Why is the vector field $X(a)\equiv (dL_a)_e X_e$ left invariant?

Question

Let $G$ be a Lie Group with $e$ as the neutral element. Taken $X_e\in T_e G$, define

$$X(a)=(dL_a)_e X_e$$

Why this vector field is left invariant? I get confused with the notation.

Thanks!

Answer 1

To show that $X$ is left-invariant, we must prove $(L_g)_{*}X = X$ for all $g\in G$. This means $(dL_g)_aX(a) = X(L_g(a))$ for all $a,g\in G$, i.e., $(dL_g)_aX(a) = X(ga)$ for all $a,g\in G$. Given $a, g\in G$,

$$(dL_g)_aX(a) = (dL_g)_a(dL_a)_eX_e = (dL_g)_{L_a(e)}(dL_a)_eX_e \underset{(*)}{=} d(L_g\circ L_a)_eX_e = (dL_{ga})_eX_e.$$

The chain rule was used to get equation $(*)$. Since $(dL_{ga})_eX_e = X(ga)$, we have $(dL_g)_aX(a) = X(ga)$.

Answer 2

In general, let $M$ and $N$ be smooth manifolds and let $f\in C^\infty(M, N)$.

$(i)$ The tangent bundle of $M$ is $$TM:=\{(p, X): p\in M, X\in T_pM\}.$$ This comes with a submersion $\pi:TM\longrightarrow M$ which is just the projection in the first component. One might show $TM$ is a smooth manifold and $\pi:TM\longrightarrow M$ is a vector bundle.

$(ii)$ $f$ induces a map $df\in C^\infty(TM, TN)$ given by $$df(p, X):=(f(p), df_p(X)),$$ where $df_p:T_pM\longrightarrow T_{f(p)}N$ is the differential of $f$ in $p$.

$(iii)$ A vector field on $M$ is just a smooth map $X:M\longrightarrow TM$ such that $\pi\circ X=id_M$ (that is, $X_p\in T_pM$ for all $p\in M$).

Now let us specialize to the case of Lie groups. We have a smooth map $L_g\in C^\infty(G, G)$ given by $$L_g(h):=gh,$$ where we're using the group multiplication on the right-hand side. We say a smooth vector field $X:G\longrightarrow TG$ is left invariant if $$dL_g\circ X=X\circ L_g,$$ for all $g\in G$. Let us examine the equality $dL_g\circ X=X\circ L_g$: $$dL_g\circ X=X\circ L_g\Leftrightarrow dL_g(X(h))=X(gh),\ \forall h\in G.$$ Using the definition of $dL_g$ one finds $$dL_g\circ X=X\circ L_g\Leftrightarrow (gh, (dL_g)_h(X_h))=(gh, X_{gh})\,\ \forall h\in G\Leftrightarrow (dL_g)_h(X_h)=X_{gh},\ \forall h\in G.$$ Above $X_h:=\textrm{pr}_2\circ X(h)$, that is, $X(h)=(h, X_h)$ where $X_h\in T_hG$. Therefore, $X:G\longrightarrow TG$ is left-invariant if and only if $$(dL_g)_h(X_h)=X_{gh},\ \forall g, h\in G.$$ The above is also equivalent to:$$(dL_g)_e(X_e)=X_g.$$ If fact, if $(dL_g)_e(X_e)=X_g$ holds then using the chain rule $$X_{gh}=(dL_{gh})_e(X_e)=d(L_g\circ L_h)_e(X_e)=(dL_g)_h((dL_h)_e(X_e))=(dL_g)_h(X_h).$$ Therefore, one might say your vector field is left-invariant by definition.