## 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