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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Richard S. Millman and George D. Parker, Elements of Differential Geometry (1977), p129 ↩︎