Yang-Mills action ($S_{\mathrm{YM}}$)

On an oriented Riemannian four-manifold, one positive Euclidean convention isfor anti-Hermitian gauge fields.

Under $g\mapsto {\Omega }^{2}g$ in dimension $d$, the Hodge star on $p$-forms scales by ${\Omega }^{d-2p}$. For gauge gauge curvature, $d=4,p=2$, so the star and the action $-g_{\mathrm{YM}}^{-2}\int \mathrm{tr}(F\wedge \ast F)$ are unchanged. Dilation therefore changes the instanton size modulus without changing the classical action. Quantum running of the coupling can break this classical scale invariance.

This gauge-invariant higher-derivative density has equationThe extra commutator comes from varying the connection inside the adjoint covariant derivative: $\delta (D_{\mu }{F}_{\nu \rho })=D_{\mu }\delta F_{\nu \rho }+[\delta A_{\mu },F_{\nu \rho }]$. Two uses of gauge-covariant integration by parts give the first term, and Killing-form invariance gives the second. In an abelian limit it reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$. Keeping the covariant derivative order is essential because their commutator is curvature.

For a semisimple gauge algebra and positive integer $p$, this gauge-invariant density yields $D_{\mu }(s^{p-1}{F}^{\mu \nu })=0$. The curvature variation $\delta F_{\mu \nu }=D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$ gives $\delta s=4\kappa (D_{\mu }\delta A_{\nu },F^{\mu \nu })$, and gauge-covariant integration by parts proves the equation. More generally a density $f(s)$ gives $D_{\mu }(f^{\prime }(s)F^{\mu \nu })=0$. The product form retains the equation at zeros of $s$; dividing by $s$ can discard solutions.