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Suppose we're in $\mathbb R^n$, and we have a function on line segments ,$\omega(I)$, with values in $\mathbb R$. Give sufficient conditions for $\omega$ to be given by a generalized 1-form (that is, an $(n-1)$ -current), which is integrable (that is, can be evaluated on line segments). Obviously $\omega$ should be additive, and it should not vary too wildly as we change $I$. What is the precise statement? Is it already sufficient?

What if we are only given a sufficiently large set of segments $\{I\}$ on which $\omega$ is defined, say those not intersecting a given hyperplane.

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  • $\begingroup$ Uniqueness of such a generalized form seems fairly obvious, although I haven't written the details. $\endgroup$
    – Dima
    Nov 14, 2013 at 12:25
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    $\begingroup$ In another MO question (link), I asked about characterizing differential forms among a densities. In response, Anton Petrunin proved that differential forms were characterized by satisfying Stokes' Theorem. A density is a generalization of a form which is different from a current (I don't know much about the latter), but a 1-density does define, via integration, a function on line segments, so maybe this is of some interest. $\endgroup$
    – Tim Campion ♦
    Nov 15, 2013 at 1:21
  • $\begingroup$ Thank you Tim, your question on densities seems to be just the infinitesimal version of my question. In particular, I immediately see that I missed another obvious necessary condition, namely Stokes theorem, and it seems that nothing else is needed. However, you seem to be ignoring smoothness questions there, that is, assuming everything is sufficiently smooth, while I am particularly interested in the case when $\omega$ is not very smooth. I'll try repeating Anton Petrunin's argument while keeping this in mind. $\endgroup$
    – Dima
    Nov 15, 2013 at 13:48
  • $\begingroup$ @ Dima The argument you outlined above is what Whitney does in his book I mentioned in my answer. Again you have to be very careful what you mean by a density. A density on $\mathbb{R}^n$ (defined in a conventional way) cannot be integrated along line segments. $\endgroup$
    – Liviu Nicolaescu
    Nov 15, 2013 at 16:27

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First of all, since you are looking for a differential form, $\omega$ should be a function defined on oriented segments. By additivity it can be extended to $1$-dimensional polygonal chains in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$.

In his beautiful book Geometric Integration theory, H. Whitney addresses a more general problem, of characterizing which linear maps defined on $k$-dimensional polyhedral chains are representable as the integration by a $k$-form. For details I refer to Chapter V of Whitney's book.

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  • $\begingroup$ Some details gleaned from Whitney: The functions on segments that the question asks about are called cochains, on which Whitney defines two norms, "sharp" and "flat". In Ch. V he shows that sharp cochains correspond to Lipschitz forms, and in Ch. IX, following Wolfe, that flat cochains correspond to flat forms, whose definition (Sec. IX.6) looks more technical. $\endgroup$
    – Tim Campion ♦
    Nov 15, 2013 at 16:55
  • $\begingroup$ There is a more modern description of flat cochains in Federer's Geometric Measure Theory, Sec. 4.1.19. Example 1.12 in www3.nd.edu/~lnicolae/Fslices.pdf may clarify the notations in Federerer. Roughly, a flat $k$-cochain is the sum between a bounded measurable form and the exterior derivative of a bounded measurable $(k-1)$-form. $\endgroup$
    – Liviu Nicolaescu
    Nov 15, 2013 at 20:47

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