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!


1 Answer 1


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.

  • $\begingroup$ Thanks! I still don't see the last part, yes we know that for every unit vector in the tangent space there is a geodesic having that vector as tangent vector; but why do we have that if the curvature $k=\langle S\alpha'(0),\alpha'(0)\rangle$ of a geodesic $\alpha$ is minimum or maximum among geodesics then $k$ is an eigenvalue of $S$? $\endgroup$
    – Manuel
    May 3, 2011 at 0:59
  • $\begingroup$ @Manuel: this follows from the expression for the second fundamental form in the first paragraph. If $v$ is a unit vector then $v_1^2 + v_2^2 = 1$, so $\kappa_1 v_1^2 + \kappa_2 v_2^2$ is minimized when $v_1^2 = 1, v_2^2 = 0$ and maximized when $v_1^2 = 0, v_2^2 = 1$ (I'm assuming $\kappa_1 \le \kappa_2$ here). This is a fairly straightforward calculation. $\endgroup$ May 3, 2011 at 2:20
  • $\begingroup$ IIRC the shape operator is not quite the second fundamental form. But it is $dN$, where $N$ is the Gauss map. $\endgroup$
    – Zhen Lin
    Jun 2, 2011 at 7:14
  • $\begingroup$ @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? $ $\endgroup$
    – Narasimham
    Sep 16, 2015 at 11:56

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