6

I am learning differential manifold and got a question.

How do we calculate the surface area? Or how to calculate the volume of a submanifold? Like for the surface area of $S^n$, if $\phi$ is the embedding map, then it seems that $S=\int\phi^*(\sum_{j=1}^{n+1}(-1)^{j-1}x_j dx_1\wedge dx_2...dx_{j-1}\wedge dx_{j+1}...\wedge dx_{n+1})$ according to some webpage I found. But where did that volume form come from? For a general case, if $(N,\phi)$ is a n-dimension submanifold embedding in a m-dimension manifold M, what is the n-form in $A(M)$ that should be pulled back and integrate on $N$?

Thank you for your patience.

Neal
  • 32,659
tomography
  • 391

2 Answers2

3

I think that in general the best approach is the following: For all this discussion, we start with a Riemannian metric $ds^2$ on $M$, and we look at the induced Riemannian metric $i^*ds^2$ on $N$. We write $$i^*ds^2 = \sum_{j=1}^n \omega^j\otimes\omega^j$$ for a suitable collection of $1$-forms $\omega^j$. Then the induced volume ("area") form on $N$ will be $\omega^1\wedge\dots\wedge\omega^n$.

For example, consider $S^2\hookrightarrow \mathbb R^3$. Considering spherical coordinates, $i(\phi,\theta) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$, we have \begin{align*} i^*ds^2_{\mathbb R^3} &= i^*\big(dx^1\otimes dx^1+ dx^2\otimes dx^2+dx^3\otimes dx^3\big) \\ &= d\phi\otimes d\phi + \sin^2\phi\, d\theta\otimes d\theta \\ &= \omega^1\otimes\omega^1 + \omega^2\otimes\omega^2\,, \end{align*} where $\omega^1 = d\phi$ and $\omega^2 = \sin\phi\,d\theta$. [We order these to give the orientation we want on the submanifold.] Then our area form on $S^2$ is $$\omega^1\wedge\omega^2 = \sin\phi\,d\phi\wedge d\theta\,.$$

Ted Shifrin
  • 115,160
2

Suppose you have an orientable manifold $M$ and a volume form $\operatorname{vol}_g$ on $M$. Then to get an induced form on a submanifold $N$ you just need to choose an outward pointing unit vector field $X$ on $N$ (this can always be done) and then do $\iota^\ast(N \lrcorner \operatorname{vol}_g)$, where $\iota : N \hookrightarrow M$ is inclusion and the upper star indicates the pullback. This also gives an induced orientation on $N$. In your case above, all we do is take the standard volume form $dx^1 \wedge \ldots \wedge dx^n$ on $\Bbb{R}^n$ and contract it with the Euler vector field $$X = x^i \frac{\partial}{\partial x^i}$$

that is outward pointing on $S^n$. For example when $n = 3$ assuming I have calculated this correctly you should get the induced form to be $x dy \wedge dz - y dx \wedge dz + z dx \wedge dy$. Then if you integate this using spherical coordinates you should get the area of $S^2$ (remember $``$volume on $S^2"$ now is area.)

  • Thanks a lot but can you explain a little more. What does this expression mean $\iota^\ast(N \lrcorner \operatorname{vol}_g)$? Does it mean the (m-1)-form obtained by putting $X$ in the first place(or the last?) of the original volume form? Do we need $X$ to be normalized, orthogonal(if there is a inner product) to the tangent space of $N$? And in a general case, if the difference between the two manifold is not 1 then just use the complementary of the tangent space of $N$? And how to interpret this? Sorry but the text book really didn't cover this... also @BenjaLim – tomography Jun 25 '13 at 04:06
  • You seem to be assuming $N$ is an orientable hypersurface and there is a typo. Moreover, you need $X$ to be a unit normal field. In general, you'd need trivial normal bundle and orthonormal framing for it. To answer the OP's question, in general there is never a form on $M$ that pulls back to give the volume element on an $n$-dim. submanifold. This does happen in the land of Kähler (complex) manifolds. – Ted Shifrin Jun 25 '13 at 04:18
  • @TedShifrin I have edited my answer, I made a mistake before by not mentioning that $M$ needs to be orientable. –  Jun 25 '13 at 05:19
  • Wow. I don't know what happened to my comment! (Would you believe me if I said the new kitten did it?) Sorry about the mess, I'll delete it! – Jason DeVito - on hiatus Jun 25 '13 at 12:35