# Equation of motion of the curvature form $F$ in Yang-Mills theory

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?

• @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 $\square_A F=0$ – Jeanbaptiste Roux Feb 12 at 12:15

## 1 Answer

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.

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