Computing the differential of the coadjoint representation of a Lie group at origin ( $ad^*$) Let $G$ be a Lie group and $Ad^*: \mathfrak{g} \rightarrow GL(\mathfrak{g}^*)$, $[Ad^*(g)(\xi)](x) \stackrel{def}{=} \xi(Ad(g^{-1})x)$, the coadjoint representation.
I am trying to compute $ad^*: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g}^*)$, $ad^* = d_eAd^*$. Supposedly, it is $ad^*(x)(\xi)(y)=-\xi([x,y]_G)$.
I know that $ad(x)(y)=[x,y]_G, \forall x,y \in \mathfrak{g}$. Intuitively, it seems like what is happening is that the differential of $Ad^*$ "passes over" each 1-form $\xi$ and simply differentiates $Ad(a^{-1})$. Since $Ad(a^{-1})=(Ad(a))^{-1} = (i \circ Ad)(a)$, where $i$ is the inverse in $GL(\mathfrak{g})$, and $d_ei=-id$, the result would follow. But I can't see how to prove this "passing over" rigorously.
Also, $[\cdot, \cdot]_G$ can be seen as the linear term of the function $\nu(x,y)$ which makes it that $exp(x)exp(y)=exp(\nu(x,y))$ on a neighborhood of $0_{\mathfrak{g}}$. Based on this and the fact that the diagram:
$$
\require{AMScd}
\begin{CD}
\mathfrak{g} @>{ad}>> \mathfrak{gl}(\mathfrak{g})\\
@VV{exp_G}V @VV{exp_{GL(\mathfrak{g})}}V \\
G @>{Ad}>> GL(\mathfrak{g})
\end{CD}
$$
commutes, it is shown that $ad(x)(y)=[x,y]_G$. (A more general proposition holds, namely that if $\Phi:G \rightarrow H$ is a lie group homeomorphism, then the diagram:
$$
\require{AMScd}
\begin{CD}
\mathfrak{g} @>{d_e\Phi}>> \mathfrak{h}\\
@VV{exp_G}V @VV{exp_H}V \\
G @>{\Phi}>> \mathfrak{h}
\end{CD}
$$ commutes.)
That is, it suffices to calculate $Ad$ and then use the theorem to get to $ad$. However, the argument begins by exploiting the definition of $Ad(a) = d_e(b \rightarrow aba^{-1})$ in the sense that $Ad(exp(x))(y)$ can be seen as $\gamma'(0)$ for $\gamma(t)=exp(x)exp(ty)exp(x)^{-1}$ and proceeds from there. I tried reproducing the argument for $Ad^*: G \rightarrow GL(\mathfrak{g}^*)$ and $ad^*: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g}^*)$ instead of $Ad: G \rightarrow GL(\mathfrak{g})$ and $ad: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})$ respectively ($Ad^*$ and $ad^*$ are also Lie group homeomorphisms), but I can't find a way to exploit the definition of $Ad^*$ in the same way.
Thank you for your time.
 A: The expression $\operatorname{ad}^*(x)(\xi)(y) = -\xi([x,y])$ may be deduced using the chain rule.
By the definition of the differential at the identity, we have
$$\operatorname{ad}^*(x) = \frac{d}{dt}\left.\operatorname{Ad}^*(\exp(tx))\right\vert_{t=0}\quad \forall x\in\mathfrak{g}.$$
View the term $A(t):=\operatorname{Ad}^*(\exp(tx))$ as a curve $t\mapsto A(t)$ of linear maps within $\mathrm{GL}(\mathfrak{g}^*)$. Then the derivative $\frac{d}{dt}\left. A(t)\right\vert_{t=0}\in\mathfrak{gl}(\mathfrak{g}^*)$ acts on $\xi\in\mathfrak{g}^*$ by
$$\Big(\frac{d}{dt}\left.A(t)\right\vert_{t=0}\Big)(\xi) = \frac{d}{dt}\left.\Big(A(t)(\xi)\Big)\right\vert_{t=0}$$
and similarly, the derivative $\frac{d}{dt}\left.\Big(A(t)(\xi)\Big)\right\vert_{t=0}\in \mathfrak{g}^*$ acts on $y\in\mathfrak{g}$ by
$$\Big(\frac{d}{dt}\left.\Big(A(t)(\xi)\Big)\right\vert_{t=0}\Big)(y) = \frac{d}{dt}\left.\Big(A(t)(\xi)(y)\Big)\right\vert_{t=0}.$$
Expanding
$$A(t)(\xi)(y) = \operatorname{Ad}^*(\exp(tx))(\xi)(y) = \xi\Big(\operatorname{Ad}(\exp(tx)^{-1})(y)\Big)$$
we deduce by the chain rule that
\begin{align*}
\operatorname{ad}^*(x)(\xi)(y) 
&= \frac{d}{dt}\left.\xi\Big(\operatorname{Ad}(\exp(tx)^{-1})(y)\Big)\right\vert_{t=0}
\\&= \xi\Big(\frac{d}{dt}\left.\operatorname{Ad}(\exp(tx)^{-1})(y)\right\vert_{t=0}\Big)
\\&= \xi\Big(\left[\frac{d}{dt}\left.\exp(tx)^{-1}\right\vert_{t=0}, y \right]\Big)
\\&= \xi\Big(\left[-x, y \right]\Big)
\\&= -\xi([x,y]).
\end{align*}
