Consider the 1-form $\alpha = xdz + ydw -(x^2 + y^2 + z^2 + w^2)dt$ on $\mathbb{R}^5$.
I'm trying to find $\int_S d\alpha \wedge d\alpha$, where $S \subset \mathbb{R}^5$ is given by $x^2 + y^2 + z^2 + w^2 =1$ and $0\leq t \leq 1$.
Restricting to $S$,we get $\alpha = xdz + ydw -dt$, so $d\alpha = dx\wedge dz + dy \wedge dw$.
Now $d\alpha \wedge d\alpha$ = $d(\alpha \wedge d\alpha)$, so by Stokes' Theorem, $\int_S d\alpha \wedge d\alpha = \int_{\partial S} \alpha \wedge d\alpha$.
I found that $\alpha \wedge d\alpha = xdz\wedge dy \wedge dw + ydw\wedge dx \wedge dz – dt \wedge dx \wedge dz – dt \wedge dy \wedge dw$.
The boundary $\partial S$ of $S$ is the disjoint union $S^3 \times \{0\} \cup S^3 \times \{1\}$, where $S^3$ is the unit sphere in the $x,y,z,w$-subspace of $\mathbb{R}^5 = \{(x, y, z, w, t)\}$.
How do I integrate this 3-form $\alpha \wedge d\alpha$ over a boundary sphere of $S$?
(This isn't homework! I've never seen explicitly how to integrate differential forms over a manifold, so a worked example would be really helpful.)
Best Answer
Since $t$ is constant over each component of $\partial S$, $dt|_{\partial S}=0$. So it suffices to integrate $\eta=x \, dz \wedge dy \wedge dw + y \, dw \wedge dx \wedge dz$; the last two terms of $\alpha \wedge d\alpha$ vanish on $\partial S$.
But $\eta$ is independent of $t$, so it is identical on the two boundary components. Since whatever orientation you choose on $S$ will determine opposite orientations on the two boundary components, it follows that $\int_{\partial S} \eta = 0$.
So that's pretty boring. Let's integrate $\eta$ over each boundary component individually, just to see what happens. Notice that: $$d\eta = dx \wedge dz \wedge dy \wedge dw + dy \wedge dw \wedge dx \wedge dz = -2 dx \wedge dy \wedge dz \wedge dw \, ;$$ that is, $d\eta$ is a constant multiple of the volume form on each $x,y,z,w$-subspace of $\Bbb{R}^5$. So, applying Stokes' theorem again and recalling the formula for the volume of a hyperball, we have $$\int_{S^3 \times \{1\}} \eta = \int_{B^4 \times \{1\}} d\eta = -2 \mathrm{Vol}(B^4)= \pm \pi^2$$ depending on your original choice of orientation for $S$. You can do the same thing with $S^3 \times \{0\}$, except that you'll have to choose the opposite sign in order to be consistent.