Let a real, smooth manifold $M$ be given. Let $\Gamma$ denote the set of all path segments on $M$, namely the set of all paths of the form $\gamma:[a,b]\to M$.

Let $Q:\Gamma\to\mathbb R$ be a functional, namely a function that assigns a real number to each such path segment. Suppose that $Q$ has the property that for any path $\gamma:[a,c]\to M$, if I divide this path into two segments $\gamma_1[a,b]\to M$ and $\gamma_2[b,c]\to M$ such that $\gamma_1(t) = \gamma(t)$ for all $t\in [a,b]$ and $\gamma_2(t) = \gamma(t)$ for all $t\in [b,c]$, then \begin{align} Q[\gamma] = Q[\gamma_1] + Q[\gamma_2]. \end{align} Under what additional assumptions (if any) can one then show that there exists a 1-form $\omega$ such that \begin{align} Q[\gamma] = \int_\gamma\omega \end{align} for all $\gamma\in \Gamma$?

  • 2
    $\begingroup$ Are you sure you want to integrate 1-forms and not 1-densities? If so, you should add "sign-change condition" under reversal of orientation. You also need some regularity conditions on $Q$ so that it depends smoothly on $\gamma$. $\endgroup$ – Moishe Kohan Jan 31 '14 at 7:46
  • $\begingroup$ @studiosus I'm not sure, but I suspect not. The motivation from this comes from physics where "infinitesimal heat transfer" along a given path is represented as a 1-form, not a 1-density (I think), but that might be physicist sloppiness speaking. So you're saying that such a representation is valid provided we replace 1-form by 1-density and the sign change and smoothness conditions are added, or just that those conditions are necessary? $\endgroup$ – joshphysics Jan 31 '14 at 7:50
  • 1
    $\begingroup$ joshphysics: Note that if $\omega$ is a 1-form and $\alpha, \beta$ are paths so that one is the reverse of the other, then $\int_\alpha \omega= -\int_\beta \omega$. Thus, if you want to use forms you also need the "sign condition". You also want integral to be independent of parameterization of the path (with the same orientation). Lastly, you want smoothness. All these are necessary conditions. I will think if they are also sufficient, this is not obvious. $\endgroup$ – Moishe Kohan Jan 31 '14 at 8:01
  • $\begingroup$ @studiosus Ok I understand thanks. In that case, I wonder if one also needs a condition that mimics the scaling property $\int_\gamma c\omega = c\int_\gamma\omega$? Hopefully someone knows the answer to the question of what set of conditions is sufficient, because I certainly agree that it's not obvious. $\endgroup$ – joshphysics Jan 31 '14 at 8:06
  • 2
    $\begingroup$ I think, the latter is the additivity property you asked for. Note that in your question, you want to have a functional on paths, not a functional on forms. If you want a linear functional on forms/densities then the right object is called "current" by mathematicians (it is a generalization of a submanifold). $\endgroup$ – Moishe Kohan Jan 31 '14 at 8:09

First of all, this is a very nice and natural question. Hassler Whitney addressed exactly this question (ok, a bit more generally, for linear functionals on $k$-chains, not just 1-chains) about 60 years ago in his book "Geometric Integration Theory". The answer is given in Theorem 10A, page 167. Whitney proves that, under suitable analytical conditions, such linear functionals are indeed represented by differential $k$-forms. (In your question, $k=1$.)

  • $\begingroup$ Wow how awesome, and thanks for the praise for the question. I'll need to study this a bit to understand it as I'm not a mathematician, but thanks for putting the time into finding this; I likely would never have found it on my own. I would be grateful if you could show in a bit more detail why the theorem is just a more general case of this, but I'm already grateful as it is. $\endgroup$ – joshphysics Jan 31 '14 at 23:30
  • $\begingroup$ I actually just found a nice paper by Jenny Harrison called "Isomorphisms of Differential Forms and Cochains" that seems pretty illuminating; I'll probably tackle it first. $\endgroup$ – joshphysics Jan 31 '14 at 23:41
  • $\begingroup$ In retrospect, there's something about this result that bothers me. As you pointed out in the comments to the original question, the integral of a 1-form over a 1-chain is independent of how the chain is parameterized. This property doesn't seem to be necessary for cochains, so it seems like this parameterization-independence property comes for free. Similarly for the "sign condition" you originally referred to since for a 1-chain $\alpha$, as far as I can tell, $-\alpha$ is not the same as $\alpha$ reversed, so linearity doesn't seem like it's enough to guarantee the sign condition. $\endgroup$ – joshphysics Jul 14 '17 at 18:12
  • $\begingroup$ @joshphysics: No, it comes for free only if you know that the functional is given by integration. I do not have access to Whitney'[s book right now, I suggest, you pick it up from the library and take a look. $\endgroup$ – Moishe Kohan Jul 17 '17 at 5:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.