Field-strength formulation of gauge theories: Transformation of the functional integral

We present a formulation of gauge field theories in which the gauge potentials $A_{\mu }\mathrm{}\left(x\right)\mathrm{}$ are eliminated in a simple way in terms of the field strengths $F_{\mu \nu }\mathrm{}\left(x\right)\mathrm{}$. Our results are closely related to, but are much simpler than, Halpern's dual variable formulation of gauge theories in the axial gauge. We work in the coordinate gauge $x^{\mu }{A}_{\mu }\mathrm{}\left(x\right)=0\mathrm{}$, and show both analytically and geometrically that the potential $A_{\mu }$ can be determined uniquely from the field strengths $F_{\mu \nu }$ for a suitable class of $F\text{'}\mathrm{s}$, $A\to A\left[F\right]\mathrm{}$. We show, furthermore, that a tensor $F_{\mu \nu }\mathrm{}\left(x\right)\mathrm{}$ is a coordinate-gauge field tensor if and only if it satisfies the restricted set of Bianchi identities ${\epsilon }^{\mu \nu \sigma \lambda }{x}_{\sigma }{D}^{\rho }\mathrm{}\left[F\right]\mathrm{}{}_{}{}^{*}{}_{}{}^{}F_{\rho \lambda }^{}{}_{}{}^{}=0\mathrm{}$, $D=\partial +\left[A\right[F], ·]$. These results permit us to transform the functional integral for the vacuum-to-vacuum amplitude $Z\left[J\right]\mathrm{}$ for the gauge theory to a form in which the potentials are completely eliminated in terms of the field strengths. When the Bianchi constraints are eliminated using a set of Lagrange multiplier fields ${\lambda }_{\sigma }\mathrm{}\left(x\right)\mathrm{}$, the $F\text{'}\mathrm{s}$ can be integrated out completely to obtain a form of the theory which appears to be useful for strong coupling.