I am reading a set of notes on the classification of simple Lie algebras, in which the claim is made that since $\mathfrak h\subset g$, a Cartan subalgebra is abelian it then follows that the adjoint action of $\mathfrak h$ on $\mathfrak g$ completely commutes. My attempt at showing this is the following:
$$(\text{ad}(h_1)\circ \text{ad}(h_2))(g)=[h_1,[h_2,g]]=[[h_1,h_2],g]=[[h_2,h_1],g]=[h_2,[h_1,g]]=(\text{ad}(h_2)\circ \text{ad}(h_1))(g).$$ However, the second equality relies on the fact that the algebra is associative, which isn't necessarily true for a Lie algebra. I'm also concerned that since $[h_1,h_2]=0$ this doesn't make sense. Is there a simple way to see this?
Best Answer
As you point out yourself, the Lie bracket is in general not associative. Instead one has the Jacobi identity which tells us that
$$[h_1, [h_2, g]] + [h_2, [g, h_1]] + [g, [h_1, h_2]]].$$
Now, that $h_1, h_2$ commute means by definition that $[h_1, h_2]=0$, hence the last term is $=0$, hence
$$[h_1, [h_2, g]] = - [h_2, [g, h_1]].$$
Now use anticommutativity and bilinearity of the Lie bracket to write this as
$$= -[h_2, -[h_1, g]] = [h_2, [h_1,g]].$$
Or, since you write "adjoint action", maybe you know already (as follows from the Jacobi identity) that the adjoint action defines a Lie algebra representation, i.e.
$$ad([x,y]) =ad(x) \circ ad(y) -ad(y) \circ ad(x) \qquad \text{for all } x,y \in g,$$
and now you apply this to $h_1, h_2$ with $[h_1, h_2] = 0$.In short, if two Lie algebra elements commute, their adjoints commute.