I try to wrap my head around the idea of integrating a k-form over a manifold. Loosely speaking, my intuition is, that for each point of the manifold, we do a projection onto the tangent space and get the "area". For example the 3 dimensional unit-sphere is a 2-dimensional manifold, consequently the tanget space is also 2-dimensional. To this point my intuition was that, we do an integral over 2-dimensial-volumes that were measured in the tanget space. This also fits together with the idea that a k-form spits out the volume of the parallelepiped, that was spanned by its k-input vectors. In case of two dimensions, this would be a plane. This idea leads me to choosing the following 2-form to measure the surface of the unitball in 3 dimensions:

$$\omega = dx_1\wedge dx_2 + dx_1\wedge dx_3 + dx_3\wedge dx_2$$

Each 2-form in the sum can measure a surface. For the innocent purpose of just measuring the surface, all the coefficients are set to 1. The integral would look like this:

$$\int_{S} \omega= \int_{S}dx_1\wedge dx_2 + dx_1\wedge dx_3 + dx_3\wedge dx_2$$

But this makes no sense to me, because those 2-forms are the derivatives of the canonical projections that are wedged together. When I plugin a point $p\in S $ into $\omega$, all I get is a sum of functionals that still need an input: $$\omega(p) = x_1\wedge x_2 + x_1\wedge dx_3 + x_3\wedge x_2$$

Where does it take the additional input from? Propably from the vectors that span the tangent space. I could do a parametrization of the manifold to get them as the columns in the jacobian-matrix.

  • $\begingroup$ As an aside, integrating a form over a manifold is not a well-defined operation; a surface consists of more information than simply a two-dimensional manifold. A parametrization is (more than) sufficient to provide that information. While this comment is a bit nit-picky, I felt it's worth posting in case the difference actually relates to your difficulties. $\endgroup$
    – user14972
    Oct 14, 2017 at 19:37
  • $\begingroup$ As a second aside, I prefer to completely ignore tangent spaces; IMO they add needless complication to the notion of integration. When given a parametrized surface, I prefer to pull the differential form back to Euclidean space, since we already know how to integrate there from multivariable calculus. $\endgroup$
    – user14972
    Oct 14, 2017 at 19:43
  • $\begingroup$ @Hurkyl: I'm not sure I understand your comment. How does a surface consist of more information than simply a two-dimensional manifold? Most definitions of "surface" I've seen boil down to defining it as a two-dimensional manifold. It's true that integration of a $k$-form over a $k$-manifold is not defined until you choose an orientation for the manifold (and so it's never defined for nonorientable manifolds). But the idea of a surface does not include orientation. Integration of 2-forms is well-defined over every oriented smooth surface, without a specific choice of parametrization. $\endgroup$
    – Jack Lee
    Oct 14, 2017 at 20:44
  • $\begingroup$ Be careful with the order of factors when you're dealing with outer products: $$\omega = dx_1 \wedge dx_2 + dx_2 \wedge dx_3 + dx_3 \wedge dx_1$$ $\endgroup$
    – md2perpe
    Oct 14, 2017 at 22:10
  • $\begingroup$ Let $\omega$ be a $1$-form and $\alpha: [a,b] \to \mathbb R^n$ a continuously differentiable curve. If I integrate $\omega$ along $\alpha$, I compute the scalar product of the vector (given by $\omega$) on $\alpha$ and the tangential vector that spans the local tangential space of $\alpha$. As this gets more complicated for higher-dimensional undermanifolds, it should be still clear the the 'differentiability' of undermanifolds provides us a natural integrator. $\endgroup$
    – Pazu
    Oct 14, 2017 at 22:28

1 Answer 1


A more natural form for the calculation of the area of the unit sphere is $$\omega = \sin\varphi \, d\varphi \wedge d\theta$$ where $\varphi \in [0, \pi]$ and $\theta \in [0, 2\pi)$ are the angles shown here.

To transform this into Cartesian coordinates we start from $$\begin{cases} x = \sin\varphi \, \cos\theta \\ y = \sin\varphi \, \sin\theta \\ z = \cos\varphi \end{cases}$$ which gives $$\begin{cases} dx = \cos\varphi \, \cos\theta \, d\varphi - \sin\varphi \, \sin\theta \, d\theta \\ dy = \cos\varphi \, \sin\theta \, d\varphi + \sin\varphi \, \cos\theta \, d\theta \\ dz = -\sin\varphi \, d\varphi \end{cases}$$ or $$\begin{cases} d\varphi = \frac{1}{z} \left( \frac{x}{\sqrt{x^2+y^2}} dx + \frac{y}{\sqrt{x^2+y^2}} dy \right) \\ d\theta = \frac{x}{x^2+y^2} dy - \frac{y}{x^2+y^2} dx \end{cases}$$

Thus, $$\omega = \sqrt{x^2+y^2} \, \frac{1}{z} \left( \frac{x}{\sqrt{x^2+y^2}} dx + \frac{y}{\sqrt{x^2+y^2}} dy \right) \wedge \left( \frac{x}{x^2+y^2} dy - \frac{y}{x^2+y^2} dx \right) \\ = \frac{1}{z} \left( \frac{x^2}{x^2+y^2} dx \wedge dy - \frac{y^2}{x^2+y^2} dy \wedge dx \right) \\ = \frac{1}{z} dx \wedge dy $$


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