Qu: How (formally) - can I extract the generators in the adjoint representation from any given Lie algebra?
Attempt at Solution:
Take a matrix lie group $G$, with lie algebra $\mathfrak{g}$. For me, this means:
$$ \mathfrak{g} = \bigg\{ X \, : \, exp(tX) \in G \; \forall t \in \mathbb{R} \bigg\} $$
Then for every element $A\in G$ we can define a linear map. $Ad_{A}: \mathfrak{g} \rightarrow \mathfrak{g}, Ad_{A}(X):= AXA^{-1}$.(and you can show from an immeadiate substitution into the exponential that $AXA^{-1}$ will always be in $\mathfrak{g}$).
Can think of the set of these maps as a map - from the Lie group $G$ to the set of invertible linear maps on $\mathfrak{g}$ $Ad: G \rightarrow GL(\mathfrak{g})$
Easily checked that $GL(\mathfrak{g})$ is a valid group, and $Ad$ is then a group homomorphism (i.e $Ad$ is a group representation of $G$ over $\mathfrak{g}$)
$Ad$ is also a lie group homomorphism because it's continuous.
There's then a theorem [see e.g Hall, Introduction to Lie Algebras] that for every Lie group homomorphism, there's a corresponding homomorphism of the respective lie algebras i.e if $\Phi$ is a LG homomorphism $\Phi: G \rightarrow F$, then $!\exists$ $\phi: \mathfrak{g} \rightarrow \mathfrak{f}$
So if:
$$ Ad: G \rightarrow GL(\mathfrak{g}) $$ Then: $$ ad: \mathfrak{g} \rightarrow gl(\mathfrak{g})$$
Where $gl(\mathfrak{g})$ is the Lie algebra of the group of invertible maps on $\mathfrak{g}$, $GL(\mathfrak{g})$
What I really want however, is to figure out what the generators are for this lie algebra. (So if for sake of argument we take $G=SL(2,\mathbb{R})$), given that I understand this - how do I go about getting to the generators?? i.e What are the generators in the adjoint representation of $\mathfrak{sl}(2,\mathbb{R})$?
I realise there are quick and dirty ways of doing this, but I'm confused as to how to do so formally.
Thanks!