Math Help - [SOLVED] Lie Bracket

1. [SOLVED] Lie Bracket

Hello!

I'm trying to determine something based off of an alternative definition of the Lie Brackect.

It is known that if $X$ and $Y$ are two smooth vector fields, then the Lie bracket $\left[X,Y\right]$ is a smooth vector field at a point $p$ defined to be

$\left[X,Y\right]_p\!\left(f\right)=X_p\!\left(Y\!\left(f\right)\ri ght)-Y_p\!\left(X\!\left(f\right)\right)$

We can express $\left[X,Y\right]$ in terms of the local coordinates $\left(v_1\!\left(p\right),\dots,v_m\!\left(p\right )\right)$ and $\left(w_1\!\left(p\right),\dots,w_m\!\left(p\right )\right)$ of $X_p$ and $Y_p$ respectively.

Since $X_p$ and $Y_p$ are directional derivatives, it turns out that:

$Y f=\sum_{j=1}^m w_j\frac{\partial f}{\partial x_j}$

This implies that

\begin{aligned}
X\left(Y f\right) &= X\left(\sum_{j=1}^m w_j\frac{\partial f}{\partial x_j}\right)\\ &= \sum_{i=1}^mv_i\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m w_j\frac{\partial f}{\partial x_j}\right]\\ &= \sum_{i=1}^m\sum_{j=1}^mv_i\left(\frac{\partial w_j}{\partial x_i}+w_j\frac{\partial^2 f}{\partial x_i \partial x_j}\right)\\ \end{aligned}

\begin{aligned}{\color{white}.}\phantom{X(Yf)}&=\s um_{i=1}^m\sum_{j=1}^mv_i\frac{\partial w_j}{\partial x_i}\frac{\partial f}{\partial x_j}+\sum_{i=1}^m\sum_{j=1}^mv_iw_j\frac{\partial^ 2f}{\partial x_i\partial x_j}\end{aligned}

By a similar process,

$Y\left(X f\right)=\sum_{i=1}^m\sum_{j=1}^mw_i\frac{\partial v_j}{\partial x_i}\frac{\partial f}{\partial x_j}+\sum_{i=1}^m\sum_{j=1}^mv_iw_j\frac{\partial^ 2f}{\partial x_i\partial x_j}$

Now, taking the difference between these two values yield the alternative definition of the Lie bracket:

$\left(XY-YX\right)f=\sum_{i=1}^m\sum_{j=1}^m\left(v_i\frac{ \partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\right)\frac{\partial f}{\partial x_j}$

In the textbook I'm using, it says that an easy consequence of this alternative definition is $\left[\partial/\partial x_i,\partial/\partial x_j\right]=0\,\forall\, 1\leq i,j\leq m$. My question is why is this the case? Given that my professor didn't go over this in class makes it quite challenging for me to grasp this idea.

I'm thinking that it boils down to finding the local coordinates for the basis vectors of $T_pM$, but I'm not sure how to come up with local coordinates for the basis vectors.

I've spent a couple hours on this and really haven't made progress. If anyone could shed some light on this, I would appreciate it!!

2. Hi, Chris

(I am currently learning differential geometry too so I may write something wrong...)
Originally Posted by Chris L T521
My question is why is this the case?
The vector field $X = \frac{\partial}{\partial x_k}$ is defined at $p$ by
$X_p = (v_1(p),v_2(p),\ldots, v_n(p)) = (0,\ldots , 0,\overbrace{1}^{v_k(p)},0,\ldots,0),$
it is a constant vector field. Hence for all $j$ such that $1\leq j\leq n$ the map $v_j$ is a constant map and we have $\frac{\partial v_j}{\partial x_i}=0$. ( $1\leq i \leq n.
$
) Similarly it can be shown that if $Y=\frac{\partial}{\partial x_l}$ then $\frac{\partial w_i}{\partial x_j}=0$ ( $1\leq i,j \leq n$). Finally we get

$\left[\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_l}\right]f=\sum_{i=1}^m\sum_{j=1}^m\left(v_i\frac{\partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\right)\frac{\partial f}{\partial x_j}=\sum_{i=1}^m\sum_{j=1}^m0\cdot \frac{\partial f}{\partial x_j} = 0$
as required.