Any smooth vector field is a linear combination of left invariant vector fields

Question

Suppose that $G$ is a lie group and that $X$ is a smooth vector field on it. Is it true that $X$ is a linear combination of left invariant vector fields with smooth maps as coefficients?

Answer 1

Let $G$ be a Lie group and ${\cal G}$ its Lie algebra. Suppose that $e_1,...,e_n$ is a basis of ${\cal G}$, you can defined the left-invariant vector field $X_i(g)=(dL_g)_I(e_i)$, $(X_1(g),...,X_n(g))$ is a basis of the tangent space $T_gG$. Let $(a_1,...,a_n)$ be the dual basis of $(e_1,...,e_n)$. Let $\alpha^i$ be the $1$-form defined by $\alpha^i_g(u)=a_i(dL_g^{-1}(u))$. For every vector field $X$, $X(g)=\alpha^1_g(X)X_1+...+\alpha^n_g(X)X_n$.

Answer 2

Yes, it is true. Take a basis $\{X_e^1,\ldots, X_e^n\}$ of the tangent space $T_eG$ to $G$ in the trivial element $e$. Extend the vectors to left-invariant vector fields: For every $h$ in $H$ consider the difeomorphism $L_h:x\in G\mapsto hx \in G$ and for every $1\leq i\leq n$ define $X^i_h=d_eL_h(X^i_e)$. The vector fields $X^1,\ldots,X^n$ are smooth and for every $h$ in $G$, $\{X_h^1,\ldots,X_h^n\}$ is a basis of $T_hG$. So if $X$ is a smooth vector field on $G$, there exists functions $f_1,\ldots,f_n:G\to \mathbb{R}$ such that $X=\sum_{i=1}^nf_iX^i$.

Let's prove that the functions $f_i$ are smooth. Let $h$ in $G$ and let $(U,\varphi)$ a chart of $G$ with $h$ in $U$. For every $1\leq i \leq n$ there exists smooth functions $f_{i1},\ldots,f_{in}:U\to\mathbb{R}$, given by $f_{ij}=X^i(\varphi_j)$, such that $X^i|_U=\sum_{j=1}^nf_{ij}\frac{\partial}{\partial\varphi_j}$.

On the other hand, there exists smooth functions $\alpha_1,\ldots,\alpha_n:U\to\mathbb{R}$, given by $\alpha_i=X|_U(\varphi_i)$, such that $X|_U=\sum_{i=1}^n\alpha_i\frac{\partial}{\partial\varphi_j}$. Consider the matrix $(f_{ij})_{i,j}$ which is a matrix of change of basis between two smooth local frames so it is smooth and invertible. The inverse matrix is smooth so you can get the coordinates $f_1,\ldots,f_n$ from $\alpha_1,\ldots,\alpha_n$ multplying by this inverse matrix.