I'm trying to understand left-invariant vector fields, but I come to a contradiction. Can you tell me where I'm wrong?
Here's the definition.
Let $G$ be a Lie group. Denote by $L_g:G\to G$ the left translation $h\mapsto gh$. Denote by $DL_g:T_hG\to T_{gh}G$ its derivative. A vector field $X$ on $G$ is called left-invariant if for every $g,h\in G$, we have $DL_g(X_h)=X_{gh}$.
Now let $g:\Bbb{R}\to G$ be a morphism of Lie groups (we think of it as a 1-dimensional subgroup, the map $g$ doesn't need to be injective). Let $V_e$ be its tangent vector at the identity $e=g(0)$. We can extend $V_e$ to a vector field $V$ on the whole of $G$ by setting $V_h:= DL_h(V_e)$. The resulting $V$ is left-invariant, and every left-invariant field arises this way. Moreover, the flow of $V$ is given by $L_{g(t)}$ (is this correct?).
Let $X$ be a left-invariant vector field. Using the definition of Lie bracket as the derivative of a flow, we get
$$
[V,X]_h = \lim_{t\to 0} \dfrac{DL_{g(-t)}(X_{g(t)h})-X_h}{t} = 0,
$$
since by left-invariance,
$$
DL_{g(-t)}(X_{g(t)h})=X_{g(-t)g(t)h}=X_{g(t)^{-1}g(t)h}=X_h.
$$
Therefore $[V,X]=0$ for all left-invariant vector fields $V$ and $X$ on $G$, so the Lie algebra of $G$ is Abelian for every $G$.
Where is the mistake?
Best Answer
The mistake is that the flow of left-invariant $V$ is $R_{g(t)}$, the right translation by $g(t)=\exp(tV_e)$. This is because $$ \frac{d}{dt}R_{g(t)}(h) = \frac{d}{dt}L_h(g(t)) = DL_h(V_{g(t)}) = V_{hg(t)}. $$ Hence the Lie bracket of two left-invariant vector fields, evaluated at the identity becomes $$ [V,X]_e = \lim_{t\to0}\frac{DR_{g(-t)}X_{g(t)}-X_e}{t} = \frac{d}{dt}\Big\vert_{t=0}\mathrm{Ad}_{g(t)}X_e =: \mathrm{ad}_{V_e}X_e $$ where $\mathrm{Ad}_{g}X_e = DR_{g^{-1}}(DL_gX_e)$.