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Write an amplitude with an external massless vector of momentum $k$ asChanging the polarization representative by ${ϵ}_{\mu }\mapsto {ϵ}_{\mu }+c,k_{\mu }$ is a gauge transformation. The physical amplitude must be unchanged, so the Ward identity isThus a longitudinal pure-gauge polarization decouples from the S-matrix. This on-shell statement is the amplitude counterpart of the off-shell Ward-Takahashi identity.

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.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

For a timelike worldline, $-{\eta }_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }>0$. Its square root is a Lorentz scalar, andThus the first term is the Lorentz-invariant free-particle action $-m{c}^2\int d\tau$. Written as a line element, it is also invariant under a change of the worldline parameter $\lambda$.

The interaction is the scalar line integralUnder an electromagnetic gauge transformationit changes only byThis endpoint term does not affect a fixed-endpoint variational problem. Hence both Lorentz invariance and gauge invariance select the displayed relativistic charged-particle action, with the expected free-particle and minimal-coupling terms.

Under the stated gauge transformation, the charged-particle electromagnetic Lagrangian becomesThusThe action changes only by $q[f(t_2)-f(t_1)]$, whose variation vanishes for fixed endpoints. By total-time-derivative invariance of a Lagrangian, the Euler-Lagrange equation is unchanged.

The equation $F_{wz}=0$ says that the partial connection $D_w,D_z$ is flat. Its compatibility condition therefore guarantees a local $G$-valued solution $g$ ofAfter the associated gauge transformation, both transformed components vanish:

The constant mode of the worldline einbein is not removed by gauge transformations whose parameters vanish at the ends of a fixed interval. This Schwinger proper-time parameter remains integrated over after gauge fixing. For $\phi =(p^2+M^2)/2$, the geometric on-shell proper length is $M\int edt$.

An Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$ changes $\partial \cdot A$ by $\square \lambda$. Consequently gauge fixing by a quadratic function of $\partial \cdot A$ retains transformations satisfying $\square \lambda =0$, subject to the same boundary conditions. Eliminating those residual transformations is part of specifying an inverse with a particular Green function prescription.

The vector Dirac current has an even temporal component and odd spatial components under time-reversal symmetry. Invariance of its real minimal coupling therefore gives the displayed transformation of the gauge field, up to a gauge transformation. Differentiation at $x_T=(-x^0,\mathbf{x})$ yields $F_{\mu \nu }(x)\mapsto -t_{\mu }{t}_{\nu }{F}_{\mu \nu }(x_T)$, with $t=(1,-1,-1,-1)$. Thus the electric field is even and the magnetic field is odd.

An einbein specifies a one-dimensional metric tensor along a worldline. In the relativistic particle phase-space action it is the Lagrange multiplier for the mass-shell condition and transforms as $\delta e=\dot{ϵ}$ under the corresponding canonical gauge transformation. After eliminating the momentum, its geometric interpretation depends on the normalization of the particle action.