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I have a bit of a problem with the following identity:

Suppose that $U, V \subset \mathbb{R}^n$, are two open sets. Let $x^1,...,x^n$ be a system of coordinates of $U$ and $y^1,...,y^n$ one on $V$. Let $f:V\rightarrow U$ such that $x^i=f^i(y^1,...,y^n)$. Than: $$df^1\wedge ... \wedge df^n=\det(\frac{\partial f^i}{\partial y^j})dy^1\wedge...\wedge dy^n.$$

Is this a legal way to prove it?

$df^1\wedge ... \wedge df^n(v_1,...,v_n)=\det(f^i(v_j))=\det(\frac{\partial y^k}{\partial y^k}f^i(v_j))=\det(\frac{\partial f^i}{\partial y^k}y^k(v_j))$ which leads to the result by applying once again the determinant formula.

I have the impression the last equality is wrong or needs to be justified more in details...

Thanks a lot for your precious help and have a nice day!

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up vote 1 down vote accepted

I suggest you try to write out $df^{i}$ in concrete terms, then you will see where the equality comes from.

You should read this article carefully.

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if I write $df^i=\frac{\partial f^i}{\partial y^j}d y^j$ I have then a problem as I can't find $dy^1\wedge...\wedge dy^n$ again –  Nre Jan 13 '13 at 13:27
    
try the 2 by 2 case first. –  Bombyx mori Jan 13 '13 at 13:38
    
I think I got what you told me; We will have all the terms with the a repeating $dy^k=0$ and then the sum of all the permutations which gives the determinant. Is there a more elegant way to put it using the determinant formula instead? thanks a lot –  Nre Jan 13 '13 at 13:50
    
Notice this is actually the definition of determinant if $f$ happens to be a linear transformation. So you are having tautology at here. –  Bombyx mori Jan 13 '13 at 13:52
    
Ok, but there is no way of getting it by writing the determinant formula as I first suggested it? –  Nre Jan 13 '13 at 13:54
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