left-invariant vector field

Left-invariant vector fields

Let $G$ be a Lie group. A vector field $X:G\to TG$ is called left-invariant if $(d{L}_g{)}_{g^{\prime }}(X_{g^{\prime }})=X_{g{g}^{\prime }}\forall g,g^{\prime }\in G.$ We denote the set of all left-invariant vector fields as ${\mathfrak{X}}_L(G)$.

Motivation

Consider the diffeomorphism of left action (multiplication) by an element $g\in G$.

$$\begin{array}{rl}{L}_g:G& ⟶G\\ h& ⟼gh\end{array}$$

We can think of this as shifting or translating elements of $G$ around by $g$. Note that the differential $(d{L}_g{)}_{g^{\prime }}:T_{g^{\prime }}G⟶T_{g{g}^{\prime }}G$ This now shifts tangent vectors around in a way related to left multiplication, $L_g$.

Therefore, we don’t want this shift to do anything extra to the vector field, so we require that
$(d{L}_g{)}_{g^{\prime }}(X_{g^{\prime }})=X_{g{g}^{\prime }}\forall g,g^{\prime }\in G.$

Note that we can look only at the identity element and see for a left-invariant vector field $X_g=(d{L}_g{)}_e(X_e).$ Thus, we can get to anywhere in the vector field starting from an element $X_e\in T_e(G)=\mathfrak{g}$ and using the differential of $L_g$ for an appropriate $g\in G$.

Lie subalgebra structure

The space ${\mathfrak{X}}_L(G)$ is a Lie subalgebra of 202405081435. In this way, it inherits the same bracket as $\mathfrak{X}(G)$.

Proof

todo

Complete vector fields

Every left-invariant vector field $X\in {\mathfrak{X}}_L(G)$ is a complete vector field.

Proof

Let $\gamma :J\to G$ be an integral curve with $\gamma (0)=h\in G$. Since $\gamma$ is an integral curve, we know that ${\gamma }^{\prime }(t)=X_{\gamma (t)}$ Consider the map

$$\begin{array}{rl}{L}_g\circ \gamma :J& ⟶G\\ t& ⟼g\gamma (t)\end{array}$$

This is now a curve in $G$, so we can look at the velocity vector

$$\begin{array}{rlr}(L_g\circ \gamma {)}^{\prime }(t)& =(d{L}_g{)}_{\gamma (t)}({\gamma }^{\prime }(t))& \text{Chain rule}\\ & =(d{L}_g{)}_{\gamma (t)}(X_{\gamma (t)})& \text{left-invariant}\\ & =X_{g\gamma (t)}& \\ & =X_{L_g(\gamma (t))}& \end{array}$$

Therefore, $L_g\circ \gamma$ is also an integral curve. If we left shift by $\gamma (t_0)$ for a small enough $t_0$ then $\gamma (t_0)\gamma (t)=\gamma (t_0+t)$ is an integral curve that will have domains that intersect. So by uniqueness, these integral curves must be the same, and we know they have the same domain (since we only want maximal ones). If we keep shifting over by $t_0$ every time, we can do that infinitely to cover the whole domain of $\mathbb{R}$.