# If $f^* \omega = 0$ for every affine transformation $f$, then $\omega = 0$

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!

• Since the condition holds for all affine transformations $f$, you are also free to choose $M$. Dec 27 '19 at 23:42
• 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. Dec 28 '19 at 0:11

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 $$b\in\Bbb R^m$$.