Exterior Covariant Derivative End(E)-Valued Forms

[ergospherical]

i) Let $\pi : E \rightarrow M$ be a vector bundle with a connection $D$ and let $D'$ be the gauge transform of $D$ given by $D_v's = gD_v(g^{-1}s)$. Show that the exterior covariant derivative of $E$-valued forms $\eta$ transforms like $d_{D'} \eta = gd_D(g^{-1}\eta)$.
ii) Show that the exterior derivative of any $\mathrm{End}(E)$-valued form $\eta$ transforms as $d_{D'} \eta = g\eta g^{-1} d_D (g^{-1}\eta g)$.

i) Write $\eta = s_I \otimes dx^I$ where the $s_I$ are sections of the bundle $\pi$. Then\begin{align*}
d_{D'} \eta &= d_{D'} s_I \wedge dx^I + s_I \otimes d^2 x^I \\
&= d_{D'} s_I \wedge dx^I \\
&= (D_{\mu} s_I) \otimes dx^{\mu} \wedge dx^I \\
&= gD_{\mu} (g^{-1}s_I) \otimes dx^{\mu} \wedge dx^I
\end{align*}Meanwhile $g^{-1} \eta = g^{-1} s_I \wedge dx^I$, so\begin{align*}
g d_D(g^{-1} \eta) &= d_D (g^{-1}s_I) \wedge dx^I \\
&= g D_{\mu}(g^{-1}s_I) \otimes dx^{\mu} \wedge dx^I
\end{align*}ii) Write the connection as $D = D^0 + A$ for some $\mathrm{End}(V)$-valued 1-form $A$, then $d_D \eta = d\eta + [A,\eta]$. Write $\eta = s_I \otimes dx^I$. Then\begin{align*}
d_{D'} \eta = d\eta + [A', \eta]
\end{align*}How do I expand the expression $d_D (g^{-1}\eta g)$ on the right hand side? Can I view $g^{-1} \eta g = (g^{-1} \eta) \wedge g$ and write something like \begin{align*}
d_D(g^{-1}\eta g) = d_D(g^{-1} \eta) g + (-1)^p (g^{-1} \eta) \wedge d_D g
\end{align*}Thanks.

 

[Ambitwistor]

To expand the expression $d_D (g^{-1}\eta g)$, we can use the product rule for exterior derivatives and the fact that $g$ is a constant with respect to the exterior derivative. This gives us:\begin{align*}
d_D(g^{-1}\eta g) &= d_D(g^{-1}) \wedge \eta g + (-1)^p g^{-1} d_D(\eta) \wedge g + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)\\
&= (-1)^p g^{-1} d_D(\eta) \wedge g + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)
\end{align*}Note that we can use the fact that $d_D(g^{-1}) = 0$ since $g$ is a constant with respect to the connection $D$. Now, using the transformation rule for the connection $D'$, we have:\begin{align*}
d_{D'} \eta &= g^{-1} d_D(\eta) \wedge g + (-1)^{p+1} g^{-1} \eta \wedge d_D(g) + [A', \eta]\\
&= g^{-1} (d_D(\eta) + [A,\eta]) \wedge g + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)\\
&= g^{-1} (d_D(\eta) + [A,\eta]) \wedge g + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)\\
&= g \eta g^{-1} d_D(g^{-1}\eta g) + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)
\end{align*}Therefore, we have shown that $d_{D'} \eta = g \eta g^{-1} d_D(g^{-1}\eta g) + (-1)^{p+1} g^{-1} \eta \wedge d_D(g)$.