Can a k-Form Be Integrated Over Lower Dimensional Manifolds?

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davi2686
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i only can integrate a k-form in a n-dimensional manifold, if k=n right?
 
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davi2686 said:
i only can integrate a k-form in a n-dimensional manifold, if k=n right?

I think you can integrate a k-form in n-dim if k<=n but its not guaranteed:

or some smooth functions fi on U.

The second idea leading to differential forms arises from the following question: given a differential 1-form α on U, when does there exist a function f on U such that α = df? The above expansion reduces this question to the search for a function f whose partial derivatives ∂f / ∂xi are equal to n given functions fi. For n > 1, such a function does not always exist: any smooth function f satisfies

[itex]\frac{\partial^2 f}{\partial x^i \, \partial x^j} = \frac{\partial^2 f}{\partial x^j \, \partial x^i} ,[/itex]

so it will be impossible to find such an f unless

[itex]\frac{\partial f_j}{\partial x^i} - \frac{\partial f_i}{\partial x^j}=0.[/itex]

for all i and j.

from the wikipedia article:

http://en.wikipedia.org/wiki/Differential_form
 
You can integrate k-form on k dimensional manifold. You can have k-form on n dimensional manifold, where n>k. This can be integrated on k-dimensional submanifold of original manifold. For example you can integrate 2-form on a surface in three dimensional space.
 
are the integral of a 2-form associate with a vector field the same thing to surface integral of that vector field?
 
The integral of a k form is defined on a smooth k chain. A smooth k chain is a formal algebraic sum of smooth oriented k simplices.
An oriented k manifold can be expressed as a smooth k chain so integration of k forms is defined. Not so for an unorientable k manifold. It can not be expressed as a smooth k chain.

Sometimes an k form can be integrated over lower dimensional manifolds. The result is a lower dimensional differential form. For example integration along the fibers of a fiber bundle reduces the dimension of the form by the dimension of the fiber.