Exactness of differential forms

July 28th, 2007 ~ Posted in: Mathematics, Programming

Remember a one-form \omega = p(x,y) dx + q(x,y) dy can be identified with a functional

\displaystyle \gamma \mapsto \int_ \gamma \omega = \int_a^b p(x(t), y(t)) \frac{dx}{dt} + q(x(t),y(t))\frac{dy}{dt} \; dt

on the set of piecewise differentiable curves \gamma(t) = (x(t), y(t)); a \leq t \leq b . It can also be identified with a vector field.

In special cases, a one-form can also represent the total differential of a function: if there is a function f(x,y) such that \frac{\partial f}{\partial x} = p and \frac{\partial f}{\partial y} = q, then since the chain rule gives \frac{d f(x(t), y(t))}{dt} = p \frac{dx}{dt} + \frac{dy}{dt}, the one-form df = p dx + q dy is called the differential of the function f. A one-form \omega is called exact if there is some f such that \omega = df.

Note that if \omega = df, then

\int_\gamma \omega = \int_a^b \frac{d f(\gamma(t))}{dt} \; dt = f(\gamma(b)) - f(\gamma(a)),

which shows the integral of an exact one-form is independent of the path taken– it depends only on the endpoints. In particular, the integral of \omega around any closed path is 0. In physics, conservative forces (fields) are exact one-forms.

What if someone gives you a one-form \omega and asks you if it is exact? How could one determine an answer, or maybe even find a function with \omega as its differential?

First, if \omega = df, (assuming f is smooth), then \frac{\partial p}{\partial y} = \frac{\partial q}{\partial x}. The differential of a one-form is the two-form d\omega = \left(\frac{\partial p}{\partial y} - \frac{\partial q}{\partial x}\right) dx dy . So, if \omega = df, then d\omega = 0. One-forms such that d\omega = 0 are called closed.

By Stokes’ theorem,

\displaystyle \int_{\partial R} \omega = \int_R d\omega

if R is a simple region, where \int_R d\omega = \int_R \left(\frac{\partial p}{\partial y} - \frac{\partial q}{\partial x}\right) dx dy . Consequently, for closed forms, \int_{\partial R} \omega = 0. Applying Stoke’s theorem to the illustration below gives that \int_{\gamma_1} \omega = \int_{-\gamma_2} \omega if \omega is a closed form, so closedness implies path independence.

un := 1in; o := (0,0); z0 = o; z1 = (1un, .5un); z2 = (1un, 2un); z3 = (.25un, .75un); path p[]; label.top(btex {\Huge R} etex, (.75un, .75un)); pickup pencircle scaled 2; p0 = z0{dir 60}..z1{right}..{dir 100}z2{dir -100}; p1 = z2{dir -100}..{dir -120}z3..{down}z0; ahlength := 6; drawarrow subpath (0,1) of p0; drawarrow subpath (0,1) of p1; draw p0; draw p1; label.rt(btex \Large $\gamma_1$ etex, z1 shifted (0, -.1un) ); label.lft(btex \Large $\gamma_2$ etex, z3 shifted (-.1un, 0) );

Clearly from an earlier comment, the first prerequisite for being exact is closedness. In fact, you can show that in star-shaped domains, closedness implies exactness. Basically, you fix a value f(p) at p, the center of the star, and define f(x) = f(p) + \int_\sigma \omega for any point x in the domain, where \sigma is a path from p to x. Taking derivatives shows any such f has \omega as a differential.

If \omega is exact on two domains U_1 and U_2, and U_1 \cap U_2 is connected, then \omega is exact on U_1 \cup U_2 . The idea is that if df_i = \omega on U_i for i=1,2, then d(f_1 - f_2) = 0 on U_1 \cap U_2 , which is connected, so f_1 = f_2 + c on U_1 \cap U_2 . Let f_2^\prime = f_2 - c , then the function f which takes the values of f_1 on U_1 and f_2^\prime on U_2 satisfies df = \omega on U_1 \cup U_2.

A powerful consequence of the fact that a closed one-form is exact in a star-shaped domain is that locally any closed one-form is exact. It then follows readily that a path integral of a closed one-form can be calculated in ‘exact patches’. More specifically, if \omega is a closed one-form on U and \gamma is a smooth path in U, then there is a subdivision a = t_0 < t_1 < \cdots < t_n = b and a collection of open subsets U_1, \ldots, U_n of U such that \gamma maps [t_{i-1}, t_i] into U_i, and the restriction of \omega to U_i is the differential of a function f_i. For any such choice of subdivision and open sets, if P_i = \gamma(t_i),

\displaystyle \int_\gamma \omega = \sum_{i=1}^n \left( f_i(P_i) - f_i(P_{i-1}) \right)

The proof of this assertion is topological. First, for each point P in \gamma([a,b]), there is a neighborhood U_P in which the form is exact. The preimages of these neighborhoods under \gamma forms an open cover of [a,b], of which a finite subcover can be chosen.

An application of the Lebesgue number lemma tells us that there’s a fine enough refinement of this cover into sets [t_i, t_{i+1}] such that the image of each interval is contained in an U_i. Applying local exactness gives the result.

Mathematics, Programming ~

2 Responses to “Exactness of differential forms”

  • 1. ObsessiveMathsFreak
    July 31st, 2007 at 6:56 am

    I really can’t say I approve of differential forms. The entire subject has so far seemed to me to me a layer of excessive formalism over fairly straightforward vector analysis. I also have objections on religious grounds to writing the likes of “pdx+qdy” and then integrating it.

    Forms are too much like distributions for my taste, in that they must be integrated in order to be evaluated. A form cannot be said to have any value at a specific point. Compare and contrast to concepts like divergence, curl and the complex derivative, which all have a value at any point before any integrals are taken.

    Add to this a swathe of books in which the authors really don’t know what they’re talking about, attested by their lack of geometry, automated symbol pushing, unmotivated definitions and total lack of any concrete examples. And this whole business with working in the counterintuitive dual space is often maddening. All in all, there’s not a lot that would convince me personally to move away from vector and tensor calculus. If this is the way to move into higher dimension, I don’t think I want to.

    Word to the wise. Find out what the exterior product of two vectors _actually is_ before getting onto outer products of forms, that includes the exterior derivative. Hint: It’s geometric. Does your book use that torturous and convoluted definition of the wedge product, with the alternating permutation babble? That thing is an embarrassment to mathematics. It’s only a way to get the outer product, not what is actually _is_

  • 2. Alex
    July 31st, 2007 at 9:27 am

    Aren’t differential k-forms just antisymmetric (0,k) tensors? I have problems understanding the motivations for both tensors and differential forms, so it doesn’t matter much to me how you define either of them. However, I do find the simplified form of Stokes’ theorem to be a big upside to the formalism of differential forms, as opposed to regular vector calculus (it’s the only way I’ve found to remember Green’s theorem).

    Exterior product == cross product in \R^3? Not sure that it generalizes to anything particularly intuitive in higher dimensions except in the sense that you can use it to get a maximally independent set from (n-1) independent elements in \R^n

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