## Some integral formulas for hypersurface in Euclidean space

Recently, Hon Leung and Zorn Leung are teaching differential geometry to some bright secondary school students. So I am affected by them to make my revision on geometry of curves and surfaces in the Euclidean space. In this post, we will study some integral formula for closed hypersurface in Euclidean space. Perhaps the most famous example is the Gauss-Bonnet theorem, which state that for a closed (compact without boundary) surface in ${\mathbb{R}^3}$,

$\displaystyle \int_M K dS=2\pi \chi(M)$

where ${K}$ is the Gaussian curvature and ${\chi}$ is the Euler’s characteristic. We will see that by applying divergence theorem in different settings, we can derive many integral formulas (e.g. Gauss-Bonnet and Wongting’s first and second theorem).

Notations
In this post ${M}$ is always an orientable hypersurface in ${\mathbb{R}^{m+1}}$ (e.g. Mobius band is not allowed), ${n}$ is a unit normal vector field of ${M}$, chosen to be “outward” whenever this makes sense (e.g. ${M}$ closed, i.e. compact without boundary). $X$ is the position vector of $M$. ${h_{ij}:=\langle \partial _in,\partial_jX\rangle=-\langle n,\partial _i\partial_jX\rangle}$ is the second fundamental form. ${H}$ always denote the mean curvature of ${M}$ w.r.t. ${n}$, i.e. ${H=\sum_{ij}g^{ij}h_{ij}}$. Alternatively ${H=\sum_i {\langle \partial_i n, e_i\rangle}}$ for an othonormal frame ${\{e_i\}_{i=1}^m}$. In our convention, the mean curvature of sphere of radius ${r}$ in ${\mathbb{R}^3}$ w.r.t. outward normal is ${2/r}$.

Remark 1

1. We will write ${\nabla}$ as the covariant derivative on ${M}$; ${\partial_i }$ or ${\frac{\partial}{\partial x^i}}$ as partial derivatives. Then for function ${f}$, ${\nabla_i f=\partial _i f}$ but

$\displaystyle \nabla^2_{ij} f:=\nabla^2 f(X_i,X_j)\neq \nabla_i(\nabla_j f)(:=\partial _i \partial _j f)$

in general.

2. A very simple but important observation which we will always use is that for any tangent vector ${V}$ of ${M}$,

$\displaystyle \boxed{\nabla_V X=V(X)=V}$

when we regard ${\nabla_V X}$ as a vector rather than ${m+1}$ functions. (You may say: no big deal! What’s the difference? This problem becomes subtle when we come to the second derivative of ${X}$. See the next point. )

3. In general, when regarding ${X}$ as a vector-valued function, and regarding ${X_k:=\frac{\partial X}{\partial x^k}}$ as a tangent vector field, then all the following three terms are different:

$\displaystyle \nabla^2_{ij}X, \nabla_{X_i}X_j, \partial_i\partial_jX.$

Actually, they are related by (denote ${\nabla_{X_i}X_j}$ by ${\nabla_{i}X_j}$)

$\displaystyle \nabla_{ij}^2 X= \partial _i\partial _j X-\nabla_iX_j. \ \ \ \ \ (1)$

In particular, given the fact that ${\nabla_i X_j=(\partial _iX_j)^T}$ (the tangential part), it follows that ${\nabla^2_{ij}X}$ is normal to ${M}$, whereas ${\nabla_i X_j}$ is tangential. (1) can be seen as follows. For tangent vector fields ${U,V}$,

$\displaystyle \begin{array}{rcl} \nabla^2 X(U,V):= (\nabla_U(\nabla X))(V)&=& U(\nabla X(V))- \nabla X(\nabla_U V)\text{\quad(product rule)}\\ &=&U(V(X))- (\nabla_U V)X\quad\quad\text{(def. of }\nabla X(\cdot))\\ &=&U(V(X))- \nabla_U V.\quad \quad\text{(2nd remark)} \end{array}$

Put ${U=\partial_iX, V=\partial _jX}$, the result follows. Using ${\partial _i\partial_jX= \nabla_iX_j-h_{ij}n}$, we also have

$\displaystyle \boxed{\nabla^2_{ij}X=-h_{ij}n.}$

Proposition 1 For a closed hypersurface ${M\subset \mathbb{R}^{m+1}}$, if ${\Delta }$ is the (intrinsic) Laplacian on ${M}$, when regarding the position vector ${X}$ as a ${\mathbb{R}^{m+1}}$-valued function (i.e. ${m+1}$ “independent” functions), then

$\displaystyle \Delta X= -Hn\in \mathbb{R}^{m+1}. \ \ \ \ \ (2)$

Remark 2

1. This is a generalization of ${\alpha''=-kn }$ for arclength-parametrized curve. E.g. ${\alpha=(\cos s, \sin s)}$, ${\alpha''=-\alpha}$ is the inward normal and ${k=1}$.

Proof: Let ${x^i}$ be a coordinates on ${M}$, then using ${\nabla^2_{ij}X=-h_{ij}n}$ as in the first remark, we have

$\displaystyle \Delta X=\sum_{ij}g^{ij}(-h_{ij}n)=-Hn.$

Here ${(g^{ij})=(g_{ij})^{-1}}$. $\Box$

By divergence theorem,

Corollary 2 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M^m\subset \mathbb{R}^{m+1}}$,

$\displaystyle \int_M Hn\;dS=0.$

Theorem 3 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M^m\subset \mathbb{R}^{m+1}}$,

$\displaystyle \int_M H(x)n(x)\cdot X(x)dS(x)= m \text{ Area}(M)$

where ${\text{Area}(M)}$ is the ${m}$-dimensional area (measure) of ${M}$.

Proof: By Proposition 1 and divergence theorem,

$\displaystyle \begin{array}{rcl} -\int_M Hn\cdot X\,dS&= &\int_M(\Delta X)\cdot X\,dS\\ &=&\int_M(\frac{1}{2}\Delta (|X|^2)-| \nabla X|^2)\,dS\\ &=&-\int_M|\nabla X|^2\,dS. \end{array}$

Let ${\{x^i\}_{i=1}^m}$ be a coordinates on ${ M}$ which is orthnormal at ${p}$, then at ${p}$,

$\displaystyle |\nabla X|^2=\sum_{i=1}^m |\nabla_i X|^2=\sum_{i=1}^m \left|\frac{\partial X}{\partial x^i}\right|^2=m. \ \ \ \ \ (3)$

Thus

$\displaystyle \int_M Hn\cdot X\,dS= m \int_M \,dS=m\text{ Area}(M).$

$\Box$

By noting that if we change the coordinates ${X\rightarrow X+X_0}$ for any fixed ${X_0}$, ${H}$ and ${n}$ and ${\text{Area}(M)}$ are unchanged, i.e. independent of translation (naturally), thus we can slightly generalize the above theorem to

Theorem 4 For a closed hypersurface ${M^m\subset \mathbb{R}^{m+1}}$, for any ${X_0\in \mathbb{R}^n}$,

$\displaystyle \int_M H(x)n(x)\cdot (X(x)-X_0)dS(x)= m \text{ Area}(M).$

From this we can also see that ${\int_M H(x)n(x)\cdot X_0\,dS(x)= 0}$ for any ${X_0}$, so this in turn implies corollary 2:

$\displaystyle \int_M H(x)n(x)dS(x)= 0.$

Proposition 5 (Codazzi equations) Under any local coordinates ${x^i}$, we have

$\displaystyle \nabla _k h_{ij}= \nabla _j h_{ik}= \nabla _i h_{kj}.$

Note that by symmetry of ${h_{ij}}$, this means that the 3 indices ${i,j,k}$ of ${\nabla _k h_{ij}}$ can indeed be interchanged in any order. Note: ${\nabla_ih_{jk}:=(\nabla_ih)(X_j,X_k)\neq \nabla_i(h_{jk})}$.

Proof: (There is a more intrinsic way of proving it in the more general Riemannian setting, but here I choose to do it in a more clumsy but explicit way. ) By (1), we have

$\displaystyle \partial_{ij}^2 X= \sum_k\Gamma_{ij}^k \partial_kX -h_{ij}n \ \ \ \ \ (4)$

where ${\Gamma_{ij}^k}$ are the Christoffel symbols: ${\nabla_i(X_j)=:\sum_k\Gamma_{ij}^k X_k.}$

As ${h_{ij}= -\langle n, \partial ^2 _{ij} X\rangle}$, by (4)

$\displaystyle \begin{array}{rcl} \partial_k h_{ij}=-\langle \partial _k n, \partial ^2 _{ij} X\rangle-\langle n, \partial ^3_{kij} X\rangle &=& -\langle \partial _k n, \sum _l\Gamma _{ij}^l \partial _l X\rangle-\langle n, \partial ^3 _{kij} X\rangle\\ &=& -\sum_l\Gamma _{ij}^l h_{kl}-\langle n, \partial ^3 _{kij} X\rangle. \end{array}$

Thus with a computation similar to (1), we have

$\displaystyle \begin{array}{rcl} \nabla_kh_{ij} &=& \partial _k h_{ij}- \sum_l\Gamma_{ki}^l h_{lj}- \sum_l\Gamma _{kj}^l h_{ik}\\ &=& -\sum_l\Gamma _{ij}^l h_{kl}-\langle n, \partial ^3 _{kij} X\rangle- \sum_l\Gamma_{ki}^l h_{lj}- \sum_l\Gamma _{kj}^l h_{il}\\ \end{array}$

Thus by interchanging ${i}$ and ${k}$ in above, noting that ${\partial ^3 _{kij} X=\partial ^3 _{ikj} X}$ and ${\Gamma_{ki}^l =\Gamma_{ik}^l }$,

$\displaystyle \begin{array}{rcl} \nabla_kh_{ij}-\nabla_ih_{kj} &=& -\sum_l\Gamma _{ij}^l h_{kl}- \sum_l\Gamma _{kj}^l h_{il}-(-\sum_l\Gamma _{kj}^l h_{il}-\sum_l\Gamma _{ij}^l h_{kl})\\ &=&0. \end{array}$

$\Box$

Definition 6 For a hypersurface ${M\subset \mathbb{R}^{m+1}}$, if ${\{\lambda_i\}_{i=1}^m}$ are the principal curvatures, i.e. eigenvalues of the second fundamental form: ${h_{ij}=\lambda_ig_{ij}}$, then the scalar curvature ${R}$ is defined as

$\displaystyle R=\sum_{i, j=1, i\neq j}^m\lambda_i\lambda_j.$

Note that $R$ is independent of our choice of $n$, so it is a well-defined notion regardless of whether $M$ is orientable or not.

Remark 3 For ${m=2}$, ${R=2\lambda_1\lambda_2=2K}$ where ${K}$ is the Gaussian curvature.

Proposition 7 Regarding ${X}$ as a ${\mathbb{R}^{m+1}}$-valued function on ${M}$, then we have

$\displaystyle \Delta n = \langle \nabla H, \nabla X\rangle -|h|^2n=\nabla H -|h|^2n\in\mathbb{R}^{m+1}.$

where ${h=h_{ij}}$ is the second fundamental form and ${\langle \nabla H, \nabla X\rangle \in \mathbb{R}^{m+1}}$(!) For a local coordiantes ${\{x^j\}^m_1}$ which is orthonormal at ${p}$, ${\langle \nabla H, \nabla X\rangle}$ is defined by

$\displaystyle \langle \nabla H, \nabla X\rangle:=\sum_{i=1}^m \frac{\partial H}{\partial x^i} \frac{\partial X}{\partial x^i}=\nabla H\in T_pM\subset \mathbb{R}^{m+1}$

at ${p}$.

Proof: For convenience we use normal coordinates around ${p}$. Then at ${p}$, using Codazzi equations (Proposition 5) and (2),

$\displaystyle \begin{array}{rcl} \Delta n= \displaystyle \sum_{i} \partial _i\partial_i n &=& \displaystyle \sum_i \partial _i(\sum_j h_{ij}\partial_jX)\\ &=& \displaystyle \sum_{i,j} \partial _i h_{ij}\partial_jX+ h_{ij}\partial _i\partial _j X\\ &=& \displaystyle \sum_{i,j} (\partial _j h_{ii}\partial_jX-h_{ij}h_{ij}n)\\ &=&\displaystyle \sum_{j} \partial _j H\partial_jX-|h|^2n\\ &=& \displaystyle \langle \nabla H, \nabla X\rangle-|h|^2n. \end{array}$

$\Box$

As a corollary, we have

Theorem 8 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M^m\subset \mathbb{R}^{m+1}}$,

$\displaystyle \int_M \nabla H dS = \int_M |h|^2 n \;dS\in \mathbb{R}^{m+1}.$

Theorem 9 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M\subset \mathbb{R}^{m+1}}$, ${m\geq 2}$,

$\displaystyle \int_M H dS =\frac{1}{m-1}\int_M RX\cdot n\,dS.$

Example 1 For example when ${m=2}$, this becomes

$\displaystyle \int_M H dS =2\int_M KX\cdot n\,dS.$

Let us verify this for the sphere of radius ${r}$. We have ${H=2/r}$, ${K=1/r^2}$, so ${LHS= \frac{2}{r}4\pi r^2= 8\pi r}$. ${RHS= 2 \frac{1}{r^2} r(4\pi r^2)=8\pi r }$.

Proof:
First by applying divergence theorem two times, and using Proposition 7: (note that ${\langle \nabla H ,\nabla X\rangle \in \mathbb{R}^{m+1}}$, so it is valid (though a bit strange in notations) to talk about ${\langle \nabla H ,\nabla X\rangle \cdot X}$, also we omit ${dS}$ for notational convenience)

$\displaystyle \begin{array}{rcl} \displaystyle\int_M H dS= \displaystyle\int_M Hn\cdot n&=& \displaystyle-\int_{M} \Delta X \cdot n\\ &=&\displaystyle\int_M \nabla X\cdot \nabla n\\ &=&\displaystyle-\int_M X\cdot \Delta n\\ &=&\displaystyle-\int_M X\cdot (\langle \nabla H ,\nabla X\rangle - |h|^2n)\\ &=&\displaystyle -\int_M \langle \nabla H ,\nabla X\rangle \cdot X+ \int_M |h|^2 X\cdot n \end{array}$

Consider the first term above, by divergence theorem, (here the ${\cdot}$ always denote the Euclidean dot product, rather than the intrinsic inner product on ${M}$)

$\displaystyle \begin{array}{rcl} -\int_M \langle \nabla H , \nabla X \rangle \cdot X&=& \int_M (H \Delta X \cdot X+ H \nabla X\cdot \nabla X) \quad \text{(why?)} \end{array}$

As in (3), ${\nabla X\cdot \nabla X= m}$, and using Proposition 1, ${\Delta X=-Hn}$, so the above becomes

$\displaystyle -\int_M \langle \nabla H , \nabla X \rangle \cdot X= -\int_M H^2 X\cdot n+ m\int_M H.$

Put this back in the first computation, we have

$\displaystyle \int_M H = -\int_M (H^2-|h|^2)X\cdot n+m\int_M H$

Noting that ${H^2-|h|^2=(\sum_i \lambda_i)^2-\sum_i \lambda_i^2=R}$, we conclude that

$\displaystyle \int_M H dS =\frac{1}{m-1}\int_M (H^2 -|h|^2)X\cdot n\,dS=\frac{1}{m-1}\int_M RX\cdot n\,dS.$

$\Box$

By noting that ${R, H}$ and ${n}$ are invariant under the change of coordinates ${X\rightarrow X+X_0}$, we have the slight generalization

Theorem 10 For a closed hypersurface ${M\subset \mathbb{R}^{m+1}}$, ${m\geq 2}$, for any ${X_0\in \mathbb{R}^{m+1}}$,

$\displaystyle \int_M H dS =\frac{1}{m-1}\int_M R(X-X_0)\cdot n\,dS.$

In particular, as ${X_0}$ is arbitrary, we have

Corollary 11 (Hon Leung’s theorem, 2011) For a closed hypersurface ${M\subset \mathbb{R}^{m+1}}$, ${m\geq 2}$,

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

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