Given a bivariate polynomial of sufficient degree $P(x,y)$, I want to calculate the principal curvature direction at every point $(x,y)$ for the surface $(x,y, P(x,y))$. So to do that i need to calculate the eigenvectors of the shape operator at point $(x,y)$, or so i think. So to conclude what i think i have to do:
Surface parameterised by $S(u,v) = (u,v,P(u,v))$
Coefficients of First fundamental form:
$$E = 1+f_u^2$$ $$F = f_u*f_v$$ $$G = 1+f_v^2$$
Coefficients of Second fundamental form:
$$L=\frac{f_{uu}}{\sqrt{1+f_u^2+f_v^2}}$$ $$M=\frac{f_{uv}}{\sqrt{1+f_u^2+f_v^2}}$$ $$N=\frac{f_{vv}}{\sqrt{1+f_u^2+f_v^2}}$$
with $f_u = P_x = \frac{\delta f(x,y)}{\delta x}$, $f_v = P_y = \frac{\delta P(x,y)}{\delta y}$, $f_{uu} = P_{xx} = \frac{\delta^2 f(x)}{\delta^2 x}$ and so on.
shape operator from weingarten equations:
$$-S = \begin{matrix} \frac{FM-GL}{EG-F^2} & \frac{FL-EM}{EG-F^2}\\ \frac{FN-GM}{EG-F^2} & \frac{FM-EN}{EG-F^2} \end{matrix}$$
eigenvectors of the shape operator for (u,v)=(x,y) are the principal curvature directions in tangent space at point (x,y)
My problem is that i now get complex eigen vectors which i cant rly interpret in this context, so where did i go wrong? The Weingarten equations are in regards to a basis, which was one thing i didnt quite fully understood, but i thought it might be the basis of the tangential space of the point and that the calculated eigenvectors are in tangent space and not in parameter space but even in tangent space complex coordinates dont make sense to me. My understanding of differential geometry might be very wrong, im thankful for evry answer.