# Inverse of gauge covariant derivative

Consider the gauge covariant derivative defined by $$D_z = d_z + \Delta_z$$ or explicitly $$(D_z)^a{}_c = \delta^a_c d_z + (\Delta_z)^a{}_c = \delta^a_c d_z + f_{bc}{}^a A_z^b$$ Here, $d_z$ is the covariant derivative w.r.t. to the metric $$ds^2 = \gamma_{z{\bar z}} \left( dz \otimes d{\bar z} + d {\bar z} \otimes d z \right)$$ $f_{ab}{}^c$ are the structure constants of a Lie algebra satisfying $$[T_a, T_b] = f_{ab}{}^c T_c$$ Let us assume that $d_z^{-1}$ is a well-defined operator on the background. Let us also assume that the gauge field satisfies $$F^a_{z{\bar z}} = d_z A^a_{\bar z} - d_{\bar z} A^a_z + f_{bc}{}^a A_z^b A_{\bar z}^c = 0$$ In matrix notation, we can write this equation as $$\begin{split} D_z A_{\bar z} = d_{\bar z} A_z,~~ D_{\bar z} A_z = d_z A_{\bar z} \end{split}$$ Under these conditions, I wish to write the inverse of the gauge covariant derivative $D_z^{-1}$ in terms of $d_z^{-1}$. Is this possible?

I have been able to determine the answer to this question myself. If $A$ is a flat connection, then $$\partial_z A_{\bar z} - \partial_{\bar z} A_z - i [ A_z , A_{\bar z} ] = 0 \implies A_z = i U^{-1} \partial_z U$$ for some scalar $U \in {\cal G}$. Now, we wish to invert the equation $$D_z \phi = f(z,{\bar z})$$ To do this, we note $$D_z \phi = U^{-1} \partial_z (U \phi U^{-1} ) U$$ Using this, we find $$\phi (z,{\bar z}) = \frac{1}{2\pi} \int d^2 w \frac{ U^{-1} (z,{\bar z}) U(w,{\bar w}) f (w,{\bar w}) U^{-1} (w,{\bar w}) U(z,{\bar z})}{{\bar z} - {\bar w} }$$ which inverts the relation, as required. $\blacksquare$