Orienting curves with differential forms

188 Views Asked by At

Consider the circle given by the equation $x^2+y^2=1$. We can orient this curve by choosing the tangent vector field $(-y,x)^T$, which defines a direction.

Supposedly we can do this with by taking an appropriate differential form $\omega$. How is this done? The idea, I thought, should be to find a 1-form such that $\omega((-y,x)^T)>0$ for all $x,y$, but this seems to be impossible since all 1-forms look like $adx+bdy$.

EDIT: If we are allowed to orient curves with form fields, then this isn't too bad! Is this allowable?

1

There are 1 best solutions below

0
On

First, a remark about "form" vs "form field": as Hurkyl said, the words "differential form", or "differential $k$-form", or simply "$k$-form" usually refer to a smooth choice of a covector at every point. Differential geometers also say "tensor" when they mean "tensor field".

Second, a choice of a top-dimensional nowhere vanishing differential form is what is usually understood as orientation: to orient an $n$-manifold is to specify a nowhere vanishing $n$-form. Any two such forms are related by multiplication by a nowhere vanishing smooth function (hence, either positive or negative). The equivalence class up to multiplication by positive functions constitutes an orientation.

Finally, the construction of nowhere vanishing $1$-form on a circle was discussed here.