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

on $R^2$

Can you help me please?

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    $\begingroup$ Do you know the definition of the Lie bracket? $\endgroup$
    – J126
    Apr 18 '14 at 22:53
  • $\begingroup$ [X,Y]f=(XY-YX)f $\endgroup$
    – user114952
    Apr 18 '14 at 22:55
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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: $$\begin{align} \frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right) = \frac{\partial}{\partial y} + x \frac{\partial^2}{\partial x\partial y} \end{align} $$

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  • $\begingroup$ @user114952: You are welcome. $\endgroup$
    – Thomas
    Apr 18 '14 at 22:57
  • $\begingroup$ doesn't it simplify to $-\frac{\partial}{\partial y}$ assuming we are dealing with a $C^\infty$ structure? $\endgroup$
    – Seth
    Apr 18 '14 at 23:05
  • $\begingroup$ @Seth: Yes, it will simplify quite a bit. Would we get $-\frac{\partial}{\partial x}\frac{\partial}{\partial y} = -\frac{\partial^2}{\partial x\partial y}$? $\endgroup$
    – Thomas
    Apr 18 '14 at 23:09
  • $\begingroup$ We should get $-\frac{\partial}{\partial y}$ $\endgroup$
    – Seth
    Apr 18 '14 at 23:12
  • $\begingroup$ @Seth: OK, you are probably right :) $\endgroup$
    – Thomas
    Apr 18 '14 at 23:12
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[−y∂/∂x+x∂/∂y,∂/∂x]=(−y∂/∂x+x∂/∂y)(∂/∂x)−(∂/∂x)(−y∂/∂x+x∂/∂y)= - ∂/∂y
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