Two different Stokes' theorems

Question

There are two different Stokes' theorems. One is used typically in analysis and can be stated as follows: for each $u\in C^1(\bar{U})$ ,with $U\subset \Bbb{R}^n$ with boundary $C^1$,and $U$ bounded we have $$\int_{U} \operatorname{div} \mathbf{u}\, d x=\int_{\partial U} \mathbf{u} \cdot \nu \,d S$$ where $\nu$ takes the outer normal direction of $U$.

Another Stokes' theorem typically appears in differential geometry and says

let $M$ be an oriented smooth $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$ \int_{M} d \omega=\int_{\partial M} \omega $$

Here are some questions about them:

1. when we consider $U\subset \Bbb{R}^n$ as manifold,there is no notion of manifold boundary if $U$ is open, why we can consider the "boundary" here?

2. the second theorem needs compact supported condition, whereas the first one does not need. What's the difference between them?

3. Can we remove the bounded condition for $U$ in theorem 1?

Answer 1

1. To apply Stokes, you cannot just have an $\omega$ that's defined only over $U$, instead $\omega$ has to be extended to $\partial U$ (at least $C^1$-ly), so the boundary is implicitly there. In practice, when we integrate an explicitly expressed function, it's usually defined almost everywhere. For example, integerate div $\textbf{u}$ on the open unit ball where $\textbf{u} = (x, y, z)$. We can instead integrate $\textbf{u}$ on the unit sphere. Here we are implicitly using the fact that the vector field $\textbf{u}$ can be extended to the sphere by the same formula.

2. Compactly supported is not the best condition, but some condition to make the integrals finite is necessary. The first version can be false if no such condition is posed. In calculus, we usually only apply Stokes when $\overline{U}$ is bounded, therefore compact. If we remove the boundedness, we have to then assume something like $\omega$ has compact support like in the second version, which is a weaker condition than $\overline{U}$ is bounded.

Answer 2

1. when we consider $U \subset \mathbb{R}^n$ as manifold,there is no notion of manifold boundary if $U$ is open, why we can consider the "boundary" here?

A "manifold with boundary" is not a particular type of manifold, it's a generalization. Instead of a set where every point has a neighborhood diffeomorphic to a ball $\{x:|x|<1\} \subset \mathbb{R}^n$, a manifold with boundary is a set where every point has a neighborhood diffeomorphic either to a ball $\{x:|x|<1\}$ or to a half-ball $\{x:|x|<1 \land \langle x, e_1 \rangle \geq 0\}$. So a manifold in the usual sense can be considered a "manifold with boundary" which has the empty set as boundary. (This implies that every exact differential form $d \omega$ on a manifold $M$ has $\int_M d \omega = \int_{\emptyset} \omega = 0$.)

2. the second theorem needs compact supported condition, whereas the first one does not need. What's the difference between them?

Your first stated theorem is the special case of the second version where $\omega$ is a $1$-form and $M \subset \mathbb{R^n}$.

A vector field on $U \subset \mathbb{R^n}$ is essentially a function ${\mathbf u}: U \to \mathbb{R^n}$. We associate this $\mathbf u$ with the $1$-form $\omega: U \to \mathbb{R}$ by $$\omega(x)(v) = \langle{\mathbf u}(x), v\rangle$$ Then it will work out that $d \omega = \mathrm{div}\, {\mathbf u}\, dx$ and $\int_{\partial U} \omega = \int_{\partial U} {\mathbf u} \cdot v\, dS$.

The conditions for the two versions are exactly the same, given that the first statement makes the domain $U$ a subset of $\mathbb{R}^n$. In $\mathbb{R}^n$ with the standard topology and metric, a set is compact if and only if the set is both closed and bounded. The second version doesn't require that the manifold is embedded in any particular $\mathbb{R}^{N}$, but since the first version is in the context of $\mathbb{R}^n$, it can restate "compact support" as the more familiar terms "closed" and "bounded".