# Integration and Stokes' theorem for vector bundle-valued differential forms?

Is there a version of Stokes' theorem for vector bundle-valued (or just vector-valued) differential forms?

Concretely: Let $E \rightarrow M$ be a smooth vector bundle over an $n$-manifold $M$ equipped with a connection. First of all, is there an $E$-valued integration defined on the space $\Omega^{n-1}(M,E)$ of smooth sections of $E\otimes \Lambda^{n-1}T^\ast M$ mimicking what you have for $\mathbb{R}$-valued forms? If so, does

$\int_{\partial M} \omega = \int_M d\omega$

hold (or even make sense) for $\omega\in\Omega^{n-1}(M,E)$ if $d$ is the exterior derivative coming from our connection on $E$?

In this setting, letting $E$ be the trivial bundle $M\times \mathbb{R}$ should give the ordinary integral and Stokes' theorem.

Sorry if the answer is too obvious; I just don't have any of my textbooks available at the moment, and have never thought about Stokes' theorem for anything other than scalar-valued forms before.

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You want a connection on $E$ as well, right? Otherwise $d\omega$ doesn't make sense, see e.g. staff.science.uu.nl/~ban00101/anman2009/lectures-10-11-12.pdf. Since connections are locally trivial and vector-valued one-forms can locally be written as vector of ordinary one-forms, I think the usual proof of Stokes by localizing with a partition of unity and treating two basic cases ($\mathbb{R}^n$ and a half plane) should work. –  skupers Aug 12 '10 at 10:32
Thanks skupers. Edited to include a connection on $M$. Also I think the rest of your comment makes sense and reassures me that everything is OK. –  gspr Aug 12 '10 at 10:39
Oh, and I meant of course "coming from our connection on $E$", not $M$. Corrected. General MO question: Is it customary to add comments (such as what I just wrote) when one edits, or should the edit list speak for itself? –  gspr Aug 12 '10 at 10:53

For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.
For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.
What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b).$$ So Stokes theorem gives $$\int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b.$$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.
TeX confusion: you wrote \mathbb{R^k} which renders as $\mathbb{R^k}$ instead of \mathbb{R}^k $\mathbb{R}^k$. –  Nate Eldredge Aug 12 '10 at 17:47