I'm reading Bröckner and Dieck's Representations of compact Lie groups and I have some problems in understanding the proof that the bracket defined via the adjoint representation indeed satisfies the properties of a Lie bracket in the abstract sense (page 18). I explain myself better, starting from the background and the work I have done so far.

Given $f:G\rightarrow H$, we define \begin{align*} L f:=T_ef:L G&\rightarrow L H\\ X &\mapsto X\circ f^* \end{align*}where $f^*(\varphi):=\varphi\circ f$. Hence \begin{equation*} L f(X)\varphi=(X\circ f^*)\varphi=X(f^*(\varphi))=X(\varphi\circ f). \end{equation*}Also, there is a canonical isomorphism between $LG$ and the vector space of left-invariant vector fields on $G$ and, in particular, $X\in LG$ can be seen as \begin{align*} X:G&\rightarrow TG\\ p &\mapsto \frac{\partial}{\partial s}\bigg|_0g\alpha^X(s), \end{align*}where $\alpha^X$ is the integral curve of $X$ starting at $\alpha^X(0)=e$.

Now, for $g\in G$, let \begin{align*} c(g):G&\rightarrow G\\ x&\mapsto gxg^{-1}. \end{align*}Then $c(g)$ is an automorphism hence we can consider \begin{align*} Lc(g):LG&\rightarrow L G\\ X&\mapsto X\circ c(g)^* \end{align*}and we have that \begin{equation*} Lc(g)(X)\varphi=X(\varphi\circ c(g)). \end{equation*}Now let \begin{align*} Ad:G&\rightarrow Aut(LG)\\ g&\mapsto Lc(g). \end{align*}Then \begin{equation*} (Ad(g)Y)\varphi=Lc(g)(Y)\varphi=Y(\varphi\circ c(g))=\frac{\partial}{\partial t}\bigg|_0(\varphi\circ c(g))\alpha^Y(t), \end{equation*}hence \begin{align*} (Ad(\alpha^X(s))Y)\varphi&=\frac{\partial}{\partial t}\bigg|_0(\varphi\circ c(\alpha^X(s)))\alpha^Y(t)\\ &=\frac{\partial}{\partial t}\bigg|_0\varphi\biggl(c(\alpha^X(s))\alpha^Y(t)\biggr), \end{align*}hence \begin{align*} Ad(\alpha^X(s))Y&=\frac{\partial}{\partial t}\bigg|_0c(\alpha^X(s))\alpha^Y(t)\\ &=\frac{\partial}{\partial t}\bigg|_0a(s,t), \end{align*}where \begin{equation*} a(s,t):=c(\alpha^X(s))\alpha^Y(t)=\alpha^X(s)\cdot\alpha^Y(t)\cdot\alpha^X(-s). \end{equation*}In particular $Ad(\alpha^X(s))Y\in LG$ as $Ad(\alpha^X(s))\in Aut(LG)$ and $Y\in LG$.

Now, let \begin{align*} ad:=LAd:LG&\rightarrow LAut(LG)=End(LG)\\ X&\mapsto X\circ Ad^*. \end{align*}Then \begin{align*} ad(X)Y&=(X\circ Ad^*)(Y)\\ &=X(Y\circ Ad)\\ &=\frac{\partial}{\partial s}\bigg|_0(Y\circ Ad)\alpha^X(s)\\ &=\frac{\partial}{\partial s}\bigg|_0Y\biggl(Ad(\alpha^X(s))\biggr). \end{align*}

So my question is: Why\begin{equation*} ad(X)Y=\frac{\partial}{\partial s}\bigg|_0\frac{\partial}{\partial t}\bigg|_0a(s,t)? \end{equation*}Can we say that\begin{equation*} Y\biggl(Ad(\alpha^X(s))\biggr)=Ad(\alpha^X(s))Y? \end{equation*}

Thank you!


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