Let $(x,y,z)$ a coordinate system, $M=\mathbb{R}^3$ and we also denote by $x$ the first coordinate function : $x:M \rightarrow \mathbb{R},\; q=(a,b,c) \mapsto a$.

We have $dx:TM \rightarrow \mathbb{R},\; (q,v=(v_1,v_2,v_3)) \mapsto dx[q](v)=v_1$.

Let $f=dx[q]$.

We have $df:TM \rightarrow \mathbb{R},\; (p,u=(u_1,u_2,u_3)) \mapsto df[p](u)=u_1$.

How to get $d(dx) = df = 0$ ?

  • $\begingroup$ $d$ is the derivative operator. So use the properties of the exterior derivative: $df = d(dx) = d(1 dx) = d(1)\wedge dx = 0$ $\endgroup$
    – user137731
    Commented Jun 21, 2016 at 13:22
  • $\begingroup$ is there a way by just manipulating the definitions that I recalled and not using the exterior calculus ? $\endgroup$
    – prolea
    Commented Jun 21, 2016 at 13:26
  • $\begingroup$ What you don't do is define the $d$ operator. So your statement for $df$ is wrong but it'd be hard to see that without some definition of $d$. To find the correct formula for $df$ you can use the properties of $d$ or just go back to the limit-integral definition. $\endgroup$
    – user137731
    Commented Jun 21, 2016 at 13:32
  • $\begingroup$ oki I see if I take the definition in local coordinates: Given a fucntion $f : \mathbb{R}^3 \rightarrow R, q\mapsto f(q)$, $df$ is defined as: $df[q](h) = \sum_i h_i \partial_i f(q) $ but then I need to define $d^2 f[q]$ with second partial derivatives and then the null function appears. $\endgroup$
    – prolea
    Commented Jun 21, 2016 at 13:39
  • 1
    $\begingroup$ $dx$ has zero differential as a $1$-form. But you differentiate $f=dx[q]$ as a $0$-form, i.e. as a function. In this case your calculation is correct and you don't get $d(dx[q])=0.$ $\endgroup$ Commented Jun 21, 2016 at 13:39

1 Answer 1


Let's use the definition of the exterior derivative to see why $d(dx)=0$:

Let $x:\Bbb R^3\to \Bbb R$ be given by $x(q) = x(q_1,q_2,q_3) = q_1$. Then by definition:

$$dx[q](v) = \lim_{t\to 0} \frac {1}{t} \int_{\{q,q+tv\}} x = \lim_{t\to 0}\frac{x(q+tv)-x(q)}{t} = v_1$$


$$d(dx)[q](u,v) = \lim_{t_1,t_2\to 0} \frac{1}{t_1t_2} \int_{C} dx$$ where $C$ is the oriented parallelogram with vertices at $q$, $q+t_1u$, $q+t_1u+t_2v$, and $q+t_2v$. But $\int_C dx=0$ for any such $C$ so $$d(dx) = 0$$

However as Valentin noted in the comments, $d(dx)$ and $d(dx[q])$ are two different things. $dx$ is a $1$-form but $dx[q]$ is a $0$-form on $T_qM$. I think that's where the disconnect is here.


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