1
$\begingroup$

Last week I considered again principal curvature (pc) and principal curvature directions (pcd) of a, for the sake of simplicity, 2-manifold embedded in 3-space. In this simple case, the pc and pcd of at a point are the eigenvalues and eigenvectors of the shape operator. The magnitude of the pc's corresponds to the minimal and maximal normal curvature at the point. My question, however, is:

What does the principal curvatures direction magnitude represents?

In the textbooks I looked up in (Kühnel and do Camro) I couldn't find a reference to the principal curvature direction's magnitude. Is there something known about this? Is it something basic (maybe even from linear algebra)?

Edit 1: A somewhat more general, but related, question is:

What is the geometrical meaning of an eigenvector's magnitude?

$\endgroup$
1
  • 2
    $\begingroup$ Sorry, this really should be over at math.stackexchange.com. Closing the gate after the horse has bolted, etc, but we've belatedly closed the question. $\endgroup$ Dec 21, 2010 at 6:34

3 Answers 3

5
$\begingroup$

Unless there's an additional constraint on the defintion of the principal curvature directions, being just defined as eigenvectors means their magnitude is arbitrary and so meaningless.

$\endgroup$
2
  • $\begingroup$ That is my understanding as well. However, is it possible that some geometrical meaning of the pcd's magnitude does exist? $\endgroup$ Dec 20, 2010 at 10:59
  • 2
    $\begingroup$ How could there be any meaning? If $\vec{v}$ is a principal curvature direction then so is $2\vec{v}$. So, in order for there to be a geometrical meaning you would have to modify the definition so that the magnitude isn't allowed to be arbitrary. $\endgroup$
    – Zhen Lin
    Dec 20, 2010 at 11:54
7
$\begingroup$

While all of the above remarks about the principal curvature magnitudes being arbitrary are correct, I think it is fairly customary for differential geometers to think of the eigenvectors of the shape operator as being the principal axes of the the curvature ellipsoid, and so giving them lengths equal to the principal curvatures. Of course this only makes sense on the complement of the umbilic points.(Recall that the umbilics are the points where the principal curvatures are equal, so that at these points the principal directions are not well-defined.)

$\endgroup$
2
$\begingroup$

In general, the concept of "eigenvector" is slightly misleading. The fundamental concept is eigenspace. However, eigenspaces of dimension greater than 1 are generally considered pathological and rather a nuisance. Mostly we would rather have one-dimensional eigenspaces, and for these any convenient nonzero element---an eigenvector--- will serve as a representative. It would be pedantically correct but encumbering to insist on talking about one-dimensional eigenspaces rather than eigenvectors. The arbitrariness of eigenvectors becomes clearer when we really have to deal with a multidimensional eigenspace.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.