Area form a differential form?

Question

I just read that $\omega_x( \eta, \zeta) := \langle x, \eta \times \zeta \rangle $ is the area form on the sphere, where $x \in \mathbb{S}^2$ and $\eta,\zeta \in T_xM.$ All I see is that this is basically a determinant which is of couse somehow related to a volume, but why is this called the area form on the sphere?

Answer 1

If you restrict the inner product on $\mathbb{R}^3$ to the sphere $\mathbb{S}^2$, the result is a Riemannian metric. Every Riemannian metric comes with a canonical top-dimensional form, the volume form, which tells you which bases for a tangent space have unit volume. This 2-form is the volume form for the metric on the sphere. It's called an "area form" because we tend to call two-dimensional volume "area."

In this case, the canonical volume form in coordinates is the square root of the determinant of the metric's matrix. If you want to crunch out the computation, take a local coordinate patch, identify the pullback of $\omega$, write the metric in those coordinates, and compute.

That's no fun, so let's see if we can do this more elegantly. The volume form is equivalently defined in terms of oriented orthonormal bases $(e_1,e_2,\ldots,e_n)$ so that $\operatorname{dVol}(e_1,e_2,e_3,\ldots,e_n) = 1$. Let us take an oriented basis $e_1,e_2$ of $T_x\mathbb{S}^2$. Now we compute: $$\omega_x(e_1,e_2) = \langle x,e_1\times e_2\rangle = \langle x,x\rangle = \|x\|^2 = 1$$ because $x = e_1\times e_2$. Now we've won! The form $\omega_x$ took an orthonormal frame and spit out $1$, so it's the volume form for this metric.

(You should prove that the volume form is unique and that this doesn't depend on choice of orthonormal framing.)