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I am somewhat confused as to whether $ds$ (line element) is actually a differential form... we have (in $\mathbb{R}^2$):

$$ds^2 = dx^2 + dy^2$$

Differential 1-forms are supposed to be linear combinations of the $dx_i$, but the $ds$ shown above is definitely not a linear combination of $\{dx, dy\}$.

So what is it?

EDIT The link given in the comments seems to indicate that $ds$ is not a differential form. The following quote is from C. H. Edwards, "Advanced Calculus of Several Variables" section V.1

Given an oriented curve $C$ in $\bf R$, its arclength form $ds$ is defined for ${\bf x} \in C$ by $$ds_{\bf x}({\bf v}) = T({\bf x}) \cdot {\bf v}$$ Thus $ds_{\bf x}({\bf v})$ is simply the component of ${\bf v}$ in the direction of the unit tangent vector $T({\bf x})$. It is clear that $ds_{\bf x}({\bf v})$ is a linear function of ${\bf v} \in {\bf R}^n$, so $ds$ is a differential form on $C$.

Is the book wrong (or does "is a differential form on $C$" not mean "is a differential form" in general)?

Ken Garoo
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1 Answers1

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On a given curve, $ds$ (as in 'integration with respect to arc length', so allowing a negative sign according to the orientation of the interval of integration) can be locally written as $f(p) dp$ where $p$ is a parameter for the curve. It is a $1$-form on the curve. If you require positivity no matter the direction of integration, so that $\int_p^q ds = \int_q^p ds$ then it is not a $1$-form but a $1$-density.

There is no $1$-form on the plane $f(x,y) dx + g(x,y) dy$ whose restriction to every differentiable curve is $ds$ of that curve. This is what is meant by the slogan that arclength is not a differential form.

zyx
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  • Ahh this seems to make some sense (see my simultaneous edit). Thanks – Ken Garoo Sep 03 '13 at 19:16
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    When I read this question on the main page, I thought "On a given curve..." when I opened this question and saw your answer, I chuckled. (+1) – robjohn Sep 03 '13 at 19:30