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Following 4.2.1 in this document (Muharrem Küskü, The Free Maxwell Field in Curved Spacetime, 2001), I tried to adapt the method used (in particular equations 4.21 and 4.32) to Yang-Mills theory rather than curved space-time by introducing a gauge covariant exterior derivative $d_A$, and its associated coderivative on $\mathfrak{g}$-valued $k$-forms: $(-1)^k \star^{-1} d_A \star$. These operators are such that \begin{equation*} \left\{ \begin{array}{ll} \delta_A F = 0\\ d_A F=0 \end{array} \right., \end{equation*} where $F$ is the curvature form of the gauge connection form $A$. My problem arises when I try to find a unified equation of motion for $F$ of the form $\square_A F=0$. Since the paper is using $\square_A = d_A \delta_A+\delta_A d_A$ I thought I have to use it, but I have a doubt because I can use either $\square_A = d_A \delta_A$ or $\square_A = \delta_A d_A$ since they lead to $0$ too. So, which one should I use in my derivation?

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  • $\begingroup$ @Nihar Karve Yes of course, but as I said I have a doubt since the other two also lead to an equation of the form $\square_A F=0$ $\endgroup$ – Jeanbaptiste Roux Feb 12 at 12:15
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The correct equation is $(d_A\delta_A+\delta_Ad_A)F_A=0$.

The reason for using the Laplacian in the (generalised) wave equation is not as subtle as one would think: there are two conditions that the curvature form of a Yang-Mills connection must satisfy - $$ d_AF_A=0 \qquad\text{Bianchi Identity}\tag{1} $$ $$ d_A \star F_A = 0 \Leftrightarrow \delta_AF_A=0\qquad\text{Equation of Motion}\tag{2} $$

So, the curvature form must be such that $F_A \in \ker(d_A\delta_A)\subseteq\bigwedge^2 T^*(M)\otimes \mathrm{ad}(P)$, and $F_A \in \ker(\delta_Ad_A)$. It is straightforward to show that $\ker(d_A\delta_A)\cap\ker(\delta_Ad_A)=\ker(d_A\delta_A+\delta_Ad_A)\equiv\ker\Delta$, in other words, $F_A$ is a (non-linear) generalisation of a harmonic form. The equations $d_A\delta_AF=0$ and $\delta_Ad_AF=0$ are insufficient to individually determine whether $A$ is a Yang-Mills connection, although they are trivially satisified for the curvature form of a Yang-Mills connection.

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    $\begingroup$ Thank you for your answer, this is what I sought. $\endgroup$ – Jeanbaptiste Roux Feb 13 at 7:44

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