Volume form on S^1

Question

I know that the volume form on $S^1$ is $\omega= ydx-xdy$. But how I can derive that? The only things that I know are the definition of differential q-form, and the fact that the vector field $v= y \frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$ never vanishes on $S^1$.

Answer 1

1. You speak of "the" volume form. There are many volume forms on any orientable smooth manifold. Given $\omega$ a volume form, and a non-vanishing smooth function $f$, then $f\omega$ is again a volume form.

2. The task of finding "a" volume form is to find a non-vanishing top degree form on the manifold in question. To verify that a one form $\omega$ is a volume form on $S^1$ you of course need to check that for any tangent vector $X$ of $S^1$, the contraction $\omega(X) \neq 0$. Now, any one form on $\mathbb{R}^2$ can be written as $\omega_x \mathrm{d}x + \omega_y \mathrm{d}y$. Since you already know that the tangent space of $S^1$ is spanned by $y \partial_x - x\partial_y$, it suffices to find any pair of real functions $\omega_x, \omega_y$ such that $$ \omega(X) = \omega_x y - \omega_y x \neq 0 $$ for any $(x,y)\in S^1$. The choice that $\omega_y = -x$ and $\omega_x = y$ is just one of many possible choices.

3. Because of the freedom described above, there isn't one method to derive your given one form as the volume form on $S^1$ ... unless you add additional requirements. For example, on the unit circle the volume form you wrote down is the "natural" one (up to choice of orientation) given by the induced Riemannian metric. And perhaps this is what you are seeking, in the end:

   The natural volume form on $\mathbb{R}^2$ is $\mathrm{d}x\mathrm{d}y$. Using the Riemannian metric on $\mathbb{R}^2$ we can write down the unit normal vector field to the unit circle, and this happens to be $x\partial_x + y\partial_y$. Since $\mathbb{R}^2$ is orientable and $S^1$ has a unit normal field, it is also orientable. And a volume form (indeed the one for the induced Riemannian metric) can be given by $$ \omega = (\mathrm{d}x\mathrm{d}y)(x \partial_x + y\partial_y, \cdot) $$

   This method is not restricted to Riemannian manifolds. Suppose $\Sigma$ is a hypersurface in a smooth manifold $M$. Assume that $M$ is orientable and so $\omega$ is a volume form for $M$. Assume also that $\Sigma$ admits a field of "normal vectors", by which I mean that there exists a vector field $n$ defined along $\Sigma$ that is never in the tangent space to $\Sigma$. Then a volume form for $\Sigma$ can be found by contracting (taking the interior product) $\iota_n\omega$.

Answer 2

See the proof of Proposition 12.6 in http://www.math.toronto.edu/mat1300/orientation.11.pdf.

EDIT: Wikipedia gives the following reference for the deduction of the generalization of your formula: Flanders, Harley (1989). Differential forms with applications to the physical sciences.

EDIT2: $$x=\cos\theta, y=\sin\theta$$ $$xdy-ydx=\cos\theta\cos\theta d\theta - \sin\theta(-\sin\theta)d\theta=d\theta$$ and now, go back.