Let $r:M\rightarrow{\mathbb{R}^{n+1}}$ be an isometric immersion and $M$ is an $n$-dimensional Riemannian Manifold. That is to say, $M$ is the hypersurface in $\mathbb{{R}^{n+1}}$.

Then we can introduce a normal vector field: $N:M\rightarrow{T\mathbb{{R}^{n+1}}}=\mathbb{R}^{n+1}\times{\mathbb{R}^{n+1}}$ satisfies $N_p\in{T_{r(p)}\mathbb{R}^{n+1}}=\{r(p)\}\times\mathbb{R}^{n+1}$. So we will have $T_{r(p)}\mathbb{R}^{n+1}=r_*(T_pM)\oplus{span\{N_p\}}$.

Before we talk about this problem, we look at the connection on $\mathbb{R}^{n+1}$. Let $\bigtriangledown$ be its connection and $X=\sum_{i=1}^{n+1}x_ie_i$, $Y=\sum_{i=1}^{n+1}y_ie_i$. So $\bigtriangledown_XY=\sum_{i=1}^{n+1}\sum_{j=1}^{n+1}x_je_j(y_i)e_i$.

When I read book, I find two different definitions of The Second Fundamental Form. I want to verify that they are the same.

Let $\bar{n}$ denote the Guass Map which is actually: $n=\pi_2\circ{N}:M\rightarrow{\mathbb{R}^{n+1}}$.

The first definition is:


The second definition is:


So I think they are the same. Then I try to prove it. But I am failed.

I want to prove $\bigtriangledown_{\frac{\partial{r}}{\partial{x_i}}}N=\frac{\partial{n}}{\partial{x_i}}$.

Proof. Let $\frac{\partial{r}}{\partial{x_i}}=\sum_{k=1}^{n+1}\frac{\partial{r_k}}{\partial{x_i}}e_k$ and $N=\sum_{k=1}^{n+1}n_ke_k$.
Then I have no idea.

How to prove??

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    $\begingroup$ Use Gauss' equation in the form $\nabla_XY=D_XY-(D_XY\cdot n)n$ where $D$ is the ${\Bbb{R}}^{n+1}$-standard connection. $\endgroup$ – janmarqz Jan 15 '14 at 2:18
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    $\begingroup$ looking at the case of surfaces in ${\Bbb{R}}^3$ will give the clue. $\endgroup$ – janmarqz Jan 15 '14 at 2:26
  • $\begingroup$ You use the Guass's Equation, so is that to say that one should use the connection's fractionation along tangent space in the inner product in The Second Fundamental Form? And could you write down the answer by using the local representation of The Second Fundmantal Form? I will be more thankful.@janmarqz $\endgroup$ – gaoxinge Jan 15 '14 at 2:40


Consider that if $\langle\partial_j,n\rangle=0$ then also $\langle D_{\partial_i}\partial_j,n\rangle+\langle\partial_j,D_{\partial_i}n\rangle=0$. Use it to correct your guessing.

One is taking $\partial_i=\frac{\partial}{\partial x^i}$ for the coordinated tangent vectors.


Very probable you are using the standard covariant derivative $\nabla$ of ${\Bbb{R}}^{n+1}$.

In this case one computes for $\nabla_XY=[JY]X$ for vector fileds $X,Y$ where $[JY]$ means the Jacobian of the field $Y$.

The covariant derivative and the pairing satisfy: $$ \frac{\partial}{\partial x^i}\langle X,Y\rangle= \langle \nabla_{\frac{\partial}{\partial x^i}}X,Y\rangle + \langle X,\nabla_{\frac{\partial}{\partial x^i}}Y\rangle .$$

When applied to $N$ and $\frac{\partial}{\partial x^j}$ which $\langle N,\frac{\partial}{\partial x^i}\rangle=0$ then \begin{eqnarray*} \frac{\partial}{\partial x^i}\langle N,\frac{\partial}{\partial x^j}\rangle&=&0\\ \langle \nabla_{\frac{\partial}{\partial x^i}}N,\frac{\partial}{\partial x^j}\rangle+\langle N,\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}\rangle&=&0\\ \langle \nabla_{\frac{\partial}{\partial x^i}}N,\frac{\partial}{\partial x^j}\rangle&=&- \langle N,\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}\rangle \end{eqnarray*}

So $$II|_U=\sum_{ij} \langle\nabla_{\frac{\partial{}}{\partial{x^i}}}N,\frac{\partial{}}{\partial{x^j}} \rangle dx^i\otimes{dx^j}, $$ is also $$\qquad =-\sum_{ij} \langle N,\nabla_{\frac{\partial{}}{\partial{x^i}}}\frac{\partial{}}{\partial{x^j}} \rangle dx^i\otimes{dx^j}. $$

  • $\begingroup$ $\nabla_{\partial_i}n=D_{\partial_i}n$ because $<\partial_i,n>=0$ $\endgroup$ – janmarqz Jan 15 '14 at 2:38
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    $\begingroup$ meanwhile check the pages 26-27 of wisdom.weizmann.ac.il/~yakov/scanlib/hicks.pdf for the form of the Gauss's eqn mentioned $\endgroup$ – janmarqz Jan 15 '14 at 3:06
  • $\begingroup$ write briefly $\partial_i=\frac{\partial}{\partial x^i}$ $\endgroup$ – janmarqz Jan 15 '14 at 3:09

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