📂Geometry

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 papers¹.

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.

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