Why do we need left invariant vector fields?

Question

Here Wikipedia Says

$1-$Vector fields on any smooth manifold $M$ can be thought of as derivations $X$ of the ring of smooth functions on the manifold, and therefore form a Lie algebra under the Lie bracket $[X, Y] = XY − YX,$ because the Lie bracket of any two derivations is a derivation.

$2-$We apply this construction to the case when the manifold M is the underlying space of a Lie group $G$, with $G$ acting on $G = M$ by left translations $L_g(h) = gh$. This shows that the space of left invariant vector fields on a Lie group is a Lie algebra under the Lie bracket of vector fields.

As we can see that for any smooth manifold $M$, the vector fields for a Lie Algebra under the Lie bracket.

When we are considering a smooth manifold $M$ which is a group as well(that is it is a lie group) why are we suddenly interested in the left invariant vector fields instead of all the vector fields?

Is the following true? The collection of all left invariant vector fields is a sub Lie Algebra of the collection of vector fields?

Answer 1

A short answer is that the Lie algebra determines the Lie group to large extent. Even when the Lie algebra does not determine the Lie group, it provides some key invariants of the Lie group. In contrast, if you consider general smooth manifolds, the tangent space at one point says almost nothing about the manifold (apart from its dimension). The rest, as Arctic Char suggested, you will find out by reading textbooks on the subject.

For instance: Suppose that $G_1, G_2$ are simply-connected Lie groups and $\phi: {\mathfrak g}_1\to {\mathfrak g}_2$ is an isomorphism of their Lie algebras. Then there exists an isomorphism of Lie groups $f: G_1\to G_2$ such that $df_e=\phi$.

Answer 2

Another useful property of the Lie algebra of left-invariant vector fields is that a basis for the latter always forms a global frame for the tangent bundle of the Lie group. While the space of all vector fields is quite large, it still has a "basis" such that any $X\in\mathfrak{X}(G)$ is a finite linear combination $X=f_1\cdot X_1+\dots+f_n\cdot X_n$ where $X_1,\dots,X_n$ is a basis of the Lie algebra and $f_1,\dots,f_n\in\mathcal{C}^\infty(G)$. So it shows how the tangent bundle of a Lie group is trivializable.