To prove Stokes's theorem, why does it suffice to prove it for $\mathbb{R}^n$ and for $\mathcal{H}^n$?

126 Views Asked by At

In Tu's An Introduction to Manifolds, he states Stokes's theorem as:

For any smooth $(n-1)$-form $\omega$ with compact support on the oriented $n$-dimensional manifold $M$, $$\int_M d\omega = \int_{\partial M}\omega.$$

In his proof he chooses an atlas $\{(U_\alpha, \phi_\alpha)\}$ for $M$ in which each $U_\alpha$ is diffeomorphic to either $\mathbb{R}^n$ or $\mathcal{H}^n$ via an orientation-preserving diffeomorphism. He then says

Suppose Stokes's theorem holds for $\mathbb{R}^n$ and for $\mathcal{H}^n$. Then it holds for all the charts $U_\alpha$ in our atlas, which are diffeomorphic to $\mathbb{R}^n$ or $\mathcal{H}^n$.

I am having trouble seeing how Stokes's theorem applying to $\mathbb{R}^n$ and $\mathcal{H}^n$ implies it also holds for all the charts $U_\alpha$. Is he using the diffeomorphism and a pull back to make some sort of change of variables that preserves the integral?

2

There are 2 best solutions below

0
On BEST ANSWER

Let $F: \mathbb R^n \to U$ be a diffeomorphism. If you know that

$$\tag{1} \int_{\mathbb R^n} d\omega =0$$

for all compactly support smooth differential $n-1$ form $\omega$ on $\mathbb R^n$, show that

$$\tag{2} \int_U d\beta = 0$$ for all compactly support smooth differential $n-1$ form $\beta$ on $U$.

Sketch of proof: For any such $\beta$, $\omega = F^*\beta$ is a compactly support differential $(n-1)$ forms on $\mathbb R^n$. Use (1) on this $\omega$ to obtain (2).

Similar for $\mathbb H^n$.

0
On

You can sort of understand it as that the properties and theorems proved for the flat space $\mathbb R ^n$ is "transferred" on your manifold by the atlas in some way that the "important properties" , for example, the topological properties and some integration properties of differential forms (using the fact that the operator $\mathrm d$ commutes with the pullback), are remained unchanged (because of the compatibility of different coordinate maps).

Some similar theorem includes the Darboux' theorem in symplectic geometry which states that the properties proved in the flat symplectic space can be transferred on any symplectic manifold.

You can also use the Stokes' theorem of integration on regular chains to prove the Stokes' theorem of regular domains on a manifold. See GTM94 Chapter 4.