Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
You want a connection on $E$ as well, right? Otherwise $d\omega$ doesn't make sense, see e.g. 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

2 Answers 2

up vote 28 down vote accepted

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, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.

share|cite|improve this answer
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
@Nate - eagle eyes! I'm so used to my own macros that I normally blunder when having to type out the whole of \mathbb. Anyhow, I've changed it now. –  Joel Fine Aug 12 '10 at 18:38
Thanks for the clarifying answer, Joel Fine! –  gspr Aug 13 '10 at 9:55

After looking at this question for a few days in the context of the Riemann curvature tensor, holonomy for a given affine connection, and the (false) conjecture that the parallel transport around the boundary curve could equal the integral of the Riemann tensor within the span of the closed curve, I've concluded that the Stokes theorem cannot be applied to this conjecture except when the connection is flat.

The reason for the failure of the conjectured relation between curvature and parallel transport is that the Stokes theorem's integrals are themselves not really well defined. But it is not that simple. I'll explain...

When even constructing a simple Riemann integral from the fundamentals, one has to add vectors at different points inside the region. Even if you have a connection, you have to decide which paths to use to connect the points of the region. You can do a kind of a "raster scan" of the image of a rectangular region of $\mathbb{R}^2$, parallel transporting the vectors back to the left of the scan to add them on the left hand side, and then you can parallel transport all of these X-scans down the Y-axis by transporting them down to the bottom left of the rectangle. But then what do you have? It's clearly not geometrically meaningful. And then you have the same problem with the boundary integral of a vector function.

A second conclusion which I came to is that if you do apply the Stokes theorem to this scenario, you get a mathematically correct identity, which has a practical value as the first iteration of the Picard iteration procedure to compute the parallel transport around the curve. This, clearly, is not extremely useful. But in my opinion, the Stokes theorem is applicable to this situation. It just doesn't give anything geometrically meaningful for a non-flat geometry, and it has limited value for relating curvature to parallel transport and holonomy. On the other hand, it does get the right answer in the limit of a shrinking region to a point, which gives the correct answer for the "Cartan characterization of curvature".

This issue is related to questions 16850 and 50051.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.