📂Geometry

The Christoffel Symbols are Intrinsic

Theorem¹

The Christoffel symbols $\Gamma_{ij}^{k}$ satisfy the following equation. In other words, they are intrinsic.

$$ \Gamma_{ij}^{k} = \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{lk} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right) $$

Explanation

Gauss proved it.

The Christoffel symbols depend only on the Riemann metric and are independent of the normal vector. Therefore, by using Christoffel symbols, one can understand the structure of a surface without leaving the surface.

Proof

First, the partial derivatives of each index of the Riemann metric coefficients are as follows.

$$ \dfrac{\partial g_{il}}{\partial u_{j}} = \dfrac{\partial}{\partial u_{j}} \left\langle \mathbf{x}_{i} , \mathbf{x}_{l} \right\rangle = \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{lj} \rangle $$

$$ \dfrac{\partial g_{ij}}{\partial u_{l}} = \dfrac{\partial}{\partial u_{l}} \left\langle \mathbf{x}_{i} , \mathbf{x}_{j} \right\rangle = \langle \mathbf{x}_{il} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{jl} \rangle $$

$$ \dfrac{\partial g_{lj}}{\partial u_{i}} = \dfrac{\partial}{\partial u_{i}} \left\langle \mathbf{x}_{l} , \mathbf{x}_{j} \right\rangle = \langle \mathbf{x}_{li} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle $$

Since $\mathbf{\mathbf{x}}_{ij} = \mathbf{\mathbf{x}}_{ji}$,

$$ \begin{align*} \dfrac{\partial g_{il}}{\partial u_{j}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{lj}}{\partial u_{i}} =&\ \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{lj} \rangle - \langle \mathbf{x}_{il} , \mathbf{x}_{j} \rangle - \langle \mathbf{x}_{i}, \mathbf{x}_{jl} \rangle + \langle \mathbf{x}_{li} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle \\ =&\ \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle \\ =&\ 2 \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle \end{align*} $$

Therefore, the Christoffel symbols are

$$ \Gamma_{ij}^{k} = \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \dfrac{1}{2} \sum \limits_{l=1}^{2} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right) g^{lk} $$

Example

Suppose we are given a Monge patch as in $\mathbf{x}(u_{1}, u_{2}) = \left( u_{1}, u_{2}, f(u_{1}, u_{2}) \right)$. Then

$$ \mathbf{x}_{1} = \partial_{u_{1}}\mathbf{x} = (1,0,f_{1}) \quad \text{and} \quad \mathbf{x}_{2} = (0,1,f_{2}) $$

At this time, $f_{i} = \partial_{u_{i}}f$ is. $\Gamma_{11}^{1}$ can be obtained in the following two ways.

Extrinsically computing

$$ \mathbf{x}_{1} \times \mathbf{x}_{2} = (-f_{1}, -f_{2}, 1) $$

The unit normal is

$$ \mathbf{n} = \dfrac{(-f_{1}, -f_{2}, 1)}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}} $$

Gauss formula

$$ \mathbf{x}_{ij} = L_{ij} \mathbf{n} + \sum \limits_{k=1}^{2} \Gamma_{ij}^{k} \mathbf{x}_{k} $$

The second derivative of $\mathbf{x}$ is as follows according to the Gauss formula.

$$ \mathbf{x}_{11} = (0, 0, f_{11}) = L_{11}\mathbf{n} + \Gamma_{11}^{1}\mathbf{x}_{1} + \Gamma_{11}^{2}\mathbf{x}_{2} $$

Therefore, the coefficients of the second fundamental form are $L_{ij} = \langle \mathbf{x}_{ij}, \mathbf{n} \rangle$

$$ L_{11} = \left\langle (0,0,f_{11}), \dfrac{(-f_{1}, -f_{2}, 1)}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}\right\rangle = \dfrac{f_{11}}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}} $$

If we break down $\mathbf{x}_{11}$ into components, we get the following.

$$ \mathbf{x}_{11} = (0, 0, f_{11}) = \dfrac{L_{11}}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}(-f_{1}, -f_{2}, 1) + \Gamma_{11}^{1}(1,0,f_{1}) + \Gamma_{11}^{2}(0,1,f_{2}) $$

Looking at the first component, we obtain the following equation.

$$ \begin{align*} && 0 =&\ \dfrac{L_{11}(-f_{1})}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}} + \Gamma_{11}^{1} \\[1em] \implies&& \Gamma_{11}^{1} =&\ \dfrac{L_{11}(f_{1})}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}} \\[1em] \implies&& \Gamma_{11}^{1} =&\ \dfrac{f_{1} f_{11}}{(f_{1})^{2} + (f_{2})^{2} + 1} \end{align*} $$

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Intrinsically computing

According to the theorem above, $\Gamma_{11}^{1}$ can be computed as follows.

$$ \begin{align*} \Gamma_{11}^{1} =&\ \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{l1} \left( \dfrac{\partial g_{l1}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{l}} + \dfrac{\partial g_{1l}}{\partial u_{1}} \right) \\ =&\ \dfrac{1}{2} \left[ g^{11} \left( \dfrac{\partial g_{11}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{1}} + \dfrac{\partial g_{11}}{\partial u_{1}} \right) + g^{21} \left( \dfrac{\partial g_{21}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{1}} \right) \right] \\ =&\ \dfrac{1}{2} \left[ g^{11} \dfrac{\partial g_{11}}{\partial u_{1}} + 2g^{21}\dfrac{\partial g_{21}}{\partial u_{1}} - g^{21}\dfrac{\partial g_{11}}{\partial u_{2}} \right] \end{align*} $$

The coefficients of the first fundamental form are $g_{ij} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle$

$$ \left[ g_{ij} \right] = \begin{bmatrix} 1+(f_{1})^{2} & f_{1}f_{2} \\[1em] f_{1}f_{2} & 1 + (f_{2})^{2} \end{bmatrix} $$

The inverse matrix is

$$ \left[ g_{ij} \right]^{-1} = \left[ g^{lk} \right] = \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\begin{bmatrix} 1+(f_{2})^{2} & -f_{1}f_{2} \\[1em] -f_{1}f_{2} & 1 + (f_{1})^{2} \end{bmatrix} $$

So, when we gather everything needed, we get the following.

$$ \begin{align*} \dfrac{\partial g_{11}}{\partial u_{1}} =&\ \dfrac{\partial }{\partial u_{1}}\left( 1+ (f_{1})^{2} \right) = 2f_{1}f_{11} \\ \dfrac{\partial g_{21}}{\partial u_{1}} =&\ \dfrac{\partial }{\partial u_{1}}\left( f_{1}f_{2} \right) = f_{11}f_{2} + f_{1}f_{21} \\ \dfrac{\partial g_{11}}{\partial u_{2}} =&\ \dfrac{\partial }{\partial u_{2}}\left( 1+ (f_{1})^{2} \right) = 2f_{1}f_{12} \end{align*} $$

And

$$ \begin{align*} g^{11} =&\ \dfrac{1+(f_{2})^{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} \\ g^{21} =&\ \dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} \end{align*} $$

Now, if we substitute, we get as follows.

$$ \begin{align*} \Gamma_{11}^{1} =&\ \dfrac{1}{2} \left[ g^{11} \dfrac{\partial g_{11}}{\partial u_{1}} + 2g^{21}\dfrac{\partial g_{21}}{\partial u_{1}} - g^{21}\dfrac{\partial g_{11}}{\partial u_{2}} \right] \\ =&\ \dfrac{1}{2} \left[ \dfrac{1+(f_{2})^{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} 2f_{1}f_{11} + 2\dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} \left( f_{11}f_{2} + f_{1}f_{21} \right)- \dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1}2f_{1}f_{12} \right] \\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ \left( 1+(f_{2})^{2} \right)f_{1}f_{11} + \left( -f_{1}f_{2} \right) \left( f_{11}f_{2} + f_{1}f_{21} \right)- (-f_{1}f_{2})f_{1}f_{12} \right] \\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ f_{1}f_{11} + f_{1}(f_{2})^{2}f_{11} - f_{1}(f_{2})^{2}f_{11} -(f_{1})^{2}f_{2}f_{21} + (f_{1})^{2}f_{2}f_{12} \right] \\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ f_{1}f_{11} \right] \\ =&\ \dfrac{f_{1}f_{11}}{(f_{1})^{2} + (f_{2})^{2} + 1} \end{align*} $$

We can see that computing Christoffel symbols intrinsically is more complicated compared to when not doing so.

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