The only reasonable way to interpret "$ds$" as a functional on tangent vectors has to be that it takes a tangent vector and spits out its length, but this is not linear. So $ds$ is not a 1form. It still seems like a nice sort of object to think about integrating. Does $ds$ fit into a larger class of gadgets generalizing differential forms? Or it there some compelling reason that I shouldn't care about $ds$?

$\begingroup$ What sort of properties of differential forms would you like this generalisation to have? $\endgroup$– Marco GollaJun 13, 2012 at 20:06

11$\begingroup$ I see a vote to close, and I find this very strange: this is a perfectly good question. (I vote to keep open) $\endgroup$– André HenriquesJun 13, 2012 at 20:12

2$\begingroup$ Contrary to Malte's comment above, the $ds$ in a scalar line integral is the same as the $ds$ in the expression of a Riemannian metric. As ÁlvarezPaiva indicates in the currently accepted answer below, this object is an even density of rank $1$; on the one hand, such a thing may be multiplied by a scalar field to produce another even density of rank $1$, and such a thing may be integrated along any (unoriented!) $1$dimensional submanifold (aka curve); on the other hand, such a thing may be multiplied (symmetrically!) by itself to produce (sometimes) a symmetric bilinear form. $\endgroup$– Toby BartelsMar 3, 2013 at 8:10

1$\begingroup$ And as long as it is understood that we are using symmetrised multiplication (not the antisymmetrised multiplication that is the wedge product of differential forms), then the equation $(ds)^2 = (dx)^2 + (dy)^2$ is literally correct (for the Euclidean metric on the $(x,y)$plane, which is literally $(dx)^2 + (dy)^2$). And even the formula $ds = \sqrt{(dx)^2 + (dy)^2}$ is literally correct and defines $ds$ as the unique positive semidefinite (which in this case is actually positive definite) square root of the Euclidean metric $(dx)^2 + (dy)^2$. $\endgroup$– Toby BartelsMar 3, 2013 at 8:15

1$\begingroup$ Here $dx$ and $dy$ are the differentials of the coordinate functions $x$ and $y$, but $ds$ is an unanalysable whole, not the differential of anything. Of course, once you have chosen a curve (complete with starting point), you can define a function $s$ along this curve, but there is no $s$ definable on the plane as a whole. In contrast, $ds$ is a density on the entire plane, defined once the metric is chosen, without the need to specify any particular curve. So any particular $s$ is merely one choice of indefinite integral of $ds$, which is the prior concept. $\endgroup$– Toby BartelsMar 3, 2013 at 8:19
4 Answers
It is not a 1form, it is a 1density: a function that is continuous and homogeneous of degree 1 on the tangent space of the manifold. It also happens to be convex and positive in the complement of the zero section (actually, its restriction to each tangent space is a Euclidean norm). If the norm is not Euclidean, you have the arclength element of a Finsler metric. The convexity is basically necessary and sufficient for the lower semicontinuity of the length functional (BusemannMayer theorem).
See my answer to this question for more on densities.


1$\begingroup$ And integrals with respect to $ds$ make sense on any rectifiable curve, even those that are not differentiable. $\endgroup$ Jun 13, 2012 at 20:24

$\begingroup$ Integrals with respect to act length are usually signed things, in that they take into account the orientation of the curve, no? That would suggest the $ds$ is a form then. $\endgroup$ Jun 13, 2012 at 21:00

3$\begingroup$ @Mariano: No, I would say that they are not signed things. This is clearly visible in the notation $\sqrt{dx^2+dy^2}$. Of course, you can make them signed things if you want: given an orientation on a 1dimensional manifold, you can turn a density into a oneform and vice versa. $\endgroup$ Jun 13, 2012 at 21:24

$\begingroup$ Gerald Edgar's comment is correct but may not be as big a generalisation as one might think, since every rectifiable curve may be parametrised by arclength, and this parametrisation is continuously differentiable almost everywhere, and the classical definition of the line integral of a continuos function using a Riemann integral and the derivative of the parametrisation therefore already applies (and gives the correct answer). $\endgroup$ Mar 1, 2013 at 19:52
It is an example of an absolute differential form, as defined by Toby Bartels here: http://ncatlab.org/nlab/show/absolute+differential+form.

$\begingroup$ Interesting. I will have to look into this closer  I know anything by Toby Bartels is good. $\endgroup$ Jun 14, 2012 at 2:25

2$\begingroup$ Actually, Bartles's definition is incomplete: at each point $p$ and for each $k$tuples of linearly independent tangent vectors $v_1$, $\dots$, $v_k$, if $A: T_p M \rightarrow T_p M$ is an invertible linear map, then one needs either $\omega_p(Av_1,\dots, Av_k) = det(A)\omega_p(v_1,\dots,v_k)$ (odd densities that need to be integrated on oriented submanifolds) or $\omega_p(Av_1,\dots, Av_k) = det(A)\omega_p(v_1,\dots,v_k)$ (even densities). If you don't have this condition, the integral will depend on the parametrization of the submanifold. $\endgroup$ Jun 14, 2012 at 6:33

1$\begingroup$ Hmm, ok. Better update the nLab then. Thanks! $\endgroup$– David Roberts ♦Jun 14, 2012 at 6:57

2$\begingroup$ I'm delighted to learn that my ‘absolute differential forms’ have been seen before! I'm also not surprised that my definition is too general; I was never really satisfied with it. I only wish that David had reported this conversation in the relevant thread on the nForum: nforum.mathforge.org/discussion/3744/…. (But that's not David's job either.) $\endgroup$ Mar 1, 2013 at 21:50

1$\begingroup$ Actually, I don't believe that my definition of absolute differential form is incomplete. It satisfies every condition (for an even density) that you (Álvarez Paiva) list in the other thread (at least if the form is continuous). It doesn't satisfy your condition above about determinants, but then neither does $ds$, so that condition must be mistaken. I believe that you mean for $A$ to be a map from and to the span of the $v_i$ (which is all of $T_p M$ only if $k = \dim M$), in which case my definition satisfies this (as given for even densities, again). $\endgroup$ Mar 3, 2013 at 10:42
If you have a curve, also known as a 1manifold, inside a Riemannian manifold, the Riemannian metric on the manifold restricts to a 1dimensional Riemannian metric on the 1manifold. The square root of this metric is a density (see alvarezpaiva's answer) that can indeed be integrated along the 1manifold.

$\begingroup$ More generally, given a regular curve, i.e. an immersion $\gamma: [a,b] \rightarrow (M,g)$, one has the pullback metric $\gamma^\ast g$. The volume of $([a,b],\gamma^\ast g)$ is the length of the curve $\gamma$, which is $\int g(\dot\gamma(s),\dot\gamma(s)) ds. However, $ds$ here denotes the volume element of $[a,b]$ with the standard euclidean metric from the real line. $\endgroup$– MalteJun 13, 2012 at 20:26

$\begingroup$ This is also not a good answer, since we want to know what $ds$ is on the ambient space, not just what it is on the curve. $\endgroup$ Mar 23, 2013 at 5:34

$\begingroup$ Toby, you have a reasonable point. It is a 1density on the ambient space that, when restricted to any curve and integrated, gives the length of the curve. Thanks. $\endgroup$ Mar 23, 2013 at 13:59
It is the volume element on the one manifold. It is a 1form.

3$\begingroup$ In what sense can you say that $ds$ = $\sqrt(dx^2+dy^2)$? This is very common in high school calculus classes, and matches my intuition well, but it doesn't seem to make any sense from a differential forms perspective. Maybe this is more what I wanted to ask about. $\endgroup$ Jun 13, 2012 at 19:58

1$\begingroup$ The volume element of the projective plane with its (let's say) standard Riemannian metric is a form ? $\endgroup$ Jun 14, 2012 at 6:40

1$\begingroup$ It should be a 1pseudoform, just as the volume element of the projective plane must be a 2pseudoform. Only with an orientation (not available for the projective plane) can one identify forms and pseudoforms. But this is still not a good answer, since we want to know what it is on the ambient space, not just what it is on the curve. $\endgroup$ Mar 23, 2013 at 5:33