I want to prove that the tangent space of a Lie group at its identity $e$ is isomorphic to the vector space of left-invariant vector fields.
Given an element $D \in T_e G$ (a derivation), the corresponding left-invariant vector field is $$X : G \rightarrow TG, \; X(g) := L_g^*(D),$$ where $L_g$ is left multiplication by $D$.
To verify that this is smooth, I take an open subset $U \subseteq G$ and a smooth function $f : U \rightarrow \mathbb{R};$ then I need to see that $Xf$ is smooth, where $$Xf : U \rightarrow \mathbb{R}, \; \; Xf(g) := X_g(f) = (L_g^*)(D)(f) = D(f \circ L_g).$$
I do not know how to show that $Xf$ is smooth.
Best Answer
Possible Outline: You know that group multiplication is smooth, implying that $L_g$ is smooth with $L_g (a) = g\cdot a$. Since $f$ is smooth and $L_g$ is smooth, then $f \circ L_g$ is smooth. Moreover, the (directional) derivative of a smooth function is smooth, so $D(f \circ L_g)$ is smooth.