Maurer-Cartan 1-form as a connection 1-form

Question

As MO questions go, this one might be borderline - I'm guessing it could be a homework problem in a suitably advanced differential geometry class. I tried asking on math.stackexchange yesterday and it has scarcely received 20 views let alone an answer, so I'm trying it here instead. If it gets 4 votes to close, I'll give the 5th.

I'm trying to decipher a differential geometric comment on page 23-24 of Berline, Getzler, and Vergne's "Heat Kernels and Dirac Operators".

Take a trivial vector bundle $E \times M$ on a manifold $M$ with connection $\nabla = d + \omega$ where $\omega$ is an $End(E)$-valued 1 form. Let $g: GL(E) \to End(E)$ be the tautological map sending a linear map in $GL(E)$ to itself as an element of $End(E)$. The claim is that the connection 1-form on the (trivial) frame bundle for $E \times M$ is given by $g^{-1} \pi^* \omega g + g^{-1} d g$. In particular, if $\omega = 0$ then we get that the trivial connection on the trivial bundle is the Maurer-Cartan 1-form. Unfortunately, I don't see how to give a convincing proof of this - can someone help?

Answer 1

A concrete explanation: If you have a trivial connection, then a constant frame has covariant derivative zero. If you have a non-constant frame, then it can be written as a $GL(n)$-valued function multiplied by the constant section. Then the covariant derivative of the non-constant frame relative to itself can be obtained by differentiating this using the product rule. The Maurer-Cartan forms appear when you do this.

Of course, after you do this, you want to translate this into a much more abstract and sophisticated proof.

(But I do consider this to be a reasonable exercise for an advanced graduate student in geometric analysis and therefore at best borderline for MathOverflow. I'm surprised that no one on math.stackexchange.com has helped.)