📂Geometry

Gauss's Theorem in Differential Geometry

정리¹

Let’s call $\mathbf{x} : U \to \R^{3}$ the coordinate patch. Let $(u_{1}, u_{2})$ be the coordinates of $U$.

Let $\mathbf{n}$ be the unit normal, $L_{ij} = \left\langle \mathbf{x}_{ij}, \mathbf{n} \right\rangle$ the coefficients of the second fundamental form, and $\Gamma_{ij}^{k} = \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk}$ the Christoffel symbols.

Then, the following are true:

(a) Gauss’s formulas:

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

(b) For any unit speed curve $\boldsymbol{\gamma}(s) = \mathbf{x}\left( \gamma^{1}(s), \gamma^{2}(s) \right)$,

$$ \kappa_{n} = \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} L_{ij} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} $$

And

$$ \kappa_{g}\mathbf{S} = \sum \limits_{k=1}^{2} \left[ u_{k}^{\prime \prime} + \sum \limits_{i,j=1}^{2} \Gamma_{ij}^{k}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime} \right] \mathbf{x}_{k} $$

Where $\kappa_{n}$ is the normal curvature, $\kappa_{g}$ is the geodesic curvature, and $\mathbf{S} = \mathbf{n} \times \mathbf{T}$.

Explanation

In fact, (a) is by itself a definition of $L_{ij}$ and $\Gamma_{ij}^{k}$.

From the result of (a), we obtain the following equation:

$$ \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle = \sum \limits_{k=1}^{2}\Gamma_{ij}^{k}g_{kl} $$

This is called the first Christoffel symbol.

Proof

(a)

The unit normal is perpendicular to the tangent space, and since $\left\{ \mathbf{x}_{1}, \mathbf{x}_{2} \right\}$ is a basis of the tangent space, $\left\{ \mathbf{n}, \mathbf{x}_{1}, \mathbf{x}_{2} \right\}$ becomes the basis of $\R^{3}$. Therefore, all vectors of $\R^{3}$ can be represented as a linear combination of these. Now, let’s represent $\mathbf{x}_{ij}$ as follows:

$$ \mathbf{x}_{ij} = a_{ij}\mathbf{n} + {b_{ij}}^{1}\mathbf{x}_{1} + {b_{ij}}^{2}\mathbf{x}_{2} $$

Therefore, since $\left\langle \mathbf{x}_{i}, \mathbf{n} \right\rangle=0$, by the definition of the coefficients of the second fundamental form,

$$ L_{ij} = \left\langle \mathbf{x}_{ij}, \mathbf{n} \right\rangle = \left\langle a_{ij}\mathbf{n} + {b_{ij}}^{1}\mathbf{x}_{1} + {b_{ij}}^{2}\mathbf{x}_{2}, \mathbf{n} \right\rangle = a_{ij} $$

Also, the coefficients of the Riemann metric are defined as follows:

$$ \begin{align*} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle =&\ \left\langle a_{ij}\mathbf{n} + {b_{ij}}^{1}\mathbf{x}_{1} + {b_{ij}}^{2}\mathbf{x}_{2}, \mathbf{x}_{l} \right\rangle \\ &= {b_{ij}}^{1} \left\langle \mathbf{x}_{1}, \mathbf{x}_{l} \right\rangle + {b_{ij}}^{2} \left\langle \mathbf{x}_{2}, \mathbf{x}_{l} \right\rangle \\ &= {b_{ij}}^{1} g_{1l} + {b_{ij}}^{2} g_{2l} \\ &= \sum \limits_{m=1}^{2} {b_{ij}}^{m} g_{ml} \\ &= {b_{ij}}^{m} g_{ml} \quad (\text{Einstein notation}) \end{align*} $$

Therefore, $[g^{lk}]$ being the inverse of $[g_{ij}]$, the following equation holds:

$$ \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \sum \limits_{m=1}^{2} {b_{ij}}^{m} g_{ml}g^{lk} $$

Summing the left-hand side over all $l$ gives the Christoffel symbols. Since $g_{ik}g^{kj} = {\delta_{i}}^{j}$ holds for the Riemann metric,

$$ \Gamma_{ij}^{k} = \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \sum \limits_{l=1}^{2} \sum \limits_{m=1}^{2} {b_{ij}}^{m} g_{ml}g^{lk} = \sum \limits_{m=1}^{2} {b_{ij}}^{m} {\delta_{m}}^{k} = {b_{ij}}^{k} $$

Therefore,

$$ \begin{align*} \mathbf{x}_{ij} =&\ a_{ij}\mathbf{n} + {b_{ij}}^{1}\mathbf{x}_{1} + {b_{ij}}^{2}\mathbf{x}_{2} \\ =&\ L_{ij} \mathbf{n} + {\Gamma_{ij}}^{1} \mathbf{x}_{1} + {\Gamma_{ij}}^{2} \mathbf{x}_{2} \\ =&\ L_{ij} \mathbf{n} + \sum \limits_{k=1}^{2} \Gamma_{ij}^{k} \mathbf{x}_{k} \end{align*} $$

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(b)

• Part 1. Calculation of $\boldsymbol{\gamma}^{\prime \prime}$

  Calculating the tangent vector of $\boldsymbol{\gamma}(s) = \mathbf{x}\left( \gamma^{1}(s), \gamma^{2}(s) \right)$ gives the following result:

  $$ \begin{align*} T(s) =&\ \dfrac{d \boldsymbol{\gamma}}{d s} \\ =&\ \dfrac{d}{ds} \mathbf{x}(\gamma^{1}, \gamma^{2}) \\ =&\ \dfrac{\partial \mathbf{x}}{\partial \gamma^{1}}\dfrac{d \gamma^{1}}{ds} + \dfrac{\partial \mathbf{x}}{\partial \gamma^{2}}\dfrac{d \gamma^{2}}{ds}& \text{by } \href{https://freshrimpsushi.github.io/posts/derivative-of-three-dimentional-scalar-vector-function}{\text{chain rule}} \\ =&\ \mathbf{x}_{1}(\gamma^{1})^{\prime} + \mathbf{x}_{2}(\gamma^{2})^{\prime} \\ =&\ \sum \limits_{i=1}^{2}\mathbf{x}_{i}(\gamma^{i})^{\prime} \\ =&\ \mathbf{x}_{i}(\gamma^{i})^{\prime} \end{align*} $$

  In the last equality, Einstein notation was used. Let’s calculate the acceleration. For someone familiar with Einstein notation and differential calculus, it can be calculated in one shot as follows:

  $$ \boldsymbol{\gamma} ^{\prime \prime} = \dfrac{d}{ds}\left( \mathbf{x}_{i}(\gamma^{i})^{\prime} \right) = \mathbf{x}_{ij}(\gamma^{j})^{\prime}(\gamma^{i})^{\prime} + \mathbf{x}_{i}(\gamma^{i})^{\prime \prime} = \sum \limits_{i=1}^{2}\left( \sum\limits_{j=1}^{2}\mathbf{x}_{ij}(\gamma^{j})^{\prime}(\gamma^{i})^{\prime} + \mathbf{x}_{i}(\gamma^{i})^{\prime \prime} \right) $$

  For those not familiar with Einstein notation, detailing the calculation process gives the following:

  $$ \begin{align*} \boldsymbol{\gamma}^{\prime \prime} =&\ \dfrac{d}{ds} \left( \mathbf{x}_{1}(\gamma^{1})^{\prime} + \mathbf{x}_{2}(\gamma^{2})^{\prime} \right) \\ =&\ \dfrac{d}{ds} \left( \mathbf{x}_{1}(\gamma^{1})^{\prime} \right) +\dfrac{d}{ds} \left( \mathbf{x}_{2}(\gamma^{2})^{\prime} \right) \\ =&\ \dfrac{d}{ds} \left( \mathbf{x}_{1} \right) (\gamma^{1})^{\prime} +\mathbf{x}_{1} \dfrac{d}{ds} \left( (\gamma^{1})^{\prime} \right) + \dfrac{d}{ds} \left( \mathbf{x}_{2} \right) (\gamma^{2})^{\prime} +\mathbf{x}_{2} \dfrac{d}{ds} \left( (\gamma^{2})^{\prime} \right) \\ =&\ \left( \dfrac{\partial \mathbf{x}_{1}}{\partial \gamma^{1}}\dfrac{d \gamma^{1}}{ds} + \dfrac{\partial \mathbf{x}_{1}}{\partial \gamma^{2}}\dfrac{d \gamma^{2}}{ds} \right) (\gamma^{1})^{\prime} + \mathbf{x}_{1} (\gamma^{1})^{\prime \prime} + \left( \dfrac{\partial \mathbf{x}_{2}}{\partial \gamma^{1}}\dfrac{d \gamma^{1}}{ds} + \dfrac{\partial \mathbf{x}_{2}}{\partial \gamma^{2}}\dfrac{d \gamma^{2}}{ds} \right) (\gamma^{2})^{\prime} + \mathbf{x}_{2} (\gamma^{2})^{\prime \prime} \\ =&\ \left( \mathbf{x}_{11}(\gamma^{1})^{\prime} + \mathbf{x}_{12}(\gamma^{2})^{\prime} \right)(\gamma^{1})^{\prime} + \mathbf{x}_{1} (\gamma^{1})^{\prime \prime} + \left( \mathbf{x}_{21}(\gamma^{1})^{\prime} + \mathbf{x}_{22}(\gamma^{2})^{\prime} \right)(\gamma^{2})^{\prime} + \mathbf{x}_{2} (\gamma^{2})^{\prime \prime} \\ =&\ \mathbf{x}_{11}(\gamma^{1})^{\prime}(\gamma^{1})^{\prime} + \mathbf{x}_{12}(\gamma^{2})^{\prime}(\gamma^{1})^{\prime} + \mathbf{x}_{1} (\gamma^{1})^{\prime \prime} + \mathbf{x}_{21}(\gamma^{1})^{\prime}(\gamma^{2})^{\prime} + \mathbf{x}_{22}(\gamma^{2})^{\prime}(\gamma^{2})^{\prime} + \mathbf{x}_{2} (\gamma^{2})^{\prime \prime} \\ =&\ \left( \mathbf{x}_{11}(\gamma^{1})^{\prime}(\gamma^{1})^{\prime} + \mathbf{x}_{12}(\gamma^{2})^{\prime}(\gamma^{1})^{\prime} + \mathbf{x}_{21}(\gamma^{1})^{\prime}(\gamma^{2})^{\prime} + \mathbf{x}_{22}(\gamma^{2})^{\prime}(\gamma^{2})^{\prime} \right) + \mathbf{x}_{1} (\gamma^{1})^{\prime \prime} + \mathbf{x}_{2} (\gamma^{2})^{\prime \prime} \\ =&\ \sum \limits_{j=1}^{2}\left( \mathbf{x}_{1j}(\gamma^{j})^{\prime}(\gamma^{1})^{\prime} + \mathbf{x}_{2j}(\gamma^{j})^{\prime}(\gamma^{2})^{\prime} \right) + \mathbf{x}_{1} u_{1}^{\prime \prime} + \mathbf{x}_{2} (\gamma^{2})^{\prime \prime} \\ =&\ \sum \limits_{i=1}^{2} \left(\sum \limits_{j=1}^{2} \mathbf{x}_{ij}(\gamma^{j})^{\prime}(\gamma^{i})^{\prime} + \mathbf{x}_{i} (\gamma^{i})^{\prime \prime} \right) \\ =&\ \mathbf{x}_{ij}(\gamma^{j})^{\prime}(\gamma^{i})^{\prime} + \mathbf{x}_{i} (\gamma^{i})^{\prime \prime} \end{align*} $$

• Part 2.

  Substituting Gauss’s formulas (a) into $\boldsymbol{\gamma}^{\prime \prime}$,

  $$ \begin{align*} \boldsymbol{\gamma}^{\prime \prime} (s) &= \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} \mathbf{x}_{ij} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} + \sum \limits_{k=1}^{2} \mathbf{x}_{k} (\gamma^{k})^{\prime \prime} \\ &= \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} \left( L_{ij} \mathbf{n} + \sum \limits_{k=1}^{2} \Gamma_{ij}^{k} \mathbf{x}_{k} \right) (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} + \sum \limits_{k=1}^{2} \mathbf{x}_{k} (\gamma^{k})^{\prime \prime} \\ &= \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} L_{ij} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} \mathbf{n} + \sum \limits_{k=1}^{2}\left( (\gamma^{k})^{\prime \prime} + \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} \Gamma_{ij}^{k} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} \right)\mathbf{x}_{k} \end{align*} $$

  However, because it is expressed as $\boldsymbol{\gamma}^{\prime \prime} = \kappa_{n}\mathbf{n} + \kappa_{g}\mathbf{S}$in this way

  $$ \kappa_{n} = \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} L_{ij} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} $$

  $$ \kappa_{g}\mathbf{S} = \sum \limits_{k=1}^{2}\left( (\gamma^{k})^{\prime \prime} + \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} \Gamma_{ij}^{k} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} \right)\mathbf{x}_{k} $$

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