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


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

\mathbf{Remark:}see the notation on

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