Gauss-Weingarten relations

Question

In the 3.4 theorem, $F:R^n\supset U \rightarrow F(U)\subset M\subset R^{n+1}$ is a local represent of Riemannian manifold $M$. $g_{ij}$ is Riemannian metric , and $h_{ij}$ is second fundamental form. $\nu$ is outward normal vector. Then, how to show the red line from the Weingarten equation ?

Answer 1

A surfaces in the euclidean space the Gauss equations and the Weingarten equations are

$$X_{uu} = \Gamma^1 _{1 1}X_u + \Gamma^2 _{ 1 1}X_v + eN, $$ $$X_{uv} =\Gamma^1_{1 2}X_u + \Gamma^2_{ 1 2}X_v + fN, $$ $$X_{vu} =\Gamma^1_{2 1}X_u + \Gamma^2 _{ 2 1}X_v + fN, $$ $$X_{vv }=\Gamma^1_{2 2}X_u + \Gamma^2_{ 2 2}X_v + gN,$$ $$N_u = aX_u + cX_v,$$ $$N_v = bX_u + dX_v$$

A case general is your question.

Answer 2

Let $\frac{\partial{F}}{\partial x_{k}} = X_k$, and

$$ \nabla_{X_{i}} X_{j} = \sum\limits_{k=1}^n \Gamma^k_{ij} X_k.$$ Because tangential component of Euclidean directional derivative is the Riemannian connection on the surface we have that
$$\frac{\partial^2{F}}{\partial{ x_{i}}\partial{x_{j}}}=D_{X_{i}} X_{j} = \nabla_{X_{i}} X_{j} + \alpha \nu. $$ Now, if we multiply( inner product ) both sides of the last equation by $\nu$, we get $$<D_{X_{i}} X_{j},\nu> = <\nabla_{X_{i}} X_{j},\nu> + <\alpha \nu,\nu>= \alpha;$$ (note that $\nabla_{X_{i}} X_{j} \perp \nu, $ so $<\nabla_{X_{i}} X_{j},\nu>=0$). Thus $\alpha= -h_{ij}$. Therefore $$\frac{\partial^2{F}}{\partial{ x_{i}}\partial{x_{j}}} = \nabla_{X_{i}} X_{j} -h_{ij}\nu = \sum\limits_{k=1}^n \Gamma^k_{ij} X_k -h_{ij}\nu.$$ Since $\left< \nu , X_k \right>= 0$, we infer from compatibility of metric with respect to the connection that $\left< \frac{\partial{\nu}}{\partial{ x_{j}} }, X_k \right>+ \left<\nu, \frac{\partial^2{F}}{\partial{ x_{k}}\partial{x_{j}}} \right>=0$. Thus $$\left< \frac{\partial{\nu}}{\partial{ x_{j}} }, X_k \right>= h_{jk}.$$ Similarly, we infer from $\left< \nu, \nu \right> = 1$ and compatibility of metric with respect to connection that $\left< \frac{\partial{\nu}}{\partial x_{j}}, \nu \right> =0 $. Thus $ \frac{\partial{\nu}}{\partial x_{j}}$ is in the tangent space. So there are scalars $\alpha_i $ such that $$ \frac{\partial{\nu}}{\partial x_{j}} = \sum\limits_{l=1}^n \alpha^j_{l} X_i.$$

We have that $$ h_{jk}= \left< \frac{\partial{\nu}}{\partial x_{j}} , X_k \right> = \sum\limits_{i=1}^n \alpha^j_{i} \left< X_i, X_k \right>= \sum\limits_{i=1}^n \alpha^j_{i} g_{ik}= (\alpha^j_{1},..., \alpha^j_{n})(g_{1k},..., g_{nk})^T.$$ If we multiply both sides by $(g^{k1},..., g^{kn})$, we get $$(\alpha^j_{1},..., \alpha^j_{n})= h_{jk}(g^{k1},..., g^{kn}). $$ Thus $$\alpha^j_{l} =h_{jk}g^{kl} .$$ Now, consider

$$Z=\sum\limits_{k=1}^n \sum\limits_{l=1}^n h_{jk}g^{kl} X_l.$$

We claim that $ \frac{\partial{\nu}}{\partial x_{j}}= Z$. To see this, we multiply $Z$ by $X_i$, we will have

$$\left< Z, X_i \right> =\left< \sum\limits_{k=1}^n \sum\limits_{l=1}^n h_{jk}g^{kl} X_l, X_i \right> = \sum\limits_{k=1}^n \sum\limits_{l=1}^n h_{jk}g^{kl} g_{li} = h_{ji}= \left< \frac{\partial{\nu}}{\partial x_{j}} , X_i \right>.$$

Since $X_i$ is a basis for the tangent space, $Z$ is uniquely determined by the values of $ \left< Z, X_i \right>$. Also, we got $\left< Z, X_i \right>= \left< \frac{\partial{\nu}}{\partial x_{j}} , X_i \right>$ for each $i$, thus $\frac{\partial{\nu}}{\partial x_{j}} =Z.$ Concluding the proof.