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!
Differential Geometry – Shape Operator and Principal Curvature
differential-geometryriemannian-geometry
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This is just a linear algebra fact.
The shape operator is known to be symmetric, so by the spectral theorem it has real eigenvalues, so there is an orthogonal basis in which this operator has a diagonal matrix, and the determinant of a diagonal matrix is easy to compute.
Of course, the validity of my "proof" depends on how you define the principal curvatures, but I rely here on the treatment given in the book of Theodore Shifrin "Differential Geometry. A first Course in Curves and Surfaces", where one can find a very accessible exposition.
The obvious example one thinks of is the $1$-parameter family of minimal surfaces interpolating between the catenoid and the helicoid. These are locally — but not globally — isometric. However, because they're minimal, having equal curvatures (and mean curvature $0$) in fact forces the principal curvatures to be equal.
Here's the next "obvious" example. If we puncture a unit sphere at the north and south poles, we can take the standard embedding or we can take various embeddings as different surfaces of revolution (still symmetric about the $z$-axis and still having $K=1$). By Minding's Theorem, these surfaces are (locally) isometric. (You can also check directly from the formulas below.) However, the principal curvatures are quite different, as you can easily check.
(Take the profile curve $(f(u),g(u))$ with $f(u)=a\cos u$, $|a|\ne 1$, $g(u) = \int\sqrt{1-a^2\sin^2u}\,du$.)
Best Answer
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.