Why is $D\omega(R_gX,R_gY)=ad g^{-1}D\omega(X,Y)$?

Question

I am reading Singer and Ambrose's paper "A Theorem on Holonomy" in which they develop some theory of principal bundles. At one point they mention that $D\omega(R_gX,R_gY)=ad g^{-1}D\omega(X,Y)$.

Given that $D\omega=d\omega\circ H$ (H the projection onto the horizontal bundle) it suffices to prove that since $\omega$ has the property that $\omega\circ R_g$=$adg^{-1}\circ\omega$ that $d\omega$ does as well. To clarify notation: $\omega$ is the Lie algebra valued connection 1-form, $R_g$ is right translation by $g$ in the fiber, $adg$ is the differential of the map $h\mapsto ghg^{-1}$ in $G$ and hence can be viewed as a map from $\mathfrak{g}$ to itself.

So far I have:

$d\omega(R_gX,R_gY)=R_gX(\omega(R_gY))-R_gY(\omega(R_gX))-\omega([R_gX,R_gY])$

$= R_gX(adg^{-1}(\omega(Y)))-R_gY(adg^{-1}(\omega(X)))-\omega(R_g[X,Y])$

and the lattermost term is $adg^{-1}\circ\omega([X,Y])$. It seems like $adg^{-1}$ can be pulled out of the first two derivatives since it is constantly the same action on $\mathfrak{g}$. If this is correct, we are at: $adg^{-1}\circ R_gX(\omega(Y))-adg^{-1}\circ R_gY(\omega(X))-adg^{-1}\omega([X,Y])$.

Any suggestions or hints for how to proceed?

Answer 1

If $M$ is a manifold $w$ a $p$-form defined on $M$ and $f:M\rightarrow M$ a diffeomorphism, $d(f^*w)=f^*dw$. This implies that

$R_g^*d\omega=d(R_g^*\omega)=d(ad(g^{-1}\omega)=ad(g^{-1})d\omega$ (1)

since $ad(g^{-1})$ is a linear map defined on the Lie algebra.

This implies that $R_g^*D\omega=R_g^*d\omega\circ H=ad(g^{-1})d\omega\circ H$. To see this, it is enough to check the previous equality for $X,Y$ where $X$, $Y$ is horizontal or vertical. If $X$ or $Y$ is horizontal both sides are zero. If $X,Y$ are vertical it reduces to (1) since $H(X)=X$ and $H(Y)=Y$ in this case.