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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:

  1. Are we only concerned about the differential at $x$?
  2. 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 $ad(X)$.


  1. Yes, so in that case it indeed becomes the definition by Fulton, if this is true: $$T_{Ad(e)} (Aut(T_eG)) = End(T_eG)$$

  2. 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.

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up vote 1 down vote accepted

I assume, $G$ is of finite dimension ($n$).

(1.) Since we can multiply any curve $\gamma$ by $(\gamma(0))^{-1}$, we can restrict ourselves to the case when the base point $x$ is $e$.

For the next 1., $T_eG\cong\mathbb R^n$ as vector space, then $Aut(T_eG)$ is $GL(n,\mathbb R)$ which is an open subset of the space of all $n\times n$ matrices (open, because contains a ball about $I$). That is, $Aut(T_eG)$ is an open submanifold of $\mathbb R^{n\times n}\cong End(T_eG)$, hence its tangent spaces are canonically isomorphic to this $End(T_eG)$.

(2.) Usually (later on) capital letters mean vector fields (continuous/smooth $X:G\to TG$ mappings such that $Xg\in T_gG$ for all $g\in G$).

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Thanks, illustrating with matrices is very helpful. – galileilei Oct 3 '12 at 19:18

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