Let $\Omega = dx^1 \wedge \ldots \wedge dx^n$ an $n$-form on $\mathbb R^n$ and
$$\alpha = \sum_{j = 1}^n (-1)^{j-1}x^j dx^1 \wedge \ldots \wedge dx^{j-1} \wedge dx^{j+1}\wedge \ldots \wedge dx^n.$$
In order to prove Brouwer fixed point, I've to show that for every smooth function $F: B^n \to \mathbb S^{n-1}$, with $B^n$ the unit ball in $\mathbb R^n$, such that $F(x) = x$ for all $x \in \mathbb S^{n-1} \subset B^n$, we have $d(F^* \circ \iota^*)(\alpha) = 0$ where $\iota: \mathbb S^n \to \mathbb R^n$ is the inclusion map. Since the exterior derivative commute with the pullback, I find
$$d(F^* \circ \iota^*)(\alpha) = (F^* \circ \iota^*)(d\alpha)= n (F^* \circ \iota^*)(\Omega).$$
Now here is my problem: As $\Omega$ is an $n$-form on $\mathbb R^n$, $\iota ^* \Omega$ is an $n$-form on $\mathbb S^{n-1}$ which is zero because the dimension $\mathbb S^{n-1}$ is $n-1$.
Is this argument correct ? It seems too easy, I didn't use anywhere the fact $F|_{\mathbb S^{n-1}} = \text{Id}_{\mathbb S^{n-1}}$. More generally, if we've $N \subset M$ with $\dim N = n < \dim M = m$, does the pullback by the inclusion map of a $k$-form (for $k > n$) always vanishes ?
Best Answer
You use the fact that $F\big|_{S^{n-1}} = \text{id}$ when you finish the proof and apply Stokes's Theorem. $\displaystyle\int_{S^{n-1}} \alpha\ne 0$ by inspection, and yet $$\int_{S^{n-1}} \alpha = \int_{S^{n-1}} F^*\alpha = \int_{D^n} dF^*\alpha = \int_{D^n} F^*(d\alpha) = 0,$$ because of your observation. (I've omitted the $\iota^*$'s, but you can put them in.)