Let $g$ be an element of Lie Group $G$, and $\gamma(t) : \mathbb{R} \rightarrow G$ be a path in $G$ such that $\gamma(0) = e$, the identity element of $G$. Denote the tangent space at $e$ as $T_eG$, the collection of all $\left.\frac{d}{dt} (\gamma(t)) \right|_{t=0}$ for $\gamma(0) = e$.
Definition 1(Adjoint Representation): The map $Ad: G \rightarrow Aut(T_eG)$ is (in gory details) $ Ad : g \rightarrow \{f_g : \left. \frac{d}{dt} (\gamma(t)) \right|_{t=0} \rightarrow \left. \frac{d}{dt} (g\cdot\gamma(t)\cdot g^{-1}) \right|_{t=0} \} $ so my understanding is that $Ad$ maps $g$ into $f_g$, an automorphism $T_eG$. Therefore $f_g$ lies in the the manifold(?) $Aut(T_eG)$, and apply this definition verbatim I get:
$ ad_{x} : T_xG \rightarrow T_{Ad(x)} (Aut(T_eG)) $
which becomes the following definition in Fulton's book: $ ad : T_eG \rightarrow End(T_eG) $
I guess I am confused because I have not learnt differntial geometry yet:
- Are we only concerned about the differential at $x$?
- If the answer to first question is "Yes", then what are X and Y in $ad(X)(Y)$? Here $ad(X)(Y)$ is defined as the image of $Y$ under the map $ad(X)$.
Edit:
Yes, so in that case it indeed becomes the definition by Fulton, if this is true: $T_{Ad(e)} (Aut(T_eG)) = End(T_eG)$
Take what we got from 1, we have ( I like to use $f$ to denote maps) $ ad: X \rightarrow \{f_X : Y \rightarrow Z\} $ where $X$, $Y$ and $Z$ are all elements of $T_eG$. so this is a billinear function $ ad(X)(Y) =f_X(Y) = Z $.
I guess I sort of get the definition now, but it would be nice if someone could explain the intuition without all the formalism of the definition, so I will leave this question open.