(Background and notation) Let $M$ be a smooth manifold. Given $p\in M$, define the tangent space $T_p M$ as the set of equivalence classes of curves sharing the same slope in some coordinate chart: $\gamma'(0)\in T_pM$ where $\gamma'(0)\subset C^\infty(I\to M)$ for some interval $I\subset \mathbb R$ is such that $\forall \gamma,\eta\in \gamma'(0)$ and local chart $\phi:M\to\mathbb R^n$ we have $(\phi\circ\gamma)'(0)=(\phi\circ\eta)'(0)$.
Another way to treat tangent vectors is via derivations. I know that there is a bijection between derivations $\mathrm{Der}(C^\infty(p))$ and equivalence classes of curves. Write said bijection as
$$\gamma'(0)\mapsto D(\gamma)\in \mathrm{Der}(C^\infty(p)) : \qquad D(\gamma)(f) \equiv (f\circ\gamma)'(0)\in T_{f(p)}\mathbb R\simeq \mathbb R.$$
If now $G$ is a Lie group with identity $e$, we define its Lie algebra as the tangent space $\mathfrak g\equiv T_e G$.
Moreover, one can show that the set of left-invariant vector fields, $X\in\Gamma(TG)$ such that $d\ell_g\circ X=X\circ\ell_g$ with $\ell_g$ the translation map, is one-to-one with $T_e G$.
(Defining commutator in $\mathfrak g$ via that in $\Gamma(TM)$)
We can use this bijection to define the commutator of $\gamma'(0),\eta'(0)\in T_e G\equiv\mathfrak g$ via the commutator of the corresponding vector fields. Denote with $X_\gamma$ the vector field such that $X_\gamma(e)=(e,\gamma'(0))\in \{e\}\times T_e G$. Then we define $[\gamma'(0),\eta'(0)]\in\mathfrak g$ via the relation
$$(e,[\gamma'(0),\eta'(0)]) \equiv [X_\gamma,Y_\eta](e).$$
(The question) While the above is all nice and well, is there a more direct way to define the commutator of two tangent vectors without having to pass through the bijection with vector fields?
I know I can write the commutator of vector fields explicitly in coordinates as
$$[X,Y] = (X^i (\partial_i Y^j) – Y^i (\partial_i X^j))\partial_j,$$
where the correspondence between $X_\gamma\equiv X^i\partial_i$ and $\gamma'(0)$ reads $\gamma'(0)=X^i(e)\partial_i|_e$. My problem is that $[X,Y]$ involves terms such as $(\partial_i X^j)$, and this information doesn't seem to exist in the tangent vectors themselves, as $\gamma'(0)$ only depend on the numbers $X^i(e)\in\mathbb R$. So how can I define $[\gamma'(0),\eta'(0)]$ without introducing the associated left-invariant vector fields?
Best Answer
For $g \in G$, define $\iota_g : G \to G$ the conjugation $\iota_g(x) = gxg^{-1}$. It is a smooth automorphism and therefore, one can consider $\mathrm{Ad}_g = \mathrm{d}_e \iota_g \in GL(\mathfrak{g})$. The map $\mathrm{Ad} \colon g \in G \mapsto \mathrm{Ad}_g \in GL(\mathfrak{g})$ is a smooth representation of the Lie group $G$ and we call it the adjoint representation of $G$.
As a smooth map, it induces a map: $$ \mathrm{ad} \colon : X \in \mathfrak{g} \mapsto \mathrm{ad}_X = \mathrm{d}_e(\mathrm{Ad})(X) \in T_{\mathrm{id}}GL(\mathfrak{g}). $$ which is called the adjoint representation of $\mathfrak{g}$. Now, the space $T_{\mathrm{id}} GL(\mathfrak{g})$ is canonically isomorphic to $End(\mathfrak{g})$ (as a linear space), hence one can define for $X,Y \in \mathfrak{g}$, the vector $[X,Y] = \mathrm{ad}_XY \in \mathfrak{g}$.
Thanks to linear group considerations, one can show that it is indeed a Lie bracket, and thus $\mathfrak{g}$ is a Lie algebra.