Gauge field strength ($F_{\mu \nu }$)

The curvature of a non-Abelian gauge field is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. It satisfies $[D_{\mu },D_{\nu }]=F_{\mu \nu }$ and transforms as $F_{\mu \nu }\mapsto g{F}_{\mu \nu }{g}^{-1}$.

In complexified four-dimensional space-time, antisymmetry giveswhere both gauge curvature two-component spinors are symmetric. The two summands are the two duality components. In the convention where the primed symmetric part is self-dual, the Anti-self-dual Yang-Mills equations are ${\phi }_{A^{\prime }{B}^{\prime }}=0$. Equivalently ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}{F}_{A{A}^{\prime }B{B}^{\prime }}=0$ for every primed two-component spinor $\pi$, since a quadratic polynomial vanishing for every two-component spinor has all symmetric coefficients zero.

For the ordinary three-dimensional cross product, $D_{\mu }\Phi ={\partial }_{\mu }\Phi +e{A}_{\mu }\times \Phi$ requires $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+e{A}_{\mu }\times A_{\nu }$. This follows from $[A\times ,B\times ]=(A\times B)\times$. The same curvature then transforms covariantly under local rotations. Using a minus sign in its quadratic term while retaining the plus sign in $D$ breaks local gauge invariance: a pure gauge for $D$ has $F_{+}=0$ but generally $F_{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Reversing both signs is an alternative consistent convention.

For ${\mathcal{F}}_{\mu \nu }={\partial }_{\mu }{\mathcal{A}}_{\nu }-{\partial }_{\nu }{\mathcal{A}}_{\mu }+ig[{\mathcal{A}}_{\mu },{\mathcal{A}}_{\nu }]$, the CP transformation of a non-Abelian gauge connection gives ${\mathcal{F}}_{\mu \nu }^{CP}(x)=-P_{\mu }{}^{\alpha }{P}_{\nu }{}^{\beta }{\mathcal{F}}_{\alpha \beta }^T(x_P)$. The commutator term transforms with the same sign as the derivative terms because $[X^T,Y^T]=-[X,Y{]}^T$. A simple component check is $F_{0i}\mapsto +F_{0i}^T$ and $F_{ij}\mapsto -F_{ij}^T$, evaluated at the reflected point.

Write $B_i=({\partial }_{i}g)g^{-1}$ and use $D=d+A$. The right Maurer-Cartan equation gives ${\partial }_x{B}_y-{\partial }_y{B}_x=[B_x,B_y]$. Thus $A_i=\alpha B_i$ has the displayed gauge curvature. Both $\alpha =0$ and $\alpha =-1$ are universally flat; the latter is a pure gauge potential. Commuting $B_x,B_y$ give additional flat cases. For the opposite convention $D=d-A$, the corresponding component formula instead has $\alpha (1-\alpha )$.

For $g=\exp (-i(x{\sigma }_1+y{\sigma }_2)/2)$, write $r=\sqrt{x^2+y^2}$ and $n=(x/r,y/r,0)$. The Pauli matrix multiplication law gives $g=\cos (r/2)I-i\sin (r/2)n\cdot \sigma$. On the circles $r=2\pi k$, the angular derivative is zero, so the two Cartesian right logarithmic derivatives are proportional to the same radial generator. They commute, giving zero gauge curvature for every scaling constant. At the origin their commutator is $-i{\sigma }_3/2$, which need not vanish.

With gauge covariant derivative $D=d+A$, transforming the zero connection by a smooth group-valued map $g$ gives $A=-dg{g}^{-1}$. This potential has zero gauge curvature. A flat potential is locally of this form; global reconstruction can also depend on topology and holonomy. The sign changes when the definition of the covariant derivative changes.

The Jacobi identity for gauge covariant derivatives gives $D_{[\mu }{F}_{\nu \rho ]}=0$. In four dimensions this is $D_{\mu }{\tilde{F}}^{\mu \nu }=0$, with ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. It holds independently of the Yang-Mills equations.