4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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.) $\endgroup$ Aug 4, 2019 at 23:05

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .