# line element as a 1 form

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.

• 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 $\|\phi'(t)\|$. – Rylee Lyman Apr 2 at 12:02
• @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 $\mathbb{R}^2$? – enochk. Apr 2 at 12:24
• 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. – Rylee Lyman Apr 2 at 19:54
• 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? – enochk. Apr 2 at 21:59
• Sorry, I think this is beyond my cobwebby knowledge of differential geometry! – Rylee Lyman Apr 3 at 1:51

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.