The coefficients of a vector field in the Lie Algebra basis are smooth functions

Question

Suppose $G$ is a Lie Group. $Lie(G)$ be its Lie Algebra. Let $\tilde{X_1},\tilde{X_2},\cdots\tilde{X_n}$ be a basis. Then in a local chart we can express every vector field $X$ as $$X= \varphi_1\tilde{X_1}+\varphi_2\tilde{X_2}+\cdots\varphi_n\tilde{X_n}$$ I do not understand why these functions $\varphi_i$'s are smooth.

Answer 1

That's nothing to do with Lie group. Just write everything in local coordinates:

$$X = \sum_j a^j(x) \frac{\partial}{\partial x^j}, \ \ \widetilde X_i = B^j_i (x) \frac{\partial}{\partial x^j}.$$

Then

$$\frac{\partial}{\partial x^j} = \sum_i C^i_j (x) \widetilde X_i,$$

where the matrix $C = (C^i_j)$ is the inverse of $B = (B_i^j)$. Since the inverse is given by determinant and cofactors, $C$ has smooth coefficients. Thus

$$X = \sum_j a^j (x) \frac{\partial}{\partial x^j} = \sum_{j,i} a^j (x) C^i_j(x) \widetilde X_i $$

and this implies $\varphi_i = \sum_j a^j C_j^i$, which are smooth functions.