Gauss's Great Theorem
정리[^1]
Gaussian curvature $K$ is intrinsic, and the following holds.
$$ K = \dfrac{\sum\limits_{l} R_{121}^{l}g_{l2}}{g} $$
Here, $R_{ijk}^{l}$ are the coefficients of the Riemann curvature tensor, and $g$ and $g_{ij}$ are the coefficients of the Riemann metric.
Corollary
Since $R_{ijk}^{l} = \dfrac{\partial \Gamma_{ik}^{l}}{\partial u^{j}} - \dfrac{\partial \Gamma_{ij}^{l}}{\partial u^{k}} + \sum_{p} \left( \Gamma_{ik}^{p} \Gamma_{pj}^{l} - \Gamma_{ij}^{p}\Gamma_{pk}^{l}\right) \text{ for }$, the following holds.
$$ K = \dfrac{1}{g}\left( \dfrac{\partial \Gamma_{11}^{1}}{\partial u^{2}} - \dfrac{\partial \Gamma_{12}^{1}}{\partial u^{1}} + \Gamma_{11}^{1} \Gamma_{12}^{1} - \Gamma_{12}^{1}\Gamma_{11}^{1} + \Gamma_{11}^{2} \Gamma_{22}^{1} - \Gamma_{12}^{2}\Gamma_{21}^{1}\right) \left( \dfrac{\partial \Gamma_{11}^{2}}{\partial u^{2}} - \dfrac{\partial \Gamma_{12}^{2}}{\partial u^{1}} + \Gamma_{11}^{1} \Gamma_{12}^{2} - \Gamma_{12}^{1}\Gamma_{11}^{2} + \Gamma_{11}^{2} \Gamma_{22}^{2} - \Gamma_{12}^{2}\Gamma_{21}^{2}\right) $$
Description
This is called Gauss’s Theorem Egregium. Theorem Egregium translates to great theorem or outstanding theorem, and this term was used by Gauss himself in his Latin papers.1
Gauss used the term egregium because Gaussian curvature $K$ is defined in terms of the product of principal curvatures, and principal curvatures are derived from the Weingarten map, which measures the rate of change of the unit normal $\mathbf{n}$. Thus, while $K$ is defined in a very extrinsic way, it is in fact intrinsic, which is why Gauss named it so.
Proof
By the Gauss equations, $R_{ijk}^{l} = L_{ik}L_{j}^{l} - L_{ij}L_{k}^{l}$ holds. Therefore, we obtain
$$ \begin{align*} \sum\limits_{l} R_{ijk}^{l} g_{lm} &= \sum\limits_{l} ( L_{ik}L_{j}^{l} - L_{ij}L_{k}^{l} ) g_{lm} \\ &= \sum\limits_{l} ( L_{ik}L_{j}^{l}g_{lm} - L_{ij}L_{k}^{l}g_{lm} ) \\ &= L_{ik}\sum\limits_{l}L_{j}^{l}g_{lm} - L_{ij}\sum\limits_{l}L_{k}^{l}g_{lm} \end{align*} $$
Here, $g_{lm}$ are the coefficients of the first fundamental form. Then, since $L_{j}^{l} = \sum\limits_{i} L_{ij}g^{il}$ and $\sum\limits_{k}g_{ik}g^{kj} = \delta_{i}^{j}$, we obtain
$$ \begin{align*} && L_{j}^{l} &= \sum_{i} L_{ij}g^{il} \\ \implies && \sum\limits_{l} L_{j}^{l}g_{lm} &= \sum\limits_{l} \sum\limits_{i} L_{ij}g^{il}g_{lm} \\ && &= \sum\limits_{i} L_{ij}\delta_{m}^{i} \\ && &= L_{mj} \end{align*} $$
Substituting this into the above equation,
$$ \sum\limits_{l} R_{ijk}^{l} g_{lm} = L_{ik}L_{mj} - L_{ij}L_{mk} $$
Let $i = k = 1$ and $j = m = 2$, then
$$ \begin{align*} \sum\limits_{l} R_{121}^{l}g_{l2} &= L_{22}L_{22} - L_{12}L_{12} = \det ([L_{ij}]) = \det ([(L_{j}^{k}) (g_{ik})]) \\ &= \det ([L_{j}^{k}]) \det([g_{ik}]) = Kg \end{align*} $$
Therefore,
$$ K = \dfrac{\sum\limits_{l} R_{121}^{l}g_{l2}}{g} $$
Since $g$ is intrinsic and $R_{ijk}^{l}$ is also intrinsic, $K$ is intrinsic.
■