Integrating 2-forms over a parameterised surface

[Rasalhague]

Not sure if this question belongs more here or in Topology & Geometry or in Math & Science Software...

I'm trying out the technique described in Bachman: A Geometric Approach to Differential forms, § 6.2 "Integrating 2-forms". To take a simple example, I tried to use this method to get the surface area of a sphere, S², with radius R, embedded in R³. I think the 2-form I need to integrate is

$$\omega = dy \wedge dz + dz \wedge dx + dx \wedge dz.$$

I've chosen the obvious parameterization

$$\psi : \mathbb{R}^2 \rightarrow \mathbb{R}^3 \; \bigg| \; \psi (\zeta, \alpha) = R \; (\sin \zeta \cos \alpha, \sin \zeta \sin \alpha, \cos \zeta)$$

where $0 \leq \zeta \leq \pi$ is the zenith angle, and $0 \leq \alpha \leq 2 \pi$ the azimuth angle. Let $\textbf{x}=\psi(\zeta, \alpha)$. Then

$$\frac{\partial \textbf{x}}{\partial \zeta} = R \; (\cos \zeta \cos \alpha, \cos \zeta \sin \alpha, -\sin \zeta)$$

$$\frac{\partial \textbf{x}}{\partial \alpha} = R \; (-\sin \zeta \sin \alpha, \sin \zeta \cos \alpha, 0)$$

Barring copying errors on my part, mathematica confirms this.

$$dy \wedge dz\left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix} R \cos \zeta \sin \alpha & R \cos \alpha \sin \zeta\\ -R \sin \zeta & 0 \end{vmatrix} = R^2 \cos \alpha \; \sin^2 \zeta$$

$$dz \wedge dx \left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix} -R \sin \zeta & 0 \\ R \cos \zeta \cos \alpha & -R \sin \zeta \sin \alpha \end{vmatrix} = R^2 \sin \alpha \; \sin^2 \zeta$$

$$dx \wedge dy \left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix} R \cos \zeta \cos \alpha & -R \sin \zeta \sin \alpha \\ R \cos \zeta \sin \alpha & R \sin \zeta \cos \alpha \end{vmatrix} = R^2 \cos \zeta \sin \zeta$$

But according to Mathematica,

$$R^2 \int_0^{2\pi} \int_0^\pi \left [ \sin^2 \zeta (\cos \alpha + \sin \alpha) + \cos \zeta \sin \zeta \right ] d \zeta \wedge d \alpha = 0$$

What went wrong?

x = R {Sin[z] Cos[a], Sin[z] Sin[a], Cos[z]};
tz = D[x, z]; ta = D[x, a];
Integrate[
 Simplify[Det[{{tz[[2]], ta[[2]]}, {tz[[3]], ta[[3]]}}] + 
   Det[{{tz[[3]], ta[[3]]}, {tz[[1]], ta[[1]]}}] + 
   Det[{{tz[[1]], ta[[1]]}, {tz[[2]], ta[[2]]}}]], {z, 0, Pi}, {a, 0, 
  2 Pi}]

Mathematica gave the same results for each of the matrices and their determinants as I got by hand.

 

[quasar987]

You are not using the right 2-form if you're hoping to get 4piR² as the answer.

Try integrating

(x/R)dydz - (y/R)dxdz + (z/R)dxdy

(See Proposition 13.24 in Lee's books which tells you which form is induced on an hypersurface)

 

[Rasalhague]

Thanks for the pointer, Quasar. I don't have a copy of Introduction to Smooth Manifolds, but I can read bits of it on Google Books, and bits on Amazon, including this proposition.

What does the symbol between N and dV_g mean?

 

[quasar987]

It is interior multiplication, also often written $\iota_NdV_g$.

More generally, if you have a k form $\Omega$ and a vector field X, then you can construct a (k-1)-form from these two ingredients by setting $(\iota_X\Omega)(\cdot,\ldots,\cdot):=\Omega(X,\cdot,\ldots,\cdot)$. It is called the interior product of X with $\Omega$.

 

[Rasalhague]

Excellent, that works perfectly. Thanks, Quasar, I understand now.

x = R {Sin[z] Cos[a], Sin[z] Sin[a], Cos[z]};
tz = D[x, z]; ta = D[x, a];
R^(-1)*Integrate[
  Simplify[x[[1]]*Det[{{tz[[2]], ta[[2]]}, {tz[[3]], ta[[3]]}}] + 
    x[[2]]*Det[{{tz[[3]], ta[[3]]}, {tz[[1]], ta[[1]]}}] + 
    x[[3]]*Det[{{tz[[1]], ta[[1]]}, {tz[[2]], ta[[2]]}}]], {z, 0, 
   Pi}, {a, 0, 2 Pi}]

gives $4 \pi R^2$, since the unit normal vector is

$$\frac{\textbf{x}}{R}$$

And a more round-about way to the same result is

x = R {Sin[z] Cos[a], Sin[z] Sin[a], Cos[z]};
tz = D[x, z]; ta = D[x, a]; n = 
 Simplify[Cross[tz, ta]/Norm[Cross[tz, ta]]]; Integrate[
 Simplify[n[[1]]*Det[{{tz[[2]], ta[[2]]}, {tz[[3]], ta[[3]]}}] + 
   n[[2]]*Det[{{tz[[3]], ta[[3]]}, {tz[[1]], ta[[1]]}}] + 
   n[[3]]*Det[{{tz[[1]], ta[[1]]}, {tz[[2]], ta[[2]]}}]], {z, 0, 
  Pi}, {a, 0, 2 Pi}, Assumptions -> R \[Element] Reals]

defining the unit normal by

$$N=\frac{\partial_\zeta \textbf{x} \times \partial_\alpha \textbf{x}}{\left \| \partial_\zeta \textbf{x} \times \partial_\alpha \textbf{x} \right \|}.$$

And we get the action of a k-form, k > 1, on a single vector, starting with the complete determinant formula, given by Lee in Riemannian Manifolds as

$$\omega^1 \wedge ... \wedge \omega^k (v_1, ..., v_k) = \text{det}(\omega^i(v_j))$$

(Bachman defines it equivalently by the transpose), and then carrying out the first stage of computing this determinant, splitting it into determinants of (empty) submatrices, each multiplied by the appropriate coefficient of the vector, then plugging into these subdeterminants the coefficients of the tangent basis to the embedded surface to be integrated over.

Incidentally, Wikipedia's remark on the terminology made me smile: "The interior product, named in opposition to the exterior product, is also called interior or inner multiplication, or the inner derivative or derivation, but should not be confused with an inner product." (Also called... Also not to be confused with the following things also called inner product...)