# The tangent bundle of a Lie group is trivial

I'm trying to recreate the proof given by Alex Youcis at https://math.stackexchange.com/a/308798/86801 and got everything except $\displaystyle (L_h)_\ast \left.\frac{\partial}{\partial x_i}\right|_e = \left.\frac{\partial}{\partial x_i}\right|_h$.

Let $f$ be a germ of functions at $h$, then $$\left( (L_h)_* \left.\frac{\partial}{\partial x_i}\right|_e \right) \ f = \left.\frac{\partial}{\partial x_i}\right|_e (f\circ L_h) = \left.\frac{\partial}{\partial x_i}\right|_0 (f\circ L_h\circ \varphi^{-1}),$$ while $$\left.\frac{\partial}{\partial x_i}\right|_h f = \left.\frac{\partial}{\partial x_i}\right|_{\psi(h)} (f \circ \psi^{-1}) = \left.\frac{\partial}{\partial x_i}\right|_{\varphi(g^{-1} h)} (f \circ L_g \circ \varphi^{-1}).$$ Here $\varphi$ is a chart of $G$ centered at $e$, $\psi=\varphi\circ L_{g^{-1}}$ is a chart centered at $g$ and $h$ is a point in the domain of $\psi$.

Can anybody explain why these derivations should be equal?

• This question is now a rather old, but anyway...That solution is not right. I posted an answer there. This statement just holds for $g$. – user40276 Oct 12 '15 at 6:40