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From Munkres:

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As I read this definition, I was wondering what it would mean if we were integrating a 0-form (a scalar function). According to this definition, it seem that if $A$ is open in $\mathbb R^k$, then we need $\omega$ to be a $k$-form. So are we even able to integrate 0-forms over $k$-dimensional manifolds, using this definition? I know that that Munkres defines the integral of a scalar function over a $k$-dimensional parametrised-manifold separately - prior to discussing manifolds - but I would think that this definition is a generalisation of the scalar variant, and should therefore match up in the case we integrate a 0-form.

I understand that only a $k$-form will actually yield a regular integral if $A\subset\mathbb R^k$, but I don't see how we would integrate an $r$-form over a $k$-dimensional manifold, when $r\neq k$ (if that's even possible at all). Could someone clarify?

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You cannot integrate an $r$-form $\omega$ over a $k$-dimensional (oriented) manifold if $r\neq k$. This includes the case $r=0$.

However, if $\xi$ is any $(k-r)$-form on an open set containing $Y$, then $\omega\wedge\xi$ is a $k$-form, and so you can integrate it. This is how you integrate functions ($0$-forms) on $\mathbb{R}^k$: if $A\subseteq \mathbb{R}^k$ is open and $f$ is a smooth function on $A$, let $\xi=dx_1\wedge dx_2\wedge\dots\wedge dx_k$, where $x_1,\dots,x_k$ are the coordinates of $\mathbb{R}^k$. Then $\int_A f\wedge\xi$ is the usual integral of $f$ as a function (with respect to Lebesgue measure). (Note that $f\wedge\xi$ is usually just written as $f\xi$, since $f$ is scalar-valued and the wedge product in this case is just the pointwise scalar product.)

In general, there is not any natural way to integrate a function on a $k$-manifold (if $k>0$). It is only possible to integrate functions by first choosing a $k$-form $\xi$ as above (a "volume form" to integrate against). On $\mathbb{R}^k$ there is a canonical choice coming from the coordinates, and more generally if you have a Riemannian metric (and orientation) on your manifold there is a canonical choice by replacing the coordinate functions with a positively oriented orthonormal frame. But if you have no structure besides a smooth manifold, there is no canonical choice.

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  • $\begingroup$ (+1) O right, I didn't realise we should think of it as "volume forms", but I understand what you mean! Thanks for the clarification, this helps a lot! $\endgroup$ – Sha Vuklia Apr 29 '18 at 11:15

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