Differential Geometry, why orthogonal eigenvectors have minimum and maximum curvature? why not they happen in other vectors without orthogonal? Hello brothers and masters, I am undergrad student, learning introduction of Differential Geometry (DG).
I am very curious that in the DG, the principal curvature appear in the eigenvectors which is orthogonal to each other (because linear algebra), given that a point of the regular surface.
But I cannot hard to understand with my heart. Because it looks also possible that the principal curvature may happen in the tangent vectors which are not orthogonal to each other.
For example, the counterexample which I am suspecting:


This surface has maximum curvature at $\theta = 0$ and $\theta = \frac{\pi}{2}$, and has minimum curvature at $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$.
Is there a something wrong with this surface, or what can be a good explanation for this surface?
Please quench this curiosity! Thank you in advance.
the wolfram alpha code of the surface
 A: Through the power of Wolfram Alpha, one can check that
$$
 \cos(4\arctan(t)) = 3 - \frac{8t^2}{(t^2+1)^2},
$$
so that
$$
 \cos(4\arctan(v/u))+2 = 5 - \frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
 (u^2+v^2)(\cos(4\arctan(v/u))+2) = 5(u^2+v)^2 - \frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
 f(u,v) = 5(u^2+v^2) - \frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) \neq (0,0)$, one can compute
$$
 \partial_1f(u,v) = 10u - \frac{16uv^4}{(u^2+v^2)^2}, \quad \partial_2f(u,v) = 10v - \frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t \neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
 \partial_2f(u,tu) = 10tu - \frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $\partial_2f(u,v)$ as $(u,v) \to (0,0)$ along the line $v = tu$ is
$$
 \lim_{u \to 0} \partial_2f(u,tu) = -\frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $\partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
