# Naive question about differential forms

1. May 16, 2010

### Manicwhale

In math, differential forms are alternating: dx^dy=-dy^dx. But in physics, we seem to exchange the order freely: dxdy=dydx. What's going on?

I am comfortable with an answer that involves tensors, differential geometry, physics, volume forms, etc. In fact, this is really something I should already know the answer to...

2. May 17, 2010

### haushofer

I think in the end it boils down to the fact that an integrand over an n-dimensional space can naturally be seen as an n-form. If you don't regard the integrand as such, you can just regard the measure d^n x as composed of commuting measures dx,dy,dz,...

3. May 17, 2010

### George Jones

Staff Emeritus
In physics, $dxdy$ often is shorthand (particularly in the context of symmetric "metric" tensors) for a symmetrized tensor product, e.g.,

$$dx^2 = dxdx = \frac{1}{2} \left(dx \otimes dx + dx \otimes dx \right) = dx \otimes dx$$

and

$$dxdy = \frac{1}{2} \left(dx \otimes dy + dy \otimes dx \right).$$

Last edited: May 17, 2010
4. May 17, 2010

### dx

This is not really a difference between physics and math. It's a difference between old math and new math. In old math, when you integrate something, like charge density, over a surface, you would write that as

$$\int_{S} \rho(x,y) dxdy$$

Where $$\rho(x,y)$$ is thought of as the density, and dxdy is thought of as the 'volume element'. How this changes between coordinate systems is encoded in the Jacobian matrix and so on.

In new math, we take advantage of the recognition that geometric objects like densities etc. have their own symmetries and orientations. For example, if we want to find out how much fluid is flowing out of a surface, i.e. calculate the flux, then we focus attention on two objects:

1. Extension
2. Density

An extension of linear size ε in the cartesian plane R x R can be represented by the bivector ε² ∂x Λ ∂y. Here the wedge product 'Λ' is used so that what is encoded in this object is nothing but the area and orientation of the area element. Things like shape etc that we are not interested in are removed by the antisymmetrization involved in using 'Λ'.

A scalar density on a surface is represented and object of the form by ρ(x,y) dx Λ dy, called a 2-form. If we want to be more careful about the symmetries of these representations matching the intrinsic symmetries of the objects themselves, then we use Weyl 2-forms and so on.

Now we have two things: the density ω2 = ρ(x,y) dx Λ dy and a small area element με = ε² ∂x Λ ∂y. The area element is a bivector and the density is a 2-form. Using these we can get the amount of substance in the area element by contracting it with the extension: ω2⋅με. Starting here, it is easy to define the integral. We simply divide the surface into many small extensions, and make them act on the densities and then sum. The sum becomes an integral in the limit. In general, we integrate n-forms over an n-surface to get a real number.

Last edited: May 17, 2010
5. May 17, 2010

### lavinia

dxdy is the product of the displacements dx and dy. This is clearly commutative since it is just a product.

dx^dy measures an oriented area spanned by the projections of two displacements onto the xy-plane. this is anti-commutative since reversing the order of dx and dy reverses the orientation.

The integral of a differential form can always be boiled down to a sum of ordinary integrals of functions times ordinary displacements. This is done with the change of variables formula - a.k.a. the Chain Rule - If you integrate this way, orientations are preserved in the sign of the function to be integrated.

Last edited: May 17, 2010
6. May 17, 2010

### zhentil

The symbol dxdy implies an iterated integral. By Fubini's theorem, you can reverse the order. If one integrates over something that's not a rectangle, the symbol dxdy is a very misleading abuse of notation (this is the reason most calculus textbooks use the notation dA).

7. Jun 15, 2010

### OB1

The wedge product is not the same as the tensor product. They do not act on the same space - the tensor product is on the tensor algebra (which is of course just the free vector space modulo a bunch of ideals), whereas the wedge product acts on the exterior algebra (which is the tensor algebra modulo an ideal). So in the end, it isn't that dx or dy are intrinsically symmetric or antisymmetric (whatever that would mean), but that the operation you use on them behaves in a particular way.
Now for the notation: physicists like to drop the tensor symbol and mathematicians like to drop the wedge symbol, so it's a miserable situation for someone just starting to learn linear algebra. I recommend starting out with a mathematical text. You can find a very nice explanation of all of this in F. Warner's Foundations of Differentiable Manifolds and Lie Groups, chapter 2 (even if you don't know any manifold theory, you can still understand this part).

8. Jun 15, 2010

### Hurkyl

Staff Emeritus
The exterior algebra is also isomorphic to the algebra of antisymmetric tensors; sometimes people use that model, rather than the model of tensors modulo symmetric tensors.

9. Jun 15, 2010

### DrRocket

With no quarrel to any of the answers above, what you are dealing with is basically the difference between a normal measure and an oriented integral.

To be simple consider just an integral over an ordinary interval, and the difference among the following:

$$\int_{[0,1]} f(x) dx$$

$$\int_{0}^{1} f(x) dx$$

$$\int_{1}^{0} f(x) dx$$

A similar thing happens when considering integrals of chains over co-chains. You have to consider the orientation of the co-chain.

The other way to think about is that dxdy is a surface element with a direction "normal to the surface", but which of the two possible directions is not clear, while dx^dy is clearly defined and is pointed in the opposite direction from dy^dx.

10. Jun 16, 2010

### LukeD

In the books that I've been reading on differential geometry, integration is only defined for forms (anti-symmetric tensors). But it's defined by referring to "the usual" n-fold integration.
What is this "usual" n-fold integration? Is it integration of symmetric tensors? Is there a separate theory of integration of symmetric tensors? I understand differential forms as being egg crates of sorts that exist on manifolds that let us measure quantities associated with flows through surfaces and the such.
So what are these symmetric tensors? I guess I think of symbols like dxdy as grids on a surface. 5dxdy would be a grid that is 5x as fine. The grids let us answer questions like "how many grid spaces are taken up by my square"

Are these things really tensors though? It seems to me that if we can work out a theory of calculus allowing integration on symmetric and anti-symmetric tensor fields, that we can can work out a theory of calculus on any sort of tensor fields. I've never head of this though, so I am skeptical.

I guess we can give integrals like $$\int_{[0,1]\times[0,1]} \frac{1}{2}(dxdy + dx \wedge dy)$$ meaning though?
With the usual orientation of the manifold, this would give 1, but with the reversed orientation, it would give 0. That's interesting. I imagine this has a lot of uses, but does it work like this?

But how do I give meaning to an integral like $$\int_{[0,1]} dx^2$$?
Edit: I realize that dx^2 is given meaning in the theory of Stochastic Calculus. I don't get the connection though.

Last edited: Jun 16, 2010
11. Jun 16, 2010

### Hurkyl

Staff Emeritus
The Riemann integral you learned in your elementary calculus courses. Or maybe something more general like Lebesgue integration with respect to the Lebesgue measure.

12. Jun 16, 2010

### LukeD

What I mean is that the "usual" integration doesn't integrate over anti-symmetric forms. It integrates over... symmetric things (are they tensors? I don't know). I usually read that "everything under the integral symbol is a differential form". Is that really true? what about the symmetric integrals? Are those technically integrals over something else or can they be interpreted as forms as well?

What about $$\int_t dW^2$$ from Stochastic Calculus? Is dW^2 a differential form on some weird space? I've heard of "vector-valued differential forms" is this one of them things?

13. Jun 16, 2010

### Hurkyl

Staff Emeritus
Usual integration doesn't have anything to do with tensors or forms; it's simply a Riemann integral or a Lebesgue integral or something similar.

Not literally. It's only true in the sense that for a "usual integral" you can generally make a trivial change to find an integral of a differential form that gives the same value.

e.g. let:
• x be the standard coordinate on the real line
• f be a continuous, real-valued function of the real line
• $\varphi$ be the scalar field corresponding to f
• a and b be real numbers with a < b
• C be any smooth increasing function from [0,1] to [a,b]
Then, the Riemann integral
$$\int_a^b f$$​
and the path integral
$$\int_C \varphi \, dx$$​
have the same value.

Of course, we would often write the Riemann integral using an indeterminate variable, making the notation look more similar:
$$\int_a^b f(\xi) \, d\xi$$​
(where $d\xi$ has nothing to do with differential forms -- it's just a formal symbol saying what variable we're integrating over)
And we might even write the interval as a region, making notation look even more similar:
$$\int_{[a,b]} f(\xi) \, d\xi$$​

Last edited: Jun 16, 2010
14. Jun 18, 2010

### LukeD

Isn't there some kind of isomorphism in terms of an exterior algebra of vector valued forms? It'd be really nice to have a consistent framework to do everything with. I suppose there has to be one. I could just assume that it exists and work with it. Nothing could go wrong right?

But it would be nice to know what the vector space of functions is that is described by the exterior algebra of completely symmetric basis forms. I suppose it would have to be a very large vector space. Now that I think about it... i suppose it would probably represent the exterior algebra of a space describable by functions that could fail to be described accurately by a Taylor series no matter how many terms. their directional derivatives probably commute though, otherwise I wouldn't expect dx^dy = dy^dx.

Edit: I should mention that I'm still learning Differential Geometry & I'm also trying to learn Stochastic Calculus. As a test of my understanding, I want to try to figure out how Stochastic Geometry works. So far I've found... a 15 page paper and the name of a book. Any recommendations?

Last edited: Jun 18, 2010
15. Jun 18, 2010

### DrRocket

I get the impression that you are seeking a single all-encompassing framework that includes everything with the name "integral". I don't think you will find it. There are all sorts of specialized integrals. In your study of the stochastic calculus you might want to take the time to look at the difference between the Ito and Stratonovich integral -- I am not particularly expert in this subject, but I do know that the two integrals do not always agree.

16. Jun 18, 2010

### LukeD

Ah but we can express stochastic calculus in the language of non-commutative geometry (which is to say, algebra valued forms). We recover differential forms that satisfy the Leibniz product rule in that case. (Though 0-forms and 1-forms no longer commute)

In that case, the difference between the Ito integral and the Stratonovich integral is the difference between expressing a 1 form as a left-martingale (for the Ito integral) or as a half-left, half-right martingale (for the Stratonovich integral)

I am absolutely positive that what I am searching for does exist in the realm of non-commutative geometry. It needs to exist for the self-consistency of the whole study.

Edit: The paper I found that explains stochastic calculus in terms of non-commutative geometry is here: http://phorgyphynance.files.wordpress.com/2008/06/blackscholes.pdf

17. Jun 18, 2010

### quasar987

One way to motivate the theory of integration of differential forms is through the desire to integrate plain old numerical functions on a manifold.

Suppose you have a manifold M and and map f:M-->R. How would you define $\int_Mf$? To simplify the problem, suppose that f has support in a coordinate nbhd (U,phi) of M (i.e. f=0 outside of U), or even better, that M can be covered by a single chart (U,phi) (where U=M). Then it is tempting to set

$$\int_Mf :=\int_{\phi(U)}f\circ \phi^{-1}$$

where the integral on the right is plain old riemannian integration.

Notice first that the value of $\int_Mf$ depends on the choice of the chart (U,phi). For instance if f is identically 1 and if phi maps U to the unit ball in R³, then we would have $\int_Mf$=4pi/3, while if phi maps U to the whole of R³, we would have $\int_Mf$=+infinity.

This is a little weird, but isn't really a problem. What is more of a problem is the fact that with this definition, it is not possible to meaningfully compare the values of the integrals of different functions. Indeed, if f,g:M-->R are maps on M and if

$$\int_Mf =\int_{\phi(U)}f\circ \phi^{-1}>\int_Mg =\int_{\phi(U)}g\circ \phi^{-1}$$

for a certain chart (U,phi), then in general, it is possible to find another chart (V,psi) (which will map the parts of M where g is greater than f to a large portion of R^n and which will map the parts of M where g is lesser than f to a tiny portion of R^n) so that with this chart, we have instead

$$\int_Mf =\int_{\psi(V)}f\circ \psi^{-1}<\int_Mg =\int_{\psi (V)}g\circ \psi^{-1}$$

The way that pleople have found to define the integral of a function in a way that the relation $\int_Mf>\int_Mg$ is an intrinsic one (i.e. is independent of the choices of charts involved) is in first observing that if M is an orientable n-manifold, there is a way of integrating an n-form on M in a way that does not depend on the charts involved. As above, sticking for simplicity to the special case of an n-form $\omega$ with support in a coordinate chart (U,phi), on U, $\omega$ has the form

$$\omega|_U=hdx^1\wedge\ldots\wedge dx^n$$

for some function h:U-->R, and we set

$$\int_M\omega:=\int_{\phi(U)}h\circ \phi^{-1}$$

And now this number does not depend on the choice of (U,phi). Indeed, if (V,psi) is another chart containing the support of $\omega$ that is coherently oriented with (U,phi) (that is to say, $\det(D(\phi\circ\psi^{-1}))>0$ everywhere), then

$$dx^1\wedge\ldots\wedge dx^n=\det(D(\phi\circ\psi^{-1}))dy^1\wedge\ldots\wedge dy^n=|\det(D(\phi\circ\psi^{-1}))|dy^1\wedge\ldots\wedge dy^n$$

And so, if we now compute $\int_M\omega$ using the chart (V,psi), we get

$$\int_M\omega=\int_{\psi(V)}h\circ \psi^{-1}|\det(D(\phi\circ\psi^{-1}))|=\int_{\phi(U)}h\circ \phi^{-1}$$

which is the same as the computation of $\int_M\omega$ using the chart (U,phi). In the second equality I invoked the change of variable formula.

In short, because of the transformation property $dx^1\wedge\ldots\wedge dx^n=|\det(D(\phi\circ\psi^{-1}))|dy^1\wedge\ldots\wedge dy^n$ of n-forms, their integral, defined in the obvious most natural sense, is independent of the choice of (coherently oriented) charts involved.

Finally, to solve our original problem concerning the integration of functions, we proceed like so: If M is an orientable n-manifold, then by definition, this means that there exists a nowhere vanishing n-form $\Omega$ on M, commonly called a "volume form". We may simply define the integral of f over M (with respect to the volume form $\Omega$) as the number

$$\int_{(M,\Omega)}f:=\int_Mf\Omega$$

Note that $f\Omega$ is an n-form and so the above is well defined and independent of the choice of coherently oriented charts involved in the computation. It is only dependent on the choice of the n-form $\Omega$.

18. Jun 18, 2010

### quasar987

Um.. when I began writing the post, I thought it addressed your question.. now that i re-read your posts, I'm not so sure, lol...

19. Jun 18, 2010

### LukeD

It almost does. The thing is at the end, your volume is defined up to a sign (the orientation of $$\Omega$$. We can just add absolute value bars or decide that at the end we multiply by the orientation of our basis or something to get rid of the sign dependency, but as far as I know, there's nothing immediately within the machinery of forms that makes
$$\int_S f(x,y)dxdy = \int_S f(x,y)dydx$$ a true statement.

20. Jun 18, 2010

### quasar987

But this is not a statement about the integration of differential forms, it is a statement about the Riemann integral of a function.

21. Jun 19, 2010

### LukeD

It would be a true statement about forms if we were on a Moebius strip! Moebius strip are non-orientable, because there isn't a difference between dx^dy and dy^dx.

22. Jun 19, 2010

### Hurkyl

Staff Emeritus
No, the wedge product of forms is always antisymmetric. In fact, it's pure algebra; the identity dx^dy = -dy^dx has nothing to do with any notion of geometry or topology.

That the Möbius strip is non-orientable means that there does not exist a global area form; every 2-form has to degenerate somewhere. It is impossible to represent any area measure by a 2-form.
(Of course, you can do it locally, since even the Möbius strip is locally Euclidean)

Incidentally, I suspect you need to pay more attention to what you mean by dx and dy. For the Möbius strip, there doesn't exist a global coordinate chart. Furthermore, I believe the set of 1-forms doesn't have a basis; any spanning set must contain at least three forms. (Of course, locally, you only need two)

23. Jun 19, 2010

### quasar987

I wonder where you got that from. As Hurkyl said, that dx^dy=-dy^dx is an algebraic fact relating to the definition of the wedge product (namely, $dx\wedge dy=dx\otimes dy - dy\otimes dx$) and it is always true.

24. Jun 19, 2010

### LukeD

Sorry, I didn't mean the Moebius Strip, that was a mistake. I meant the Moebius strip with the front and back identified with each other. In that case, i can take the 2 form described by dx^dy and I translate it across the Moebius strip until I get back to the some point on the other side. Now it's flipped to its mirror image, which would have been dy^dx". And I'm aware of the algebraic definition of ^ (though I'm a bit confused. Can I wedge together any x&y as long as $$x \otimes y - y \otimes x$$ makes sense? I've only seen this done with forms, but it could be done with vectors, right?)

Locally of course dx^dy and dy^dx look different... I'd need something where dxdy and dydx look the same locally. After reading some more about non-commutative geometry, I see that there is a more general meaning for df than $$\sum_i \partial_{x^i} f dx_i$$ and there seems to also be a more general meaning for dxdy than dx^dy. I'm still trying to grapple with the definition of d over the smooth functions on a commutative algebra.

Last edited: Jun 19, 2010
25. Jun 19, 2010

### Hurkyl

Staff Emeritus
Mathematical surfaces don't have a front and a back; if you have a path on the Möbius strip that stays to the middle and goes around once, the it's a closed loop: the initial and final points are the same.

The notion of a surface having a front and back has to do with embedding it into a three-dimensional space, rather than being a notion intrinsic to the surface itself.

The fact that dx^dy and dy^dx are unequal* is not an issue of local vs global; they are simply unequal. And if you had a parametrized surface that covered the strip, you'd find that the integrals of the two 2-forms would have opposite sign.

Instead, the relationship you describe is a matter of symmetry. Up to diffeomorphism, I believe the only property of a global 2-form is the shape (up to homotopy) of its vanishing set.

P.S. I really wish you'd stop using dx and dy unless you clarify their meaning -- I'm very worried you are making some implicit assumption that is true for the Euclidean plane but not for the Möbius strip.

*: I'm assuming they're nonzero