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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.

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1 Answer 1

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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:

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

I hope that answers your question.

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