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.

Theorem 1 [Reilly’s formula]

\displaystyle  \int_\Omega (\overline \Delta f)^2-|\overline \nabla ^2f|^2= \int_\Omega Rc(\overline \nabla f, \overline \nabla f)+\int_{\partial \Omega} 2u \Delta z+H u^2 +A(\nabla z,\nabla z). \ \ \ \ \ (1)

 

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<n},

\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<n}(X_j \overline \nabla _j X_n- X_n \overline \nabla_j X_j) \end{array}

Consider

\displaystyle  \begin{array}{rcl}  \sum_{j<n} \overline \nabla_j X_j =\sum_{j<n} \langle \overline \nabla_j X, e_j\rangle =\sum_{j<n} \langle \overline \nabla_j (Z+ u \nu), e_j\rangle &=&\sum_{j<n} \langle \nabla_j Z, e_j\rangle+u\langle \overline \nabla_j \nu, e_j\rangle \\ &=& \mathrm{div} \,Z +Hu. \end{array}

and for {j<n}

\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<n}(X_j \overline \nabla _j X_n- X_n \overline \nabla_j X_j) &=& \int_{\partial \Omega} \sum_{j<n}(Z_j(u_j -\langle Z, \overline \nabla_j \nu\rangle) - u(\mathrm{div} \,Z +Hu))\\ &=& \int_{\partial \Omega} (\langle Z,\nabla u\rangle-A( Z,Z) -u\, \mathrm{div} \,Z -Hu^2)\\ &=& -\int_{\partial \Omega} (A( Z,Z) +2u\, \mathrm{div} \,Z +Hu^2) \end{array}

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

  1. 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}.
  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

Let’s return to the application of Theorem 1.

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

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