I am trying to understand how the Hodge decomposition is affected by gauge transformations in non-abelian in gauge theory (eg $SU(N)$). In particular, I am searching for a way to generalise the below splitting of the connection in the abelian case. Please correct me if I'm wrong about any of the below.

In abelian gauge theory, eg $U(1)$, we know that our connection $A$ transforms by an exact shift $A\to A+d\lambda$, or (non-infinitesimally) $A\to A+gdg^{-1}$. Because we have $d^2=0$, on a closed surface we can perform a Hodge decomposition into exact and co-exact parts. For the Laplacian $\Delta=\delta d+d\delta$, one can in theory extract the exact part by $$A_{exact}=\int_\Sigma \Delta^{-1} \delta A.$$ Under gauge transformation, only $A_{exact}$ is modified, and $A_{coexact}=A-A_{exact}$ is unchanged.

My question is then how this is modified in the non-abelian case. Now, gauge transformations act as $A\to A+d_A\lambda$, or $A\to gAg^{-1}+gdg^{-1}$. We can formally define $\delta_A$ I suppose, however in general $d_A^2\neq 0$ so we don't have a complex such that we can form a Hodge decompositon. Is there an analogous decomposition to split $A$ into the part that is unchanged by gauge transformations, and a part that captures all of the changes? Is there anything interesting at all we can say about the non-Abelian case?

I am trying to understand how the Hodge decomposition is affected by gauge transformations in non-abelian in gauge theory (eg $\mathrm{SU}(N)$). In particular, I am searching for a way to generalise the below splitting of the connection in the abelian case. Please correct me if I'm wrong about any of the below.

In abelian gauge theory, e.g. $\mathrm{U}(1)$, we know that our connection $A$ transforms by an exact shift $A\to A+d\lambda$, or (non-infinitesimally) $A\to A+gdg^{-1}$. Because we have $d^2=0$, on a closed surface we can perform a Hodge decomposition into exact and co-exact parts. For the Laplacian $\Delta=\delta d+d\delta$, one can in theory extract the exact part by $$A_{\mathrm{exact}}=\int_\Sigma \Delta^{-1} \delta A.$$ Under gauge transformation, only $A_{exact}$ is modified, and $A_{coexact}=A-A_{\mathrm{exact}}$ is unchanged.

My question is then how this is modified in the non-abelian case. Now, gauge transformations act as $A\to A+d_A\lambda$, or $A\to gAg^{-1}+gdg^{-1}$. We can formally define $\delta_A$ I suppose, however in general $d_A^2\neq 0$ so we don't have a complex such that we can form a Hodge decompositon. Is there an analogous decomposition to split $A$ into the part that is unchanged by gauge transformations, and a part that captures all of the changes? Is there anything interesting at all we can say about the non-Abelian case?