23
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ That's funny; I thought measurable functions were the things you can integrate... $\endgroup$ Aug 22, 2010 at 21:36
  • 7
    $\begingroup$ @Qiaochu: evidently, there's more than one kind of thing you can integrate. $\endgroup$ Aug 22, 2010 at 22:11
  • 1
    $\begingroup$ 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 $dy$... $\endgroup$ Aug 23, 2010 at 2:00
  • $\begingroup$ @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. $\endgroup$ Aug 24, 2010 at 18:22
  • 2
    $\begingroup$ 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 $ds$. $\endgroup$ Dec 4, 2019 at 3:39

2 Answers 2

29
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 mathoverflow.net/questions/90455/… $\endgroup$ Dec 4, 2019 at 15:32
2
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2019 at 3:43
  • $\begingroup$ @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 ... $\endgroup$ Dec 4, 2019 at 9:25
  • $\begingroup$ ... 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. $\endgroup$ Dec 4, 2019 at 9:25
  • $\begingroup$ 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.) $\endgroup$ Dec 4, 2019 at 13:32
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 4, 2019 at 15:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .