Given a Lie Group $G$ with $\dim G = n$, I can take a basis of $T_e G$ and left translate it to obtain a basis at any point $g$ in $G$. This way I can construct $n$ left translated vector fields
$\underline{E}_i, i=1,…,n$ that span $L(G)$.
My question is, is it possible to express any smooth vector field as a linear combination:
$\underline{X} = f^i\underline{E}_i$, where $f^i$ smooth functions ?
More abstract, if a smooth manifold $M$ can be turned into a Lie Group, is the basis of the Lie algebra of vector fields also a basis for the module of all smooth vector fields ?
Thanks
Best Answer
Yes, because the frame $(\underline{E}_i)$ spans the tangent space $T_g M$ at each $g \in G$; this follows immediately from the left-invariance of the frame.
Explicitly, let $(\underline{e}^i)$ denote the (again, left-invariant and global) coframe dual to $(\underline{E}_i)$. For any smooth vector field $X$ on $G$ we can recover the coefficients $f^j$ explicitly: $$X = e^j (\underline{X} ) \underline{E}_j = e^j \left(f^i \underline{E}_i \right) \underline{E}_j = f^i e^j(\underline{E}_i) \underline{E}_j = f^i \delta^j{}_i \underline{E}_j= f^j \underline{E}_j .$$ So, $f^j = e^j(X)$; in particular, $f^j$ is the contraction of smooth tensors and so is itself smooth.