= Solution
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 gives
$$
[D_x,D_y]v=(\partial_xA_y-\partial_yA_x+[A_x,A_y])v=F_{xy}v.
$$
Compatibility means local solvability for arbitrary initial vectors, equivalently the existence of an invertible parallel <bundle frame> $S$ with $D_xS=D_yS=0$. 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 $A_x=0$, $A_y=x\operatorname{diag}(1,0)$ and $v=(0,1)^T$ have $D_xv=D_yv=0$ although $F_{xy}=\operatorname{diag}(1,0)\ne0$. This is why arbitrary-data compatibility is the necessary interpretation of “consistent.”
Conversely assume $F_{xy}=0$. On $x=0$, solve $\partial_yS=-A_yS$ with $S(0,0)=1$. For each $y$, extend it along $x$ by $\partial_xS=-A_xS$. Linear <ordinary differential equations> with smooth coefficients give global invertible <matrices> on these finite coordinate paths. Put $R=D_yS$. Since $D_xS=0$,
$$
D_xR=[D_x,D_y]S+D_yD_xS=F_{xy}S=0,\qquad R(0,y)=0.
$$
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 $A=A_xdx+A_ydy$ 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 $\mathbb R^2$:
$$
\boxed{A_x=g^{-1}\partial_xg,\qquad A_y=g^{-1}\partial_yg,\qquad
g:\mathbb R^2\to GL(n,\mathbb R).}
$$
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 equations>
$$
\boxed{L_0(\lambda)=D_1+iD_2-\lambda(D_3+i\Phi),\qquad
L_1(\lambda)=D_3-i\Phi+\lambda(D_1-iD_2).}
$$
The auxiliary equations are $L_0(\lambda)\psi=L_1(\lambda)\psi=0$. To check their compatibility, use $[D_i,D_j]=F_{ij}$ and $[D_i,\Phi]=D_i\Phi$. Put
$$
a=F_{12}-D_3\Phi,\quad b=F_{31}-D_2\Phi,\quad c=F_{23}-D_1\Phi.
$$
The constant, linear and quadratic terms in the <matrix commutator> are respectively
$$
[L_0,L_1]=-b+ic-2i\lambda a-\lambda^2(b+ic).
$$
Thus vanishing for every $\lambda$ forces $a=0$, $b=0$, $c=0$, and those conditions also make the <matrix commutator> zero. They are exactly $\tfrac12\epsilon_{ijk}F_{jk}=D_i\Phi$. 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.
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