# Is there a Stokes theorem for covariant derivatives?

A $V$-valued differential form on $M$ is a smooth map $\omega : TM \to V$ such that $\omega$ restricted to any tangent space $T_p M$ is an element of the $V$-valued exterior algebra $\Lambda^n (T_p M, V)$ of $T_p M$. That is, the restriction $\omega_p$ is a completely antisymmetric map $\omega_p : T_p M \times T_p M \times \cdots \times T_p M \to V$.

In some situations (I forgot which ones), one can define a flat covariant derivative $\mathrm{d}$ which is just the exterior derivative. It fulfills the Stokes theorem.

For a $V$-valued differential form $\omega$, there is also a covariant derivative ( principal connection) that acts on all $W$-valued differential forms $\phi$ where there is a representation $R$ of $V$ on $W$ by the formula $\mathrm{d}_\omega \phi := \mathrm{d} \phi + \omega \wedge_R \phi$.

Is there a Stokes theorem for $\mathrm{d}_\omega$? Maybe something like $\int_M \mathrm{d}_\omega \phi = \int_{\partial M} \phi$ up to terms proportional to the curvature of $\mathrm{d}_\omega$?

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I would like to tag this question with "principal-connection" and "stokes-theorem", which are new tags. If there is someone with high reputation who finds this appropriate, could he/she tag my question accordingly? –  Turion Jul 5 '11 at 9:27
I think these two tags are far too specific to be of real use. –  t.b. Jul 5 '11 at 9:32
There are $\sim 10^1$ questions about Stokes theorem, I think that justifies a tag. –  Turion Jul 5 '11 at 9:37
As I said, I think it's too specific and (differential-forms) is a good enough match in my opinion, so I'm not going to create this tag. But if someone else wants to, feel free to do so. –  t.b. Jul 5 '11 at 9:51
Very interesting question, BTW. The usual generalization of Stokes' theorem to the case of a principal connection is in the context of non-Abelian gauge theories. The generalization, however, is to generalize the relation between holonomy and curvature. (For example, this.) If you don't get an answer here after a few days, I encourage to ask on MathOverflow. –  Willie Wong Jul 5 '11 at 12:03

I am no mathematician, but I have studied differential forms and covariant derivatives enough that I think I have a good foundation on the subject. I see a vector (or p-form) in curved space as a tensor with a basis that changes from point to point. Therefore, when you take the derivative, you should not just take the derivative of the tensor, but apply the product rule and take the derivative of the basis as well. This leads to the Misner, Thorn and Wheeler definition of the Christoffel symbols: $\nabla_ie_j=\Gamma^k_{ji}e_k$ (with $e_i$ and $e_k$ being basis vectors). One can define the exterior covariant derivative as the exterior derivative plus the product rule applied to the basis. So the exterior covariant derivative acts on magnitudes and directions (or dual directions) while the exterior derivative just acts on magnitudes. The exterior covariant derivative reduces to the covariant derivative when the bases are constant.