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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 $\mathfrak{g}$ $Ad: G \rightarrow GL(\mathfrak{g})$

  1. 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 $\mathfrak{g}$)

  2. $Ad$ 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!

thesundayscientist
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  • If you say "generators" of Lie algebra, do you mean basis? A Lie algebra is a vector space with a Lie bracket. Have a look here – Dietrich Burde Nov 12 '17 at 14:14
  • Sure :) I thought the set of generators was technically the any set of elements of the LA such that the smallest subalgebra which contained them was the LA... but that seems like the same thing as a basis set to me.. – thesundayscientist Nov 12 '17 at 15:04
  • So then you know how to come from $G=SL(2,R)$ to the "generators" of the Lie algebra - a basis for the trace zero matrices of size $2$. The vector space has dimension $3$, therefore you get the three "generators" from the link. – Dietrich Burde Nov 12 '17 at 16:07
  • By result that $det(exp(tX)) = exp(tTr(X))$ then the lie alebgra of $SL(2,R)$ is the set of $2\times 2$ traceless matrices. $$ \mathfrak{sl}(2,R) = \bigg{ \begin{pmatrix} \alpha & \beta \ \delta & -\alpha \end{pmatrix} \alpha,\beta,\delta\ \in \mathbb{R} \bigg} $$ and it's then obvious how to get the basis matrices out of that - but I really don't understand how those adjoint matrices have been calculated. I think it's something obvious, but it's not clear to me, sorry. – thesundayscientist Nov 12 '17 at 17:06
  • It's just linear algebra. The images of the linear map $f$ of the basis are the column vectors of the matrix. So if $(e_1,e_2,e_3)$ is a basis, then $0=[e_1,e_1]={\rm ad}(e_1)(e_1)$, so that the first column vector of the matrix ${\rm ad}(e_1)$ is $(0,0,0)^T$. – Dietrich Burde Nov 12 '17 at 19:04

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