a left-invariant vector field

On a Lie group, every point looks like the identity, because left multiplication slides the identity anywhere. So a single tangent vector at the identity can be 'spread out' over the whole group in a canonical way: at each point g, place the vector you get by sliding the identity vector over to g via left translation. The result is a vector field that the group's own left-multiplications never disturb — a left-invariant vector field.

Precisely, for a Lie group G let L_g(x) = g*x be left translation, a diffeomorphism, with differential (L_g)_* mapping tangent spaces. A vector field X on G is left-invariant if (L_g)_* X = X for every g, meaning X at the point g*h equals the pushforward by (L_g)_* of X at h. Such an X is completely determined by its value at the identity: given any v in T_e G, set X_g = (L_g)_* v. This gives a vector-space isomorphism between left-invariant vector fields and the tangent space T_e G. The decisive fact is that the Lie bracket [X, Y] of two left-invariant fields is again left-invariant — invariance is preserved by the bracket — so the left-invariant fields form a finite-dimensional Lie algebra under [.,.], and through the isomorphism this bracket lands on T_e G. That algebra is the Lie algebra g of G.

Left-invariant fields are how the Lie algebra is born from the Lie group: they convert the global, possibly complicated group into the linear-algebraic data of one tangent space with a bracket. They are also automatically complete (their flows exist for all time, and the time-1 flow is the exponential map). A small caveat: one could equally use right translations to get right-invariant fields and another copy of the same abstract Lie algebra; the left convention is just a choice. And a left-invariant field is generally not invariant under right translation, nor a Killing field for a chosen metric unless the metric, too, is left-invariant.