Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to prove the following property of the map between differential forms: (Spivak's book ''Calculus on manifolds'' p.91)

$$f^{\star}\;\Lambda^{k}(\mathbb{R}^{m}_{f(p)})\to \Lambda^{k}(\mathbb{R}^{n}_{p}) $$

which is defined as $$(f^{\star}\omega)(p)=f^{\star}(\omega(f(p)))$$ where $\omega$ is a $k$-form on $\mathbb{R}^{m}$ and $f^{\star}\omega$ is a $k$-form on $\mathbb{R}^{n}$, and  $p\in\mathbb{R}^{n}$ for the mapping $$f_{\star}\;:\mathbb{R}^{n}_{p}\to \mathbb{R}^{m}_{f(p)}$$

Now if $\omega$ is a $k$-form and $\eta$ is a $l$-form then $$f^{\star}(d\omega)=d(f^{\star}\omega)$$ The comment for the $0$-form case is ''it is clear'' (just the chain rule).

why is this clear when $\omega$ is a $0$-form? A $0$-form is a function of zero variables right? then its differential will be a $1$-form $dx^{i}$?

But then the lhf of the equation will have form $$f^{\star}(dx^{i})=\displaystyle\sum_{j}^{n}\frac{\partial f^{i}}{dx^{j}}\cdot dx^{j}$$

on the other hand


Any help appreciated

share|cite|improve this question

You seem to be a bit confused on a couple of points. For example, I'm not sure what you mean by $d(f^\ast f)$. Remember that $f$ is a function $\mathbb{R}^n \to \mathbb{R}^m$, and we're using $f$ to pull back forms on $\mathbb{R}^m$ to obtain forms on $\mathbb{R}^n$.

A $0$-form on $\mathbb{R}^m$ is simply a function $\mathbb{R}^m \to \mathbb{R}$ (I don't know what you mean by a "function of zero variables"). Let's call that function $g$. We need to show that $f^\ast (dg) = d(f^\ast (g))$. If you write down the left and right sides of that equation using the definitions of $d$ and $f^\ast$, you'll find that they are equal. I can elaborate if you like, but this is a good exercise for you.

share|cite|improve this answer
Thank you, you are right, this is a bit confusing, but you made it a bit clearer. I am afraid I still don't see it... there is the following theorem in the book: If $g\;\mathbb{R}^{m}\to\mathbb{R}$ is differentiable then $dg=\displaystyle\sum_{i=1}^{m}\frac{\partial g}{dx^{i}}dx^{i}$ is a $1$-form. But then again the lhs : $f^{\star}(dg)=f^{\star}\displaystyle\sum_{i=1}^{m}\frac{\partial g}{dx^{i}}dx^{i}$ can anything more be done here? how to calculate the rhs? I still obtain exactly the same expression as in the original post... – user124471 May 16 '14 at 19:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.