# Lie bracket of left-invariant vector fields, wrong reasoning

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?

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 $$\mathrm{Ad}_{g}X_e = DR_{g^{-1}}(DL_gX_e)$$.