Area form from volume form

Question

In $\mathbb{R}^3$ the volume form is defined as $$ \omega = dx\land dy \land dz $$

Such differential form is exact, indeed it can be written as

$$ \omega = d\left(\frac{1}{3}\left(x dy \land dz - y dx \land dz + z dx\land dy \right)\right) = d \alpha $$

I wonder if $\alpha$ does correspond to the area form. The question is if there's a relationship between the $\alpha$ form and the area of a surface.

I am also aware of this question where the given differential form is the same as mine, however the factor $\frac{1}{3}$ is missing. I therefore wonder what's the relationship between the two.

Also can this be generalized to any volume form defined by a Riemannian metric $g$?

Answer 1

• Yes, $\mu=dx\wedge dy\wedge dz$ is exact as you have seen. It is volume forms on compact boundaryless manifolds which cannot be exact (due to Stokes’ theorem).
• $\alpha$ is a $2$-form for sure, but it doesn’t make sense to call it an area form. An area form for which submanifold? In most cases, no it is not the (‘standard’) area-form of every submanifold. However, apart from the factor of $3$, if you pullback $\alpha$ to the unit sphere $S^2$, then you do indeed get the area 2-form of the sphere.
• The above remark about spheres is special, it is because the unit outward normal on the sphere is $\nu=x\frac{\partial}{\partial x}+ y\frac{\partial}{\partial y} + z\frac{\partial}{\partial z}$, and the interior product of this vector field with the ambient volume form $\mu$ equals $3\alpha$: $\nu\lrcorner\,\mu=3\alpha$. In a general Riemannian manifold $(M,g)$, if $\mu_g$ denotes the volume form of $M$, and $S\subset M$ is any submanifold with unit normal $\nu$, then $\sigma_{S,g}:=\nu\lrcorner\,\mu_g$ (well technically its pullback by $\iota_S:S\hookrightarrow M$) gives the area form on $S$. See here for more details.

Answer 2

Notice that for the Euler vector field on $\Bbb R^3$, $$X := x \partial_x + y \partial_y + z \partial_z ,$$ we have $$\iota_X \omega = x \,dy \wedge dz + y \,dz \wedge dx + z \,dx \wedge dy ,$$ where again $\omega := dx \wedge dy \wedge dz$. It's true that $\omega$ is exact: In the language of the divergence operator, we have $$d(\iota_X \omega) = (\operatorname{div} X) \omega = 3 \omega,$$ so $$d\left(\frac{1}{3} \iota_X \omega\right) = \omega .$$

It's perhaps misleading to call $\alpha := \frac{1}{3} \iota_X \omega$ an area form: After all, it's defined on a $3$-manifold, not a $2$ manifold. But we can say this: Given an oriented surface $S \subset \Bbb R^3$, say, with inclusion map $i_S : S \hookrightarrow \Bbb R^3$, either unit normal vector field $N$ determines a volume (area) form $i_S^*(\iota_N \omega)$ for the induced metric on $S$. In particular, if the restriction $X \vert_S$ of $X$ to $S$ is a unit normal vector field for some $S$, $i_S^*(\iota_X \omega) = 3 i_S^* \alpha$ is an area form for the induced metric on $S$---but the only surfaces $S$ for which $X\vert_S$ is a unit normal vector field are the open subsets of the unit sphere $S^2 \subset \Bbb R^3$.

Some of the above, but not all of it, can be generalized to general oriented Riemannian manifolds: The definition of the divergence works just as well for any manifold equipped with a volume form (including an oriented Riemannian manifold), and the definition of the area form induced on a hypersurface works just as well for oriented hypersurfaces of oriented Riemannian manifolds. But the volume form $\Omega$ on an oriented manifold need not be exact. In fact, if $M$ is a (nonempty) compact, oriented $n$-manifold (without boundary) and $\eta$ is a smooth $(n - 1)$-form on $M$, Stokes' Theorem gives $$\int_M d\eta = \int_{\partial M} \eta = \int_\varnothing \eta = 0 ,$$ whereas $\int_M \Omega > 0$, so no volume form on $M$ is exact.