hermitian and trace free, differentiating $u^{\ast }u$ and $\det u$ shows that $u(x,y)\in SU(2)$. Under the bundle gauge transformation $u$, the transformed connection hasIts vector-bundle curvature is still zero. The remaining vector-bundle curvature equation is ${\partial }_x{A}_y^{\prime }=0$, so $A_y^{\prime }=B(y)$ is independent of $x$. Now solve $v^{\prime }(y)=-B(y)v(y)$ with $v(0)=I$. Again $v\in SU(2)$; this second bundle gauge transformation, independent of $x$, leaves $A_x^{\prime }=0$ and makes $A_y^{\prime \prime }=0$. Thus the original connection is gauge equivalent to the trivial connection. Both transformations exist on the whole square, not just in a small coordinate neighborhood.

For a connected base with a chosen frame at a base point, flat connections modulo the based unitary gauge group correspond to flat holonomy representations ${\pi }_1(X,x)\to G$. Removing the frame, or allowing the full unitary bundle gauge group, also divides by conjugation in $G$. On a prescribed underlying bundle, retain only those representations whose associated flat bundle has that bundle type.

The correspondence can be seen directly. Zero vector-bundle curvature makes parallel transport invariant under homotopies of paths with fixed endpoints, giving a group homomorphism on loops. Conversely, a representation $\rho$ gives the flat bundle $(\tilde{X}\times G)/{\pi }_1(X)$, with the deck action on the first factor and $\rho$ on the second, and the product horizontal distribution descends. If two framed connections have identical holonomies, compare their parallel transports along a path from the base point to each point. The comparison is path independent and defines a bundle gauge transformation equal to the identity at the base point. These constructions are inverse. This is framed flat connections and holonomy.

The punctured torus retracts onto a wedge of two circles, so its fundamental group is the free group $F_2$. A representation into $SU(2)$ is specified freely by the two generator images $(A,B)$, with no relation. Every resulting principal $SU(2)$ bundle is trivial, since a bundle with connected structure group over a graph is trivial. Thus no representation is excluded by the fixed trivial-bundle condition. Holonomy depends continuously on a connection, and the flat-bundle construction gives locally continuous choices of representatives as the two matrices vary; local choices of paths in $SU(2)$ give this continuity without requiring a global logarithm. ConsequentlyThe last identification is the description of $SU(2)$ by unit quaternions. The unbased quotient used next is simultaneous conjugation, not independent conjugation of the two generator images.

Use $D_x={\partial }_x+A_x$, $D_y={\partial }_y+A_y$, acting on columns. Expanding the two compositions and cancelling their ordinary derivative terms givesCompatibility means local solvability for arbitrary initial vectors, equivalently the existence of an invertible parallel bundle frame $S$ with $D_{x}S=D_{y}S=0$. Such a bundle frame gives $F_{xy}S=0$, hence $F_{xy}=0$. A single specified parallel vector only implies $F_{xy}v=0$: for example $A_x=0$, $A_y=x\mathrm{diag}(1,0)$ and $v=(0,1{)}^T$ have $D_{x}v=D_{y}v=0$ although $F_{xy}=\mathrm{diag}(1,0)\ne 0$. This is why arbitrary-data compatibility is the necessary interpretation of “consistent.”

Conversely assume $F_{xy}=0$. On $x=0$, solve ${\partial }_{y}S=-A_{y}S$ with $S(0,0)=1$. For each $y$, extend it along $x$ by ${\partial }_{x}S=-A_{x}S$. Linear ordinary differential equations with smooth coefficients give global invertible matrices on these finite coordinate paths. Put $R=D_{y}S$. Since $D_{x}S=0$,Uniqueness of the homogeneous transport equation gives $R=0$. Thus $S$ is a full parallel bundle frame. This proves the parallel-frame compatibility criterion without assuming a nonzero vector spans the whole fiber.

Geometrically $A=A_{x}dx+A_{y}dy$ is a gauge connection on the trivial real rank-$n$ vector bundle, and $F_{xy}dx\wedge dy$ is its gauge curvature. The displayed nonlinear equation is flat connection. Writing $g=S^{-1}$ gives its general solution on ${\mathbb{R}}^2$:Conversely substitution, or the Maurer-Cartan equation, proves that every such gauge potential is flat. The parallel sections are $v=g^{-1}c$ for constant columns $c$. Multiplying $g$ on the left by a constant matrix leaves $A$ unchanged. On a nonsimply connected domain a flat connection can have nontrivial holonomy, so a single-valued global pure-gauge expression needs an extra holonomy condition; the specified plane has no such obstruction.

For the Bogomolny equations, complexify the auxiliary vector space and introduce the spectral parameter $\lambda$. Let $\Phi$ denote multiplication by the Higgs matrix, while $D_i\Phi$ denotes its adjoint gauge covariant derivative. Define the Lax pair for the Bogomolny equationsThe auxiliary equations are $L_0(\lambda )\psi =L_1(\lambda )\psi =0$. To check their compatibility, use $[D_i,D_j]=F_{ij}$ and $[D_i,\Phi ]=D_i\Phi$. PutThe constant, linear and quadratic terms in the matrix commutator are respectivelyThus vanishing for every $\lambda$ forces $a=0$, $b=0$, $c=0$, and those conditions also make the matrix commutator zero. They are exactly $\frac{1}{2}{ϵ}_{ijk}{F}_{jk}=D_i\Phi$. No spectral-parameter derivative is needed. Homogenizing in the two projective coordinates of $\lambda \in {\mathbb{CP}}^1$ includes the point at infinity.

The dimensional-reduction interpretation is equally explicit. On Euclidean ${\mathbb{R}}^4$ with orientation $1234$, take fields independent of $x^4$ and set $A_4=-\Phi$. Then $F_{i4}=-D_i\Phi$. Anti-self-duality of gauge curvature says $F_{12}=-F_{34}$, $F_{23}=-F_{14}$, and $F_{31}=-F_{24}$, which reduce to the same three Bogomolny equations. Translation symmetry reduction therefore turns the four-dimensional ASDYM zero-curvature system into the parameter-dependent three-dimensional system above.

On complexified Minkowski spacetime, use the twistor incidence relationThe harmless factor $i$ is a conventional choice; omitting it corresponds to rescaling the complexified coordinates. For fixed $x$, projectivizing the nonzero two-component spinors $\pi$ gives a twistor line $L_x\simeq {\mathbb{CP}}^1$. The two stated patches cover all these incidence lines; they describe the affine-space-time twistor domain with the line $\pi =0$ removed from full projective twistor space.

Since $E{|}_{L_x}$ is holomorphically trivial, choose a global bundle frame on that line. Comparing its columns with the bundle frames in the two patch trivializations gives invertible matrices $H(x,\pi )$ and $\tilde{H}(x,\pi )$, holomorphic on their respective patches, withThis is the holomorphic splitting of a twistor patching matrix. It follows directly from existence of the global bundle frame, rather than from an arbitrary factorization of an arbitrary matrix: nontrivial line bundles would obstruct such a nonsingular splitting.

The factors can be chosen holomorphically in $x$ near any given $x_0$. An explicit local argument is useful. Normalize the known splitting at $x_0$ and set $Q(x,\lambda )=\tilde{H}(x_0,\lambda {)}^{-1}F(x,\lambda )H(x_0,\lambda )$ on a fixed circular contour in the overlap. It is close to the identity when $x$ is close to $x_0$. Let $P_{+}$ select strictly positive Laurent powers, and seek $q=1+$ positive powers satisfyingIn a holomorphic contour-function norm the operator has norm below one for sufficiently small variation, so its Neumann series converges and is holomorphic in $x$. Then $Qq=\tilde{q}$ has only nonpositive powers and is holomorphic on the exterior patch including infinity. Both factors are close to the identity and invertible. They extend over their full patches through $Q$ and its inverse. Taking $H=H(x_0)q$ and $\tilde{H}=\tilde{H}(x_0)\tilde{q}$ gives the required local family. This is also the elementary Laurent series reason that $H^1({\mathbb{CP}}^1,\mathcal{O})=0$ leaves no infinitesimal obstruction.

Raise primed two-component spinor indices with the antisymmetric two-component spinor form and define $V_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$, holding $\pi$ fixed. Incidence givesThus $V_{A}F=0$ for a pulled-back function depending only on twistor coordinates. Differentiating the splitting and multiplying by the inverse factors givesThe left expression is holomorphic on one two-component spinor patch, the right on the other, and each is homogeneous of degree one in $\pi$. They therefore glue to a global matrix-valued section of $\mathcal{O}(1)$ on the line.

Such a section is linear in its homogeneous coordinates. To see this explicitly, set $\lambda ={\pi }_{0^{\prime }}/{\pi }_{1^{\prime }}$ on $U$. The normalized expression is an entire matrix function of $\lambda$; on the other patch its quotient by $\lambda$ is holomorphic at infinity. Hence it has at most linear growth. Cauchy estimates force all its derivatives of order at least two to vanish. Entry by entry it is $a_A(x)+\lambda b_A(x)$, or invariantlyThe coefficients are holomorphic in $x$. This proves existence of the desired space-time one-form, with no remaining dependence on $\pi$ in its coefficients.

A second splitting must differ by right multiplication $H\mapsto Hg(x)$ and $\tilde{H}\mapsto \tilde{H}g(x)$. Indeed the comparison matrix is holomorphic on both patches of a compact complex projective line and therefore constant along the line. The resulting transformation isthe gauge connection transformation law. Local coefficient matrices consequently glue to a holomorphic gauge connection on the bundle whose fiber at $x$ is the space of holomorphic sections of $E{|}_{L_x}$. If a single global matrix-valued one-form is wanted on all $M_{\mathbb{C}}\simeq {\mathbb{C}}^4$, choose a basis at the origin and parallel-transport it along the straight radial paths to each $x$. Linear transport exists on every finite path and varies holomorphically with $x$; this gives a global holomorphic bundle frame. The assertion does not require flat connection in all four directions, only a fixed choice of paths for a trivialization. On more general space-time domains the intrinsic object is a gauge connection with these gauge-related local one-forms.

Finally put $D_{A{A}^{\prime }}={\partial }_{A{A}^{\prime }}+{\Phi }_{A{A}^{\prime }}$ and ${\mathcal{D}}_A(\pi )={\pi }^{A^{\prime }}{D}_{A{A}^{\prime }}$. From the defining identity,Thus the columns of $H^{-1}$ are an invertible parallel bundle frame in the two alpha-plane directions. Their matrix commutator must annihilate that bundle frame, soUse the spinor decomposition of gauge curvaturewhere both gauge curvature two-component spinors are symmetric. The second term vanishes under contraction with ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}$. The first is a quadratic polynomial in the two two-component spinor components; vanishing for every two-component spinor forces each of its coefficients to vanish. Thereforewhich is precisely the ASDYM equation in the convention where the primed symmetric two-component spinor is the self-dual part. Equivalently, the gauge connection is flat when restricted to every alpha-plane. This completes the forward Penrose-Ward correspondence, including both the existence of the gauge connection and its anti-self-duality of gauge curvature.

On an oriented pseudo-Riemannian vector space $({\mathbb{R}}^n,\eta ,\mathrm{vol})$, the Hodge star operator is the unique linear mapsatisfyingfor all $p$-forms $\alpha ,\beta$. If $s$ is the number of negative metric directions, thenon $p$-forms under this convention. For the two-forms in the question,whereasThe induced inner product is symmetric, so these expressions are negatives of one another. Hence a self-dual differential form and an anti-self-dual differential form are orthogonal and

Write $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The metric is conformal to the standard Euclidean metric, and the Hodge star on middle-degree differential forms is conformally invariant. Taking $e^{ij}=d{x}^i\wedge d{x}^j$, the three real forms arebecause ${\omega }_1+i{\omega }_2=dw\wedge dz$. For the orientation specified by
$dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$, one hasThe corresponding relations for the complementary basis forms immediately giveThus these forms give a real basis of the self-dual two-forms.

Let $D_{\alpha }={\partial }_{\alpha }+A_{\alpha }$ be the gauge covariant derivative. In these complex coordinates the Anti-self-dual Yang-Mills equations areIntroduce the spectral parameter $\lambda$ and the linear operatorsTheir commutator isTherefore the Lax pair for the anti-self-dual Yang-Mills equationsis compatible for every $\lambda$ exactly when the ASDYM equations hold.

In particular, $F_{wz}=0$ says that the connection restricted to each $(w,z)$ surface is a flat connection. On a simply connected coordinate patch, the compatible equationshave an invertible solution $g$. Applying the associated gauge transformation setsThis conclusion is local; global topology can obstruct a single such gauge over the whole space.