Left- and right-invariant vector fields

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$, and let $X$ be a smooth vector field on a Lie group $G$. The field $X$ is left invariant if

$d(L_g{)}_h(X_h)=X_{gh}$

for every $g,h\in G$. It is right invariant if

$d(R_g{)}_h(X_h)=X_{hg}$

for every $g,h\in G$, where $R_g(h)=hg$ is the ordinary right translation.

In the left-invariant case, setting $h=e$ gives $X_g=d(L_g{)}_e(X_e)$. Conversely, if that identity-value formula holds for every $g$, then the chain rule and $L_g\circ L_h=L_{gh}$ give

$d(L_g{)}_h(X_h)=d(L_g{)}_{h}d(L_h{)}_e(X_e)=d(L_{gh}{)}_e(X_e)=X_{gh}\mathrm{.}$

Thus left invariance is equivalent to the displayed identity-value formula. Likewise, $R_g\circ R_h=R_{hg}$ shows that right invariance is equivalent to $X_g=d(R_g{)}_e(X_e)$. This agrees with Kirillov's Definition 2.26 after translating the source's inverse-parametrized right action into the ordinary right-translation convention used here.

The assumption ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through the supplied definition of a smooth vector field on the canonical smooth tangent bundle and through the supplied translation trivializations. The chain-rule calculation is pointwise and makes no further choice. A Lie group is nonempty; in dimension zero the tangent values are all zero, and the same definition applies in dimension one. No metric or nondegeneracy hypothesis occurs, and Lie groups are boundaryless by the page convention.