Gauge field with flat connection Consider a gauge field $A_z^a$ with a flat connection 
$$F_{z{\bar z}}^a = \partial_z A_{\bar z} ^a - \partial_{\bar z} A_z^a + f_{bc}{}^a A_z^b A_{\bar z}^c = 0$$
where $f_{bc}{}^a$ is the structure constant satisfying $[T_a, T_b] = f_{ab}{}^c T_c$. We define the gauge covariant derivative
$$
(D_z)^a{}_c = \delta^a_c \partial_z + f_{bc}{}^a A_z^b
$$
This satisfies
$$
(D_z)^a{}_c (D_{\bar z})^c{}_b - (D_{\bar z})^a{}_c (D_z)^c{}_b = f_{cb}{}^a F_{z{\bar z}}^c = 0
$$
Now, consider the following equation

$$
(D_z)^a{}_b B^b_{\bar z} - (D_{\bar z})^a{}_b B^b_z = 0
$$
  What is the most general solution to the above equation?

Clearly, one solution is (owing to the flatness of the connection)
$$
B_z^a = (D_z)^a{}_c \phi^c
$$
for any $\phi^c$. Are there more solutions?
PS - $(z,{\bar z})$ are coordinates on $S^2$.
 A: I've figured out the answer to the question. I write the answer here to help others who might be interested. Also, there is an assumption here that I'm not totally comfortable with, so it might be interesting to get some insight there. 
Let us start with the equation
$$
\partial_z B_{\bar z} - \partial_{\bar z} B_z = 0
$$

The most general solution to this is $B_z = \partial_z \phi$.

I'm not 100% sure this assumption is true. However, if we impose reality conditions on $B_z$, i.e. $B_z^* = B_{\bar z}$, then it seems to be true. Any comments on this?
We assume this to be true. We now write everything in matrix notation. Firstly, the vanishing of the field strength tensor implies
$$
F_{z {\bar z}} = 0 \implies A_z = U^{-1} \partial_z U
$$
We then find
$$
D_z B_{\bar z} = U^{-1} \partial_z \left( U B_{\bar z} U^{-1} \right) U
$$
The equation then reduces to
$$
\partial_z \left( U B_{\bar z} U^{-1} \right) - \partial_{\bar z} \left( U B_z U^{-1} \right)  = 0
$$
Now, using our earlier assumption, we find
$$
U B_z U^{-1} = \partial_z \phi \implies B_z = U^{-1} \partial_z \phi U = D_z \left( U^{-1} \phi U\right)
$$
QED
