This is a sequel to the previous post on some integral formulas for hypersurface in Euclidean space. As in the previous post, always denotes the outward normal of whenever this makes sense.

Theorem 1 (Hon Leung’s theorem, 2011)For a closed hypersurface ,

*Proof:* Let be the standard coordinate vectors, let be the region enclosed by , then by divergence theorem,

Since is arbitrary, .

For , the above theorem is trivial because and where is the anticlockwise rotation by . It turns out that there is another way to prove the above theorem when . The observation is the following

Theorem 2 (Hon Leung-Raymond’s theorem, 2011)For a closed surface with outward unit normal , suppose is a domain with smooth one-dimensional boundary (not necessarily connected). Suppose is the positive tangent vector of (i.e. if we arclength-parametrize such that always lies on the left hand side of , then ), and is the position vector, then

*Proof:* Note that

(as a vector-valued 2-form) for any positively oriented local coordinates . We will simply denote by , by , and so on. Note that on one hand, by product rule,

On the other hand, we have

Combining, we have

(In a coordinate-free language, the above is just , which is understood as for differential forms . Note that in this setting !) So by Stokes theorem,

But for positively oriented parametrized by arclength, , so

In particular, for a closed surface , since , we have recovered Theorem 1: .

~~I do not know the answer to the following~~

**Question**: Is there a higher dimensional analogue of the above result?

As remarked by Edward, the above result can be generalized to the higher dimension.

Theorem 3 (Hon Leung-Raymond-Edward Theorem, 2011)For a closed hypersurface , if is its unit outward normal and is a domain in with smooth -dimensional boundary with induced orientation, thenwhere is a positively oriented orthonormal basis for the tangent space of and is the position vector.

*Proof:* As in the previous proof, it is not hard to see that

For example for (),

Thus by Stokes theorem,

With a similar computation as before, it can be seen that

Therefore we have

Let me also state the discrete version of Theorem 1:

Theorem 4 (Hon Leung’s theorem, 2011, discrete version)If is a –polytope and are all its -dimensional faces, denote to be the unit outward normal on , thenHere is the -dimensional volume of .

*Proof:* Clearly we only have to prove the identity for to be a –simplex (triangle for , tetrahedron for ), for, given a polytope we can triangulate into finitely many -simplices, since the result is true for these simplices, summing up these identities, and noting that the terms involving a pair of adjacent faces of two adjacent simplices cancel each others, we can get the result.

The simplex case is very easy. Let us take for notational simplicity. Without loss of generality the 4 vertices are where are linearly independent. Let us denote the normal of simply by . Then

Theorem 2 above can be generalized directly, with the -fold cross product instead of the standard vector cross product. The theorem becomes

where is a -dimensional submanifold of , is a domain in with smooth -dimensional boundary , is the positive coordinate tangent vectors when you fix a positively oriented local parametrization.

The -fold cross product on of vectors is defined as the unique vector, denoted by such that

or in a more fancy way, the Hodge dual of the -form on under the standard dot product.

[Yes indeed! I’ve updated the note accordingly, thanks! -KKK]

I think in Theorem 3, the RHS of the equality should have constant instead of .

Hmm…. I don’t really think so. The reason is the following: In , in normal coordinates,

Thus Stokes theorem gives

But

by a similar computation as before. So

By the way, to type in latex, you have to put the word “latex” (without the quotation mark “) in between two $’s. The format is

$latex your-latex-code-here$