# Lie brackets of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

Let $f$ be a smooth function, $f\colon\mathbb{R}^2\to \mathbb{R}$.

What is $\left[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right]$ ?

I want to say it's $0$ since $\frac{\partial f^2}{\partial x\partial y}=\frac{\partial f^2}{\partial y\partial x}$ but I am unsure about why for the two functions : $\frac{\partial f}{\partial x}$ ,$\frac{\partial f}{\partial y}$ the composition is the same...

Can someone please help with this ? [I didn't know what tag to give this post, hope it's ok]

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Maybe you should give some context? There are different Lie algebra structures that may be put on the space of smooth functions. If you want to think of them as differential operators then they do commute. If you're using a Poisson structure, it may not be zero. – Eric O. Korman Apr 23 '12 at 18:48
I think $[\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}}]$ makes no sense since $\frac{\partial{f}}{\partial{x}}$ is a function, and you most likely are looking at the Lie bracket that is defined for two vector fields. So, think of $[\frac{\partial}{\partial{x}},\frac{\partial}{\partial{y}}]$ – Yuri Vyatkin Apr 23 '12 at 18:49
Yeah, sorry, I implicitly assumed (and didn't notice the $f$, and even copied it into the first draft of my answer!) that the OP was talking about Lie bracket of vector fields. – user29743 Apr 23 '12 at 18:51

## 1 Answer

Your first answer is correct for the reason that you said. There is no composition of functions involved in Lie bracketing vector fields, only composition of derivations (= composition of operators). It may help to look at some nontrivial Lie brackets, e.g., compute $$\left[\frac{\partial}{\partial x}, x\frac{\partial}{\partial y} \right].$$

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Actually, I am thw one with the mistake. I didn't understand the question and it's interpreted like you said! thank you – Belgi Apr 23 '12 at 18:53