# Orthogonality of principal curvature directions

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 $$3D$$:

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:

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

• 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. – Ted Shifrin Nov 13 '18 at 18:41
• 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. – Anthony Carapetis Nov 14 '18 at 5:20
• 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. – Wazowski Nov 14 '18 at 13:17
• 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. – Anthony Carapetis Nov 14 '18 at 14:04
• I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations. – Wazowski Nov 15 '18 at 8:09