📂Geometry

Gaussian Curvature and Mean Curvature

Definition¹

Let’s consider the principal curvature at a point $p$ on a surface $M$ be denoted as $\kappa_{1}, \kappa_{2}$. Let $L$ be referred to as the Weingarten map. The Gaussian curvature $K$ is defined as follows:

$$ K := \kappa_{1} \kappa_{2} = \det L = \det ([{L^{i}}_{j}]) $$

where ${L^{i}}_{j} = \sum \limits_{k} L_{kj}g^{ki}$ applies.

Formula

• The product of the principal curvatures

$$ K = \kappa_{1} \kappa_{2} $$

• Gaussian curvature can be expressed with the Riemann curvature tensor.

$$

$$

• Gaussian curvature can be expressed with Christoffel symbols.

$$ K = $$

• The Gaussian curvature at point $p$ can be shown using the Gauss map and area of the domain.

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

Theorems

1. $H^{2} \ge K$ holds.

2. Let $\mathbf{X}, \mathbf{Y}$ be the orthonormal vectors at point $p$. Then, the following is true:

$$ H = \dfrac{1}{2}\left( II(\mathbf{X}, \mathbf{X}) + II(\mathbf{Y}, \mathbf{Y}) \right) $$

3. Let $\mathbf{Y} \in T_{p}M$ be the unit tangent vector, $\kappa_{n}$ be the normal curvature, and $\theta$ be the angle between the principal directions $\mathbf{X}_{1}$ and $\mathbf{Y}$. The following holds:

$$ H = \dfrac{1}{2\pi}\int_{0}^{2\pi}\kappa_{n}d\theta $$

Proof

3.

Given that $\kappa_{n} = II(\mathbf{Y}, \mathbf{Y})$ and $II(\mathbf{Y}, \mathbf{Y}) = \kappa_{1}\cos^{2}\theta + \kappa_{2}\sin^{2}\theta$,

$$ \begin{align*} \dfrac{1}{2\pi}\int_{0}^{2\pi}\kappa_{n}d\theta &= \dfrac{1}{2\pi}\int_{0}^{2\pi} \kappa_{1}\cos^{2}\theta + \kappa_{2}\sin^{2}\theta d\theta \\ &= \dfrac{1}{2\pi} \left( \kappa_{1} \int_{0}^{2\pi} \cos^{2} d\theta + \kappa_{2} \int_{0}^{2\pi}\sin^{2}\theta d\theta \right) \\ &= \dfrac{1}{2\pi} \left( \kappa_{1} \pi + \kappa_{2} \pi \right) \\ &= \dfrac{\kappa_{1} + \kappa_{2}}{2} \\ &= H \end{align*} $$

(Refer to trigonometric integral table $(2), (3)$)

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