📂Geometry

Bingarten Equation

Theorem¹

On the surface $M$, the following equation holds.

$$\mathbf{n}_{j} = - \sum_{k} {L^{k}}_{j}\mathbf{x}_{k}$$

Here, $\mathbf{x} : U \to M$ is a coordinate chart mapping, $\mathbf{n}$ is the unit normal, and ${L^{k}}_{j} = \sum\limits_{i}L_{ij}g^{ik}$ is.

Explanation

Consider the Frenet-Serret frame $\left\{ \mathbf{T}, \mathbf{N}, \mathbf{B} \right\}$ of a curve. Since these are three mutually orthogonal vectors, they form a basis of $\mathbb{R}^{3}$. Moreover, the derivative of each is expressed as a linear combination of the other vectors, which is called the Frenet-Serret equations.

$$\begin{align*} \mathbf{T}^{\prime}(s) =&\ \kappa (s) \mathbf{N}(s) \\ \mathbf{N}^{\prime}(s) =&\ - \kappa (s) \mathbf{T}(s) + \tau (s) \mathbf{B}(s) \\ \mathbf{B}^{\prime}(s) =&\ - \tau (s) \mathbf{N}(s) \end{align*}$$

Now, consider the set $\left\{ \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{n} \right\}$. $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ generate the tangent space, and since $\mathbf{n}$ is perpendicular to these two, this set also forms a basis of $\mathbb{R}^{3}$. Therefore, from the Gauss equation and the Weingarten equations, we can obtain the following formula that serves a similar function to the Frenet-Serret equations for the surface $M$.

$$\begin{align*} \mathbf{x}_{ij} &= L_{ij}\mathbf{n} + \sum_{k} \Gamma_{ij}^{k}\mathbf{x}_{k} \\ \mathbf{n}_{j} &= - \sum_{k} {L^{k}}_{j}\mathbf{x}_{k} \end{align*}$$

Proof

Since $\mathbf{n}_{j} = - L(\mathbf{x}_{j})$ holds,

$$\mathbf{n}_{j} = - L(\mathbf{x}_{j}) = - \sum_{k} {L^{k}}_{j}\mathbf{x}_{k}$$

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