Lie bracket on the tangent space of a Lie group

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$, let $G$ be a Lie group with identity $e$, and write $\mathfrak{g}=T_{e}G$. For $u,v\in \mathfrak{g}$, let $u^L$ and $v^L$ be their unique left-invariant smooth extensions. Define the Lie-group tangent bracket by

$[u,v{]}_G=[u^L,v^L{]}_e\mathrm{.}$

The evaluation isomorphism Left-invariant vector fields evaluate isomorphically at the identity makes both extensions unique, so the definition is unambiguous. Moreover, The Lie bracket of left-invariant fields is left invariant makes $[u^L,v^L]$ left invariant, and hence its identity value determines it:

$[u^L,v^L]=[u,v{]}_G^L\mathrm{.}$

The bracket on fields is the library's fixed commutator $[X,Y]f=X(Yf)-Y(Xf)$ from The Lie bracket of smooth vector fields. This explicitly fixes the sign for every later occurrence of $\mathrm{ad}$, the Maurer--Cartan equation, and right-invariant fields. In particular, no opposite vector-field commutator convention is being imported from a source.

The assumption ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through the supplied invariant-extension and bracket-closure results; evaluating the supplied vector-field bracket at $e$ adds no choice. A Lie group is nonempty. If $\dim G=0$, then $\mathfrak{g}=0$ and this is the unique zero bracket; the definition applies unchanged in dimension one. No metric or nondegeneracy hypothesis occurs, and the group is boundaryless by convention.