# Using coordinates to show left-invariant vector fields form Lie algebras

I came across a simple Lie group question while reading Bernard Schutz's Geometrical Methods of Mathematical Physics. I am brand new to group theory, and this is my first attempt at grasping its physical significance.

He writes (p. 93) that if $\overline{V}$ and $\overline{W}$ are any two left-invariant vector fields on a manifold $G$, then the left translation map $L_g$ (which maps tangent vectors at the identity $e$ to those at $g$) maps $[\overline{V},\overline{W}]$ at $e$ to $[\overline{V},\overline{W}]$ at $g$. Then he draws a figure of the Lie bracket, and states that the field $[\overline{V},\overline{W}]$ is also left-invariant. He says, "The reader who is not convinced by the [diagrammatic proof] is invited to use coordinates on $G$ to prove the result".

I am wondering how one uses coordinates to prove this result. The Lie bracket, of course, can be expressed $$[\overline{V},\overline{W}]=\left(V^i\frac{\partial W^j}{\partial x^i}-W^i\frac{\partial V^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}$$ (summed over the repeated indices). But how would the operation $L_g$ interact with each of these component terms? I only know that $L_g(\overline{V}(e))=\overline{V}(g)$, and likewise for $\overline{W}$.

Let $M$ be a smooth manifold and $\varphi: M\to M$ a diffeomorphism. For a vector field $\xi: M\to TM$ one can define a new vector field $\varphi_*\xi$ by the relation:

$$(\varphi_*\xi)|_q = T_p\varphi(\xi|_p)$$

where $q = \varphi(p)$. This lead to an automorphism $\varphi_*: \Gamma(TM)\to \Gamma(TM)$ of the $C^{\infty}(M)$-module of vector fields called the pushforward by $\varphi$.

The heart of your question rely on the following relation:

$$\varphi_*[\xi_1, \xi_2] = [\varphi_*\xi_1, \varphi_*\xi_2]$$

Indeed if $M$ is a Lie group $G$, a vector field $\xi$ is left-invariant if and only if $(L_g)_*\xi = \xi$ for every $g\in G$. So using the previous relation it's obvious that the Lie bracket of two left-invariant vector fields is left-invariant.

Now since this relation works for any diffeomorphisms I don't think it's easier to prove it in the very particular case $\varphi = L_g$. Also, prooving this relation depends on how you define the Lie bracket and it's usually not done in coordinates. My own opinion is coordinates are not suitable to prove this relation; it would be tedious. For a standard proof I recommand John. M. Lee's Introduction to smooth manifolds.

• I agree that coordinates aren't the ideal way to prove this statement, but my question is, how would one use coordinates to prove this statement. – Doubt Oct 2 '17 at 1:54
• It's indeed possible to prove the relation I mentionned in coordinates. But it's neither usefull, nor instructive and it's extremely tedious. So I don't think it's the answer you desire. If you really want to do so, compute $\varphi_*[\xi_1, \xi_2]$ and $[\varphi_*\xi_1, \varphi_*\xi_2]$ in coordinates is a way. – paeolo Oct 3 '17 at 8:42
• Thanks, I'm almost there! I see the last piece I'm missing: given $\xi = \xi^i\frac{\partial}{\partial x^i}$, how does one write out (using the coordinate basis) the quantity $\varphi_*\xi$? – Doubt Oct 13 '17 at 3:26
• Use the definition $(\varphi_*\xi)|_q = T_p\varphi(\xi|_p)$ and the relation $T_p\varphi(\frac{\partial}{\partial x_i}) = \frac{\partial\varphi^j}{\partial x_i}\frac{\partial}{\partial x_j}|_{\varphi(p)}$. – paeolo Oct 14 '17 at 15:12