# For a cordinate system $(U,x^1,\ldots , x^d)$ show that $[\partial / \partial x_i,\partial / \partial x_j]=0$ on $U$.

I'm working through Warner's Foundations of Differentiable Geometry and stuck on this question. Let $$(U,x^1,\ldots x^d)$$ be a coordinate system on $$M$$, show that $$[\partial / \partial x_i,\partial /\partial x_j]=0$$ on $$U$$.

I'm confused on what this question is asking. Are we supposed to apply $$[\partial / \partial x_i,\partial /\partial x_j]$$ to an arbitrary function $$f$$ defined on $$U$$ and use that partial derivatives commute? Or since $$\partial /\partial x_i$$ is a basis for a vector field on $$M$$ should we interpret $$[\partial / \partial x_i,\partial /\partial x_j]$$ a vector field and show that is it is zero on $$U$$.

If $$[\partial / \partial x_i,\partial /\partial x_j]$$ is not a vector field could you explain what kind of object it is in your answer.

• You achieve the latter by doing the former. (Of course, in general, the reason the bracket of two arbitrary vector fields is again a vector field is that mixed partials are equal.) Aug 4, 2019 at 23:05

You can prove it with both ways, as you described above. Let $$M$$ be a differentiable manifold with a coordinate chart $$\big(U,\phi=(x_1,...,x_n)\big)$$ and $$X:=[\frac{\partial }{\partial x_i}, \frac{\partial }{\partial x_j}]$$. Let's begin with your second way.

1. First Solution

Of course any Lie bracket is always a vector field. Moreover, if $$\xi, \eta$$ are smooth vector fields, then $$[\xi, \eta]$$ is a smooth vector field too. As you said before, $$\{\frac{\partial }{\partial x_i}\}_{i=1}^n$$ is a basis for the space of vector fields, and so $$X=\sum\limits_{i=1}^n a_i \frac{\partial }{\partial x_i}$$ , where of course $$a_i=X(x_i)$$ are the coefficient functions on $$U$$.

In order to show that $$X\rvert_U\equiv 0$$, you have to show that $$a_1=...=a_n=0$$. Βut for $$m \in \{1,..,n\}$$ we have \begin{align*} a_m=X(x_m)=\bigg[\frac{\partial }{\partial x_i}, \frac{\partial }{\partial x_j}\bigg](x_m)&=\frac{\partial }{\partial x_i}\bigg(\frac{\partial }{\partial x_j}(x_m)\bigg)-\frac{\partial }{\partial x_j}\bigg(\frac{\partial }{\partial x_i}(x_m)\bigg)\\ &=\frac{\partial }{\partial x_i}(\delta_{jm})-\frac{\partial }{\partial x_j}(\delta_{im})=0 \end{align*}

2. Second solution

Another way to prove that $$[\frac{\partial }{\partial x_i}, \frac{\partial }{\partial x_j}]\equiv 0$$ is to show that $$X_p=0$$ for every $$p\in U$$, or, in other words, $$\begin{equation*}\tag{1} \frac{\partial }{\partial x_i}\Big\rvert_p\bigg(\frac{\partial }{\partial x_j}\Big\rvert_p\bigg)-\frac{\partial }{\partial x_j}\Big\rvert_p\bigg(\frac{\partial }{\partial x_i}\Big\rvert_p\bigg)=0, \end{equation*}$$

If $$f\in C^\infty(M)$$, then $$f\circ \phi^{-1}:\mathbb{R}^n \to \mathbb{R}$$ is differentiable, and $$\frac{\partial}{\partial x_j}\Big\rvert_{p}(f)=\frac{\partial f}{\partial x_j}\Big\rvert_{p}=\frac{\partial (f\circ \phi^{-1})}{\partial u_j}\Big\rvert_{\phi(p)},$$ hence $$\frac{\partial }{\partial x_i}\Big\rvert_p\bigg(\frac{\partial }{\partial x_j}\Big\rvert_p(f)\bigg)-\frac{\partial }{\partial x_j}\Big\rvert_p\bigg(\frac{\partial }{\partial x_i}\Big\rvert_p(f)\bigg)=\frac{\partial^2 (f\circ \phi^{-1})}{\partial u_iu_j}\Big\rvert_{\phi(p)}-\frac{\partial^2 (f\circ \phi^{-1})}{\partial u_ju_i}\Big\rvert_{\phi(p)}=0.$$ Since $$f$$ was arbitrary, $$(1)$$ holds.