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Proposition: Let $U\subset\mathbb{R}^n$ and $V\subset\mathbb{R}^n$ be open sets and $\phi:U\to V$ be differentiable. For all $k\in\mathbb{N}_0$ and $\omega\in \Lambda^k(V)$ it is true that

$$d(\phi^*\omega)=\phi^*(d\omega)$$

I am trying to understand its prove. But there are some steps I do not understand. Here are the first lines of the

Proof: At first let $f\in \mathcal{C}^\infty (V)$ be a differentialform of degree $0$. Then $\phi^*(f)=f\circ\phi$. Hence \begin{eqnarray*} d(\phi^*(f)) &=&d(f\circ\phi)\\ &=&\sum_{j=1}^{n}\frac{\partial(f\circ\phi)}{\partial x_j}dx_j\\ &\overset{?}{=}&\sum_{i,j}\frac{\partial f}{\partial y_i}\circ\phi(\frac{\partial (\phi_i)}{\partial x_j}dx_j)\\ &=&\sum_{i=1}^{m}\frac{\partial f\circ\phi}{\partial y_i}d\phi_i\\ \end{eqnarray*}

I marked the position I don't understand with a question mark. What exactly happens here?

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2 Answers 2

up vote 4 down vote accepted

As mentioned by Daniel, this is just the chain rule. As for your comment, I think what is written in your post is not correct. I claim it should be $$\sum_{i,j}\left(\frac{\partial f}{\partial y_i}\circ\phi\right)\cdot\left(\frac{\partial\phi_i}{\partial x_j}\right)dx_j$$ where $\cdot$ denotes multiplication of functions. Now you can see how it is just the chain rule. For two differentiable functions $f, g : \mathbb{R} \to \mathbb{R}$, the composition $f \circ g :\mathbb{R} \to \mathbb{R}$ is differentiable and $$(f\circ g)'(a) = f'(g(a))\cdot g'(a)$$ where $\cdot$ denotes multiplication of real numbers. Without the value $a$, this is just the statement that $$(f\circ g)' = (f'\circ g)\cdot g'$$ where $\cdot$ denotes multiplication of functions.

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The step you don't understand is simply chain rule, $y_i$ denotes $\phi_i(x)$.

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I see, but where does the $\phi$ in $\sum_{i,j}\frac{\partial f}{\partial y_i}\circ\phi(\frac{\partial (\phi_i)}{\partial x_j}dx_j)$ come from? –  Aufwind Jun 9 '13 at 13:32
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