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I'm studying differential forms and I know how to manipulate all the equations. On trying to find a pictorial understanding, I am a bit stuck on the following. A one-form is suppose to assign a measurement to a line and it must also be linearly dependent on the tangent vectors at that point.

Consider the line integral $\int f(s)ds$. At a point $p$, consider a path to be integrated over, say $p +\phi(t)$. We have $ds=\|\phi'(t)\|dt$. Obviously, $ds$ does not depend linearly on the tangent vectors at point $p$, i.e., the paths that go through $p$. Does this mean $ds$ is not a one-form?

Thanks in advance for any explanation.

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    $\begingroup$ this is a $1$-form. $dt$ is the $1$-form that eats the tangent vector $\partial_t$ and spits out $1$, so $ds$ eats $\partial_t$ and spits out $\|\phi'(t)\|$. $\endgroup$ – Rylee Lyman Apr 2 at 12:02
  • $\begingroup$ @RyleeLyman thanks. this would mean that the tangent is at $t$ in the domain of $\phi$ instead of the tangent of $\phi$ in $\mathbb{R}^2$, correct? Whereas the integral $\int F(r) dr$ uses the tangent of $r(t)$ as seen from $\mathbb{R}^2$? $\endgroup$ – enochk. Apr 2 at 12:24
  • $\begingroup$ No. $\frac{d \phi}{dt}$ should live in $T_{\phi(t)}\mathbb R^2$. Why should this be true? Well, the tangent vector at a point is really some information about the image, not the domain. $\endgroup$ – Rylee Lyman Apr 2 at 19:54
  • $\begingroup$ What I meant was that $dr$ is a one form, linear with respect to the tangent of the curve $\phi$ as seen in $\mathbb{R}^2$. On the other hand, $ds=\|\phi'(t)\|dt$ is not linear with respect to the tangent of $\phi$, but to the tangent of $t$ as seen from $\mathbb{R}^1$. Is that correct? $\endgroup$ – enochk. Apr 2 at 21:59
  • $\begingroup$ Sorry, I think this is beyond my cobwebby knowledge of differential geometry! $\endgroup$ – Rylee Lyman Apr 3 at 1:51
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No, $ds$ is not a $1$-form. It is technically a density (or the "absolute value" of a $1$-form). Note that if you reverse the orientation of the curve, you get the same integral, whereas for the integral of a $1$-form you will get the negative of the integral.

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