📂Geometry

Euler's Theorem in Differential Geometry

Theorem¹

Let’s define the unit tangent vector to the surface $M$ at point $p$ as $\mathbf{Y}$.

$$ \mathbf{Y} \in T_{p}M \quad \text{and} \quad \left\| \mathbf{Y} \right\| = 1 $$

Let $\kappa_{1} \ge \kappa_{2}$ represent the principal curvature at $p$. Then, the following equation holds:

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

In this context, $II$ represents the second fundamental form, $\mathbf{X}_{1}$ represents the principal direction corresponding to $\kappa_{1}$, and $\theta$ represents the angle between $\mathbf{Y}$ and $\mathbf{X}_{1}$.

Proof

By the definition of principal curvature, the following is true:

$$ L(\mathbf{X}_{i}) = \kappa_{i} \mathbf{X}_{1},\quad i=1,2 $$

Since $\mathbf{Y}$ is a unit vector and $\left\{ \mathbf{X}_{1}, \mathbf{X}_{2} \right\}$ is the orthonormal basis for $T_{p}M$, the angle between $\mathbf{X}_{1}$ can be expressed as $\theta$. Hence, we can express it as follows:

$$ \mathbf{Y} = \cos \theta \mathbf{X}_{1} + \sin \theta \mathbf{X}_{2} $$

Therefore, we obtain the following. Since $II (\mathbf{Y}, \mathbf{Y}) = \left\langle L(\mathbf{Y}), \mathbf{Y} \right\rangle$,

$$ \begin{align*} II (\mathbf{Y}, \mathbf{Y}) =&\ \left\langle L(\mathbf{Y}), \mathbf{Y} \right\rangle \\ =&\ \left\langle L(\cos \theta \mathbf{X}_{1} + \sin \theta \mathbf{X}_{2}), \cos \theta \mathbf{X}_{1} + \sin \theta \mathbf{X}_{2} \right\rangle \\ =&\ \left\langle \cos \theta L(\mathbf{X}_{1}) + \sin \theta L(\mathbf{X}_{2}), \cos \theta \mathbf{X}_{1} + \sin \theta \mathbf{X}_{2} \right\rangle \\ =&\ \left\langle \kappa_{1} \cos \theta \mathbf{X}_{1} + \kappa_{2} \sin \theta \mathbf{X}_{2}, \cos \theta \mathbf{X}_{1} + \sin \theta \mathbf{X}_{2} \right\rangle \\ =&\ \kappa_{1} \cos^{2} \theta + \kappa_{2} \sin^{2} \theta \end{align*} $$

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