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.

Induced orientation of the boundary from the surface

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
    $latex your-latex-code-here$

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