Left-invariant vector field

Remarks

Example: on $G = GL_n(\R)$, open in the space $M_n(\R)$ of matrices, fix $A \in M_n(\R)$ and put $X_g = gA$. Left translation by $g$ is the linear map $h \mapsto gh$, so its differential is again $v \mapsto gv$, and $(dL_g)_h(X_h) = g h A = X_{gh}$.

Non-example: on the same group, the field $Y_g = Ag$ with $A \ne 0$ is not left-invariant in general. For $n = 2$, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $g = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$, one gets $(dL_g)_e(Y_e) = gA \ne Ag = Y_g$. It is right-invariant instead.

The map $X \mapsto X_e$ is a linear isomorphism from the space of left-invariant vector fields onto the tangent space $T_eG$; the inverse sends $v$ to the field $g \mapsto (dL_g)_e(v)$. The Lie bracket of two left-invariant vector fields is left-invariant. So $T_eG$ inherits a bracket, and with it becomes the Lie algebra of $G$, written $\mathfrak{g}$ or $\operatorname{Lie}(G)$. Every left-invariant vector field is complete: its integral curves are defined on all of $\R$.