All left invariant vector fields are right invariant: What's my failure in reasoning

Question

I understand that the conclusion is absurd. But here is my reasoning.

Let $G$ be a lie group, let $x \in \mathfrak{g}=T_e(G)$, and let $\gamma(t)$ be a homomorphism $\mathbb{R} \to G$ such that $\gamma'(0)=x$. Then for any $t_0, t$, $$L_{\gamma(t_0)}(\gamma(t))=\gamma(t+t_0):=\delta(t),$$ $$ (L_{\gamma(t_0)})_* (\gamma'(0))=\delta'(0)=\gamma'(t_0).$$ By the same argument $(R_{\gamma(t_0)})_*(\gamma'(0))=\gamma'(t_0)$.

This shows $\gamma(t)$ is the unique flow of the left invariant vector field $v_x^L$ defined by $v_x^L(g)=dL_g x$ such that $\gamma(0)=e$. Similarly $\gamma(t)$ is also the flow of the vector field $v_x^R$ defined by $v_x^R(g)=dR_g x$ such that $\gamma(0)=e$. Therefore $v_x^L(\gamma(t))=\gamma'(t)=v_x^R(\gamma(t))$.

Since there is a one parameter subgroup that goes through any group element $g \in G$ this apparently shows that $v_x^L(g)=v_x^R(g)$ for all $g \in G$.

What is wrong?

Answer 1

Your argument establishes the following (true) theorem.

Let $v\in T_e G$ and let $\gamma:\mathbb{R}\rightarrow G$ be the unique homomorphism for which $\gamma'(0) = v$. Let $X_v$ denote the unique left invariant vector field with $X_v(e) = v$, and similarly, let $Y_v$ denote the corresponding right invariant vector field. Then, for any point of the form $\gamma(t_0)$, $X_v(\gamma(t_0)) = Y_v(\gamma(t_0))$.

However, this does not mean every left invariant vector field is right invariant because it never establishes that $X_v(g) = Y_v(g)$ for $g$ not of the form $\gamma(t_0)$.