Lie bracket of vector fields on $R^2$ [duplicate]

Compute the Lie bracket$$\Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big]$$

on $R^2$

• Do you know the definition of the Lie bracket? – Joe Johnson 126 Apr 18 '14 at 22:53
• [X,Y]f=(XY-YX)f – user114952 Apr 18 '14 at 22:55

I am not 100% sure that this is what you want, but I think that \begin{align} \Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big] &= \left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\frac{\partial}{\partial x} - \frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right) \\ &= -y\frac{\partial^2}{\partial x^2}+x\frac{\partial}{\partial y}\frac{\partial}{\partial x} +y\frac{\partial^2}{\partial x^2} - \frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right) \\ &= \dots \end{align}
Use the product rule on the last part: \begin{align} \frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right) = \frac{\partial}{\partial y} + x \frac{\partial^2}{\partial x\partial y} \end{align}
• doesn't it simplify to $-\frac{\partial}{\partial y}$ assuming we are dealing with a $C^\infty$ structure? – Seth Apr 18 '14 at 23:05
• @Seth: Yes, it will simplify quite a bit. Would we get $-\frac{\partial}{\partial x}\frac{\partial}{\partial y} = -\frac{\partial^2}{\partial x\partial y}$? – Thomas Apr 18 '14 at 23:09
• We should get $-\frac{\partial}{\partial y}$ – Seth Apr 18 '14 at 23:12
[−y∂/∂x+x∂/∂y,∂/∂x]=(−y∂/∂x+x∂/∂y)(∂/∂x)−(∂/∂x)(−y∂/∂x+x∂/∂y)= - ∂/∂y