Right-Invariant Vector Field

Let $G$ be a Lie group . Write $R_g:G\to G$ for right translation by $g$.

Definition (right invariance).
A smooth vector field $X$ on $G$ is right-invariant if for all $g,h\in G$,

where $(dR_g)_h$ is the differential (pushforward) of $R_g$ at $h$. Equivalently, $(R_g)_*X=X$ for every $g\in G$.

Proposition (determined by the value at the identity).
Let $e$ be the identity of $G$. The assignment

defined by

is a vector space isomorphism from $T_eG$ onto the space of right-invariant vector fields on $G$.

Bracket and the “opposite” Lie algebra.
Right-invariant vector fields are closed under the Lie bracket of vector fields. If one identifies $\mathfrak g=T_eG$ using left-invariant vector fields (the usual convention), then the correspondence $v\mapsto X^R_v$ is a Lie algebra anti-homomorphism:

Equivalently, right-invariant vector fields realize the opposite Lie algebra structure on the same underlying vector space.

In an abelian Lie group, left and right translations coincide, so left- and right-invariant vector fields are the same.

Examples

1. $(\mathbb R^n,+)$: constant vector fields.
   Since $R_a(x)=x+a$ and $(dR_a)_x=\mathrm{id}$, right-invariant vector fields are again precisely constant fields $X(x)=v$.

2. Matrix groups: $X(g)=Ag$.
   For $G\subseteq GL(n,\mathbb R)$ and $A\in\mathfrak g$, the right-invariant vector field associated to $A$ is

   $$X(g)=Ag,$$

   because $(dR_g)_e(A)=Ag$.

3. Circle group $S^1$.
   As $S^1$ is abelian, right-invariant and left-invariant fields agree; a standard right-invariant vector field is $X(z)=iz$.