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.

$\displaystyle \overline{\nabla}_XY=\nabla_XY+h(X,Y)\nu$,

$\nu$ is the outer normal vector.

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

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

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

Extend $\partial_iF=F_*(\frac{\partial}{\partial u^i})$ 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

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

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

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

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