The claim here is Every Lie group admits a left-invariant smooth global frame.
To prove this we use the vector space isomorphism the evaluation map $\epsilon:\text{Lie}(G)\to T_eG$, which is defined as $\epsilon(X) = X_e$.Where $\text{Lie}(G)$ is the Lie algebra of all smooth left-invariant vector field on Lie group $G$.
The problem here is since we can map back the basis of $T_eG$ to basis of $\text{Lie}(G)$.Why the basis for $\text{Lie}(G)$ forms the global frame,the global frame is the set of vector field $(E^1,E^2,…,E^n)$ that for each $p\in G$. we have $(E^1_p,E^2_p,…,E^n_p)$ linear independent.
Here We only find a vector field level basis,how to show for each point the choosen vector filed at this point are linear independent?
Best Answer
The basic argument is that every point "looks the same" as the identity by left invariance, so linear independence are the identity suffices.
Let $L_g:G\to G$ be the left translation operator $L_g(h):=gh$. These are all diffeomorphisms.
Since the frame $(E_1,\dots,E_n)$ is a basis for $\mathfrak{g}$, $(E_1(e)\dots,E_n(e))$ is a (linearly independent) basis for $T_eG$. Since $E_i(g)=d_eL_gE_i(e)$, and the differential $d_eL_g:T_eG\to T_gG$ is a linear isomorphism, $(E_1(g),\dots,E_n(g))$ is linearly independent as well.