Codazzi-Mainardi equation from the integrability condition for the Weingarten equation

Question

Here In my textbook , the author states that the integrability condition

$$ 0 = \frac{\partial^2\nu}{\partial u^i \partial u^j} - \frac{\partial^2\nu}{\partial u^j \partial u^i}$$

together with the Weingarten equation

$$ \frac{\partial \nu}{\partial u^i} = - \sum_{j,k} h_{ij}g^{jk} \cdot \frac{\partial f}{\partial u^k} = - \sum_k h_i^k \cdot \frac{\partial f}{\partial u^ k}$$

yield a "reformulation" of the Codazzi-Mainardi equation:

$$\frac{\partial}{\partial u^k}h_{ij} - \frac{\partial}{\partial u^j}h_{ik} + \sum_r \left( \Gamma_{ij}^r h_{rk} - \Gamma_{ik}^rh_{rj}\right) = 0 \quad \text{for all} \; i, j, k.$$

(Here, $h_{ij}$ are the coefficients of the second fundamental form, $h_i^j$ are the coefficients of the Weingarten map, $g^{ij}$ are the coefficients of the inverse of the first fundamental form, $f$ is the local parametrization of the surface and $\Gamma_{ij}^k$ are the Christoffel symbols of the second kind).

I presume the proof will be somewhat similar to the one above in the text (see above link), and will use Gauss' formula

$$ \frac{\partial^2 f}{\partial u^i \partial u^j} = \sum_k \Gamma_{ij}^k \cdot \frac{\partial f}{\partial u^k} + h_{ij} \cdot \nu$$

but I've tried and cannot see how to rearrange the terms so that something meaningful comes out. Could someone do the derivation?

Answer 1

In Ricci calculus

$$e_{i,jk}=e_{i,kj}$$ $$(\Gamma^l_{ij}e_l+b_{ij}N),_k=(\Gamma^l_{ik}e_l+b_{ik}N),_j$$ $$\Gamma^l_{ij,k}e_l+\Gamma^l_{ij}e_{l,k}+b_{ij,k}N+b_{ij}N,_k= \Gamma^l_{ik,j}e_l+\Gamma^l_{ik}e_{l,j}+b_{ik,j}N+b_{ik}N,_j$$

$$\Gamma^l_{ij,k}e_l+\Gamma^l_{ij}(\Gamma^m_{lk}e_m+b_{lk}N)+b_{ij,k}N-b_{ij}b^l_ke_l = \Gamma^l_{ik,j}e_l+\Gamma^l_{ik}(\Gamma^m_{lj}e_m+b_{lj}N)+b_{ik,j}N-b_{ik}b^l_je_l$$

$$\Gamma^l_{ij,k}e_l+\Gamma^m_{ij}(\Gamma^l_{mk}e_l+b_{mk}N)+b_{ij,k}N-b_{ij}b^l_ke_l = \Gamma^l_{ik,j}e_l+\Gamma^m_{ik}(\Gamma^l_{mj}e_l+b_{mj}N)+b_{ik,j}N-b_{ik}b^l_je_l$$

$$(\Gamma^l_{ij,k}+\Gamma^m_{ij}\Gamma^l_{mk}-b_{ij}b^l_k)e_l+(\Gamma^m_{ij}b_{mk}+b_{ij,k})N = (\Gamma^l_{ik,j}+\Gamma^m_{ik}\Gamma^l_{mj}-b_{ik}b^l_j)e_l+ (\Gamma^m_{ik}b_{mj}+b_{ik,j})N$$

The vectors $e_l$ and $N$ are linearly independent, then Codazzi-Mainardi equation

$$b_{ij,k}-b_{ik,j}=\Gamma^m_{ik}b_{mj}-\Gamma^m_{ij}b_{mk}$$

And Gauss equation

$$b_{ij}b^l_k-b_{ik}b^l_j=\Gamma^l_{ij,k}-\Gamma^l_{ik,j}+\Gamma^m_{ij}\Gamma^l_{mk}-\Gamma^m_{ik}\Gamma^l_{mj}=R_{ikj}^l$$