Computing a Lie Bracket: General Questions

Question

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 :)

Answer 1

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

Answer 2

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.

Answer 3

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

$$[V,W]=VW-WV.$$

I.e.,

\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 $W(-y)\partial_x=0$,

Answer 4

I originally stumbled upon this post while trying to understand how to compute the Lie bracket, but I'll post the way of doing it that helped me the most. It's similar to @anonymous' answer, but I will write everything out in detail.

Recall that if $M$ is a smooth manifold with or without boundary, and $X,Y\in \mathfrak{X}(M)$ are smooth vector fields on $M$, then the Lie bracket $$[X,Y]=\sum_{j=1}^n\left(X\left(Y^j\right)-Y\left(X^j\right)\right)\frac{\partial}{\partial x^j}=\sum_{j=1}^n\sum_{i=1}^n\left(X^i\frac{\partial Y^j}{\partial x^i}-Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}$$ where $X^j,Y^j:U\to \mathbb{R}$ are the vector field coefficients of $X$ and $Y$ for some smooth chart $(U,(x^i))$ on $M$ (it's a good exercise to prove this, but if you can't check out John Lee's Introduction to Smooth Manifolds). So in this case, our vector fields are \begin{align*} X&=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\\ Y&=\frac{\partial}{\partial x}. \end{align*} The above theorem states that our Lie bracket is going to have the following formula: $$[X,Y]=X(1)\frac{\partial}{\partial x}-Y(-y)\frac{\partial}{\partial x}-Y(x)\frac{\partial}{\partial y}=\left(\sum_{i=1}^2X^i\frac{\partial 1}{\partial x^i}\right)\frac{\partial}{\partial x}-\left(\sum_{i=1}^2Y^i\frac{\partial (-y)}{\partial x^i}\right)\frac{\partial}{\partial x}-\left(\sum_{i=1}^2Y^i\frac{\partial x}{\partial x^i}\right)\frac{\partial}{\partial y}. $$ Computing each of these sums tells us that \begin{align*} X(1)&=\left(\sum_{i=1}^2X^i\frac{\partial 1}{\partial x^i}\right)=X^1\frac{\partial1}{\partial x}+X^2\frac{\partial 1}{\partial y}=-y\frac{\partial 1}{\partial x}+x\frac{\partial 1}{\partial y}=0+0=0\\ Y(-y)&=\left(\sum_{i=1}^2Y^i\frac{\partial (-y)}{\partial x^i}\right)=Y^1\frac{\partial(-y)}{\partial x}+Y^2\frac{\partial (-y)}{\partial y}=1\frac{\partial (-y)}{\partial x}+0\frac{\partial (-y)}{\partial y}=0+0=0\\ Y(x)&=\left(\sum_{i=1}^2Y^i\frac{\partial x}{\partial x^i}\right)=Y^1\frac{\partial x}{\partial x}+Y^2\frac{\partial x}{\partial y}=1\frac{\partial x}{\partial x}+0\frac{\partial x}{\partial y}=1+0=1.\\ \end{align*} So, after computing all of the above, our Lie bracket is going to be $$[X,Y]=X(1)\frac{\partial}{\partial x}-Y(-y)\frac{\partial}{\partial x}-Y(x)\frac{\partial}{\partial y}=0\frac{\partial}{\partial x}-0\frac{\partial (-y)}{\partial x}-1\frac{\partial}{\partial y}=\boxed{-\frac{\partial}{\partial y}}.$$ There's no reason to use derivations, or introduce any test function here. It's just a formula.