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 :)

  • 1
    $\begingroup$ Expand it. Looks like some things will cancel. $\endgroup$
    – JEM
    Apr 13, 2014 at 18:13

3 Answers 3


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: $$\left[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\right]=-\frac{\partial}{\partial y}.$$


The idea is to operate on a (sufficiently nice) function. In your case, "sufficiently nice" means at least twice continuously differentiable (since you will have mixed partials). Here's an example that can guide you through the computation. Suppose we want to evaluate $\left[x,\frac{d}{dx}\right]$. Derivatives don't make sense on their own. It isn't until they operate on something that they have real meaning. With this in mind, let's see what happens:

$$\left[x,\frac{d}{dx}\right]f \equiv x\frac{d}{dx}f - \frac{d}{dx}(xf).$$

By chain rule, we get that

$$\left[x,\frac{d}{dx}\right]f = xf'-xf'-f,$$

or equivalently

$$\left[x,\frac{d}{dx}\right]f = -f.$$

As an operator, we would then say that $$\left[x,\frac{d}{dx}\right] = -I,$$

where $I$ is the identity operator on whatever "sufficiently nice" space of functions you are considering. Can you see how to proceed here? Note that JEM's comment about "expanding" and "canceling" is full of potential errors. It can work sometimes but it is not guaranteed to do so. If we took his advice and applied it to the example above, we would not get the correct answer and the reason is ultimately because chain rule is lost in his approach.

  • $\begingroup$ Perfect. I guess it would make a lot of sense to stick a function in there haha. $\endgroup$
    – Calculus08
    Apr 13, 2014 at 18:16

This of if like this, if $V=-y \partial_x+x \partial_y,W=\partial_x$., then



\begin{align} [V,W]&=V(1)\partial_x - [W(-y)\partial_x+W(x)\partial_y]\\ &=0-[0+(1)\partial_y]\\ &=-\partial_y \end{align}

We read $W(-y)$ as $W$ acts on $(-y)$ in which case $W$ takes the $x$ partial, we have $W(-y)\partial_x=0$,


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