I Vector valued integrals in the theory of differential forms

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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##)?

[EDITED; I first typed div(rho) instead of grad(rho) by accident]
 
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Orodruin

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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.
 
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pasmith

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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 [tex]
\Lambda^0 \to \Lambda^3 : f \mapsto f\,dx \wedge dy \wedge dz[/tex] 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 [itex]\rho \mathbf{c}[/itex]. Because of the special way in which calculus in [itex]\mathbb{R}^3[/itex] works, we can say that [itex]\mathbf{c}[/itex] is an arbitrary "constant" vector field, which means that (1) its divergence is zero, so that [itex]\nabla \cdot (\rho \mathbf{x}) = \mathbf{c}\cdot \nabla \rho[/itex], and (2) taking the dot product with [itex]\mathbf{c}[/itex] commutes with integration (over both [itex]V[/itex] and [itex]\partial V[/itex]). Then, because [itex]\mathbf{c}[/itex] was arbitrary, we could let it be each standard basis vector in turn to conclude that the components with respect to that basis of [itex]\int_{\partial V} \rho\,d\mathbf{S}[/itex] and [itex]\int_V \nabla \rho\,dV[/itex] were equal, allowing us to cancel [itex]\mathbf{c}[/itex].

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 [itex]\mathbb{R}^3[/itex] 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.
 

fresh_42

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

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