# Proof that vector area is a boundary integral?

Let $M \subset \mathbb{R}^2$ be a closed topological disk and let $f: M \rightarrow \mathbb{R}^3$ be a smooth embedding; let $N$ be the corresponding unit normal field on $M$. The vector area is defined as

$$A := \int_M N\ dA.$$

It is well-known that $A$ can also be expressed via the boundary integral

$$\frac{1}{2}\int_{\partial M} f \times df.$$

But in all the references I've seen the author merely invokes Stokes' theorem without providing further details. (See, for instance, Sullivan, "Curvatures of Smooth and Discrete Surfaces," section 3.)

I'm not seeing it. My question is: in excruciating detail, how do you get from point A to point B?

Thanks!

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I'm a little lost in the notation here. In particular, what does $f \times df$ mean? $df$ is a 1-form, and $f$ is a map from the disk to $\mathbb{R}^3$. – Sam Lisi Oct 25 '11 at 9:50
I'm with you -- I have never seen a formal definition of the cross product of $\mathbb{R}^3$-valued differential forms, though I'd love to see one! However this is not my notation, but really the notation used by John Sullivan and others. I can only assume that, for instance, $$(f \times df)(u) = f \times (df(u))$$ and $$(df \times df)(u,v) = df(u) \times df(v) - df(v) \times df(u),$$ in which case everything works out (even if the notation is a bit abusive...). But what I mean by "excruciating detail" is that I also want a solid definition for the cross product of differential forms. – PolyKnowMeAll Oct 27 '11 at 15:53
There are many ways of proving that $A=B$, depending on the desired level of sophistication. But there is no way around Stokes' or Green's formula. – Christian Blatter Oct 27 '11 at 19:06

## 1 Answer

Ok, finally figured it out -- the confusion basically arises from slightly nonstandard notation. First, however, let me review some important facts about differential forms (mostly following the discussion on Wikipedia). The application of a 2-form to a pair of vectors is often written as

$$\alpha \wedge \beta(u,v) = \alpha(u)\beta(v) - \alpha(v)\beta(u).$$

Differential forms are often introduced in the scalar setting (e.g., $\alpha$ and $\beta$ are $\mathbb{R}$-valued) and so this expression makes sense: $\alpha(u)$ and $\beta(v)$ are both real-valued, so we can simply multiply them to get another real number.

More generally, however, suppose that $\alpha$ and $\beta$ take values in some vector space $E$. Then there may no longer be a natural or obvious way to "multiply" vectors, and so in general the best we can do is write something like

$$\alpha \wedge \beta(u,v) = \alpha(u) \otimes \beta(v) - \alpha(v) \otimes \beta(u),$$

where $\otimes$ denotes the outer or tensor product (which is defined for any vector space $E$). Note, then, that the wedge product of two $E$-valued forms is not $E$-valued, but in fact $E \otimes E$ valued!

Logically, however, an element of $E \otimes E$ can be thought of as a pair of elements of $E$. So if we happen to have some binary product $\cdot: E \times E \rightarrow E$, then we can adopt the convention that any $(E \otimes E)$-valued 2-form "automatically" gets mapped to the $E$-valued 2-form

$$\alpha \wedge \beta(u,v) = \alpha(u) \cdot \beta(v) - \alpha(v) \cdot \beta(u).$$

In the case of the vector area, the most (only?) natural binary product on $\mathbb{R}^3$ is given by the cross product, since it takes a pair of 3-vectors to a 3-vector. However (and here's the confusion), standard notation in this case would be

$$\alpha \wedge \beta(u,v) = \alpha(u) \times \beta(v) - \alpha(v) \times \beta(u),$$

and not

$$\alpha \times \beta(u,v) = \alpha(u) \times \beta(v) - \alpha(v) \times \beta(u).$$

(Personally I prefer to stick with the $\wedge$ notation because it is the only thing that gives the reader a clue that antisymmetrization occurs.)

So, with all that pedantic garbage out of the way, we can write a really pedantic proof of the statement from the question:

Consider the algebra bundle $E = (\mathbb{R}^3, \times)$ over a disk-like region $M \subset \mathbb{R}^2$ and let $f$ be an $E$-valued 0-form representing a smooth immersion; let $N$ be the unit normal field on $M$ induced by $f$ (also viewed as an $E$-valued 0-form). Finally, let $dA$ be the standard volume form on $M$. Noting that

$$df \wedge df(u,v) = df(u) \times df(v) - df(v) \times df(u) = 2 df(u) \times df(v) = 2 N dA(u,v)$$

for all pairs of vectors $u,v \in \mathbb{R}^2$ (since the cross product of two tangents points in the normal direction!), we have

$$\int_M N dA = \frac{1}{2} \int_M df \wedge df = \frac{1}{2} \int_M d(f \wedge df) = \frac{1}{2} \int_{\partial M} f \wedge df,$$

as desired (up to a change in notation).

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I like this. Could you please explain how to extend your idea to higher (>3) dimension? – G. Fougeron Jul 15 '15 at 9:47