I am trying to use differential forms to determine the surface area element for a sphere. For a sphere of radius $r=1$. I think I am loosing something in the algebra (tried to check symbolic calculations on computer, still don't know how to proceed)

In terms of Cartesian coordinates the surface of the sphere is: $x^2+y^2+z^2=1$. The spherical coordinates relate to Cartesian coordinates in the standard way:

$$ \begin{align} x=&\sin\theta\cos\phi \\ y=&\sin\theta\sin\phi \\ z=&\cos\theta \\ \end{align} $$

The area element in Cartesian coordinates is: $$ d^2S = dx \wedge dy - dx \wedge dz + dy \wedge dz $$

Computing the equivalents in spherical coordinates (not quite equivalent since radius is fixed):

$$ \begin{align} dx=& \cos\theta\cos\phi\,d\theta - \sin\theta\sin\phi\,d\phi \\ dy=& \cos\theta\sin\phi\,d\theta + \sin\theta\cos\phi\,d\phi \\ dz=& -\sin\theta\,d\theta \\ \end{align} $$


$$ \begin{align} d^2 S =\quad&\left(\cos\theta\sin\theta\cos^2\phi+\cos\theta\sin\theta\sin^2\phi\right)d\theta\wedge d\phi-\\ -&\left(-\sin^2\theta\sin\phi\right)d\theta\wedge d\phi+ \\ +&\left(\sin^2\theta\cos\phi\right)d\theta\wedge d\phi \\ \\ d^2 S =&\sin\theta\cdot\left(\cos\theta + \sin\theta\cdot\left(\cos\phi+\sin\phi\right)\right)d\theta\wedge d\phi \end{align} $$

I know the correct result should be $d^2S=\sin\theta \, d\theta\wedge d\phi$, and that it certainly should not depend on $\phi$. But I can't quite see where I went wrong. I suppose I am looking at using a push-forward from $\theta\phi$ space to the surface of a 3D sphere, and then I am trying to pull-back the area element from the 3d space, but this statement will still lead to the same calculations.


2 Answers 2


Your cartesian area element is quite wrong. Try $$x\, dy\wedge dz + y\, dz\wedge dx + z\, dx\wedge dy.$$

  • $\begingroup$ Thanks. I started suspecting that too $\endgroup$
    – Cryo
    Sep 29, 2020 at 4:20
  • $\begingroup$ Schifrin. Thanks for your pointer. I was wondring if you could comment on the approach I have outlined in the answer post. Does it make sense? $\endgroup$
    – Cryo
    Oct 1, 2020 at 5:43

As pointed out by @Ted Schifrin, the differential form I have used at the outset was wrong. Here's how I think one can arrive at the correct form. I would appreciate any comments.

General approach

Start with $m$-dimensional Eucledian space with Cartesian coordinates $\{x^{(i)}\}_{i=1,\dots,m}$. There is a single equation that defines an $\left(m-1\right)$-dimensional surface $f\left(x^{(1)},\dots,x^{(m)}\right)=\mbox{const}$.

One can define: $$ \begin{align} df =& d(f)=\partial_if\,dx^{(i)} \\ dn=& df/\sqrt{\langle df,\,df\rangle} \end{align} $$

The former is a 1-form dual to vector that is normal to the surface (in the sence that it would give zero when applied to any vector in the tangent space of the surface). The latter is the normalized 1-form: $\langle df,\, df \rangle=g^{\alpha\beta}\,\partial_\alpha f\,\partial_\beta f$, where $g^{\alpha\beta}$ is the inverse metric tensor.

From $dn$ one can extract a Hodge dual:

$$ \star dn=\frac{\partial_\mu f\,g^{\mu\nu}}{\sqrt{\langle df,\,df\rangle}} \: \,\frac{\sqrt{g}\,\epsilon_{\nu\alpha_2\dots\alpha_m}}{\left(m-1\right)!}\:dx^{(\alpha_2)}\wedge\dots\wedge dx^{(\alpha_m)} $$

Where $g$ is the determinant of the metric tensor and $\epsilon$ is the Levi-Civita relative tensor. Given the volume form: $d^m V=dx^{(1)}\wedge\dots\wedge dx^{(m)}$, one can check that:

$$ dn\wedge\star dn=\langle dn,\,dn\rangle d^m V=d^m V $$

As it should be.

Next, I come do definition. Since $\star dn$ is the form that together with the 1-form 'perpendicular' to the surface $f=const$ gives the volume element, I define $\star dn$ to be the area element (for integrating on $f=const$ surface).

Specific to sphere

With Cartesian coordinates the metric is trivial (diagonal), so $g=1$. $f=x^2+y^2+z^2=r^2$, so:

$$ \begin{align} df =& 2x\,dx + 2y\,dy + 2z\,dz \\ dn =& \frac{x}{r}\,dx + \frac{y}{r}\,dy + \frac{z}{r}\,dz \\ \star dn =& \frac{x}{r} dy \wedge dz + \frac{y}{r} dz \wedge dx + \frac{z}{r} dx \wedge dy \\ \end{align} $$

Now, finally its time to sub in the spherical coordinates:

$$ \begin{align} \star dn =& \quad\sin\theta\cos\phi \cdot \left(\sin^2\theta\cos\phi\right)r^2\,d\theta\wedge d\phi +\\ &+\sin\theta\sin\phi\cdot\left(\sin^2\theta\sin\phi\right)r^2\,d\theta\wedge d\phi +\\ &+\cos\theta\cdot \left(\cos\theta\sin\theta\cos^2\phi+\cos\theta\sin\theta\sin^2\phi\right)r^2\,d\theta\wedge d\phi \end{align} $$

Pulling things together:

$$ \begin{align} d^2 S = \star dn =& \left(\sin^3\theta\cos^2\phi+\sin^3\theta\sin^2\phi+\cos^2\theta\sin\theta\right)r^2\,d\theta\wedge d\phi \\ =& \sin\theta\left(\sin^2\theta+\cos^2\theta\right)d\theta\wedge d\phi \\ =& r^2\,\sin\theta\,d\theta\wedge d\phi \end{align} $$


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