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

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

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.

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

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 $xexp(tX)$ passes through $x$ and that $\frac{d}{dt}(xexp(tX)) = xX = \tilde{X}(x)$ 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

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

I hope that answers your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.