# Obstruction to this gauge choice of the connection of a vector bundle

Let $M$ be a compact manifold with a nowhere-vanishing vector field $R$. Consider principal $G$-bundle $P$ over $M$, and $\mathcal{A}$ being the space of irreducible connections.

Let me denote a subspace $\mathcal{A}_R$ of all irreducible connections $\mathcal{A}$ as $$\mathcal{A}_R \equiv \{{A \in \mathcal{A}} | \mathcal{{L}}_RA = {d_A}\left( {{\iota _R}A} \right) \}$$ where $\mathcal{L}_R$ denotes the Lie-derivative. The condition implies an infinitesimal action of diffeomorphism generated by $R$ is equivalent to infinitesimal gauge transformation with parameter $\iota_R A$, and it is equivalent to $\iota_R F_A = 0$.

My goal is to understand the space $\mathcal{A}_R$. One naive guess is that one could choose a gauge such that $\mathcal{L}_R A = 0$, since infinitesimally Lie-derivative can be compensated by gauge transformation. Note that by irreducibility it implies $\iota_R A = 0$.However this is not true in general: if some orbits $\mathcal{C}$ of $R$ is closed and $A$ has non-zero holonomy along $\mathcal{C}$, then one cannot have $\iota_R A = 0$

So my question is: Is there a way to classify the space $\mathcal{A}_R$ (modding out gauge equivalence $\mathcal{G}$), namely some decomposition (some Fourier decomposition according to behavior of $A$ under action of $R$) $$\mathcal{A}_R/\mathcal{G} = \{ A|\mathcal{L}_R A=0\}\cup \{...\}\cup\{...\}...?$$ In particular, is there obstruction to the existence of connection $\mathcal{L}_R A = 0$?

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Somehow, a short computation shows me that your equation $\mathcal L_RA=d_A(I_R A)$ is not gauge invariant, and therefore not even well-defined on an arbitrary vector bundle – Sebastian Oct 28 '13 at 9:05
I think it is gauge invariant: it is equivalent to the condition ${\iota _R}{F_A} = 0$. – Lelouch Oct 28 '13 at 20:36
Sorry, my mistake. Of course, you are right, as I should have been aware of if I had read the whole question. – Sebastian Oct 29 '13 at 7:54