I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal. Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.

I construct the surface as follows. I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:

$$ \left [X = rcos(\theta), \quad \quad Y = rsin(\theta), \quad \quad Z = ar^2 \right],$$ $$a(\theta)=0.5(1 + sin(4\theta)),$$

where the parametric domain is $$ \theta \in [0,\pi], \quad \quad r \in [-0.5,0.5].$$

The "problematic" point is the origin $(0,0,0)$. See a Matlab rendering of the surface in $3D$:

enter image description here

For $\theta \in \{3\pi/8,7\pi/8\}$ the $\sin(4\theta)$ in the expression for $a(\theta)$ is at minimum ($-1$) and we get $a=0$. The restriction of $\theta$ to these two values corresponds to two orthogonal straight lines on the surface (green in the figure). Hence $\kappa_2 = 0$. For $\theta \in \{\pi/8,5\pi/8\}$ the $\sin(4\theta)$ in the expression for $a(\theta)$ is at maximum ($1$) and we get $a=1$. The restriction of $\theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $\kappa_1=2$.

Here is a top view of the $XY$ plane: enter image description here

The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.

  • $\begingroup$ First of all, $\theta=-\pi/8$ and $\theta=15\pi/8$ are the same (mod $2\pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $\theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is. $\endgroup$ – Ted Shifrin Nov 13 '18 at 18:41
  • $\begingroup$ Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization. $\endgroup$ – Anthony Carapetis Nov 14 '18 at 5:20
  • $\begingroup$ The Hessian of $Z$ is $ \left[ \begin {array}{cc} 1+\sin(4\theta) & 4\cos( 4\theta) r\\ 4\cos(4\theta) r & -8\sin(4\theta)r^2 \end {array} \right]$ so $Z$ is twice differentiable. $\endgroup$ – Wazowski Nov 14 '18 at 13:17
  • $\begingroup$ I meant $Z$ as a function of $X,Y$, not $r,\theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense. $\endgroup$ – Anthony Carapetis Nov 14 '18 at 14:04
  • $\begingroup$ I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations. $\endgroup$ – Wazowski Nov 15 '18 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.