1. 协变基矢对曲线坐标的导数与Christoffel符号

由于 $\frac{\partial {\vec{g}}_i}{\partial x^j}(i,j=1,2,3)$ 为一阶张量，也可在协变基或逆变基上就地展开，即：
$$\frac{\partial {\vec{g}}_i}{\partial x^j}={\Gamma }_{ij}^k{\vec{g}}_k={\Gamma }_{ij,k}{\vec{g}}^k(a)$$
其中，组合系数 ${\Gamma }_{ij}^k、{\Gamma }_{ij,k}$ 分别称作第二类Christoffel符号与第一类Christoffel符号。

2. Christoffel符号的计算与性质

2.1. Christoffel符号的计算

由 $(a)$知 Christoffel符号的计算表达式为：
$$\{\begin{array}{l}{\Gamma }_{ij}^k=\frac{\partial {\vec{g}}_i}{\partial x^j}\cdot {\vec{g}}^k=\frac{{\partial }^2{\bar{x}}_p}{\partial x^i\partial x^j}\frac{\partial x^k}{\partial {\bar{x}}_p}\\ \\ {\Gamma }_{ij,k}=\frac{\partial {\vec{g}}_i}{\partial x^j}\cdot {\vec{g}}_k=\frac{{\partial }^2{\bar{x}}_p}{\partial x^i\partial x^j}\frac{\partial {\bar{x}}_p}{\partial x^k}\end{array}(b)$$
其中，${\bar{x}}_1(x^1,x^2,x^3)、{\bar{x}}_2(x^1,x^2,x^3)、{\bar{x}}_3(x^1,x^2,x^3)$ 为仿射坐标。

2.2. Christoffel符号的对称性与指标升降关系

根据Christoffel符号的计算表达式，假设混合偏导与求导次序无关，则 ${\Gamma }_{ij}^k、{\Gamma }_{ij,k}$关于指标 $i,j$ 均具有对称性。 根据 $(a)$ 式有：
$$\begin{array}{rl}& \frac{\partial {\vec{g}}_i}{\partial x^j}={\Gamma }_{ij}^k{\vec{g}}_k={\Gamma }_{ij}^k{g}_{kl}{\vec{g}}^l={\Gamma }_{ij,l}{\vec{g}}^l\\ & \\ & ⟹({\Gamma }_{ij}^k{g}_{kl}-{\Gamma }_{ij,l}){\vec{g}}^l=\vec{0}\\ & \\ & ⟹{\Gamma }_{ij,l}={\Gamma }_{ij}^k{g}_{kl}\\ & \\ & ⟹{\Gamma }_{ij,l}{g}^{lm}={\Gamma }_{ij}^k{g}_{kl}{g}^{lm}={\Gamma }_{ij}^k{\delta }_k^m={\Gamma }_{ij}^m\end{array}$$
即，Christoffel符号关于第三指标 $k$ 可以通过度量张量的协变/逆变分量实现指标升降：
$$\{\begin{array}{l}{\Gamma }_{ij,l}={\Gamma }_{ij}^k{g}_{kl}\\ \\ {\Gamma }_{ij}^l={\Gamma }_{ij,k}{g}^{kl}\end{array}(c)$$

2.3. 度量张量的协变分量与第一类Christoffel符号的关系

$$\begin{array}{rl}& \frac{\partial g_{ij}}{\partial x^k}=\frac{\partial }{\partial x^k}({\vec{g}}_i\cdot {\vec{g}}_j)=\frac{\partial {\vec{g}}_i}{\partial x^k}\cdot {\vec{g}}_j+{\vec{g}}_i\cdot \frac{\partial {\vec{g}}_j}{\partial x^k}\\ & \\ & ={\Gamma }_{ik,m}{\vec{g}}^m\cdot {\vec{g}}_j+{\vec{g}}_i\cdot {\Gamma }_{jk,m}{\vec{g}}^m\\ & \\ & ={\Gamma }_{ik,j}+{\Gamma }_{jk,i}(d)\end{array}$$
同理，
$$\begin{gathered} \frac{\partial g_{jk}}{\partial x^i}={\Gamma }_{ji,k}+{\Gamma }_{ki,j} \\ \frac{\partial g_{ki}}{\partial x^j}={\Gamma }_{kj,i}+{\Gamma }_{ij,k} \end{gathered}$$
则
$${\Gamma }_{ij,k}=\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{jk}}{\partial x^i}-\frac{\partial g_{ij}}{\partial x^k}\right)(e)$$

2.4. 第二类Christoffel符号与协变基矢的混合积$\sqrt{g}$ 的关系

$$\begin{array}{rl}& \frac{\partial \sqrt{g}}{\partial x^l}=\frac{\partial }{\partial x^l}[{\vec{g}}_1\cdot ({\vec{g}}_2\times {\vec{g}}_3)]\\ & \\ & =\frac{\partial {\vec{g}}_1}{\partial x^l}\cdot ({\vec{g}}_2\times {\vec{g}}_3)+{\vec{g}}_1\cdot (\frac{\partial {\vec{g}}_2}{\partial x^l}\times {\vec{g}}_3)+{\vec{g}}_1\cdot ({\vec{g}}_2\times \frac{\partial {\vec{g}}_3}{\partial x^l})\\ & \\ & ={\Gamma }_{1l}^k{\vec{g}}_k\cdot ({\vec{g}}_2\times {\vec{g}}_3)+{\vec{g}}_1\cdot ({\Gamma }_{2l}^k{\vec{g}}_k\times {\vec{g}}_3)+{\vec{g}}_1\cdot ({\vec{g}}_2\times {\Gamma }_{3l}^k{\vec{g}}_k)\\ & \\ & ={\Gamma }_{1l}^1{\vec{g}}_1\cdot ({\vec{g}}_2\times {\vec{g}}_3)+{\vec{g}}_1\cdot ({\Gamma }_{2l}^2{\vec{g}}_2\times {\vec{g}}_3)+{\vec{g}}_1\cdot ({\vec{g}}_2\times {\Gamma }_{3l}^3{\vec{g}}_3)\\ & \\ & ={\Gamma }_{kl}^k\sqrt{g}(f)\end{array}$$
由于，$\sqrt{g}\ne 0$
$${\Gamma }_{kl}^k=\frac{1}{\sqrt{g}}\frac{\partial \sqrt{g}}{\partial x^l}=\frac{\partial ln(\sqrt{g})}{\partial x^l}(g)$$

2.5. 正交曲线坐标系中的Christoffel符号

在正交曲线坐标系中，
$${\vec{g}}_i\cdot {\vec{g}}_j=0(i\ne j)$$
故根据（e）式知：(不对指标求和)

(1) 当 Christoffel 符号的三个指标均不相同时，$i\ne j\ne k$
$${\Gamma }_{ij,k}=0{\Gamma }_{ij}^k=0(i\ne j\ne k)$$
(2) 只有前两指标相同时，$i=j\ne k$
$$\begin{gathered} {\Gamma }_{ii,k}=-\frac{1}{2}\frac{\partial g_{ii}}{\partial x^k}=-A_i\frac{\partial A_i}{\partial x^k} \\ {\Gamma }_{ii}^k=\sum _l{g}^{kl}{\Gamma }_{ii,l}=g^{kk}{\Gamma }_{ii,k}=-\frac{A_i}{A_k^2}\frac{\partial A_i}{\partial x^k} \end{gathered}$$
(3) 第一个指标或第二个指标与第三个指标相同时，$i=k\ne j或j=k\ne i$
$$\begin{gathered} {\Gamma }_{ji,i}={\Gamma }_{ij,i}=\frac{1}{2}\frac{\partial g_{ii}}{\partial x^j}=A_i\frac{\partial A_i}{\partial x^j} \\ {\Gamma }_{ji}^i={\Gamma }_{ij}^i=\sum _k{g}^{ik}{\Gamma }_{ij,k}=g^{ii}{\Gamma }_{ij,i}=\frac{1}{A_i}\frac{\partial A_i}{\partial x^j}=\frac{\partial ln(A_i)}{\partial x^j} \end{gathered}$$
(4) 三指标均相同时，$i=j=k$
$$\begin{gathered} {\Gamma }_{ii,i}=\frac{1}{2}\frac{\partial g_{ii}}{\partial x^i}=A_i\frac{\partial A_i}{\partial x^i} \\ {\Gamma }_{ii}^i=\sum _k{g}^{ik}{\Gamma }_{ii,k}=g^{ii}{\Gamma }_{ii,i}=\frac{1}{A_i}\frac{\partial A_i}{\partial x^i}=\frac{\partial ln(A_i)}{\partial x^i} \end{gathered}$$
其中，$A_i$ 为 Lame 常数。

2.6. Christoffel符号不是三阶张量的分量

首先，在仿射坐标系中基矢量不随坐标改变，因此仿射坐标系的Christoffel符号均为0，而在一般坐标系中Christoffel符号并不一定等于零，这说明Christoffel符号必定不是三阶张量的分量。下面通过Christoffel符号的坐标转换关系来说明它不是三阶张量的分量：
$$\begin{array}{rl}& {\Gamma }_{i^{\prime }{j}^{\prime }}^{k^{\prime }}=\frac{\partial {\vec{g}}_{i^{\prime }}}{\partial x^{j^{\prime }}}\cdot {\vec{g}}^{k^{\prime }}\\ & \\ & =\frac{\partial ({\beta }_{i^{\prime }}^m{\vec{g}}_m)}{\partial x^p}\frac{\partial x^p}{\partial x^{j^{\prime }}}\cdot {\beta }_n^{k^{\prime }}{\vec{g}}^n\\ & \\ & ={\beta }_{i^{\prime }}^m{\beta }_n^{k^{\prime }}\frac{\partial x^p}{\partial x^{j^{\prime }}}\frac{\partial {\vec{g}}_m}{\partial x^p}\cdot {\vec{g}}^n+{\beta }_n^{k^{\prime }}\frac{\partial {\beta }_{i^{\prime }}^m}{\partial x^p}\frac{\partial x^p}{\partial x^{j^{\prime }}}{\vec{g}}_m\cdot {\vec{g}}^n\\ & \\ & ={\beta }_n^{k^{\prime }}{\beta }_{i^{\prime }}^m{\beta }_{j^{\prime }}^p\frac{\partial {\vec{g}}_m}{\partial x^p}\cdot {\vec{g}}^n+\frac{\partial x^{k^{\prime }}}{\partial x^n}\frac{{\partial }^2{x}^n}{\partial x^{i^{\prime }}\partial x^{j^{\prime }}}\\ & \\ & ={\beta }_n^{k^{\prime }}{\beta }_{i^{\prime }}^m{\beta }_{j^{\prime }}^p{\Gamma }_{mp}^n+\frac{\partial x^{k^{\prime }}}{\partial x^n}\frac{{\partial }^2{x}^n}{\partial x^{i^{\prime }}\partial x^{j^{\prime }}}\end{array}$$
同理，
$${\Gamma }_{ij}^k={\beta }_{n^{\prime }}^k{\beta }_i^{m^{\prime }}{\beta }_j^{p^{\prime }}{\Gamma }_{m^{\prime }{p}^{\prime }}^{n^{\prime }}+\frac{\partial x^k}{\partial x^{n^{\prime }}}\frac{{\partial }^2{x}^{n^{\prime }}}{\partial x^i\partial x^j}$$
另外
$$\begin{array}{rl}& {\Gamma }_{i^{\prime }{j}^{\prime },k^{\prime }}=g_{l^{\prime }{k}^{\prime }}{\Gamma }_{i^{\prime }{j}^{\prime }}^{l^{\prime }}={\beta }_{l^{\prime }}^m{\beta }_{k^{\prime }}^n{g}_{mn}\left[{\beta }_p^{l^{\prime }}{\beta }_{i^{\prime }}^q{\beta }_{j^{\prime }}^h{\Gamma }_{qh}^p+\frac{\partial x^{l^{\prime }}}{\partial x^p}\frac{{\partial }^2{x}^p}{\partial x^{i^{\prime }}\partial x^{j^{\prime }}}\right]\\ & \\ & ={\delta }_p^m{\beta }_{k^{\prime }}^n{\beta }_{i^{\prime }}^q{\beta }_{j^{\prime }}^h{g}_{mn}{\Gamma }_{qh}^p+g_{l^{\prime }{k}^{\prime }}\frac{\partial x^{l^{\prime }}}{\partial x^p}\frac{{\partial }^2{x}^p}{\partial x^{i^{\prime }}\partial x^{j^{\prime }}}\\ & \\ & ={\beta }_{k^{\prime }}^n{\beta }_{i^{\prime }}^q{\beta }_{j^{\prime }}^h{\Gamma }_{qh,n}+g_{l^{\prime }{k}^{\prime }}\frac{\partial x^{l^{\prime }}}{\partial x^p}\frac{{\partial }^2{x}^p}{\partial x^{i^{\prime }}\partial x^{j^{\prime }}}\\ & \\ & {\Gamma }_{ij,k}={\beta }_k^{n^{\prime }}{\beta }_i^{q^{\prime }}{\beta }_j^{h^{\prime }}{\Gamma }_{q^{\prime }{h}^{\prime },n^{\prime }}+g_{lk}\frac{\partial x^l}{\partial x^{p^{\prime }}}\frac{{\partial }^2{x}^{p^{\prime }}}{\partial x^i\partial x^j}\end{array}$$

3. 逆变基矢对曲线坐标的导数

$$\begin{array}{rl}& \frac{\partial {\vec{g}}_i}{\partial x^j}=\frac{\partial }{\partial x^j}(g_{ik}{\vec{g}}^k)=\frac{\partial g_{ik}}{\partial x^j}{\vec{g}}^k+\frac{\partial {\vec{g}}^k}{\partial x^j}{g}_{ik}\\ & \\ & =({\Gamma }_{ij,k}+{\Gamma }_{jk,i}){\vec{g}}^k+\frac{\partial {\vec{g}}^k}{\partial x^j}{g}_{ik}\\ & \\ & ={\Gamma }_{ij,k}{\vec{g}}^k\\ & \\ & ⟹\frac{\partial {\vec{g}}^k}{\partial x^j}{g}_{ik}=-{\Gamma }_{jk,i}{\vec{g}}^k\\ & \\ & ⟹\frac{\partial {\vec{g}}^k}{\partial x^j}{g}_{ik}{g}^{im}=-{\Gamma }_{jk,i}{g}^{im}{\vec{g}}^k\\ & \\ & ⟹\frac{\partial {\vec{g}}^m}{\partial x^j}=-{\Gamma }_{jk}^m{\vec{g}}^k=-{\Gamma }_{jk}^m{g}^{nk}{\vec{g}}_n(h)\\ & \end{array}$$