Category Archives: Mean curvature flow

Unit normal to a radial graph over sphere

Consider \Omega\subset \mathbb{S}^n is a domain in the sphere. S is a radial graph over \Omega.

\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}

What is the unit normal to this radial graph?

Suppose \{e_1,\cdots,e_n\} is a smooth local frame on \Omega. Let \nabla be the covariant derivative on \mathbb{S}^n. Tangent space of S consists of \{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n which are

\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x

In order to get the unit normal, we need some simplification. Let us assume \{e_i\} are orthonormal basis of the tangent space of \Omega and \nabla v=e_1(v)e_1. Then

\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2

Then we obtain an orthonormal basis of the tangent of S

\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}

We are able to get the normal by projecting x to this subspace

\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.

After normalization, the (outer)unit normal can be written

    \frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.

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Principle curvature of translator

Suppose {\Sigma\subset\mathbb{R}^{3}} is a translator in {e_{3}} direction. Denote {V=e_{3}^T} and {A} is the second fundamental form of {\Sigma}. Then it is well know that

\displaystyle \Delta A+\nabla_VA+|A|^2A=0 \ \ \ \ \ (1)

Choose a local orthonormal frame {\{\tau_1,\tau_2\}} such that {A(\tau_1,\tau_1)=\lambda}, {A(\tau_2,\tau_2)=\mu} and {A(\tau_1,\tau_2)=0} in the neighborhood of some point {p}. We want to change (1) to some expression on {\lambda} or {\mu}. To do that, we need to apply both sides of (1) to {\tau_1,\tau_1}, getting

\displaystyle (\nabla_VA)(\tau_1,\tau_1)=\nabla_V \lambda-2A(\nabla_V \tau_1,\tau_1)=\nabla_V\lambda \ \ \ \ \ (2)

where we have used {\nabla_V\tau_1\perp \tau_1}. Similarly

\displaystyle (\nabla_{\tau_k} A)(\tau_1,\tau_1)=\nabla_{\tau_k}\lambda

\displaystyle (\nabla_{\tau_k}A)(\tau_1,\tau_2)=(\lambda-\mu)\langle \nabla_{\tau_k}\tau_1,\tau_2\rangle

Now let us calculate the Laplacian of second fundamental form

\displaystyle (\nabla_{\tau_k}^2A)(\tau_k,\tau_1)=\tau_k(\nabla_{\tau_k}\lambda)-2(\nabla_{\tau_k}A)(\nabla_{\tau_k}\tau_1,\tau_1)-(\nabla_{\tau_k}\tau_k)\lambda

\displaystyle =\tau_1(\nabla_{\tau_1}\lambda)-2\frac{(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}{\lambda-\mu}.

Then

\displaystyle (\Delta A)(\tau_1,\tau_1)=(\nabla_{\tau_k}^2A)(\tau_k,\tau_1)= \Delta \lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}.

Combining all the above estimates to (1), we get

\displaystyle \Delta \lambda+\nabla_V\lambda+|A|^2\lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}=0.

where we write {A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}. Using this equation and the other one on {\mu}, one can derive that if {\Sigma} is mean convex then it is actually convex.

Laplacian on graph

Suppose {\Omega\subset \mathbb{R}^n} is a domain and {\Sigma=graph(F)\subset \mathbb{R}^{n+1}} is a hypersurface, where {F=F(u_1,\cdots,u_n)} is a function on {\Omega}. Define {f=f(u_1,\cdots,u_n)} on {\Omega}. Then {f} also can be considered as a function on {\Sigma}. How do we understand {\Delta_\Sigma f}?

Denote {\partial_i=\frac{\partial}{\partial{u_i}}} for short. If we pull the metric of {\mathbb{R}^{n+1}} back to {\Omega}, denote as {g}, then

\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2

where {W=\sqrt{1+|\nabla F|^2}} and {F_{u_i}=\frac{\partial F}{\partial u_i}}. Then one can use the local coordinate to calculate {\Delta_\Sigma f}

\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)

\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}

also one can see from another definition of Laplacian

\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]

\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f

By using the expression of {g^{ij}} stated above, we can calculate

\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}

It follows from the definition of tangential derivative on {\Sigma}, see, that

\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}

then

\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

where {H} is the mean curvature of the {\Sigma}. Combining all the above calculations,

\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

Frist and second variation of translating soliton

Suppose {F(x,t):\Sigma\times(-\epsilon,\epsilon)\rightarrow \mathbb{R}^{n+1}} be a variation of {\Sigma} with compact support and fixed boundary. Consider the weighted area functional

\displaystyle \mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}dv

What is the critical point of this area functional? To see that,

\displaystyle \frac{\partial}{\partial t}\mathcal{A}(\Sigma)=\int_\Sigma F_t(x_{n+1})e^{x_{n+1}}dv+e^{x_{n+1}}\partial_t dv

From some basic calculation in minimal surface(see colding minicozzi’s book)

\displaystyle \partial_t dv=-\langle F_t,\mathbf{H}\rangle+div_{\Sigma}F_t^T

Stokes’ theorem implies

\displaystyle \int_\Sigma div_\Sigma F_t^Te^{x_{n+1}}dv=-\int_{\Sigma}\langle F_t^T,\nabla_\Sigma x_{n+1}\rangle e^{x_{n+1}}dv=-\int_\Sigma F_t^T(x_{n+1})e^{x_{n+1}}dv

Combining these above fact, we get

\displaystyle \partial_t\mathcal{A}(\Sigma)=\int_\Sigma F_t^{\perp}(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma \langle F_t^{\perp},N\rangle N(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv

\displaystyle =\int_\Sigma \langle F_t,N\rangle \langle N,e_{n+1}\rangle-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma\langle F_t,\langle N,e_{n+1}\rangle N-\mathbf{H}\rangle dv

where {N} is the unit normal of {\Sigma}. Therefore the critical point of {\mathcal{A}} will satisfy

\displaystyle \mathbf{H}=\langle N, e_{n+1}\rangle N=e_{n+1}^{\perp}

This is so called translating soliton.

Now consider the second variation at a translation soliton,

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+2F_t(x_{n+1})^2dv+2F_t(x_{n+1})\partial_tdv+\partial_{tt}dv\right]

\displaystyle =\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+\partial_{tt}dv\right]

While

\displaystyle \partial_{tt}dv=-|\langle A(\cdot,\cdot),F_t\rangle|^2+|\nabla_\Sigma^NF_t|^2+div_\Sigma(F_{tt})

Suppose {F_t=\eta N}, where {\eta\in C_c^\infty(\Sigma)}, then {|\nabla_\Sigma^NF_t|^2=|\nabla \eta|^2}

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma [|\nabla\eta|^2-|A|^2\eta^2]e^{x_{n+1}}dv

Self-shrinker and polynomial volume growth

Proposition: If M is an entire graph of at most polynomial volume growth and H=\langle X,\nu\rangle, namely M is a self-shrinker. Then M is a plane.

Proof: Suppose

\displaystyle v=\frac{1}{\langle \nu, w\rangle}

Then one can derive the following equation

\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2

Multiplying both sides by e^{-|X|^2/2} and integration on M, which makes sense because of the polynomial volume growth, we get

\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu

However, integration by parts shows the LHS is zero. Thus A\equiv v\equiv 0, M must be a plane.

Parallel surfaces and Minkowski formula

Suppose {X:M^n\rightarrow \mathbb{R}^{n+1}} is an immersed orientable closed hypersurface. {N} is the inner unit normal for {X(M^n)} and denote by {\sigma} the second fundamental form of the immersion and by {\kappa_i}, {i=1,\cdots,n} the principle curvatures at an arbitrary point of {M}. The {r-}th mean curvature of {H_r} is obtained by applying {r-}elementary symmetric function to {\kappa_i}. Equivalently, {H_r} can be defined through the identity

\displaystyle P_n(t)=(1+t\kappa_1)\cdots(1+\kappa_n)=1+\binom{n}{1}H_1 t+\cdots+\binom{n}{n}H_n t^n

for all real number {t}. One can see that {H_1} represents the mean curvature of {X}, {H_n} is the gauss-Kronecker curvature. {H_2} can reflect the scalar curvature of {M} on the condition that the ambient manifold is a space form.

We want to study the consequence of moving the hypersurface parallel. Namely, define {X_t} to be

\displaystyle X_t= X-tN.

When {t} is small enough, {X_t} is well defined immersed hypersurface. Suppose {e_1,\cdots, e_n} are principle directions at a point {p} of {M}, then

\displaystyle \quad(X_t)_*(e_i)=(1+\kappa_it)e_i

here we identify {X_*(e_i)=e_i} as abbreviation. This implies that {N_t= N\circ X_t^{-1}} is also an unit normal field of {X_t}. The area element {dA_t} will be

\displaystyle dA_t=(1+t\kappa_1)\cdots(1+t\kappa_n)dA=P_n(t)dA.

The second fundamental form of {X_t} with respect to {N} will be

\displaystyle \sigma_t(v,w)=\langle N_t,\nabla^{\mathbb{R}^{n+1}}_vw\rangle=-\langle \nabla^{\mathbb{R}^{n+1}}_vN_t,w\rangle

for all {v,w} tangent vector fields on {X_t(M)}. Plugging in {v=(X_t)_*(e_i)} and {w=(X_t)_*(e_j)}, we get

\displaystyle (\nabla^{\mathbb{R}^{n+1}}_vw)(X_t(p))=(\nabla^{\mathbb{R}^{n+1}}_{e_i}e_j)(X(p))

\displaystyle \nabla_{v}^{\mathbb{R}^{n+1}}N_t=-\frac{\kappa_i}{1+t\kappa_i}v

So {e_1,\cdots, e_n} are also principle directions for {X_t} and principle curvatures are

\displaystyle \frac{\kappa_i}{1+t\kappa_i}

Another way to see this is by choosing a geodesic local coordinates such that {\partial_iX} are the principle directions of {X} at {p}. Then

\displaystyle \partial_j\partial_iX=\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij}

\displaystyle \partial_iN=-\kappa_i\partial_iX

\displaystyle \partial_i X_t=\partial_i X-t\partial_i N=\partial_i X+t\kappa_i\partial_iX

\displaystyle \partial_j\partial_iX_t=(1+t\kappa_i)\partial_j\partial_iX=(1+t\kappa_i)(\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij})

Since {g^{ij}_t=(1+\kappa_it)^{-2}\delta_{ij}} at {p}. Therefore we get the principle curvature are {\frac{\kappa_i}{1+t\kappa_i}}.

Therefore the mean curvature {H(t)} for {X_t} is

\displaystyle H(t)=\frac{1}{n}\frac{P_n'(t)}{P_n(t)}

Since we have identity

\displaystyle \Delta|X_t|^2=2n(1+H\langle X_t,N\rangle)

which implies

\displaystyle \int_M\left(1+H(t)\langle X_t,N\rangle\right)dA_t=0

Plugging in all the information,

\displaystyle \int_M\left(nP_n(t)+P_n'(t)\langle X,N\rangle-tP_n'(t)\right)dA=0

Reorder the terms in the above identity by the order of {t}, we get

\displaystyle \int_M (H_{r-1}+H_r\langle X,N\rangle )dA=0

One can use this to prove Heintze-Karcher inequality. There are Minkowski formula in Hyperbolic space and \mathbb{S}^n also.

Remark: S. Montiel and Anotnio Ros, compact hypersurfaces: the alexandrov theorem for higher order mean curvatures. Differential Geometry, 52, 279-296

Mean curvature of sphere cap

Suppose one has the disc {B_1=\{x\in \mathbb{R}^n||x|\leq 1\}}, {n\geq 3}, prescribe the ball with metric

\displaystyle g_{ij}=4u^{\frac{4}{n-2}}\delta_{ij}, \quad u=\left(\frac{\epsilon}{\epsilon^2+|x|^2}\right)^{(n-2)/2}

What is the mean curvature of the boundary? As we all know that under the Euclidean metric, the boundary of unit ball has mean curvature {h=1}. We want to use the formula of mean curvature under the conformal transmformation. Namely, suppose the {(M,g_0)} has mean curvature {h_0}, then under metric {g=v^{\frac{4}{n-2}}g_0}, the mean curvature of {(M,g)} will be

\displaystyle h_g=\frac{2}{n-2}v^{-\frac{n}{n-2}}\left(\frac{\partial v}{\partial \eta}+\frac{n-2}{2}h_0 v\right)

where {\eta} is the normal outer unit vector under {g_0}. Using the above principle, let {v=2^{\frac{n-2}{2}}u}, {g_0} be the Euclidean flat metric, then {h_0=1}.

\displaystyle \frac{\partial v}{\partial \eta}=(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)

\displaystyle \frac{n-2}{2}h_0 v=\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}

\displaystyle h_g=\frac{2}{n-2}\frac{(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)+\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}}{(2\epsilon)^{-\frac{n}{2}}(\epsilon^2+1)^{-\frac{n}{2}}}=\frac{\epsilon^2-1}{2\epsilon}

Remark: Escobar. Conformal Defromation of a Riemannnian metric to a constant scalar curvature metric with constant mean curvature on the boundary. Indiana University Mathematics Journal 1996.

A note on Huisken’s paper 1984

\mathbf{Thm(Myers):} If a Riemannian manifold has Ric\geq (n-1)K, then diam(M)\leq \frac{\pi}{K}.

Suppose we know a function h:M\to \mathbb{R}^+, M is compact and the gradient estimate

|\nabla h|\leq \eta^2 h^2_{\max} on M      (1)

and Ric_M\geq (n-1)c^2h^2 . Consider the relation \displaystyle h_{\min}\geq (1-\eta)h_{\max}.

\mathbf{Proof:}

Let x is a point on M, h achieves the maximum. Since (1), then

h(x)>(1-\eta)h_{\max} in B=B(x,\eta^{-1}h^{-1}_{\max}),

By the Myers thm, we get

\displaystyle diam(B)\leq \frac{\pi}{(1-\eta)h_{\max}}

But we know if \eta is small enough, then \displaystyle \eta^{-1}h^{-1}_{\max}>\frac{\pi}{(1-\eta)h_{\max}}. This forces diam(M)<\eta^{-1}h^{-1}_{\max}, then

\displaystyle h_{\min}\geq (1-\eta)h_{\max}.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Gerhard Huisken. Flow by mean curvature of convex surfaces into spheres.1984. p258 lemma 8.6.

The parameter \eta has relation with M and h_{\max}. We can not apply Myers thm directly. Guo Bin told me this method.

Second fundamental form of hypersurface

\mathbf{Problem:} Suppose F:\Omega\to \mathbb{R}^{n+1} is an embedding map. \Omega\subset\mathbb{R}^nis an open set. Consider the Gauss equation of hypersurface

\displaystyle \overline{\nabla}_XY=\nabla^M_XY+h(X,Y)\nu\quad (1),

\nu is the outer normal vector. The Codazzi equation

\displaystyle \overline{\nabla}_X\nu=-A(X)\quad (2)

Then the second fundamental form of M is B=h_{ij}\partial_iF\otimes\partial_j F, where

\displaystyle h_{ij}=h(\partial_iF,\partial_jF)

There are two ways to express the second fundamental form, one is from (1), the other is from (2). Firstly, let us consider from (2)

\displaystyle h(\partial_iF,\partial_jF)\langle A(\partial_iF),\partial_jF\rangle_g=-\langle \displaystyle \overline{\nabla}_{\partial_iF}\nu,\partial_jF\rangle_g

Extend \displaystyle \partial_iF=F_*(\frac{\partial}{\partial u^i}) and \nu to a vector field X and Y on \mathbb{R}^{n+1}

\displaystyle X^A\big|_{F(p)}=\frac{\partial F^A}{\partial u^i}\bigg|_p, \displaystyle Y^A\big|_{F(p)}=\nu^A\big|_{F(p)}

\displaystyle \overline{\nabla}_{\partial_iF}\nu\big|_{F(p)}=\overline{\nabla}_{X}Y\big|_{F(p)}=X^A\frac{\partial}{\partial x^A}\left(Y^B\right)\frac{\partial}{\partial x^B}\bigg|_{F(p)}=\frac{\partial F^A}{\partial u^i}\bigg|_{p}\frac{\partial}{\partial x^A}\left(\nu^B\right)\bigg|_{F(p)}\frac{\partial}{\partial x^B}\bigg|_{F(p)}

Note the fact that \displaystyle \partial_i\nu^B=\frac{\partial}{\partial u^i}\nu^B=\frac{\partial}{\partial x^A}\left(\nu^B\right)\bigg|_{F(p)}\frac{\partial F^A}{\partial u^i}\bigg|_{p}, we get

\displaystyle \overline{\nabla}_{\partial_iF}\nu\big|_{F(p)}=\partial_i\nu^B\frac{\partial}{\partial x^B}=\partial_i \nu

\displaystyle h_{ij}=-\langle \partial_i\nu,\partial_jF\rangle_g=-\partial_i\nu\cdot \partial_jF

Secondly, let us consider from (1). Using the same extension process, one can prove

\displaystyle \overline{\nabla}_{\partial_iF}\partial_jF=\partial_i\partial_jF,    \displaystyle \nabla^M_{\partial_iF}\partial_jF=\Gamma_{ij}^k\partial_{k}F

So (1) is equivalent to

\partial_i\partial_jF=\Gamma_{ij}^k\partial_{k}F+h_{ij}\nu

\text{Q.E.D}\hfill \square

\mathbf{Remark:}see the notation on https://sunlimingbit.wordpress.com/2013/02/10/tangential-gradient-on-the-hypersurface/

Tangential gradient on the hypersurface

\mathbf{Problem:}

Let the (u^1,u^2,\cdots,u^n) be the coordinates of \Omega and (x^1,x^2,\cdots,x^n,x^{n+1}) be the coordinates of \mathbb{R}^{n+1}. Denote \displaystyle\frac{\partial }{\partial x^i}=E_i. Suppose F:\Omega\to \mathbb{R}^{n+1} is an embedding map with F(\Omega)=M where \Omega\subset\mathbb{R}^n.

Then \displaystyle F_*\left(\frac{\partial }{\partial u^i}\right)=\frac{\partial F^l}{\partial u^i}\frac{\partial }{\partial x^l}=\partial_iF, 1\leq i\leq n form a basis for T_{F(p)}M at every p\in \Omega. Define the metric on M as

\displaystyle g_{ij}=\partial_iF\cdot \partial_j F

h:M\to \mathbb{R} is a function, we can define the tangential gradient of h as

\displaystyle \nabla^Mh=g^{ij}\partial_jh\partial_iF\quad (1)

Here we view h as a function on \Omega and \displaystyle \partial_jh=\frac{\partial }{\partial u^j}h(u^1,u^2,\cdots,u^n)

Consider the case when F is a graph of some function.

\displaystyle F:\Omega\mapsto \mathbb{R}^{n+1}

\displaystyle (u^1,u^2,\cdots,u^n)\rightarrow (u^1,u^2,\cdots,u^n, f(u_1,u_2,\cdots,u_n))

Then \partial_iF=E_i+p_iE_{n+1}, here \displaystyle p_i=\frac{\partial f}{\partial u^i}.

\displaystyle g_{ij}=\delta_{ij}+p_ip_j and W=\sqrt{1+\sum p_i^2}. \displaystyle g^{ij}=\delta_{ij}-\frac{1}{W^2}p_ip_j

The unit normal vector of M is e_{n+1}=\frac{1}{W}(\sum_ip_iE_i-E_{n+1}).

Naturally, the tangential vector of h should be defined as

\overline{\nabla} h-(\overline{\nabla} h\cdot e_{n+1})e_{n+1}\quad (2)

here \overline{\nabla} is respect to \mathbb{R}^{n+1}. Verify that this definition of tangential gradient is consistent with (1)

\mathbf{Proof:} Let us find the explicit expression of (2).

\displaystyle \overline{\nabla}h =(h_{x^1}, h_{x^2},\cdots,h_{x^{n+1}}) here \displaystyle h_{x^A}=\frac{\partial }{\partial x^A}h(x^1,x^2,\cdots,x^{n+1})

\displaystyle \overline{\nabla}h\cdot e_{n+1}=\frac{1}{W}\sum_ip_ih_{x^i}-\frac{1}{W}h_{x^{n+1}}

\displaystyle (\overline{\nabla}h\cdot e_{n+1})e_{n+1}=\left(\frac{1}{W}\sum_ip_ih_{x^i}-\frac{1}{W}h_{x^{n+1}}\right)\frac{1}{W}\left(\sum_ip_iE_i-E_{n+1}\right)

\displaystyle (3)\quad \overline\nabla h-(\overline \nabla h\cdot e_{n+1})e_{n+1}=\left(h_{x^i}-\frac{1}{W^2}\sum_{j}p_ip_jh_{x^j}+\frac{1}{W^2}p_ih_{x^{n+1}}\right)E_i+\left(h_{n+1}+\frac{1}{W^2}\sum_jp_jh_{x^j}-\frac{1}{W^2}h_{x^{n+1}}\right)E_{n+1}

Next, let us calculate (1)

\partial_ih=h_{x^i}+p_ih_{x^{n+1}}, \partial_j F=E_j+p_jE_{n+1}

\displaystyle (1)=\sum_{ij}\left(\delta_{ij}-\frac{1}{W^2}p_ip_j\right)\left(h_{x^i}+h_{x^{n+1}}p_i\right)\left(E_j+p_jE_{n+1}\right)

Expand this by using the fact W=\sqrt{1+\sum_ip^2_i}, one will see this is equal to (3)

\text{Q.E.D}\hfill \square

But there should have a simple way to understand how (1) is defined. Actually, this can be seen from the identity

\langle\nabla^Mh, X\rangle_g=dh(X)=X(h)\quad (4)

holds for every vector field X on M. The definition of \nabla^Mh should satisfies (4), which can also be seen from chern’s book.

Plugging in  X=\partial_iF in (4), we get

\displaystyle \langle\nabla^Mh, \partial_i F\rangle_g=\partial_iF(h)=F_*(\frac{\partial}{\partial u^i})h=\partial_ih

considering g_{ij}=\partial_iF\cdot\partial_iF=\langle\partial_iF,\partial_jF\rangle_g, we get (1)

\mathbf{Remark:} Klaus Ecker, Regularity theory for mean curvature flow. Appendix A