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I read that the arc length is not a differential form. But I don't understand why it isn't. I understand that differential forms are integrands and arc length is an expression which is integrable. What property of differential form does it not satisfy?

zyx
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2 Answers2

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I understand that differential forms are integrands and arc length is an expression which is integrable.

On a given parametrized curve, any local expression that can be integrated can be written (locally in the parameter of the the curve) as $f(p) dp$ by differentiating the integral of the expression. This is a differential $1$-form, and accordingly the integral is invariant under changes of parameter. That is not a surprise, considering that $f(p)$ was defined by differentiating an invariant quantity.

What property of differential form does it not satisfy?

Arc length is integrated only along specified curves, and is a differential form on any one curve. However, the infinitesimal arc length (Riemannian metric line-element) $ds = \sqrt{\sum (dx_i)^2}$ is defined without reference to any curve and is a local expression in the coordinates of the space in which the curves are to be drawn. This local expression is not a differential form on the space, and it is not possible to re-write it as a differential form. If it were equal to a form, it could be restricted to any particular curve to obtain the arclength form on that curve. But there is no such form, because (through a given point, in a small neighborhood) there must pass curves along which integration of a form gives zero and this is not the case for arc length.

zyx
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For simplicity I will explain the case of the standard arclength in $\mathbb{R}^n$. Suppose $\alpha$ is a differential 1-form on $\mathbb{R}^n$: then there exist smooth functions $f_1, \ldots, f_n : \mathbb{R}^n \to \mathbb{R}$ such that, for any smooth curve $\gamma : (0, 1) \to \mathbb{R}^n$, $$\int_\gamma \alpha = \int_0^1 \sum_{i=1}^{n} f_i (\gamma_i (t)) \dot{\gamma}_i (t) \, d t$$ Thus, by the fundamental theorem of calculus, the functions $f_1, \ldots, f_n$ are uniquely determined by the assignment $\gamma \mapsto \int_\gamma \alpha$: $$f_i (x) = \left. \frac{d}{d s} \right|_{s = 0} \int_{\gamma^s} \alpha $$ where $\gamma^s$ is the line in the $i$-th direction of length $s$ starting from $x$. Thus, if arclength were determined by a differential form, it would have to be the differential form $\mathrm{d} x_1 + \cdots + \mathrm{d} x_n$, which it clearly is not.

Zhen Lin
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  • In R^2, for the curve c(t)=(t,0), dx is clearly a differential form that determines its arclength, but not of the form dx+dy, as you proved. So, the problem is not that "there cannot be a form to determine arclength". The problem is described here: http://mathoverflow.net/questions/99488/the-ds-which-appears-in-an-integral-with-respect-to-arclength-is-not-a-1-form. It is a density, or an "absolute differential form". – Xipan Xiao Jul 31 '13 at 21:36
  • You misunderstand. I said we can recover $\alpha$ as long as we know its integral with respect to all smooth curves. – Zhen Lin Jul 31 '13 at 21:39
  • Thank you for your explanation. But I still didn't get your point: yes we can recover a form if we knows its integral on all curves. But why is this related to arc length? – Xipan Xiao Jul 31 '13 at 21:45
  • Because arclength is a function ${ \text{smooth curves} } \to \mathbb{R}$ as well, so we can ask whether it is of the form $\gamma \mapsto \int_\gamma \alpha$ for some fixed differential 1-form $\alpha$. – Zhen Lin Jul 31 '13 at 22:35
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    However this is not what the OP is asking. You proved that: there doesn't exist a universal differential form $\alpha$ that can act as the arc length function for all curves. And I agree with this statement. While I believe the OP is asking: for a given curve, is its arclength ds a 1-form? – Xipan Xiao Aug 01 '13 at 13:44