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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

$$d(f^{\star}f)=$$?

Any help appreciated

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1 Answer 1

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.

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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 at 19:08

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