I'm asked to compute the following Lie Bracket:
$\left [ -y \dfrac{\partial}{\partial x} + x\dfrac{\partial}{\partial y} , \dfrac{\partial}{\partial x} \right] $ on $\mathbb{R}^2$.
Just writing it out, I get
$\left( -y \dfrac{\partial}{\partial x} + x\dfrac{\partial}{\partial y} \right) \dfrac{\partial}{\partial x} – \dfrac{\partial}{\partial x} \left(-y \dfrac{\partial}{\partial x} + x\dfrac{\partial}{\partial y} \right)$.
How can I simplify this? I know this is a very trivial question, but I'm getting stuck for some stupid reason. Any help would be greatful 🙂
Best Answer
Your vector fields really are derivation, so think that they will be applied to functions!
Take a test function $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ of class $C^\infty$. Then apply your vector field to this function (and you'll have to use the product rule and Schwarz' theorem at some point): $$\begin{align*} &\left(\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\right)f(x,y)\\ &\qquad=-y\frac{\partial^2f}{\partial x^2}(x,y)+x\frac{\partial^2f}{\partial y\partial x}(x,y)+\frac{\partial}{\partial x}\left(y\frac{\partial f}{\partial x}(x,y)\right)-\frac{\partial}{\partial x}\left(x\frac{\partial f}{\partial y}(x,y)\right)\\ &\qquad=-y\frac{\partial^2f}{\partial x^2}(x,y)+x\frac{\partial^2f}{\partial y\partial x}(x,y)+y\frac{\partial^2f}{\partial x^2}(x,y)-\frac{\partial f}{\partial y}(x,y)-x\frac{\partial^2f}{\partial x\partial y}(x,y)\\ &\qquad=-\frac{\partial f}{\partial y}(x,y) \end{align*}$$ Hence: $$\left[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\right]=-\frac{\partial}{\partial y}.$$