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I apologize for this question which is not really research-level, but I don't get any answer on mathStackExchange, and I asked a professor in my university... which couldn't find the solution.

I am writing a memoir about gauge theory, and on my way I have trouble with a small proof which should be simple and have the feeling that I am missing something obvious. I want to show that the set of connections on a vector bundle $E \to M$ of rank $p$ with a reduction of the the structure group to $H$ (a closed subset of $GL_p(\mathbb{R})$) is non-empty.

Obviously, I want to glue via a partition of unity local trivial connections $\nabla^{i}$ over $U_i \subset M$ an open trivializing set. But I need first to prove that a trivial local connection $d$ for the trivialization $\Phi_i$, whose connection matrix is null in this trivialization, is compatible with the reduction of the structure group to $H$, which means its matrix is a 1-form with matrix values in $\mathfrak{h} = Lie(H)$ for any compatible trivialization.

But if $\Phi_j$ is such a trivialization, with the transition fonctions $\phi_{ji,x} \in H$, the connection matrix in $\Phi_j$ is
$\Gamma^{\nabla^i}_{\Phi_j}(x) = - (d_x \phi_{ji}) \phi_{ji,x}^{-1}$ but this doesn't seem to be a 1-form with matrix value in $\mathfrak{h}$ !

Since $\phi_{ji} : U_i \cap U_j \to H$ we have $d_x \phi_{ji} : T_x U_i \cap U_j \to T_{\phi_{ji,x}} H \simeq \mathfrak{h}$, so $(d_x \phi_{ji})$ is a 1-form with matrix value in $\mathfrak{h}$. If I had a $\phi_{ji,x}$ in front of the expression of $\Gamma^{\nabla^i}_{\Phi_j}(x)$, I would get the adjoint representation of $H$ in its Lie algebra and it would be perfect, but where could this extra $\phi_{ji,x}$ come from ?

What am I doing wrong ?
Thanks for your answers...

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    $\begingroup$ This really belongs in MathStackExchange, but, you asked and didn't get an answer there, so here is a hint: Your confusion is this: A (smooth) map $\phi:U\to H$, satisfies $\mathrm{d}\phi:TU\to TH$, not $\mathrm{d}\phi:TU\to T_eH = {\frak{h}}$, i.e., $\mathrm{d}\phi$ does not take values in $\frak{h}$. The identification $T_hH\simeq {\frak{h}}$ that you are making in your last paragraph is an isomorphism, yes, but it is not (usually) the identity map for $h\not=e$. Making that isomorphism explicit will clear up your confusion. $\endgroup$
    – Robert Bryant
    Commented May 31, 2015 at 10:42
  • $\begingroup$ To show that there are $H$-compatible connections on $E$, it seems to be a bit easier to take an arbitrary connection on the reduced bundle and then extend it to the original bundle. $\endgroup$
    – Tobias Diez
    Commented Jun 11, 2015 at 21:04

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