Left invariant vector field

Consider a Lie group $G$ acting on itself. Every vector $v \in T_{e} G$ gives rise to a vector field $V$ associated to $v$ :

V_{g} = d (L_{g})_{e} (v)

which satisfies

\begin{matrix} (1) & d (L_{g})_{h} (V_{h}) = V_{g h} . \end{matrix}

All the vector fields satisfying (1) this are called left invariant vector fields, and the set of all of them is the Lie algebra $g$ of $G$ . Observe that there is a one-to-one relation

T_{e} G \to g

The left invariant vector fields, or the right invariant ones, are the fundamental vector fields in the case of $G$ acting on $G$ .

Matrix groups

In this case of, for example $G L (2)$ , they take the form

X_{A} = A \cdot B,

being $A \in G L (2)$ (an invertible matrix) and $B \in T_{I d} G L (2)$ (an arbitrary matrix).
Proof. Given $B \in T_{I d} G L (2)$ , the corresponding left invariant vector field is, by definition

X_{A} := d (L_{A})_{I d} (B) .

If $B$ is associated to the curve $γ (t)$ such that $γ (0) = I d, γ^{'} (0) = B$ then

X_{A} = \frac{d}{d t} |_{t = 0} L_{A} \cdot γ (t) = L_{A} \cdot B = A \cdot B .

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