I am learning gauge theory, so I tried to understand what happens in the case of a trivial principal bundle. However I have some problems understanding how a connection looks like in that case. Here's what I did:
$M$ compact manifold, $G$ compact Lie group, $P=M\times G$ the trivial principal bundle. We have the identification $$\Omega^1(P)\cong\Omega^1(M)\oplus\Omega^1(G)\cong\Omega^1(M)\oplus\mathfrak{g}^*.$$ An element $A\in\Omega^1(P)\otimes\mathfrak{g}$ is a connection if it satisfies the following two properties:
- $\iota_{X_\xi}A=\xi$, where $X_\xi$ is the vector field associated to $\xi\in\mathfrak{g}$. In our case it is given by $X_\xi=(0,\xi)$ under the identification $TP\cong TM\times\mathfrak{g}$. Thus the part of $A$ in $\mathfrak{g}^*\otimes\mathfrak{g}$ must be given by $d\xi^i\otimes\xi_i$, where $\xi_i\in\mathfrak{g}$ is a basis, and $d\xi^i$ the dual basis.
- $A$ must be invariant under the action of $G$, that is $R_g^*A=\operatorname{ad}_{g^{-1}}\circ A$, where $R^*_g$ acts on $\Omega^1(P)$.
My problem lies in the second condition. As I see it, $R^*_g$ acts trivially on $\Omega^1(P)$, while $\operatorname{ad}_{g^{-1}}$ acts non-trivially on $\mathfrak{g}$, thus the second condition is impossible to satisfy (if $G$ is not abelian, of course). Am I missing something or did I commit some stupid error?
Here's why I think $R^*_g$ acts trivially on $\Omega^1(P)$:
We can use the right shifts $(R_g)_*$ to identify $TG\cong G\times\mathfrak{g}=G\times T_eG$, and we obtain thus a basis of vector fields for $TG$ given by $$\xi_i(g)=(R_g)_*\xi_i$$ (with a small abuse of notation.) A basis for $\Omega^1(G)\subset\Omega^1(P)$ is then given by the dual basis to $\xi_i$, which we will denote by $d\xi^i$. On these elements we have: $$[R_g^*d\xi^i(hg)](v_m,\xi_j(h))=d\xi^i(hg)(v_m,(R_g)_*\xi_j(h))=d\xi^i(hg)(v_m,\xi_j(hg))=\delta^i_j$$ (sorry for the poor notation, $v_m\in T_mM$). Thus we conclude $R^*_gd\xi^i=d\xi^i$.
Elements in $\Omega^1(M)$ give zero on the part in $TG$ of vector fields, and the action of $(R_g)_*$ on vectors in $TM$ is trivial, so also those forms are invariant under $R^*_g$.
Is the mistake somewhere in here?
$R_g^*$ does not act trivially on $\Omega^1(P)$, instead we have $$R_g^*\omega|_h \cdot X = \omega|_{h\cdot g} \cdot dR_g|_h \cdot X.$$ It turns out that connections are in bijection with $\mathfrak{g}$-valued one-forms on $M$ in this case, i.e. every connection $\hat{\omega}$ has the form $$ \hat{\omega}|_g \cdot X = \omega \cdot d\pi|_e \cdot d R_{g^{-1}}|_{g} \cdot X.$$ for some $\omega \in \Omega^1(M, \mathfrak{g})$.