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It is often said that differential forms (sections of an exterior power of the cotangent bundle) are the things that you can integrate. But unless I'm being thoroughly dense differential forms are not the only things that you can integrate, c.f. the arclength form (on a 2d manifold) $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms $|f(x,y)dx+g(x,y)dy|$, or the unsigned area forms $|h(x,y)dx\wedge dy|$.

My question is:

Where do the arclength form $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms |f(x,y)dx+g(x,y)dy|, and the unsigned area forms $|h(x,y)dx\wedge dy|$ live relative to the differentials $dx$ and $dy$, which I understand to live in the cotangent bundle of some 2-dimensional manifold?

Vladimir Sotirov
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    That's funny; I thought measurable functions were the things you can integrate... – Qiaochu Yuan Aug 22 '10 at 21:36
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    @Qiaochu: evidently, there's more than one kind of thing you can integrate. – Pete L. Clark Aug 22 '10 at 22:11
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    The notation used in the right hand side of «$ds=\sqrt{dx^2+dy^2}$» is just a notation; in particular, it is not something that is built out of $dx$ and $dy$... – Mariano Suárez-Álvarez Aug 23 '10 at 02:00
  • @Mariano, I understand ds as a continuous function on the tangent space at a point. My limited understanding tells me that it is a non-linear form because c ds(v)=ds(c v) for positive constants c. I suspect that if you apply 'positively' homogeneous function of degree 1 in n variables to (dx_1, dx_2,..., dx_n), you would get a form. – Vladimir Sotirov Aug 24 '10 at 18:22
  • I guess these non-linear forms are taken from David Bachmann's book "A Geometric Approach to Differential Forms", aren't they? – shuhalo Feb 22 '12 at 09:57
  • @MarianoSuárez-Álvarez : It's not just notation. The simplest way (IMO) to see how $ds$ is built out of $dx$ and $dy$ is to think of the latter as operators on tangent vectors. If $x$ and $y$ are the standard coordinates on $\mathbb{R}^2$, then $dx$ and $dy$ map a tangent vector to its first and second components. If you take these components, square them, add them, then take the square root, you get the (Euclidean) norm of the vector. And so that's what $ds=\sqrt{dx^2+dy^2}$ is, as an operator on vectors. That seems to be what Vladimir's comment is saying too. – Toby Bartels Dec 04 '19 at 03:29
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    There is a systematic way to integrate such a thing (an arbitrary expression built out of $dx$ and $dy$) along an oriented curve: divide the curve into pieces, evaluate the expression along the vector from each endpoint to the next, add these up, and take a limit as the sizes of the pieces approach zero in an appropriate sense. This integrates any ordinary linear $1$-form correctly, but it also integrates $ds$ correctly, to get the arclength of the curve (if it is rectifiable, or infinity if it is not). So it really is $ds$. – Toby Bartels Dec 04 '19 at 03:39

2 Answers2

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The answer to "what kinds of things can you integrate" depends on the context.

  • Measurable functions are things you can integrate over measure spaces, which includes in particular measurable subsets of R^n.
  • Differential forms are things you can integrate over oriented smooth manifolds -- the key thing about them is that their integrals are invariant under smooth, orientation-preserving changes of coordinates.
  • Densities are things that can be integrated in a coordinate-independent way on any smooth manifold, regardless of whether it has an orientation or not.
  • Coming full circle, every Riemannian manifold (i.e., smooth manifold endowed with a Riemannian metric) has a naturally-defined density dV, so in that context you can integrate measurable functions again: the integral of the function f is defined to be the integral of the density f dV.

All three of the expressions you asked about are examples of densities. For details, see my book Introduction to Smooth Manifolds, pp. 375-382.

Jack Lee
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  • If you look up densities on manifolds, then the first thing that you find are, essentially, top-rank pseudo-forms, which only covers the last of the OP's three examples. If you look further, you usually find weighted $s$-densities, which don't help here. You have to instead find the $k$-densities of Gelfand and Gindikin, according to which an area form is a $2$-density and the other two examples are $1$-densities. Hopefully these are in your book, or see https://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714 – Toby Bartels Dec 04 '19 at 15:32
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In my opinion, you're looking for the notion of a cogerm.

If I understand correctly, the fact that such things act on paths (and not just vectors) allows for "higher order" forms like $d^2 x$, and the fact that such things aren't assumed linear allows for "non-linear" forms like $ds := \sqrt{dx^2+dy^2}$. And yes, there is indeed a notion of integration for such forms; see the link.

goblin GONE
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  • I of course like the theory of cogerm forms, but that's probably overkill in this case, and at the same time inadequate (since the example with the wedge product is not a cogerm form, although it is a coflare form). A better nLab page would be the one on absolute differential forms, which include all of the OP's examples. And these are really just a very explicit way of looking at the densities in Jack Lee's answer. – Toby Bartels Dec 04 '19 at 03:43
  • @TobyBartels, honestly, I don't really know what I'm talking about here. Why don't you write an answer explaining why cogerms aren't sufficiently general and why absolute differential forms are really the correct answer. In fact I think this would be a good opportunity to explain the basics of the theory, the big picture of what kind of thing can be integrated over what kind of thing, and also to include some references so that people (like myself) who wish to learn more can proceed to do so. I'll also remark that the failure to treat these objects in graduate math programs (at least in ... – goblin GONE Dec 04 '19 at 09:25
  • ... Australia where I was educated) is imo a major deficiency in present-day higher math education. I'm also very interested (and confused about) the relationship between measure theory and various notions of (generalized) differential form. In particular, since measures push forward and forms pull back, so when probability theorists say that $dx$ can be viewed as the Lebesgue measure and when differential topologists say that $dx$ can be viewed as the exterior derivative of a projection map, they can't both be right. I think it would be great if you could briefly sketch an explanation. – goblin GONE Dec 04 '19 at 09:25
  • They are both right. You may object that Lebesgue measure is a function from measurable sets to extended real numbers, while the exterior derivative of a projection map (or to keep things in one dimension for simplicity, of the identity map) is not. But this is like saying that a real number can't be both a Dedekind cut and an equivalence class of Cauchy sequences. They're two different formalizations of the same underlying concept. (And I don't know about other countries, but American mathematicians don't usually learn this stuff either.) – Toby Bartels Dec 04 '19 at 13:32
  • (Also, identifying Lebesgue measure with $dx$ uses the standard orientation on the real line, so I usually prefer to say that it's the absolute value, $|dx|$. They also both rely on the convention that $x$ is the variable for a real number; we're really talking about the identity function in both cases.) – Toby Bartels Dec 04 '19 at 13:33
  • Anyway, I don't want to write a new answer, because Jack's is correct, although I do want to make one comment on it to ensure that it's sufficiently general. Your answer is also correct for most of the OP's examples (and correct for all of them if you change cogerms to coflares, but there isn't an nLab page on coflare forms yet), and I voted it up too. (It's also nice because, out of all the concepts in the answers, cogerm forms are the only one that also includes things like $dx^2$ that you use on the way to constructing $ds$.) – Toby Bartels Dec 04 '19 at 13:47
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    The OP's last example is not a cogerm form because an area form takes two vectors as input instead of just one, and so you integrate it on a surface instead of a curve. Cogerm forms include cojet forms, which you can think of as taking multiple vectors as input; in particular, a co-$2$-jet form takes two vectors as input. But these two vectors are interpreted as velocity and acceleration, and you integrate such a form (like any cogerm form) along a curve. In contrast, an area form takes two independent velocity vectors (or the oriented parallelogram that they span) as input. – Toby Bartels Dec 04 '19 at 15:52
  • One more thing: why measures push forward but forms pull back. Instead of two ways to define real numbers, a better example is two ways to define subsets. A subset of $X$ is a function from $X$ to $2$; these pull back (giving the preimage). And a subset of $X$ is so an equivalence class on injective functions to $X$; these push forward (giving the image). Both are correct. – Toby Bartels Dec 04 '19 at 16:06
  • The situation here is subtler. If you want to pull back an absolutely continuous measure by turning it into a form and pulling that back, then you need the map to be smooth (not just measurable), and the spaces must have the same dimension (so that the pulled back form has the correct rank). Similarly, if you want to push forward a smooth top-rank pseudo-form by turning it into a measure and pushing that forward, then you need to ensure that the pushed-forward measure is absolutely continuous and Radon. I'm not sure, but I think that the requirement is a submersion with finite fibres. – Toby Bartels Dec 04 '19 at 16:41