I was trying to compute the area of the sphere using calculus and my knowledge of differential form as follow :

Consider the two form $\omega = dx \wedge dy$, we want to use this form to find out the area of a disk. We define a parametrization of the sphere as follows $F(r,\phi) \rightarrow (rcos\phi, rsin\phi)$ So we have :

\begin{align} \int_{\mathbb{S}^2} \omega = \int_{[0,R]}\int_{[0,2\pi]} F^* \omega = \\ \int_{[0,R]}\int_{[0,2\pi]}rdr \wedge d\phi = \\ \int_{[0,R]}\int_{[0,2\pi]} rdrd\phi = \pi R^2 \end{align}

I was hoping to apply the same principle to find the circumference of the circle but I think I ultimately miss some technicalities.

The circumference of the circle is a 1-dimensional manifold so I am trying to define a one-form on it.I do not know how to proceed from here. Do I have to find a 1-d parametrization of the circle ? Is the one form $dx$ the one to integrate ?

I'm trying to compute the circumference of the circle using pullback an forms. So I can understand the mechanics of integrating forms to find volumes ( and hopefully move to more exotic manifolds like the area or volume of a torus ).

EDIT : My reasoning is as follows. To find the circumference of a circle, let's define a mapping $F:\mathbb{R} \rightarrow \mathbb{R}$ that parametrize it. I was thinking of defining $F:[0,\frac{\pi}{2}] \rightarrow \mathbb{R}$ and multiply the result by 4 ( since the mapping parametrize a quarter of the circle ). Then proceed to integrate $\int_{[0,\frac{\pi}{2}]} F^*(dx)$ Unfortunately the few mapping I tried fail to provide me the right answer.

EDIT2 : Using stereo coordinates (attempt)

Let's consider the map $F:[0,\frac{\pi}{4}] \rightarrow \mathbb{R}$ the stereo projection of $\frac{1}{8}$ of the circumference of the circle, defined as $F(\alpha)=Rtan(\alpha)$ Now we have $\int_{\mathbb{\frac{S^1}{8}}} dx = \int_{[0,\frac{\pi}{4}]} F^*(d\alpha) = R\frac{\pi}{4}$ which when multiplied by 8 gives us the result $2\pi R$

  • $\begingroup$ Symbols in mathematics do not have universal, context-independent meaning. When you write $dx \wedge dy$ as "the area form on a sphere" and subsequently use polar coordinates $(r, \phi)$, it appears you're working in a disk in the Euclidean plane with Cartesian coordinates $(x, y)$, not on a sphere. <> Separately, if you're familiar with line and surface integrals in multivariable calculus, you've already learned how to integrate differential forms, and may pleasantly discover you know more than you think. $\endgroup$ – Andrew D. Hwang Jan 18 '17 at 14:12
  • $\begingroup$ I tagged the post with differential geometry so I was guessing it would be evident I was talking about the wedge product of differential forms. Also yes the mapping $F$ maps from $\mathbb{R^2}$ to $\mathbb{R^2}$ with the Image being the cartesian coordinate. This is the form I am pulling back. $\endgroup$ – user149705 Jan 18 '17 at 15:35
  • $\begingroup$ It's perfectly evident you're speaking of wedge products; the potential points of confusion are 1. You appear to be calling a "disk" a "sphere" (a sphere of radius $R$ does not have area $\pi R^{2}$, for starters); 2. You appear to be using $x$ and $y$ both as ambient Cartesian coordinates (on the disk), and as manifold coordinates (on the circle). $\endgroup$ – Andrew D. Hwang Jan 18 '17 at 15:47
  • 1
    $\begingroup$ @AndrewD.Hwang I got myself confused in my previous comment ( that I deleted ). Indeed I am talking about a disk. I am trying to integrate a 2-form on the disk. Since the area of the disk is a 2D manifold, I find a parametrization in 2D $F$ and pull back the form where I can integrate. $\endgroup$ – user149705 Jan 18 '17 at 16:15
  • $\begingroup$ I see; thank you. The disk is (aside from its boundary) an open submanifold of a Cartesian space, while the circle isn't. You might, generally, try stereographic coordinates, which are tractable in arbitrary dimension (for both balls and spheres). When time permits, I'll write up a sketch. $\endgroup$ – Andrew D. Hwang Jan 18 '17 at 16:28

$\newcommand{\Reals}{\mathbf{R}}$Stereographic coordinates give a more or less uniform way to approach the calculation of the $n$-volume of the unit $n$-sphere. If $x = (x_{1}, \dots, x_{n})$ denotes the general element of $\Reals^{n}$, the stereographic parametrization of the $n$-sphere is $$ f(x) = \frac{(2x, \|x\|^{2} - 1)}{\|x\|^{2} + 1} = \left(\frac{2x_{1}}{1 + \sum_{j} x_{j}^{2}}, \dots, \frac{2x_{n}}{1 + \sum_{j} x_{j}^{2}}, \frac{-1 + \sum_{j} x_{j}^{2}}{1 + \sum_{j} x_{j}^{2}}\right). $$ A straightforward (and edifying) calculation shows that the induced metric on $\Reals^{n}$ is $$ g = \frac{4(dx_{1}^{2} + \cdots + dx_{n}^{2})}{1 + \sum_{j} x_{j}^{2}}. $$ The volume form is therefore $$ dV = \left(\frac{2}{1 + \sum_{j} x_{j}^{2}}\right)^{n} dx_{1} \wedge \cdots \wedge dx_{n}. $$ The volume of the unit $n$-sphere $S^{n} \subset \Reals^{n+1}$ is the integral of this volume form over $\Reals^{n}$.

For instance,

  • The volume (i.e., length) of the unit circle is $$ \int_{-\infty}^{\infty} \frac{2\, dx}{1 + x^{2}}. $$ The value of this integral is

    $2(\arctan \infty - \arctan(-\infty)) = 2\pi$.

  • The volume (i.e., area) of the unit $2$-sphere is $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{4\, dx\, dy}{(1 + x^{2} + y^{2})^{2}}, $$ whose value (by converting to polar coordinates) is

    $\displaystyle\int_{0}^{2\pi} \int_{0}^{\infty} \frac{4r\, dr\, d\theta}{(1 + r^{2})^{2}} = 4\pi \int_{1}^{\infty} \frac{du}{u^{2}} = 4\pi$.

To calculate these for arbitrary $n$, it may be easier to calculate the volume of the ball of radius $r$ (using Cavalieri's theorem and recursion, see for example Where does this formula for the volume of a $n$-dimensional ball come from? or Wikipedia's page on the $n$-sphere), then use the fact that if the $n$-ball has volume $V_{n}(r) = V_{n}(1)r^{n}$, then the $(n-1)$-sphere has $(n-1)$-dimensional volume $$ \frac{d}{dr} V_{n}(r) = nV_{n}(1)r^{n-1}. $$

You may also enjoy attempting the calculation in generalized spherical coordinates, a nice application of the reduction formula for integrating powers of sine.

  • $\begingroup$ Nice answer ;-). The problem is, so it seems, that PO is still struggling with basic concepts. I done a pull-back of volume form yet he couldn't reflect on it. $\endgroup$ – Troy Woo Jan 18 '17 at 23:51
  • $\begingroup$ Thank you, that has been of great help ! $\endgroup$ – user149705 Jan 20 '17 at 19:01

Area of sphere

You should use spherical coordinate for the sphere: $$ F:(\phi,\psi)\in[0,2\pi]\times\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\mapsto(r\cos\phi\cos\psi,r\sin\phi\cos\psi,r\sin\psi)\in\mathbb R^3 $$ Now you want to integrate the volume form induced from $\mathbb R^3$ by the immersion $F$, which is given by: $$ \Omega=\sqrt{\det g}~\mathrm d\phi\wedge\mathrm d\psi $$ Now you need to compute the induced $2\times 2$ Riemannian metric matrix $g$, $$ g=\begin{pmatrix} g_{\phi\phi} & g_{\phi\psi}\\ g_{\psi\phi} & g_{\psi\psi} \end{pmatrix} $$ where $g_{*\star}=\langle X_{*},X_{\star}\rangle$ and $X_{*}$,$X_{\star}$ are the tangent vectors $\frac{\partial F}{\partial(*)}$, $\frac{\partial F}{\partial(\star)}$. Now $$ \begin{aligned} X_{\phi}&=(-r\sin\phi\cos\psi,r\cos\phi\cos\psi,0)\\ X_{\psi}&=(-r\cos\phi\sin\psi,-r\sin\phi\sin\psi,r\cos\psi) \end{aligned} $$ and therefore: $$ g=\begin{pmatrix} r^2\cos^2\psi & 0\\ 0 & r^2 \end{pmatrix} $$ This leads to $$ \begin{aligned} A_{S^2}&=\int_{S^2}\Omega=\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\det g}~\mathrm d\phi\wedge\mathrm d\psi\\ &=\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos\psi\mathrm d\phi\mathrm d\psi\\ &=r^2\int_{0}^{2\pi}\mathrm d\phi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\psi\mathrm d\psi=r^2\cdot 2\pi\cdot 2=4\pi r^2 \end{aligned} $$

Circumference of circle

This is nothing different from calculating area of the sphere.

$$ F:\phi\in[0,2\pi]\mapsto(r\cos\phi,r\sin\phi)\in\mathbb R^2 $$

$$ \Omega=\sqrt{\det g}~\mathrm d\phi $$

$$ g=g_{\phi\phi}=\langle X_{\phi},X_{\phi}\rangle $$

$$ X_{\phi}=(-r\sin\phi,r\cos\phi)\Rightarrow g=r^2 $$

$$ A_{S^1}=\int_{S^1}\Omega=\int_{0}^{2\pi}r d\phi=2\pi r $$

  • $\begingroup$ Thanks so much. how would I have to proceed if I want to calculate the circumference using a pullback of a form ? Here you are using a 3D parametrization for the area and a 2D parametrization for the circumference. Isn't there a way to parametrize the manifold with charts (2D chart for the area and 1D charts for the circumference) that will also us to pullback a form to the respective charts and compute the integral in question ? ( I am trying to find more or less an intrinsic way of doing it ) because if I have an exotic manifold that I know only the dim of it, I can start from there ! $\endgroup$ – user149705 Jan 18 '17 at 15:43
  • $\begingroup$ For example,to calculate the volume of the sphere do you have to embed it in 4D space ? $\endgroup$ – user149705 Jan 18 '17 at 15:44
  • $\begingroup$ @user149705 You have to understand the pullback is simply defined by the pushforward of tangent vectors. In the end, you have to decide where your Riemannian metric and volume form comes from. Here I use the usual isometric immersion, which is obvious: the areas are in terms of the Riemannian metric and volume form of $\mathbb R^3$. If you don't have this immersion (embedding), then you will have to decide in another way. $\endgroup$ – Troy Woo Jan 18 '17 at 15:46
  • $\begingroup$ @user149705 To calculate the volume of the sphere, I don't have to embed it in 4D space. It is simply a compact subset of $\mathbb R^3$. I simply use the volume form of $\mathbb R^3$. Maybe you should start with a standard textbook. $\endgroup$ – Troy Woo Jan 18 '17 at 15:47
  • $\begingroup$ The RIemannian metric and volume forms comes from the euclidean space locally homeomorphic to the manifold. Why is that not possible ? ( I am following Lee's textbook, smooth manifold which define integration on manifold using pullbacks ) $\endgroup$ – user149705 Jan 18 '17 at 15:54

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