Proof of an identity about Cartan gauge Let $G$ be a Lie group with its Lie algebra at the identity $\frak{g}$ and let $\theta$ be a $\frak g$-valued 1-form on the manifold $M$ and smooth function $g:M\to G$. Show that
$$d(Ad_{g^{-1}}\theta)=Ad_{g^{-1}}d\theta+[g^*\omega,Ad_{g^{-1}}\theta]$$
where $Ad_g:\frak g\to\frak g$ is an Lie algebra isomorphism, $\omega$ is Maurer-Cartan form on $M$.
My proof is as follow. Let $X,Y$ be two tangent vectors on $M$, then
$$\text{LHS}=d(Ad_{g^{-1}}\theta)(X,Y)=X(Ad_{g^{-1}}\theta(Y))-Y(Ad_{g^{-1}}\theta(X))-Ad_{g^{-1}}\theta([X,Y])$$
$$\text{RHS}=Ad_{g^{-1}}d\theta(X,Y)+[g^*\omega,Ad_{g^{-1}}\theta](X,Y)$$
And
$$Ad_{g^{-1}}d\theta(X,Y)=Ad_{g^{-1}}(X(\theta(Y))-Y(\theta(X))-\theta([X,Y]))$$
$$[g^*\omega,Ad_{g^{-1}}\theta](X,Y)=[g^*\omega(X),Ad_{g^{-1}}\theta(Y)]-[g^*\omega(Y),Ad_{g^{-1}}\theta(X)]$$
I don't know how to expand the last identity, no matter in extrinsic or intrinsic way. I have been stuck here for a whole day. Any helpful suggestions?
 A: Here's a proof that is different in style from yours. (I'm writing out the calculus for a matrix group, but you can recast it in terms of $\text{Ad}_{g^{-1}}$. Recall that $\omega=g^{-1}\cdot dg$ is the left-invariant Maurer Cartan form. Recall that $d(g^{-1}) = -g^{-1}dg\cdot g^{-1}$. So we have
\begin{align*}
d(g^{-1}\theta\,g) &= dg^{-1}\wedge \theta\,g + g^{-1}d\theta\,g -g^{-1}\theta\wedge dg = g^{-1}d\theta\,g - g^{-1}dg\wedge g^{-1}\theta\, g - g^{-1}\theta g\wedge g^{-1}dg \\
&= g^{-1}d\theta\,g - \omega\wedge (g^{-1}\theta\, g) - (g^{-1}\theta\, g)\wedge \omega.
\end{align*}
Perhaps there's a sign issue in your formula. 
A: Ted Shifrin's answer is clear, but here are some details which might be of use to others. This is too long to be a comment, so I made a post.
Ted Shifrin's calculations can be understood as using the concept of vector-valued forms with coefficient multiplication. This appears in chapter 1.5 of Sharpe's Differential Geometry, or chapter 21 of Loring Tu's Differential Geometry.
I also believe the question has a sign error, because this identity appears in exercise 3.4.14 of Sharpe as
$$d\{\mathrm{Ad}(k^{-1}) \theta \} = \mathrm{Ad}(k^{-1}) d\theta - [\mathrm{Ad}(k^{-1}) \theta, k^{\ast} \omega_H].$$
Sharpe defines the Lie bracket of $\mathfrak{g}$-valued $1$-forms as the coefficient-multiplied wedge product, using the Lie bracket as the coefficient multiplier. I.e., where $m : \mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ is the Lie bracket,
\begin{align*}
[\omega_1 ,\omega_1](X, Y) &:= m_\ast (\omega_1 \wedge \omega_2) (X, Y) \\
&= m(\omega_1(X) \otimes \omega_2(Y) - \omega_1(Y) \otimes \omega_2(X)) \\
&= [\omega_1(X), \omega_2(Y)] - [\omega_1(Y), \omega_2(X)] \\
&= [\omega_1(X), \omega_2(Y)] + [\omega_2(X), \omega_1(Y)].
\end{align*}
(This is Lemma 1.5.21 of Sharpe.)
This formula shows that the Lie bracket of $1$-forms is symmetric, so the version given by Sharpe does disagree with the version given in the question.
The exterior derivative is an anti-derivation with respect to any coefficient multiplied wedge product of vector-valued forms. I.e., for a bilinear map $m : V \otimes V \to V$, there's the formula
\begin{align*}
d (m_\ast (\omega_1 \wedge \omega_2)) = m_\ast (d\omega_1 \wedge \omega_2) + (-1)^{\mathrm{rank}(\omega_1)} m_\ast (\omega_1 \wedge d\omega_2).
\end{align*}
Sharpe lists this as Exercise 1.5.18.
Something I find confusing is that from the perspective of matrices, two different maps could naturally be used as the coefficient multiplication. There's the Lie bracket, used in order to define the Lie bracket of $\mathfrak{g}$-valued $1$-forms, and then there's the matrix wedge product. The matrix wedge product is the coefficient multiplied wedge product that uses matrix multiplication $\mu : \mathfrak{gl}(n) \otimes \mathfrak{gl}(n) \to \mathfrak{gl}(n)$ as the coefficient multiplier. I.e.,
\begin{align*}
\mu_\ast (\omega_1 \wedge \omega_2) (X, Y) = \omega_1(X)\omega_2(Y) - \omega_1(Y) \omega_2(X).
\end{align*}
Both are apparently at play here.
We calculate $d(g^{-1}) = -g^{-1} dg\ g^{-1}$ by applying the anti-derivation property in the case of the matrix wedge product, and then solving the resulting equation:
\begin{align*}
0 = d(\mathrm{Id}) = d(g g^{-1}) = dg\ g^{-1} + g\ d(g^{-1}).
\end{align*}
Ted Shifrin's centered equation is valid using the matrix wedge product, but not using the Lie bracket multiplied wedge product, because it depends on associativity of the coefficient multiplier. The first equality is an application of the anti-derivation property. The second equality depends on associativity. (Exercise 1.5.20 of Sharpe does state at least that the Lie bracket multiplied wedge product satisfies a graded version of the Jacobi identity (see Lie superalgebra).)
That is okay, however, because (denoting the Lie bracket by $m$ and matrix
multiplication by $\mu$), we have for $\mathfrak{gl}(n)$-valued $1$-forms $\omega_1$ and $\omega_2$,
\begin{align*}
(\mu_\ast (\omega_1 \wedge \omega_2) + \mu_\ast (\omega_2 \wedge \omega_1))(X, Y) &= \omega_1(X)\omega_2(Y) - \omega_1(Y) \omega_2(X) + \omega_2(X)\omega_1(Y) - \omega_2(Y) \omega_1(X) \\\
&= \omega_1(X)\omega_2(Y) - \omega_2(Y) \omega_1(X) + \omega_2(X)\omega_1(Y) - \omega_1(Y) \omega_2(X) \\\
&= [\omega_1(X), \omega_2(Y)] + [\omega_2(Y), \omega_1(X)] \\\
&= [\omega_1, \omega_2](X, Y) = m_\ast(\omega_1 \wedge \omega_2) (X, Y).
\end{align*}
Hence Ted Shifrin's answer does conclude
\begin{align*}
d\{\mathrm{Ad}(g^{-1}) \theta\} = \mathrm{Ad}(g^{-1}) d\theta - [g^\ast \omega_G, \mathrm{Ad}(g^{-1}) \theta].
\end{align*}
($g^{-1} dg$ actually equals $g^{\ast} \omega_G$, where $\omega_G$ denotes the Maurer-Cartan form on $G$.)

I also want to include this as an aside, since all of the details are present, and because the identity proved on this page is used to prove a transformation rule for curvature $2$-forms associated to gauge equivalent Cartan gauges (see Lemma 5.1.9 of Sharpe), or also to show right multiplication pulls the curvature back by $R_h^\ast \Omega = \mathrm{Ad}(h^{-1}) \Omega$ (this is Lemma 5.3.9). The curvature $2$-form $\Omega$ associated to a $\mathfrak{g}$-valued $1$-form, in the case $\mathfrak{g} = \mathfrak{gl}(n)$, is often-times written
$$\Omega = d\omega + \omega \wedge \omega.$$
Here $\wedge$ denotes the matrix wedge product, not the Lie bracket wedge product. This follows from our calculations, which show
\begin{align*}
\Omega := d\omega + \frac{1}{2} [\omega, \omega] = d \omega + \frac{1}{2} m_\ast(\omega \wedge \omega) = d \omega + \mu_\ast (\omega \wedge \omega).
\end{align*}
