# Relation between definitions of Lie bracket via adjoint representation and via left-invariant vector fields

I’ve been reading two books that touch on the Lie theory: Representation Theory by Fulton & Harris and Introduction to Manifolds by Loring Tu. They define Lie bracket on tangent space at identity in two different ways, and I want to know how these two approaches can be conciliated. Short description follows.

Let $$G$$ be a Lie group and $$\mathfrak g = T_e G$$.

In the first book, Lie bracket is defined using adjoint representation. For every $$g \in G$$ conjugation by $$g$$, $$c_g: G \to G$$ is a diffeomorphism that fixes identity. We define adjoint representation $$\operatorname{Ad}: G \to \operatorname{Aut} \mathfrak g$$ by differentiating $$c_g$$ at identity, $$\operatorname{Ad}(g) = c_{g*}$$.

We differentiate once more at identity to obtain $$\operatorname{ad}: \mathfrak g \to \operatorname{End}(\mathfrak g)$$, $$\operatorname{ad} = \operatorname{Ad}_{*}$$. Then, Lie bracket on $$\mathfrak g$$ is defined as $$[X, Y] = \operatorname{ad}(X)(Y)$$.

In the second book, we first define Lie bracket on smooth vector fields and left-invariant vector fields on $$G$$ (these are vector fields invariant to differential of left multiplication by $$g$$, for every $$g$$ in $$G$$). Next, we notice that vector space of left-invariant vector fields is Lie bracket-closed and isomorphic to $$\mathfrak g$$. Lie bracket on $$\mathfrak g$$ is transported from Lie bracket on left-invariant vector fields by that isomorphism.

I gather that you want to prove that $$L_\tilde{X} \tilde{Y}(e) = \operatorname{ad}(X)(Y)$$, where $$\tilde{X}$$ and $$\tilde{Y}$$ are the left-invariant vector fields corresponding to $$X$$ and $$Y$$ respectively. First observe that from definition we have

$$\operatorname{ad}(X)(Y) = \lim_{t \rightarrow 0}\frac{\operatorname{Ad}(\exp(tX))_*Y-Y}{t}$$

Now we want to compute $$L_\tilde{X} \tilde{Y}$$ and thus we need to compute an expression for $$\phi_{-t,*}\tilde{Y}(e)$$, where $$\phi_{t}$$ is the flow of the vector field $$\tilde{X}$$. Now we make the computation.

$${\phi_{-t,*}\tilde{Y}(e)} = {DR_{\exp(-tX)}\tilde{Y}(\exp(tX))} = {DR_{\exp(-tX)} (DL_{\exp(tX)}Y)} = {D(\operatorname{Ad}(\exp(tX))Y} = {\operatorname{Ad}(\exp(tX))_*Y}$$

Here for $$\psi$$ a map from $$G$$ to itself by $$D\psi$$ we mean the map on tangent bundles. Here the first equality follows from the fact that the curve $$x\exp(tX)$$ passes through $$x$$ and that $$\frac{\Bbb d}{\Bbb dt}(x\exp(tX)) = xX = \tilde{X}(x)$$ and hence this is the unique curve passing though $$x$$ with the given derivative and hence it is the flow of $$\tilde{X}$$. Second equality follows from the left invariance of vector field. Now, using this we have beginning from the second definition of Lie brackets:

$$[X,Y] = [\tilde{X}\tilde{Y}](e) = L_\tilde{X} \tilde{Y} = \lim_{t \rightarrow 0}\frac{\phi_{-t,*}Y - Y}{t} = \lim_{t \rightarrow 0}\frac{\operatorname{Ad}(\exp(tX)_{*}Y - Y}{t} = \operatorname{ad}(X)(Y).$$