# Vector valued integrals in the theory of differential forms

• I
• Hiero
In summary: Copied from the comments.)In summary, the conversation discusses the basics of differential forms and their integration over different regions. It also touches on the divergence theorem and the generalized Stokes theorem, and how they relate to differential forms. The conversation concludes by mentioning that while scalar quantities are naturally regarded as 0-forms, there exist isomorphic relations between different forms.
Hiero
TL;DR Summary
I’m informally learning the basics of differential forms from various sources. If anyone has book recommendations please share!
So I heard a k-form is an object (function of k vectors) integrated over a k-dimensional region to yield a number. Well what about integrals like pressure (0-form?)over a surface to yield a vector? Or the integral of gradient (1-form) over a volume to yield a vector?

In particular I’m thinking of this variation of the divergence theorem:
$$\int_{\partial V} \rho d\vec A = \int_V (\nabla \rho )dV$$
Which seems to fit so perfectly into the generalized stokes theorem:
$$\int_{\partial V} \rho = \int_V d\rho$$
Except that it seems to go against what a couple sources said; 1-forms (like ##d\rho = \nabla \rho##) are supposed to be integrated over lines, not volumes.

Another random question; can physical things be given an absolute classification as a k-form? E.g. is pressure always a 0-form? Or can it also be viewed as a 3-form (on ##R^3##)?

Last edited:
In a general manifold, expressions that are integrals of vectors do not make much sense due to the fact that you cannot add vectors from different points as they belong to different tangent spaces.

dextercioby
Hiero said:
Summary: I’m informally learning the basics of differential forms from various sources. If anyone has book recommendations please share!

So I heard a k-form is an object (function of k vectors) integrated over a k-dimensional region to yield a number. Well what about integrals like pressure (0-form?)over a surface to yield a vector? Or the integral of gradient (1-form) over a volume to yield a vector?

In particular I’m thinking of this variation of the divergence theorem:
$$\int_{\partial V} \rho d\vec A = \int_V (\nabla \rho )dV$$
Which seems to fit so perfectly into the generalized stokes theorem:
$$\int_{\partial V} \rho = \int_V d\rho$$
Except that it seems to go against what a couple sources said; 1-forms (like ##d\rho = \nabla \rho##) are supposed to be integrated over lines, not volumes.

Another random question; can physical things be given an absolute classification as a k-form? E.g. is pressure always a 0-form? Or can it also be viewed as a 3-form (on ##R^3##)?

Scalar quantities are naturally regarded as 0-forms. However there is the obvious isomorphism $$\Lambda^0 \to \Lambda^3 : f \mapsto f\,dx \wedge dy \wedge dz$$ by which an 0-form can be replaced by a 3-form if you want to integrate it over a volume.

Similarly there is an isomorphism between 1-forms (integrated over curves) and 2-forms (integrated over surfaces). In general, k-forms and (n-k)-forms are isomorphic. (For completeness one should note that some vector quantities are more naturally regarded as tangent vectors rather than 1- or 2-forms. Again, there exist isomorphisms.)

When you calculate the curl of a vector field, you calculate the exterior derivative of a 1-form, which gives a 2-form. When you calculate the divergence, you calculate the exterior derivative of a 2-form, which gives a 3-form.

With regard to the variation of the divergence theorem, what you are actually applying the divergence theorem to is the 2-form isomorphic to $\rho \mathbf{c}$. Because of the special way in which calculus in $\mathbb{R}^3$ works, we can say that $\mathbf{c}$ is an arbitrary "constant" vector field, which means that (1) its divergence is zero, so that $\nabla \cdot (\rho \mathbf{x}) = \mathbf{c}\cdot \nabla \rho$, and (2) taking the dot product with $\mathbf{c}$ commutes with integration (over both $V$ and $\partial V$). Then, because $\mathbf{c}$ was arbitrary, we could let it be each standard basis vector in turn to conclude that the components with respect to that basis of $\int_{\partial V} \rho\,d\mathbf{S}$ and $\int_V \nabla \rho\,dV$ were equal, allowing us to cancel $\mathbf{c}$.

Most of these steps are not possible in integrating over an arbitrary manifold, because the underlying vector space structure does not exist. Also, since in $\mathbb{R}^3$ we by definition integrate vector fields componentwise with respect to the standard basis, we can get away with writing a single vector equation where in terms of forms we would need three separate equations which, without that underlying standard basis, would have nothing to connect them.

pasmith said:
Scalar quantities are naturally regarded as 0-forms. However there is the obvious isomorphism $$\Lambda^0 \to \Lambda^3 : f \mapsto f\,dx \wedge dy \wedge dz$$ by which an 0-form can be replaced by a 3-form if you want to integrate it over a volume.
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Monomorphisms, not isomorphisms!

## 1. What is a vector valued integral?

A vector valued integral is an integral that involves a vector-valued function as its integrand. This means that the function being integrated returns a vector instead of a scalar value.

## 2. How is a vector valued integral different from a regular integral?

A regular integral involves a scalar function as its integrand, while a vector valued integral involves a vector-valued function. This means that the result of a vector valued integral will also be a vector, while the result of a regular integral will be a scalar.

## 3. What is the significance of vector valued integrals in the theory of differential forms?

Vector valued integrals play a crucial role in the theory of differential forms as they allow for the integration of vector fields, which are essential in the study of differential forms. Vector fields represent the direction and magnitude of change at each point in a space, making them useful in understanding the behavior of differential forms.

## 4. How are vector valued integrals related to line integrals?

Line integrals are a special case of vector valued integrals, where the vector-valued function being integrated is a vector field along a given curve. This means that line integrals can be seen as a specific type of vector valued integral, with the added constraint of being integrated along a curve.

## 5. What are some applications of vector valued integrals in real life?

Vector valued integrals have various applications in fields such as physics, engineering, and economics. For example, they are used to calculate work done by a force field, electric and magnetic flux, and fluid flow. They are also used in optimization problems and in the study of vector calculus and differential equations.

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