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I am studying analysis by using Elon Lages Lima's book "Curso de Análise (vol. 2)" (the book is written in portuguese). I am trying to solve the following problem about differential forms and would like some hint on how to advance.

Fix $k, m \in \mathbb{N}$, with $k \ge r$. Let $\omega$ be a form of degree $r$ in $\mathbb{R^m}$, such that $f^*\omega = 0$ for every affine transformation $f: \mathbb{R^k} \rightarrow \mathbb{R^m}$. Prove that $\omega = 0$.

Here's my attempt:

Since $f^* \omega = 0$, for every $x \in \mathbb{R^k}$ and for every $r$-list of vectors $w_1, \ldots, w_r \in \mathbb{R^k}$ we have:

$$ [(f^*\omega)(x)](w_1, \ldots, w_r) = \omega(f(x)) \cdot (f'(x) \cdot w_1, \ldots, f'(x) \cdot w_r) = 0 $$

An affine transformation $f$ is of the form $f(x) = Mx + b$, where $M$ is a linear transformation. Therefore, we have:

$$ [(f^*\omega)(x)](w_1, \ldots, w_r) = \omega(Mx + b) \cdot (M \cdot w_1, \ldots, M \cdot w_r) = 0 \tag{1} $$ In order to prove $\omega = 0$ we must prove that $\omega(y)(v_1, \ldots, v_r) = 0$, for an arbitrary $y \in \mathbb{R^m}$ and arbitrary vectors $v_1, \ldots, v_r \in \mathbb{R^m}$. I think I can choose $b$ conveniently in equation $(1)$ to obtain $Mx + b = y$. My next thought was to choose $w_1, \ldots, w_r$ conveniently to obtain $M \cdot w_i = v_i$, for $i = 1, \ldots, r$. But since $M$ is not necessarily surjective, I don't think I can do this.

I would like to know what to do from here. Thanks in advance!

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    $\begingroup$ Since the condition holds for all affine transformations $f$, you are also free to choose $M$. $\endgroup$
    – Berci
    Dec 27 '19 at 23:42
  • $\begingroup$ Yes, I agree. But my only thought was to choose $M$ surjective (which I don't know if I can guarantee) so that I could find, for each $v_i$, the corresponding $w_i$ such that $M \cdot w_i = v_i$. Maybe I should try another strategy... but i couldn't think of any other. Can you help me? Thanks in advance. $\endgroup$ Dec 28 '19 at 0:11
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Let $v_1,\dots,v_r\in\Bbb R^m$ arbitrary, and define a linear map $M:\Bbb R^k\to\Bbb R^n$ on the standard basis $e_1,\dots,e_k$ of $\Bbb R^k$ such that $Me_i=v_i$ if $i\le r$, and let $f:=x\mapsto Mx+b$ an affine map.

Then we get $\omega(Mx+b)(v_1,\dots,v_r)=\omega(Mx+b)(Me_1,\dots,Me_r)=[(f^*\omega)(x)](e_1,\dots,e_r)=0$.
Since $b$ is arbitrary (and we can take e.g. $x=0$), it follows that $\omega(b)=0$ for all $b\in\Bbb R^m$.

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