## Reilly type formula and its applications

In this note, I will prove two Reilly type formula and show a few applications of it. Of course, these results may not be optimal and the applications I show merely give a taste of it.

Suppose ${\Omega^n}$ is an orientable Riemannian manifold with smooth boundary ${\partial \Omega}$. Let ${f}$ be a smooth function on ${\Omega}$ and ${z=f|_{\partial \Omega}}$. Let ${\overline \nabla }$ (resp. ${\nabla}$ ) be the connection on ${\Omega}$ (resp. ${\partial \Omega}$) and ${\overline \Delta}$ be the Laplacian on ${\Omega}$ (resp. ${\partial \Omega}$). Let ${ \nu}$ be the unit outward normal on ${\partial \Omega}$ and ${ u=\frac{\partial f}{\partial \nu}}$. Let ${A(X,Y)= g(\nabla _X \nu,Y)}$ be the second fundamental form and ${H=tr A}$ be the mean curvature.

Proof: Let ${\{e_i\}_{i=1}^n}$ be a local orthonormal frame on ${\Omega}$ such that on ${\partial\Omega}$, ${e_n=\nu}$. We denote ${\overline \nabla ^2 f(e_i,e_j)}$ by ${f_{ji}}$ etc. Then by Ricci identity, (repeated index are summed from ${1}$ to ${n}$ unless otherwise stated)

$\displaystyle f_j f_{iji}- f_j f_{iij}= f_j(-R_{ijil}f_l)= R_{jl}f_jf_l = Rc(\overline \nabla f, \overline \nabla f).$

On the other hand, Stokes theorem gives

$\displaystyle \begin{array}{rcl} \int_\Omega (f_j f_{iji}- f_j f_{iij})&=&\int_\Omega (-f_{ij}f_{ij}+f_{ii}f_{jj})+ \int_{\partial\Omega} f_j f_{ij}\nu_i -f_j f_{ii}\nu_j\\ &=&\int_\Omega ((\overline \Delta f)^2- |\overline \nabla ^2 f|^2)+ \int_{\partial\Omega} f_j f_{nj} -f_n \overline \Delta f \end{array}$

We have ${ \overline \Delta f= f_{nn} + H f_n +\Delta f = f_{nn}+ Hu + \Delta z }$ and for ${j,

$\displaystyle f_{jn} =\overline \nabla ^2 f(e_j, \nu)= e_j \nu f- \overline \nabla_{\overline \nabla _{e_j}\nu}f =u_j - \langle \overline \nabla_{e_j }\nu, (\overline \nabla f )^T\rangle = u_j - A(e_j ,\nabla z).$

Thus

$\displaystyle \begin{array}{rcl} \int_{\partial\Omega} f_j f_{nj} -f_n \overline \Delta f &=&\int_{\partial\Omega} \left(\sum^{n-1}_{j=1}\left[z_j (u_j - A(e_j ,\nabla z))\right]-Hu^2 -u\Delta z\right)\\ &=&\int_{\partial\Omega} \left(\langle \nabla u, \nabla z\rangle-A(\nabla z,\nabla z)-Hu^2 -u\Delta z\right)\\ &=&\int_{\partial\Omega} \left(-A(\nabla z,\nabla z)-Hu^2 -2u\Delta z\right). \end{array}$

Substitute this into (1), we have

$\displaystyle \int_\Omega Rc(\overline \nabla f, \overline \nabla f ) = \int_\Omega ((\overline \Delta f)^2- |\overline \nabla ^2 f |^2 )- \int_{\partial\Omega} (2u\Delta z + Hu^2 + A(\nabla z, \nabla z)).$

i.e.

$\displaystyle \int_\Omega ((\overline \Delta f)^2- |\overline \nabla ^2 f |^2 )=\int_\Omega Rc (\overline\nabla f, \overline \nabla f)+\int_{\partial\Omega} (2u\Delta z + Hu^2 + A(\nabla z, \nabla z))$

$\Box$

 Corollary 2 (Choi-Wang) Let ${(M^n, g)}$ be a compact orientable Riemannian manifold such that its Ricci curvature ${Rc\geq (n-1)kg}$ for some ${k> 0}$ and ${\Sigma}$ be an embedded compact orientable minimal hypersurface of ${M}$, then the first eigenvalue of ${\Sigma}$ $\displaystyle \lambda_1(\Sigma)\geq \frac{(n-1)k}{2}.$

Proof: From a result of Lawson [Theorem 2], ${\Sigma}$ divides ${M}$ into two components ${M_1}$, ${M_2}$ such that ${\partial M_i=\Sigma}$.

Let ${z}$ be a first eigenfunction of ${\Sigma}$, i.e. ${\Delta z=-\lambda_1 z}$. Let ${f}$ be the solution of the Dirichlet problem

$\displaystyle \begin{cases} \overline \Delta f = 0 \quad\text{on }\Omega,\\ f = z\quad \text{on }\Sigma. \end{cases}$

By Cauchy-Schwarz inequality, ${(\overline \Delta f)^2\leq n |\overline \nabla ^2 f|^2 }$, thus by (1), as ${H=0}$, we have

$\displaystyle \begin{array}{rcl} 0=\frac {n-1}n\int_\Omega (\overline \Delta f)^2 &\geq &\int_{\Omega_1} (n-1)k|\overline \nabla f|^2 -2\lambda_1\int_{\Sigma}u z+ \int_{\Sigma}A(\nabla z, \nabla z)\\ &= &\int_{\Omega_1} (n-1)k|\overline \nabla f|^2 -2\lambda_1\int_{\Sigma}\frac{\partial f}{\partial \nu} f+ \int_{\Sigma}A(\nabla z, \nabla z)\\ &= &\int_{\Omega_1} (n-1)k|\overline \nabla f|^2 -2\lambda_1\int_{\Omega_1}(|\overline\nabla f|^2+f\overline\Delta f)+ \int_{\Sigma}A(\nabla z, \nabla z)\\ &= &\int_{\Omega_1}( (n-1)k -2\lambda_1)|\overline\nabla f|^2+ \int_{\Sigma}A(\nabla z, \nabla z). \end{array}$

Note that the same inequality can be obtained by reversing the choice of ${\nu}$ (and thus ${A}$ and ${H}$ change sign), so we can assume that ${\int_\Sigma A(\nabla z, \nabla z)\geq 0}$. So we have

$\displaystyle 0\geq ((n-1)k-2\lambda_1)\int_{\Omega _1} |\overline\nabla f|^2.$

Since ${f}$ is not constant, we have ${\lambda_1(\Sigma)\geq \frac{(n-1)k}2}$. $\Box$

 Remark 1 Actually the above proof shows that $\displaystyle 2\lambda_1(\Sigma)\geq (n-1)k + \frac{|\int_\Sigma A(\nabla z,\nabla z)|}{\int_{\Omega_1}|\overline \nabla f|^2}.$

It is a standard result that (p.265 Example 2)

 Lemma 3 Suppose ${u}$ is a smooth function on ${\partial \Omega}$ and ${h}$ is a smooth function on ${\Omega}$, then there exists ${f \in C^2(\Omega)\cap C(\overline \Omega)}$ which solves the following Neumann boundary value problem $\displaystyle \begin{cases} \overline \Delta f = h \quad \text{on }\Omega\\ \frac{\partial f}{\partial \nu} =u \quad \text{on }\partial \Omega. \end{cases}$ if and only if $\displaystyle \int_\Omega h =\int_{\partial \Omega}u.$

 Theorem 4 [A Reilly type formula] Suppose ${X}$ is a vector field on ${(\Omega,g)}$, let ${\overline{\mathrm{div}}X}$ be the divergence of ${X}$ w.r.t. ${g}$. Let ${u=\langle X, \nu\rangle}$ and ${Z= X^T}$ be the tangential component of ${X}$ along ${\partial \Omega}$ and let ${\mathrm{div} Z }$ be the divergence of ${Z}$ along ${\partial \Omega}$. We have $\displaystyle \int_{\Omega} ((\overline {\mathrm{div}}X)^2- \overline \nabla_i X^j\, \overline \nabla _j X^i)= \int_\Omega Rc(X,X)+ \int_{\partial \Omega}(2u \,\mathrm{div} Z +Hu^2 +A(Z,Z)).$

Proof: The proof is very similar to that of Theorem 1. Let ${e_j}$ be a local orthonormal frame on ${\Omega}$ such that ${e_n=\nu}$ on ${\partial \Omega}$. We have

$\displaystyle (\overline \nabla _{ij}X_i -\overline \nabla_{ji}X_i)X_j=R_{ijli}X_l X_j=R_{jl}X_jX_l=Rc(X,X). \ \ \ \ \ (2)$

On the other hand,

$\displaystyle \begin{array}{rcl} \int_\Omega (\overline \nabla _{ij}X_i -\overline \nabla_{ji}X_i)X_j &=& \int_\Omega (-\overline \nabla _{j}X_i \overline \nabla _i X_j+\overline \nabla_{i}X_i\overline \nabla_j X_j) + \int_{\partial \Omega} (X_j \overline \nabla _j X_n- X_n \overline \nabla_i X_i)\nonumber \\ &=& \int_\Omega (-\overline \nabla _{j}X_i \overline \nabla _i X_j +(\overline {\mathrm{div}} X)^2) + \int_{\partial \Omega} (X_j \overline \nabla _j X_n- X_n \overline \nabla_i X_i)\nonumber \\ &=& \int_\Omega (-\overline \nabla _{j}X_i \overline \nabla _i X_j +(\overline {\mathrm{div}} X)^2) + \int_{\partial \Omega} \sum_{j

Consider

$\displaystyle \begin{array}{rcl} \sum_{j

and for ${j

$\displaystyle \overline \nabla _j X_n = \langle \overline \nabla_j X, \nu\rangle = e_j \langle X, \nu\rangle - \langle X, \overline \nabla _j \nu\rangle = u_j -\langle Z, \overline \nabla_j \nu\rangle .$

Thus

$\displaystyle \begin{array}{rcl} \int_{\partial \Omega} \sum_{j

Substitute into (2), (2), we have

$\displaystyle \int_{\Omega} ((\overline {\mathrm{div}}X)^2- \overline \nabla _{j}X_i \overline \nabla _i X_j)= \int_\Omega Rc(X,X)+ \int_{\partial \Omega}(2u \,\mathrm{div} Z +Hu^2 +A(Z,Z)).$

$\Box$

 Remark 2 In general, ${\overline \nabla _i X^j \neq \overline \nabla ^j X_i}$, thus ${\overline \nabla_i X^j \overline \nabla_j X^i\neq |\overline \nabla X|^2 = \overline \nabla _i X^j \,\overline \nabla ^i X_j}$. But in some special case, we can simplify the term ${\overline \nabla_i X^j \overline \nabla_j X^i}$. E.g. when ${X= \overline \nabla f}$ is the gradient vector field of some function ${f}$, then we have ${\overline\nabla _{ij}f= \overline\nabla _{ji}f}$, thus ${\overline \nabla_i X^j \overline \nabla_j X^i= |\overline\nabla^2 f|^2}$. Actually, it is not hard to see that $\displaystyle \overline{\mathrm{div}}\left((\overline {\mathrm{div}}X)X-\overline \nabla _XX \right)=(\overline {\mathrm{div}}X)^2- \overline \nabla_i X^j\, \overline \nabla _j X^i - Rc(X,X).$ Applying divergence theorem to this equation, we can prove the above result.

 Corollary 5 For a Killing vector field ${X}$ (i.e. the Lie derivative ${L_X g=0}$) on a compact orientable Riemannian manifold ${M}$, we have $\displaystyle \int_M (Rc(X,X)-|\overline \nabla X|^2) =0. \ \ \ \ \ (3)$

Proof: We have the formula ${\langle \overline \nabla _Y X, Z\rangle + \langle \overline \nabla _Z X, Y\rangle=0}$. In fact,

$\displaystyle \begin{array}{rcl} 0=(L_X g)(Y, Z) &=& L_X(g(Y, Z))- g(L_X Y, Z)- g(Y, L_XZ)\\ &=& g(\overline \nabla _X Y, Z)+ g(Y, \overline \nabla _X Z)- g ([X,Y], Z)- g(Y, [X,Z]) \\ &=& \langle \overline \nabla _YX, Z\rangle + \langle Y, \overline \nabla _Z X\rangle. \end{array}$

From this we have ${\overline \nabla _i X_j = - \overline \nabla _j X_i}$ and thus ${\overline {\mathrm{div}}X= g^{ij}\overline \nabla_i X_j=0}$. We also have

$\displaystyle -\overline \nabla_i X^j \overline \nabla _j X^i = \overline \nabla ^j X_i \overline \nabla _j X^i= |\overline \nabla X|^2.$

In view of Theorem 4, we have

$\displaystyle \int_M (Rc (X,X )- |\overline \nabla X|^2)=0.$

$\Box$

 Remark 3 The above result can also be proved by checking that for a Killing vector field ${X}$, $\displaystyle \overline \Delta f= |\overline \nabla X|^2 - Rc(X,X)$ where ${f= \frac 1 2 |X|^2}$.

By using this Reilly type formula, we recover a result of Bochner:

 Corollary 6 (Bochner) If ${(M,g)}$ is a compact oriented manifold with ${Rc\leq 0}$, then every field is parallel (i.e. ${\overline \nabla X=0}$). If, furthermore, ${Rc<0}$ at one point (i.e. ${Rc}$ is strictly negative definite at that point), then there are no nontrivial Killing vector fields.

Proof: The first statement follows directly from (3). The second statement is true because parallel vector field has constant length. $\Box$

 Lemma 7 Let ${\Sigma^n\subset \mathbb R^{n+1}}$ be an embedded compact hypersurface into the Euclidean space. Let ${\Omega}$ be the region enclosed by ${\Sigma}$ (by generalized Jordan curve theorem) and ${Vol(\Omega)}$ be its volume. Let ${x}$ be the position vector on ${\mathbb R^{n+1}}$. With the notation as in Theorem 1, we have $\displaystyle \int_\Sigma H\langle x,\nu\rangle ^2 = n(n+1) Vol(\Omega)+ \int_\Sigma A(x^T,x^T).$ Here ${x^T}$ is the tangential component of ${x}$ (regarding it as a vector at ${p\in \Sigma}$ by parallel translation).

Proof: Define ${f=\langle x,x\rangle}$ on ${\overline\Omega}$. First of all, for the standard orthonormal frame ${e_i}$

$\displaystyle \overline \nabla f= \sum_i e_i \langle x,x\rangle e_i= 2\sum_i \langle \overline \nabla_{e_i}x,x\rangle e_i =2\sum_i \langle e_i,x\rangle e_i=2x$

and

$\displaystyle \overline \nabla_{ij}f = 2\overline \nabla_i \langle x,e_j\rangle = 2 \langle \overline \nabla _i x, e_j\rangle = 2\langle e_i, e_j\rangle = 2\delta _{ij}.$

Thus

$\displaystyle u =\overline \nabla f \cdot \nu =2\langle x, \nu\rangle , \quad \overline \Delta f= 2(n+1), \quad|\overline \nabla ^2 f|^2= 4 (n+1)\quad \text{and}\quad \nabla z = 2x^T.$

Thus

$\displaystyle 4n(n+1)Vol(\Omega)= \int_\Omega ((\overline \Delta f)^2- |\overline \nabla ^2 f|^2) = \int_\Sigma 2u \Delta z + Hu^2 +4A(x^T, x^T). \ \ \ \ \ (4)$

Let ${\{E_i\}_{i=1}^n}$ be an orthonormal frame on ${\Sigma}$ such that ${\nabla_{E_i}E_j (p)=0}$, then at ${p\in \Sigma}$,

$\displaystyle \begin{array}{rcl} \Delta z= \sum^{n}E_i E_i f - {\nabla_{E_i}E_i}(f) &=&\sum^{n}2E_i \langle \overline \nabla _{E_i}x, x\rangle \\ &=&2\sum^{n}E_i \langle E_i, x\rangle \\ &=&2\sum^{n}\langle \overline \nabla_{E_i }E_i, x\rangle+\langle E_i, \overline \nabla_{E_i }x\rangle \\ &=&2\sum^{n}\langle \overline \nabla_{E_i }E_i, \langle x,\nu\rangle \nu\rangle+\langle E_i, E_i \rangle \\ &=&2(-\langle x,\nu\rangle H+n)\\ &=& 2n - Hu \end{array}$

where we have used ${\overline \nabla_{E_i}E_i (p)= (\overline \nabla_{E_i}E_i)^\perp}$. Substitute this into (4),

$\displaystyle 4n(n+1)Vol(\Omega)= \int_\Sigma( 4nu -Hu^2 +4A(x^T,x^T)).$

By divergence theorem,

$\displaystyle \int_\Sigma u =2\int _\Sigma x\cdot \nu= 2\int_\Omega \overline {\mathrm{div}}(x) = 2(n+1 )Vol(\Omega).$

Thus

$\displaystyle 4n(n+1)Vol(\Omega)= 8n(n+1)Vol(\Omega)- 4\int H\langle x,\nu\rangle ^2 +4\int_\Sigma A(x^T,x^T)$

i.e.

$\displaystyle \int_\Sigma H\langle x,\nu\rangle ^2 = n(n+1) Vol(\Omega)+ \int_\Sigma A(x^T,x^T).$

$\Box$

 Corollary 8 (Garay) If ${\Sigma \subset \mathbb R^{n+1}}$ is an embedded compact hypersurface which is convex in the sense that ${A\geq 0}$ and its mean curvature satisfies $\displaystyle H\leq \frac{n(n+1)Vol (\Omega)}{r^2 Area (\Sigma)}$ where ${r}$ is the radius of a sphere circumscribed around ${\Sigma}$, then ${\Sigma}$ is a sphere of radius ${r}$.

Proof: We can assume the circumscribed sphere is centered at ${0}$. Then ${\langle x,\nu\rangle ^2\leq |x|^2\leq r^2}$. Thus

$\displaystyle \int_\Sigma H\langle x, \nu\rangle ^2\leq n(n+1)Vol(\Omega).$

On the other hand, by Lemma 7, ${\int_\Sigma H\langle x, \nu\rangle ^2 \geq n(n+1)Vol(\Omega)}$ since ${A\geq 0}$. We conclude that ${H=C/r^2}$ is constant and thus ${\langle x,\nu\rangle ^2\equiv r^2}$. This implies ${|x|\equiv r}$. $\Box$

 Corollary 9 For any hypersuface ${\Sigma \subset \mathbb R^{n+1}}$ which is convex in the sense that ${A\geq 0}$, then $\displaystyle \frac{(n+1)Vol(\Omega)}{Area (\Sigma)}\leq r$ where ${r}$ is the radius of the smallest sphere which circumscribe ${\Sigma}$. Equality holds if and only if ${\Sigma }$ is a sphere of radius ${r}$.

Proof: We can assume ${0}$ is in the interior of ${\Sigma}$, then by a result of Hadamard, as ${\Sigma }$ is convex, we have ${\langle x, \nu\rangle \geq 0}$. So

$\displaystyle \int_\Sigma H \langle x,\nu\rangle ^2 \leq r\int _\Sigma H \langle x,\nu\rangle$

By Minkowski’s formula (Hon Leung’s theorem), we have

$\displaystyle \int_\Sigma H \langle x, \nu\rangle = n Area (\Sigma).$

Thus by Lemma 7, we have

$\displaystyle nr Area (\Sigma)\geq n(n+1 )Vol (\Omega).$

If the equality holds, then ${r\geq |x|\geq \langle x,\nu\rangle =r}$ as ${\int _\Sigma H\langle x,\nu\rangle >0}$. Therefore ${|x|\equiv r}$.

$\Box$

 Remark 4 Using Minkowski formula, we can also characterize when ${\Sigma }$ is a sphere as follows. We have the two formulas $\displaystyle \int_\Sigma H \langle x,\nu\rangle = n Area (\Sigma ) \quad\text{and} \quad \int_\Sigma \langle x, \nu\rangle = (n+1 ) Vol(\Omega ).$ If ${H \leq \frac{n Area (\Sigma)}{(n+1)Vol(\Omega)}=c}$, then $\displaystyle n Area(\Sigma)= \int_\Sigma H \langle x, \nu\rangle \leq \frac{n Area (\Sigma)}{(n+1)Vol(\Omega)}\int_\Sigma \langle x,\nu\rangle = n Area(\Sigma ).$ So $\displaystyle \int _\Sigma (H-c) \langle x,\nu\rangle =0.$ However, if ${\Sigma }$ is convex, then ${\langle x, \nu\rangle>0}$ if ${0}$ is enclosed by ${\Sigma}$, we must have ${H\equiv c}$ and thus ${\Sigma }$ is a sphere (by Aleksandrov’s theorem).

By the above remark, we conclude that

 Theorem 10 If ${\Sigma \subset \mathbb R ^{n+1}}$ is a compact convex hypersurface and $\displaystyle H \leq \frac{n Area (\Sigma )}{(n+1) Vol (\Omega )},$ then ${\Sigma}$ is a sphere.

 Lemma 11 For a hypersurface ${\Sigma\subset \mathbb R ^{n+1}}$ whose center of mass is ${0}$, i.e. ${\int_\Sigma x=0}$. Then its first eigenvalue ${\lambda_1}$ satisfies $\displaystyle \lambda_1 \leq \frac{n Area (\Sigma )}{ \int_\Sigma |x|^2}.$

Proof: Let ${x=(x^1,\cdots, x^{n+1})}$, then the gradient of ${x^i}$ (as a function on ${\Sigma}$) on ${\Sigma }$ is easily seen to be the tangential component of ${e_i}$, where ${e_i}$ are the standard basis of ${\mathbb R^{n+1}}$: ${\langle \nabla x^i , v\rangle = dx^i(v)=v^i }$. By the minimum principle, we have

$\displaystyle \lambda _1= \min _{\int _\Sigma f=0}\frac{\int _\Sigma |\nabla f|^2}{\int _\Sigma f^2}.$

Thus ${\int _\Sigma |\nabla x^i |^2 \geq \lambda _1 \int_\Sigma (x^i)^2}$ and so

$\displaystyle \int_\Sigma \sum _{i=1}^{n+1}|\nabla x^i|^2 \geq \lambda_1 \int _\Sigma \sum _{i=1}^{n+1} (x^i)^2 = \int _\Sigma |x|^2.$

Let ${\{v_j\}_{j=1}^n}$ be an orthnormal basis for ${T_p\Sigma}$, then ${\displaystyle\sum _{i=1}^{n+1}|\nabla x^i|^2=\sum _{i=1}^{n+1}\sum _{j=1}^n\langle e_i, v_j\rangle ^2=\sum _{j=1}^n\sum _{i=1}^{n+1}\langle v_j ,e_i\rangle ^2=n.}$ Therefore ${\lambda_1 \leq \frac{n Area(\Sigma)}{\int _\Sigma |x|^2}}$. $\Box$

 Corollary 12 (Garay) For a convex hypersurface ${\Sigma\subset \mathbb R ^{n+1}}$ with ${H \leq \frac{\lambda _1 (n+1)Vol(\Omega)}{Area (\Sigma)}}$, ${\Sigma }$ is a sphere.

Proof: By Lemma 11, ${\int_\Sigma |x|^2 \leq \frac {n Area (\Sigma)}{\lambda _1 }}$.

$\displaystyle \int_\Sigma H|x|^2 \geq \int_\Sigma H\langle x,\nu\rangle ^2 = n(n+1)Vol(\Omega) +\int_\Sigma A(x^T,x^T)\geq n(n+1) Vol(\Omega).$

If ${H\leq \frac{\lambda _1 (n+1)Vol(\Omega)}{Area (\Sigma)}}$, then

$\displaystyle \int_\Sigma H |x|^2 \leq \frac{\lambda _1 (n+1)Vol(\Omega)}{Area (\Sigma)}\int_\Sigma |x|^2= n(n+1)Vol(\Omega).$

As the equality holds, ${H}$ must be constant and thus ${\Sigma }$ is a sphere.

$\Box$