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.

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. $\Omega\subset\mathbb{R}^n$is 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/

$\mathbf{Problem:}$

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