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I am trying to make geometric sense of the definition of principal curvatures as the eigenvalues of the shape operator, but I am a bit stuck. Could I have some help in showing that the principal curvatures $\kappa_1$,$\kappa_2$ at a point $p$ of a surface $M$ in $\mathbb{R}^3$ correspond to the maximum and minimum of euclidean curvatures of geodesics in $M$ through $p$? Thanks!

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The shape operator $S$ defines a quadratic form on the tangent space $T_p(M)$, the second fundamental form $\langle Sv, v \rangle$. If $e_1, e_2$ are unit eigenvectors of $S$ associated to the eigenvalues $\kappa_1, \kappa_2$, then write $v = v_1 e_1 + v_2 e_2$. Then it's not hard to see that $\langle Sv, v \rangle = \kappa_1 v_1^2 + \kappa_2 v_2^2$. This implies that $\kappa_1$ is the minimum value of the second fundamental form as $v$ ranges over all unit vectors and $\kappa_2$ is the maximum value.

On the other hand, the normal curvature of a curve $\alpha : I \to M$ parameterized by arc-length at $\alpha(0) = p$ is precisely $\langle S \dot{\alpha}(0), \dot{\alpha}(0) \rangle$. Once you know that associated to every unit vector of the tangent space is a geodesic whose tangent vector is that vector, you're done.

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    @QiaochuYuan: If $v_1 = \cos \psi, v_2 = \sin \psi $, then from it follows the Euler normal curvature formula $ \kappa_n= \kappa_1 \cos^{2} \psi + \kappa_2 \sin^{2} \psi, right? $2015-09-16