## Some integral formulas for hypersurface in Euclidean space 2

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

Theorem 1 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M^m\subset \mathbb{R}^{m+1}}$,

$\displaystyle \int_M ndS=0\in \mathbb{R}^{m+1}.$

Proof: Let ${e_i}$ be the standard coordinate vectors, let ${\Omega}$ be the region enclosed by ${M}$, then by divergence theorem,

$\displaystyle \int_M n\cdot e_i dS= \int_\Omega \text{div}(e_i) dV=0.$

Since ${e_i}$ is arbitrary, ${\displaystyle\int_M ndS=0}$. $\Box$

For ${m=1}$, the above theorem is trivial because ${M=\gamma(s), s\in [0,l]}$ and ${n=-J\gamma'(s)}$ where ${Je_1=e_2, Je_2=-e_1}$ is the anticlockwise rotation by ${\pi/2}$. It turns out that there is another way to prove the above theorem when ${m=2}$. The observation is the following

Theorem 2 (Hon Leung-Raymond’s theorem, 2011) For a closed surface ${M\subset \mathbb{R}^3}$ with outward unit normal $n$, suppose $\Omega\subset M$ is a domain with smooth one-dimensional boundary ${\partial \Omega=\Gamma}$ (not necessarily connected). Suppose ${T}$ is the positive tangent vector of ${\Gamma}$ (i.e. if we arclength-parametrize ${\Gamma(s)}$ such that ${\Omega}$ always lies on the left hand side of ${\Gamma(s)}$, then ${T=\Gamma'(s)}$), and ${X}$ is the position vector, then

$\displaystyle \int_\Omega ndS= \frac{1}{2}\int_{\Gamma} (X\times T )ds\in \mathbb{R}^3.$

Proof: Note that

$\displaystyle ndS= (\frac{\partial X}{\partial u}\times \frac{\partial X}{\partial v})du\wedge dv$

(as a vector-valued 2-form) for any positively oriented local coordinates ${\{u,v\}}$. We will simply denote ${\frac{\partial X}{\partial u}}$ by ${X_u}$, ${\frac{\partial ^2X}{\partial u\partial v}}$ by ${X_{uv}}$, and so on. Note that on one hand, by product rule,

$\displaystyle (X_u \times X_v )du\wedge dv=d\left((X\times X_v)dv\right)- (X\times X_{vu})du\wedge dv.$

On the other hand, we have

$\displaystyle (X_u \times X_v) du \wedge dv= d\left(-(X_u\times X )du\right)-(X_{uv}\times X)du\wedge dv.$

Combining, we have

$\displaystyle 2ndS= d((X\times X_u) du+ (X\times X_v)dv).$

(In a coordinate-free language, the above is just $d(X\times dX)$, which is understood as $\vec {a}\alpha \times \vec{b}\beta:=(\vec a\times \vec b)\alpha\wedge \beta$ for differential forms $\alpha, \beta$. Note that in this setting $dX\times dX\neq 0$!) So by Stokes theorem,

$\displaystyle \int _\Omega ndS = \frac{1}{2}\int_{\Gamma} (X\times X_u)du+(X\times X_v)dv.$

But for positively oriented ${\Gamma(s)}$ parametrized by arclength, ${T=\Gamma'=u'X_u+ v'X_v}$, so

$\displaystyle \int_\Omega ndS= \frac{1}{2} \int_{\Gamma }X\times T ds.$

$\Box$

In particular, for a closed surface ${M}$, since ${\partial M=\emptyset}$, we have recovered Theorem 1: ${\int_M ndS=0}$.

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 ${M^m\subset \mathbb{R}^{m+1}}$, if ${n}$ is its unit outward normal and ${\Omega}$ is a domain in ${M}$ with smooth ${(m-1)}$-dimensional boundary ${\Gamma}$ with induced orientation, then

$\displaystyle \int_\Omega n\,dS=\frac{1}{m} \int_{\Gamma}(X\times T_1\times\cdots \times T_{m-1}) ds$

where ${T_1, \cdots, T_{m-1}}$ is a positively oriented orthonormal basis for the tangent space of ${\Gamma}$ and ${X}$ is the position vector.

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

$\displaystyle m!n dS= \overbrace{dX\times \cdots \times dX}^{ m}=d(X\times \overbrace{dX\times \cdots \times dX}^{m-1}).$

For example for ${m=3}$ ($X_1:=\frac{\partial X}{\partial u^1}$),

$\displaystyle \begin{array}{rcl} dX\times dX\times dX &= &(X_1\times X_2\times X_3) du^1\wedge du^2\wedge du^3+(X_1\times X_3\times X_2 )du^1\wedge du^3\wedge du^2+\cdots\\ &=&(X_1\times X_2\times X_3) du^1\wedge du^2\wedge du^3+(X_1\times X_2\times X_3 )du^1\wedge du^2\wedge du^3+\cdots\\ &=& 3! (X_1\times X_2\times X_3 )du^1\wedge du^2\wedge du^3. \end{array}$

Thus by Stokes theorem,

$\displaystyle \int _\Omega ndS=\frac{1}{m!}\int_{\Gamma} X\times \overbrace{dX\times\cdots \times dX}^{m-1}.$

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

$\displaystyle X\times \overbrace{dX\times \cdots dX}^{m-1}= (m-1)! (X\times T_1\times \cdots \times T_{m-1})ds.$

Therefore we have

$\displaystyle \int_\Omega n\,dS= \frac{1}{m} \int_\Gamma (X\times T_1\times \cdots \times T_{m-1} )ds.$

$\Box$

Let me also state the discrete version of Theorem 1:

Theorem 4 (Hon Leung’s theorem, 2011, discrete version) If ${P\subset \mathbb{R}^{m+1}}$ is a ${(m+1)}$polytope and ${\{F_i\}}$ are all its ${m}$-dimensional faces, denote ${n_{F_i}}$ to be the unit outward normal on ${F_i}$, then

$\displaystyle \sum_i \text{Area}(F_i)n_{F_i}=0\in \mathbb{R}^{m+1}.$

Here ${\text{Area}(F)}$ is the ${m}$-dimensional volume of ${F}$.

Proof: Clearly we only have to prove the identity for ${P}$ to be a ${(m+1)}$simplex (triangle for ${m+1=2}$, tetrahedron for ${m+1=3}$), for, given a polytope ${P}$ we can triangulate into finitely many ${(m+1)}$-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 ${m+1=3}$ for notational simplicity. Without loss of generality the 4 vertices are ${o, a, b, c}$ where ${a,b,c}$ are linearly independent. Let us denote the normal of ${\triangle abc}$ simply by ${n_{abc}}$. Then

$\displaystyle \begin{array}{rcl} \text{Area}(\triangle abc)n_{abc}&=& \frac{1}{2}(c-b)\times (a-b)\\ &=&\frac{1}{2}(c\times a-b\times a- c\times b)\\ &=& -\text{Area}(\triangle oac)n_{oac}-\text{Area}(\triangle oba)n_{oba}-\text{Area}(\triangle ocb)n_{ocb}. \end{array}$

$\Box$

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### 4 Responses to Some integral formulas for hypersurface in Euclidean space 2

1. Edward Fan says:

Theorem 2 above can be generalized directly, with the $(m-1)$-fold cross product instead of the standard vector cross product. The theorem becomes

$\displaystyle\int_{\Omega} n \, dS = \frac{m-1}{m} \int_{\Gamma} X \times T_1 \times \cdots \times T_{m-1} ds$

where $M$ is a $m$-dimensional submanifold of $\mathbb{R}^{m+1}$, $\Omega$ is a domain in $M$ with smooth $m-1$-dimensional boundary $\Gamma$, $T_1, \ldots, T_{m-1}$ is the positive coordinate tangent vectors when you fix a positively oriented local parametrization.
The $(m-1)$-fold cross product on $\mathbb{R}^m$ of $m-1$ vectors $v_1, \ldots, v_{m-1}$ is defined as the unique vector, denoted by $v_1 \times \cdots \times v_{m-1}$ such that
$det(v_1, \ldots, v_{m-1}, w) = v_1 \times \cdots \times v_{m-1} \cdot w,$
or in a more fancy way, the Hodge dual of the $(m-1)$-form $v_1 \wedge \cdots \wedge v_{m-1}$ on $\mathbb{R}^m$ under the standard dot product.

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

2. Edward Fan says:

I think in Theorem 3, the RHS of the equality should have constant ${\frac{(m-1)!}{m}}$ instead of ${\frac{1}{m}}$.

3. KKK says:

Hmm…. I don’t really think so. The reason is the following: In $\mathbb{R}^{m+1}$, in normal coordinates,

$\begin{array}{rcl} d(X\times \overbrace{dX\times \cdots \times dX}^{m-1})&=&\overbrace{dX\times \cdots \times dX}^{m}\\ &=& m! (X_1\times \cdots \times X_m) du^1\wedge \cdots \wedge du^m\\ &=&m!n\,dS. \end{array}$

Thus Stokes theorem gives

$\displaystyle \int_\Omega n\,dS=\frac{1}{m!} \int_\Gamma (X\times \overbrace{dX\times \cdots \times dX}^{m-1}).$

But

$X\times \overbrace{dX\times \cdots \times dX}^{m-1}= (m-1)! (X\times T_1\times \cdots \times T_{m-1})ds$

by a similar computation as before. So
$\displaystyle \int_\Omega n\,dS= \frac{(m-1)!}{m!}\int _\Gamma (X\times T_1\times \cdots \times T_{m-1})\,ds=\frac{1}{m}\int _\Gamma (X\times T_1\times \cdots \times T_{m-1})\,ds.$

4. KKK says:

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\$