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For an anti-Hermitian SU(2) connection on ${\mathbb{R}}^4$ tending to $g^{-1}dg$ at infinity, the Chern-Simons 3-form and Maurer-Cartan equation give $(8{\pi }^2{)}^{-1}\int \mathrm{tr}(F\wedge F)=-(24{\pi }^2{)}^{-1}{\int }_{S^3}\mathrm{tr}(g^{-1}dg{)}^3=\mathrm{deg}g$. This is the second-Chern convention with the boundary orientation. Defining instanton charge with the opposite trace sign reverses this integer.

For a smooth finite-energy static planar Skyrme model field, Derrick scaling gives zero quartic energy, hence $[R_1,R_2]=0$ pointwise. The quartic first variation then vanishes, leaving the sigma-model equation ${\partial }_i{R}_i=0$. The right-current Maurer-Cartan equation gives ${\partial }_1{R}_2-{\partial }_2{R}_1=[R_1,R_2]=0$. Each current component is therefore harmonic on the plane and square-integrable. The mean-value estimate on expanding disks makes each component zero, so $U$ is constant. This excludes smooth finite-energy planar static solutions of this potential-free SU(2) model, without misapplying the same conclusion to a pure curved-target sigma model.

For a connected simply connected source Lie group, every Lie algebra homomorphism integrates uniquely to a Lie group homomorphism. Along a path $\gamma$ from the identity, transport its left logarithmic velocity by $\theta$ and solve the corresponding left-invariant differential equation in the target. The Maurer-Cartan equation and preservation of brackets make the endpoint invariant under fixed-endpoint homotopies. Simple connectedness removes path dependence, and path concatenation proves multiplicativity.

The Maurer-Cartan form of the translation-dilation group of the plane has these coefficients in its dilation–translation basis, with $a=1,2$. Left multiplication rescales each coordinate differential and $\rho$ equally, proving they are left-invariant differential forms. Their dual left-invariant vector fields are $\rho {\partial }_{\rho }$ and $\rho {\partial }_{x^a}$. The Maurer-Cartan equation is $d{\lambda }^0=0$, $d{\lambda }^a=-{\lambda }^0\wedge {\lambda }^a$. The tensor ${\sum }_j{\lambda }^j\otimes {\lambda }^j$ is a positive definite left-invariant metric.

For a matrix Lie group, the left-invariant Maurer-Cartan form is $\rho =g^{-1}dg$. It identifies each tangent space with the Lie algebra and obeys $d\rho +\rho \wedge \rho =0$.

Use an SU(2) connection one-form $A=A_{\mu }d{x}^{\mu }$ with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, so $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Absorb the gauge coupling into $A$, and write its gauge curvature as $F=dA+A\wedge A$. The Yang-Mills gauge transformation convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.

In four dimensions the Second Chern form and Second Chern number of the associated fundamental rank-two bundle areHere $M$ is a closed oriented four-manifold. This sign follows by expanding the total Chern class $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. The first Chern class vanishes for SU(2); the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual instanton number isBoth sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.

The form is gauge-invariant by conjugation and cyclicity of the matrix trace. The gauge-theory Bianchi identity $D_{A}F=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:On a closed $M$ the integral of this exact form vanishes. This is Chern-Weil connection transgression, showing that $k_C$ depends on the bundle's topology rather than on a particular gauge field.

Locally the same four-form is an exterior derivative. The Chern-Simons 3-formsatisfiesFor verification, differentiating gives $\mathrm{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\mathrm{tr}{A}^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\mathrm{tr}{A}^4=0$. This local exactness does not make the Second Chern number zero on every closed manifold: on a nontrivial bundle the gauge potentials and ${\omega }_3$ are only patchwise defined.

For the usual finite-action sector on ${\mathbb{R}}^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The Maurer-Cartan equation gives $d(h^{-1}dh)=-(h^{-1}dh{)}^2$, henceWith the outward boundary orientation, Stokes theorem therefore givesThese are opposite choices of the winding number, and either becomes the topological degree of $h$ after specifying the orientation of the target SU(2) sphere. Thus integrality also follows from ${\pi }_3(SU(2))=\mathbb{Z}$. Smooth deformations preserving the asymptotic sector cannot change it.

The positive Euclidean Yang-Mills action isDefine $\Vert X{\Vert }^2=-\int \mathrm{tr}(X\wedge \ast X)$. Since the Hodge star has square one on two-forms in Euclidean dimension four,Choosing $s=\mathrm{sgn}Q$ gives the Yang-Mills instanton Bogomolny boundThe self-dual Yang-Mills equations and their anti-self-dual version imply the Yang-Mills equations $D_A\ast F=0$ by the Bianchi identity. Their finite-action solutions are Yang-Mills instantons.

For a concrete self-dual example, the BPST instanton with center $a$ and size $\rho >0$ haswhere ${\eta }_{\mu \nu }^a$ are self-dual 't Hooft symbols. Their contraction is ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$. Thus $F_{\mu \nu }^a{F}_{\mu \nu }^a=192{\rho }^4/(|x-a{|}^2+{\rho }^2{)}^4$, andThis gives $S=8{\pi }^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the instanton moduli space.

On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the Chern-Simons number of a gauge field, in the instanton-charge conventionIt is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta =h^{-1}dh$ and $\bar{\theta }=dh{h}^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta =-{\theta }^2$ and graded trace cyclicity, gives the gauge change of the Chern-Simons three-formThe mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,Thus $N_{\mathrm{CS}}\mathrm{mod}\mathbb{Z}$, or $\exp (2\pi i{N}_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.

For a Yang-Mills vacuum on ${\mathbb{R}}^3$, zero magnetic energy requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the theta vacuum. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact no static finite-energy lump in three-dimensional pure Yang-Mills theory exists with the usual decay conditions: $A_i^{(s)}(x)=s{A}_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.

Finally, on an oriented spacetime slab $[t_{-},t_{+}]\times \Sigma$, with no side-boundary contribution, the transgression identity givesA Yang-Mills instanton therefore interpolates between vacuum representatives whose Chern-Simons numbers differ by its integer charge. In a four-dimensional quantum theory a Yang-Mills theta term weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi \ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.

Use the index convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$. For the left- and right-invariant coframes, the Maurer-Cartan equations areTheir dual invariant vector fields satisfyFor example, evaluating $d{\lambda }^b$ on two dual fields gives $-{\lambda }^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.

For a semisimple Lie algebra, the Killing form $B(X,Y)=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\mathrm{ad}[Z,X]=[\mathrm{ad}Z,\mathrm{ad}X]$ and cyclicity of trace givesGroup conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a bi-invariant pseudo-Riemannian metric.

Lower the indicated structure index by $C_{abc}=B_{be}{C}_a{}^e{}_c$. ThenThis changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence $C_{abc}=C_{[abc]}$.

Let $\eta (X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a left-invariant differential form. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the exterior derivative on four left-invariant fields. Its six bracket terms combine by invariance intoThe final equality is the Jacobi identity. Therefore the Cartan three-formis both bi-invariant and closed.

A Derrick scaling variation changes a configuration's size without changing its topological sector. Use $U_L(\mathbf{x})=U(\mathbf{x}/L)$, so $L>1$ expands the configuration. An integrated nonnegative term homogeneous in $s$ spatial derivatives scales as $L^{d-s}{E}_s$. For a scalar gradient-plus-potential energy, for example,A static solution is a stationary point of the full energy functional, so it must in particular satisfy $E^{\prime }(1)=0$. In this example the Derrick virial identity is $(d-2)E_2+d{E}_0=0$. If both energies are nonnegative and the coefficients cannot balance, a nontrivial localized static solution is impossible. If balance is possible, it fixes relations among energy contributions and may select a characteristic size. A negative $E^{\prime \prime }(1)$ proves an instability under size changes; a positive second derivative gives stability only against that one deformation. Passing the scale test is a necessary condition, not an existence theorem or a proof of full stability. It does not directly rule out time-dependent or charge-constrained solitons outside the static variational problem.

For the specified model, each spatial current is anti-Hermitian: differentiating $U{U}^{†}=1$ gives $R_i^{†}=-R_i$. Thus $-\mathrm{Tr}(R_i^2)$ is nonnegative, and the static energy of the nonlinear sigma model isThe boundary condition $U\to 1_2$ identifies spatial infinity to one point, so one-point compactification turns the field into a based map $S^d\to SU(2)$. By SU(2) as the three-sphere, its sectors are classified by the homotopy group ${\pi }_d(S^3)$. In particular,There is no corresponding topological protection in one or two spatial dimensions. In three dimensions there is an integer degree of a map between oriented manifolds, the usual topological baryon number in the Skyrme model when the same field is interpreted in that theory. With a compatible orientation convention it can be writtenConjugation from the left current to the right current preserves the trace in this formula. Higher dimensions are governed by their corresponding homotopy groups, which need not all be trivial; no assumption of their triviality is needed in the scaling obstruction below.

For the pure sigma model, $E(L)=L^{d-2}{E}_2$, so stationarity requires $(d-2)E_2=0$. In every $d\ne 2$ a finite-energy nonconstant static solution is excluded by Derrick scaling. In $d=1$ spreading lowers its energy; in $d>2$ shrinking lowers it. In particular, a degree-carrying three-dimensional sigma configuration can collapse toward zero size and energy within its sector, with a singular limiting field. The topological charge prevents smooth unwinding, but not this singular collapse.

In $d=2$, the sigma energy is scale invariant. The scaling identity supplies no obstruction, but ${\pi }_2(S^3)=0$ gives no topological soliton sector either. One must not infer nonexistence of every nonconstant stationary harmonic map from these two facts: scale-invariant sigma models can have non-topological harmonic-map saddles, for example using a totally geodesic two-sphere in the target three-sphere. The conclusion is absence of a topologically protected sigma soliton here, with no positive scale-restoring force supplied by the quadratic term alone.

For the Skyrme model, the added commutator term gives the nonnegative static energyAnti-Hermiticity of the commutators fixes its sign. The target and boundary condition are unchanged, so the topological classification is unchanged. The Derrick dimension test for the Skyrme model instead becomesIn $d=3$, the opposite exponents permit a balance:For an arbitrary reference shape with both energies nonzero, its preferred scale is $L_{\ast }=(E_4/E_2{)}^{1/2}$ and its optimized energy is $2\sqrt{E_2{E}_4}$. Expansion costs quadratic-gradient energy and contraction costs quartic-gradient energy. Integer-degree Skyrmions are therefore topologically possible and evade the pure sigma collapse. This mechanism permits finite-size stable solitons, although existence and stability in all field directions require more than the scaling argument itself.

The other dimensions behave differently. In $d=1$, all spatial commutators vanish identically, so the added term changes nothing and nonconstant static finite-energy fields remain excluded. In $d=2$, stationarity forces $E_4=0$; any configuration with nonzero commutator energy can lower it by spreading. In fact a smooth finite-energy static field of this particular potential-free model must be constant. To prove this commuting-current obstruction in a two-dimensional Skyrme model, $E_4=0$ forces $[R_1,R_2]=0$ pointwise. The quartic first variation then vanishes, leaving the sigma equation ${\partial }_i{R}_i=0$. The right-current Maurer-Cartan equation isDivergence-free and curl-free currents have componentwise $\Delta R_i=0$. Their components belong to L2 space by finite $E_2$. An entire square-integrable harmonic function on the plane vanishes: the mean-value formula and Cauchy–Schwarz bound its value by $(\pi R^2{)}^{-1/2}$ times its full L2 norm, tending to zero as the disk radius $R$ grows. Thus $R_i=0$, and the boundary condition fixes $U=1_2$. This additional argument uses the vanishing commutator; it does not apply to the pure sigma model's general noncommuting currents.

In $d=4$, the virial relation gives $2E_2=0$, hence a constant field and zero $E_4$. In $d>4$ both coefficients are positive and the same conclusion follows. Only three spatial dimensions allow a nontrivial finite-size balance of these two positive derivative energies. Higher-dimensional topological classes, where present, do not evade this obstruction.

Use the supertranslation-invariant superspace coframeIt is a basis because the change from $(dX,d\theta )$ is triangular with identity diagonal blocks. Under a rigid supertranslation, $\delta \theta =ϵ$, $\delta X^m=-i\overline{ϵ}{\gamma }^m\theta$ and $dϵ=0$. Therefore $\delta {\Pi }^m=-i\overline{ϵ}{\gamma }^{m}d\theta +i\overline{ϵ}{\gamma }^{m}d\theta =0$, and $\delta d\theta =0$.

Writing wedge products explicitly, the three-form isBoth factors are supertranslation invariant. Under a Lorentz transformation, ${\Pi }^m$ and the bilinear in $d\theta$ transform as vectors, so their contraction is invariant. Thus $H_3$ is invariant under the Super-Poincaré group.

The Maurer-Cartan equation is $d{\Pi }^m=id\bar{\theta }{\gamma }^{m}d\theta$. Since the other factor is closed,The required four-dimensional Fierz rearrangement identity, which may be stated without proof, isDifferentials of odd Grassmann variables commute in the graded exterior algebra. Thus the coefficient of the product of four $d\theta$ factors is the fully symmetric spinor-index part appearing in this identity. It vanishes, provingThis is the closed three-form on four-dimensional superspace; it should not be confused with the different closed four-form used for a membrane coupling in Question 10.

Choose a basis $T_a$ of the Lie algebra, with $[T_b,T_c]=c^a{}_{bc}{T}_a$. On the Matrix Lie group setLeft multiplication by a constant matrix preserves $\lambda$, while right multiplication preserves $\rho$. Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Apply the graded Leibniz rule to obtainBecause the wedge product antisymmetrizes the matrix products, these Maurer-Cartan equations becomeFollow the vector-field naming used in the question: $L_a(g)=T_{a}g$ is right-invariant and generates left multiplication, whereas $R_a(g)=g{T}_a$ is left-invariant and generates right multiplication. They are dual to ${\rho }^a$ and ${\lambda }^a$, respectively. For any one-form $\eta$, $d\eta (X,Y)=X(\eta (Y))-Y(\eta (X))-\eta ([X,Y])$. Evaluating the two Maurer-Cartan equations on the corresponding invariant fields yieldsFor the last identity, the left and right multiplication flows commute. The opposite sign for the right-invariant vector fields is therefore intrinsic; the letters attached to the two families are only a convention.

A Lie algebra homomorphism $\theta :{\mathfrak{g}}_1\to {\mathfrak{g}}_2$ always defines a local group homomorphism near the identities by the exponential charts:for sufficiently small $X$. Preservation of the BCH multiplication law makes this a local homomorphism. The issue is extending it consistently around all paths and periods in the source.

Suppose first that $G_1$ is connected and simply connected. For a piecewise smooth path $\gamma :[0,1]\to G_1$ from $e_1$ to $g$, letbe its left logarithmic velocity. Solve in $G_2$ the differential equationThe equation exists throughout the finite path interval: bounded smooth body velocity gives a uniform local existence interval at the identity, and left translation repeatedly continues the solution. Define the proposed map by the endpoint $\Phi (g)=\eta (1)$.

Bracket preservation makes this endpoint invariant under fixed-endpoint homotopy. To see the mechanism, for a two-parameter path write $A={\gamma }^{-1}{\partial }_t\gamma$ and $B={\gamma }^{-1}{\partial }_s\gamma$, using the Maurer-Cartan form rather than literal multiplication for nonmatrix groups. The Maurer-Cartan equation givesAfter applying $\theta$, the same identity holds for $a=\theta A$ and $b=\theta B$. If $c={\eta }^{-1}{\partial }_s\eta$, the transport equation impliesBoth have zero initial value because the path starts at a fixed identity. Uniqueness gives $c=b$. At the fixed endpoint, $B(s,1)=0$, hence ${\partial }_s\eta (s,1)=0$. The endpoint is therefore unchanged by the homotopy. Simple connectedness then makes it independent of the chosen path.

For paths to $g$ and $h$, concatenate the first with the left translate by $g$ of the second. Its transported endpoint is $\Phi (g)\Phi (h)$, because left logarithmic velocity is unchanged by left translation. It follows that $\Phi (gh)=\Phi (g)\Phi (h)$. Near the identity, choose $\gamma (t)=\exp (tX)$; transport yields $\Phi (\exp X)=\exp (\theta X)$. Thus $\Phi$ is smooth near the identity and, by translation, everywhere, and its differential is $\theta$. The uniqueness argument from the root solution givesThis is integration of a Lie algebra homomorphism.

For a connected source that is not simply connected, first integrate $\theta$ on its universal cover $p:{\tilde{G}}_1\to G_1$. The resulting map $\tilde{\Phi }:{\tilde{G}}_1\to G_2$ descends exactly whenIndeed two lifts of one element differ by a kernel element, so this condition is both necessary and sufficient for their images to agree. It is the period obstruction to integration of a Lie algebra homomorphism. For $G_1=G_2=S^1$, identify the Lie algebras with $\mathbb{R}$ and take $\theta (t)=ct$. The integrated map on the cover is $t\mapsto e^{ict}$, which respects the source period $2\pi$ precisely when $c\in \mathbb{Z}$. Therefore half-scaling is a valid Lie algebra homomorphism but does not integrate to a circle homomorphism. If both groups are written as simply connected covers modulo central subgroups, the equivalent descent condition is that the integrated cover map send ${\Gamma }_1$ into ${\Gamma }_2$.

For disconnected $G_1$, first solve this problem on $G_1^{\circ }$, obtaining ${\varphi }_0$. One must then assign images $a_{\gamma }$ to representatives $s_{\gamma }$ of the component group. These assignments must satisfy the conjugation and multiplication relations. Explicitly, with $s_e=e$ and $s_{\gamma }{s}_{\delta }=s_{\gamma \delta }{c}_{\gamma ,\delta }$ for $c_{\gamma ,\delta }\in G_1^{\circ }$, the conditions areWhen these can be solved, $\varphi (s_{\gamma }h)=a_{\gamma }{\varphi }_0(h)$ is a smooth global homomorphism. They may prevent existence or allow several extensions with the same differential. Thus a bracket-preserving linear map gives local data; global integration requires the source's periods and, when present, its component relations to be respected.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

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.

The given equation is a Maurer-Cartan equation, with the sign appropriate to the chosen invariant coframe. Write $\lambda =({\lambda }_{ij})$ as an antisymmetric matrix of one-forms, so $d\lambda =\lambda \wedge \lambda$. Using the graded product rule for the exterior derivative,The two sums coincide after exchanging their dummy indices $k,\ell$. This is also associativity of matrix-valued wedge products of differential forms.

To make the relation to the Jacobi identity explicit, let ${\theta }^a$ be any independent coframe and write $d{\theta }^a=-\frac{1}{2}{c}^a{}_{bc}{\theta }^b\wedge {\theta }^c$. Its dual invariant vector fields satisfy $[E_b,E_c]=c^a{}_{bc}{E}_a$, by the exterior derivative of a one-form evaluated on vector fields. Evaluating $d^2{\theta }^a=0$ on three basis fields gives the cyclic coefficient identityThese are exactly the components of $[[E_b,E_c],E_d]+[[E_c,E_d],E_b]+[[E_d,E_b],E_c]=0$. Thus the structure constants obey the Jacobi identity.

For the real four-dimensional case, abbreviate $a={\lambda }_{12}$, $b={\lambda }_{23}$, $c={\lambda }_{31}$, $d={\lambda }_{43}$, $e={\lambda }_{41}$, $f={\lambda }_{42}$; here $d$ without an argument is a one-form, whereas $d(\cdot )$ denotes the exterior derivative. Directly expanding the given equations givesDefine ${\alpha }_1^{\pm }=a\pm d$, ${\alpha }_2^{\pm }=b\pm e$, and ${\alpha }_3^{\pm }=c\pm f$. For example,The minus combination and the other two cyclic combinations giveThe six forms are independent, since the original pairs are their half-sums and half-differences. Their dual vector fields $E_i^{+},E_i^{-}$ consequently obeyEach triple is a three-dimensional Lie algebra ideal with the SO(3) Lie algebra brackets; the two ideals commute and span the original six-dimensional algebra. This proves the real four-dimensional rotation algebra splittingThis is a Lie-algebra statement. Globally $SO(4)$ is a central quotient of $SU(2)\times SU(2)$; a direct product of the corresponding $SO(3)$ groups is not asserted.