Tangential gradient on the hypersurface


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

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