# Lie bracket of coordinate frame

Suppose I have a local nbd $U$ in a manifold centered around $p$ and a chart on it. The chart is given by $X=(x_1,x_2,\cdots,x_n):U\rightarrow V\subset\mathbb{R}^n$. I consider the vector fields $\frac{\partial}{\partial x_i}$, ($i=1,\cdots,n$) which are given by the curve $c_i:(-\epsilon,\epsilon):\rightarrow U$, $c_i(t)=X^{-1}(0,0,\cdots,t,\cdots,0)$ ($t$ at the $i$-th coordinate).

I want to ask whether $\bigg[\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\bigg]$ is always $0$ ?

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Hint: Compute the action of your $\bigg[\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\bigg]$ on an arbitrary smooth function $f$ – Yuri Vyatkin Mar 1 '13 at 11:03

This is Schwarz's theorem on symmetric second derivatives.

http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

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I knew it as Young's theorem.. – Berci Mar 1 '13 at 11:17

Also, there is the lemma that any $f:X \longrightarrow Y$ in differentiable class preserves the bracket: Taking bracket and then pushing forward to $Y$ is same as going to $Y$ and doing bracket there. (See classical differential geometry books.)

SO, if our vectors come from local coordinates, since in $R^n$ bracket of coordinate directions vanish, we get the result.

Now, more interesting question is: Given a set of linearly independent vector fields, is the vanishing of the lie bracket sufficient for them to be realizable as vector coming from directions of a local coordinate?

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