# Volume of a manifold

Throughout this post, I am presuming $M$ to be an $2$-dimensional manifold that is parametrized by one chart $\varphi$, and I presume $\omega$ be a $2$-form on $M$.

Apparently, there is no natural way to define the volume of a manifold, if it's not a pseudo-Riemannian manifold - i.e., we don't have a metric on it of some kind. Here is a question I have, based on that:

1

My "argument" against this a few days ago was this - why does it not make sense to define the volume as $\int_M 1 dx_1 \wedge .. \wedge dx_n$?

To this I got the reply, that I can "scale" that form, and it still makes perfect sense to define that as volume - so it's arbitrary to choose $1 dx_1 \wedge .. \wedge dx_n$ as opposed to $2 dx_1 \wedge .. \wedge dx_n$. I didn't undestand this argument, but I think I do now, and I'd like to make sure I undestand it correctly.

I think the issue here stems from my previous understanding of $dx_i$ primarily as symbols, as opposed to actual linear maps from the tangent space, that are induced by some map. I thought $1 dx_1 \wedge .. \wedge dx_n$ is the always the same thing as $1 dy_1 \wedge .. \wedge dy_n$, that it's just a different notation of the coordinates - but I see now, that in some sense, that's not the case.

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As a linear map from the tangent space, the expression $1 dx_1 \wedge .. \wedge dx_n$ by itself doesn't really make sense on it's own, and it's not something I can integrate on $M$ - first, I need to pick a certain chart, to know what $dx_i$ actually are - and this is arbitrary.

More formally:

If for a chart $\varphi$ I denote the 'induced' forms $dx_i$ as those forms, for which $dx_i(\frac{\partial }{\partial x_i})=1$, where $(\frac{\partial }{\partial x_i})$ is the tangent vector induced by $\varphi$. I take a "scaled" version of $\varphi$ and call this new chart $\varphi'$. If I now consider two differential forms :

$\omega = 1 dx_1 \wedge .. \wedge dx_n$ (where $dx_i$ are induced by $\varphi$)

$\omega'= 1 dx_1' \wedge .. \wedge dx_n'$ (where $dx_i'$ are induced by $\varphi'$),

it's actually true that $\omega' = \alpha dx_1 \wedge .. \wedge dx_n$ , for some $\alpha$.

Thus, their integrals will give something different, depending on αα, and choosing one as opposed to the other is arbitrary, and so it doesn't make sense to prefer one as opposed to the other for defining volume.

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Is my answer to 1 correct?

2

Based on all of this, does it then make sense to think of differential forms as some type of "measure", that gives me volumes of tangent vectors, parallelograms formed by a choice of two tangent vectors, or generalizations of the previous?

Essentially I found that your answer is correct: $dx_1\wedge...\wedge dx_n$ depend by the parametrisation chosen, as in your case, where $\phi$ and $\phi'$ give different differential forms.

2: In general differential forms are not "measures", because they are not always positive. For example there is no $1$-form on $\mathbb{R}^2$ which give you the length of a curves.

But you could think to $n$-form as "functional" on differentiable $n$-subvarieties with satisfies a supplementary condition: you could describe it locally as a alternate form on $n$ copies of the vector space.

So, $1$-forms could be integrated on curves, $2$-forms on surfaces, etc.

For example, the integral $\int_a^bf(x) dx$ could be seen as integrating the differential form $\omega=f(x) dx$ on the interval $[a,b]$, which is a curve of $\mathbb{R}$.

Anyway there is one case where you could consider a differential form as a measure: when $\omega$ is a $n$-differential form on a $n$-dimensional variety, so you could measure a subvarieties $V$ of dimension $n$ calculating $|\int_V \omega|$

Personally, I've found these notes pretty useful in understanding differential forms: http://www.math.uqam.ca/~powell/Bachman_Geometric_Approach_to_Differential_Forms.pdf