Interior Product of the normal with the volumeform on the euclidean sphere

Question

Let $n:S^{n-1}\rightarrow \mathbb{R}^n$ be the outer normal $n(x)=x$, and $\alpha:=n\lrcorner (dx_1\wedge \dots dx_n)$ a $(n-1)$-form on $S^{n-1}$.

I want to proof that $\int_{S^{n-1}} \alpha >0$.

It is a step in a proof of Brouwer's fixed point theorem with differential function and with the use of Stokes theorem. All is really logical, i just cant see why that form is a positive meassure, in particular why the integral over the sphere is bigger than 0. Would appreciate any hint!

Answer 1

Hint: As you defined $n$, it makes sense on all of $\mathbb R^n$ and inserting this into $dx^1\wedge\dots\wedge dx^n$ you get an extension of $\alpha$ to a $(n-1)$-form defined on all of $\mathbb R^n$. Then you can apply Stokes to compute your integral as the integral of $d\alpha$ over the unit ball. But $d\alpha$ can be easily computed directly.

Answer 2

$dx_1\wedge\dots\wedge dx_n$ is a (positive) volume n-form of the ambient space which induces a (positive) volume $(n-1)$-form on $S^{n-1}$ using its unit normal vector field $\nu$, $\alpha:=\nu\lrcorner (dx_1\wedge \dots dx_n)$ and the volume (the hyper-surface "area") of the sphere is $\int_{S^{n-1}} \alpha > 0$.