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The Stokes theorem states: $$\int_\mathcal M d\omega =\int_{\partial \mathcal M} \omega $$

If we have that $\mathcal M$ is a one dimensional manifold with two extreme points, like a closed interval of $\mathbb R$, and $d\omega$ is a one-form, how could the integral in the second term be done? Does that make sense? Because $\partial\mathcal M$ is just two isolated points. We are integrating a function to a single point? The only thing that makes some sense would be the value of the function at those points, and only because it coincides in a specific problem with what I should get by integrating $d\omega$ to the curve, but I don't see that clear at all.

I know this is actually what gives:

$$\int_a^b df=\left. f\right|_a^b$$

but I can't prove it.

MyUserIsThis
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    Think of Barrow's rule $\int_a^b f'(x)dx=f(b)-f(a).$ You can consider $\int_{\partial M}\omega$ as evaluating it at the end points of the interval. – mfl May 19 '14 at 22:39

1 Answers1

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You are correct that this integral is a special case. In Spivak's Calculus on Manifolds, he writes:

A special definition must be made for $k$ = $0$. A $0$-form $\omega$ is a function; if $c : \{ 0 \} \to A$ is a singular $0$-cube$^\ast$ in $A$ we define $$\int_c \omega = \omega(c(0))$$

Since the boundary of the interval $[a,b]$ (oriented from $a$ to $b$) is $b$ (positively oriented) and $a$ (negatively oriented), Stokes' theorem tells us that: $$ \int_{[a,b]} d\omega = \int_b \omega - \int_a \omega = \omega(b) - \omega(a) $$ as is expected from the Fundamental Theorem of Calculus.

$^\ast$ by $k$-cube, Spivak means "a continuous function $c : [0,1]^k \to A \subseteq \mathbb{R}^n$."

Henry Swanson
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