📂Geometry

Definition and Relationship between the Gaussian Map and Gaussian Curvature

Definition¹

A function $\nu$ that maps each point $p$ on a surface $M$ to a unit normal is called the Gaussian map.

$$ \nu : M \to \mathbb{S}^{2} \quad \text{and} \quad \nu (p) = \mathbf{n}_{p} $$

Description

The Gaussian map is also referred to as the normal spherical image.

Theorem

Let us call the area of any region $\mathscr{R}$ on the surface $A(\mathscr{R})$ as the area of $\mathscr{R}$. Then the following holds.

$$ K = \lim \limits_{\mathscr{R} \to p} \dfrac{A(\nu (\mathscr{R}))}{A(\mathscr{R})} $$

Here, $K$ is the Gaussian curvature.

Proof

Assume that $\mathbf{n} : \mathbf{x}^{-1}(\mathscr{R}) \to \mathbb{S}^{2}$ is regular. Then it becomes a coordinate chart mapping.

$$ \dfrac{\partial \mathbf{n}}{\partial u_{1}} \times \dfrac{\partial \mathbf{n}}{\partial u_{2}} \ne 0 $$

Since $\mathbf{n}$ itself is a coordinate chart mapping, it can be written as follows.

$$ A( \nu (\mathscr{R})) = \int\int_{\mathbf{x}^{-1}(\mathscr{R})} \left[ \mathbf{n}_{1}, \mathbf{n}_{2}, \mathbf{m} \right] du^{1}du^{2} $$

$$ \mathbf{m} = \dfrac{\mathbf{n}_{1} \times \mathbf{n}_{2}}{\left\| \mathbf{n}_{1} \times \mathbf{n}_{2} \right\|} $$

But $\mathbf{m}$ is the normal of $\mathbb{S}^{2}$, so in fact $\mathbf{n} = \mathbf{m}$ is.

$$ \lim \limits_{\mathscr{R} \to p} \dfrac{A(\nu (\mathscr{R}))}{A(\mathscr{R})} = \dfrac{\left[ \mathbf{n}_{1}, \mathbf{n}_{2}, \mathbf{n} \right]}{\left[ \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{n} \right]} $$

By calculating the scalar triple product,

$$ \begin{align*} (\mathbf{n}_{1} \times \mathbf{n}_{2}) \cdot \mathbf{n} &= \left( L(\mathbf{x}_{1})\times L(\mathbf{x}_{2}) \right) \cdot \mathbf{n} \\ &= \left( \left( {L^{1}}_{1}\mathbf{x}_{1} + {L^{2}}_{1}\mathbf{x}_{2} \right) \times \left( {L^{1}}_{2}\mathbf{x}_{1} + {L^{2}}_{2}\mathbf{x}_{2} \right) \right) \cdot \mathbf{n} \\ &= \left( {L^{1}}_{1}{L^{2}}_{2} - {L^{2}}_{1}{L^{1}}_{2} \right) (\mathbf{x}_{1} \times \mathbf{x}_{2}) \cdot \mathbf{n} \end{align*} $$

In this case, ${L^{i}}_{j} = \sum \limits_{k} L_{kj}g^{ki}$ is. Therefore

$$ \begin{align*} \lim \limits_{\mathscr{R} \to p} \dfrac{A(\nu (\mathscr{R}))}{A(\mathscr{R})} &= \dfrac{\left[ \mathbf{n}_{1}, \mathbf{n}_{2}, \mathbf{n} \right]}{\left[ \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{n} \right]} \\[1em] &= \dfrac{\left( {L^{1}}_{1}{L^{2}}_{2} - {L^{2}}_{1}{L^{1}}_{2} \right) (\mathbf{x}_{1} \times \mathbf{x}_{2}) \cdot \mathbf{n}}{(\mathbf{x}_{1} \times \mathbf{x}_{2}) \cdot \mathbf{n}} \\[1em] &= \left( {L^{1}}_{1}{L^{2}}_{2} - {L^{2}}_{1}{L^{1}}_{2} \right) \\ &= \det ([L_{j}^{i}]) \\ &= K \end{align*} $$

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