Computing Lie bracket on $\mathbb{R}^3$

Question

Let us view $\mathbb{R}^3$ as a smooth manifold. Let us define vector fields $$ X_1 = \frac{\partial}{\partial x} - y\frac{\partial}{\partial z} $$ and $$ X_2 = \frac{\partial}{\partial y} - x\frac{\partial}{\partial z}. $$ I am not seeing how $$ [X_1, X_2] = z \frac{\partial}{\partial z}... $$ any explanation would be appreciated. thank you.

Answer 1

That doesn't seem right indeed. Since, given a smooth map $f\colon\mathbb R^3\longrightarrow\mathbb R$,\begin{align}X_2\bigl(X_1(f)\bigr)&=X_2\left(\frac{\partial f}{\partial x}-y\frac{\partial f}{\partial z}\right)\\&=\frac{\partial^2f}{\partial y\partial x}-\frac{\partial f}{\partial z}-y\frac{\partial^2f}{\partial z^2}-x\frac{\partial^2f}{\partial z\partial x}+xy\frac{\partial^2f}{\partial z^2}\end{align}and\begin{align}X_1\bigl(X_2(f)\bigr)&=X_1\left(\frac{\partial f}{\partial y}-x\frac{\partial f}{\partial z}\right)\\&=\frac{\partial^2f}{\partial x\partial y}-\frac{\partial f}{\partial z}-x\frac{\partial^2f}{\partial x\partial z}-y\frac{\partial^2f}{\partial z\partial y}+xy\frac{\partial^2f}{\partial z^2},\end{align}we have $[X_1,X_2]=0.$

Answer 2

$[X_1,X_2]=-DX_2(X_1)+DX_1(X_2)$

$X_1(x,y,z)=(1,0,-y)$, $X_2=(0,1,-x)$.

$DX_1(x,y,z)=(0,0,-dy)$, $DX_1(X_2)=(0,0,-1)$.

$DX_2(x,y,z)=(0,0,-dx)$, $DX_2(X_1)=(0,0,-1)$

implies that $[X_1,X_2]=0$.