Given a polynomial surface $M =\{(x,y,z): z = p(x,y))\}$ where $p(x,y)$ is a polynomial. I want compute the principal curvatures and directions at some point in $M$. But I do not have idea to compute the principal directions.

For the part of principal curvatures, I can formulate $E、F、G、L、M、N、W$, which are coefficients in first and second fundamental form, in function of $p,p_x,p_{xx},p_y,p_{yy},p_{xy}$. Then the Mean curvature and Gaussian curvature can be evaluate by $$ H = \frac{-1}{2}(EN - 2FM + GL)/W^2 \\ K = (LN - M^2)/W^2~~~~~~~~~~~~~~~~~~~~~ $$ Therefore, the principal curvatures can be formulated in function of $p,p_x,p_{xx},p_y,p_{yy},y_{xy}$ by $$ \kappa_1 = H - \sqrt{H^2 - K}\\ \kappa_2 = H + \sqrt{H^2 - K} $$ for any point in surface $M$.

But now I do not have idea to compute the principal directions corresponding to $\kappa_1$ and $\kappa_2$. Should I compute them by solving eigenvector of shape operator $S$? Thank you for your help.


Yes, you should compute the eigenvectors of the shape operator. Keep in mind you already know the eigenvalues, that's what the principal curvatures are. You can use the Weingarten equations to compute the shape operator in terms of the first and second fundamental form.

  • $\begingroup$ How could I do that? I am so confusing about the shape operator $S$. $\endgroup$ – sinoky Apr 1 '17 at 9:54
  • $\begingroup$ The best way to do that is to use the Weingarten equations, given here: en.m.wikipedia.org/wiki/… which can be found from the first and second fundamental forms. $\endgroup$ – phunfd Apr 1 '17 at 10:07
  • $\begingroup$ Why the shape operator is not a 3 by 3 matrix in wiki page? I think that the shape operator is a mapping form $R^3$ to $R^3$. I am so confusing... $\endgroup$ – sinoky Apr 1 '17 at 10:14
  • $\begingroup$ The shape operator should be thought of as mapping the tangent plane of your surface to the tangent plane of the sphere. Although both sit inside $\mathbb{R}^3$ they are both two dimensional. $\endgroup$ – phunfd Apr 1 '17 at 10:18
  • $\begingroup$ I use the formula of shape operator $S$ in wiki and compute the solution of $ (S - \kappa_i I)v = 0$, then I can get two 2 dimensional vectors $v = (v_1,v_2)$ corresponding $\kappa$. But our space is 3 dimensional, the principal direction must be 3D vector. Where is my mistake $\endgroup$ – sinoky Apr 1 '17 at 13:37

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