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I am confused about the following concerning the second fundamental form.
Consider a surface $S$ $\subset R^3$
If we consider a chart at a point $p \in S$, $f$: $R^2$$\to S$ and suppose $\partial f/\partial x$ and $\partial f/\partial y$ are orthonormal, Does it then follow that the eigenvalues of the second fundamental form are the principal curvatures and the eigenvectors are the principal directions?
Moreover If $\partial f/\partial x$ and $\partial f/\partial y$ are not orthonormal this may not be true?
Thanks in advance

TheGeometer
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  • The second fundamental form depends only on the geometry of the surface, not on a choice of coordinates. As you say, its eigenvalues are the principal curvatures and its eigenvectors the principal directions.

    If that's not what you're asking, could you please explain the purported dependence on the coordinate vectors of a parametrization?

     – Andrew D. Hwang May 27 '15 at 15:56
  • Well I found two exersices on the second fundamental form in the first exersice the solution evaluated the second fundamental form using some chart without orthonormal derivatives and got a diagonal matrix therefore the diahgonal entries were the eigenvalues but eventually it turned out that those eigenvalues were not the principal curvatures. However the second exersice I found evaluated the second fundamental form using one chart with orthonormal derivatives and then claimed that the eigenvalues of the second fundamental form are the principal curvatures. That why I am confused – TheGeometer May 27 '15 at 16:14
  • Something sounds odd with the first exercise solution. Are you sure the solution didn't say "the coordinate directions are not the principal directions"? – Andrew D. Hwang May 27 '15 at 16:52
  • The exercise I am talking about can be found if you google the following:Assume the Gauss map composed with the coordinate chart is given by This should bring you to a past paper and the question 1c of this past paper – TheGeometer May 27 '15 at 17:03
  • Did you find it? – TheGeometer May 27 '15 at 17:08
  • My oops: When I typed "second fundamental form" earlier my brain was thinking "shape operator". I'll type up an answer shortly.... – Andrew D. Hwang May 27 '15 at 17:18

1 Answers1

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Let $S$ be a regular surface, $f$ a local parametrization in some neighborhood of a point $p$, and $f_{x}$, $f_{y}$ the coordinate vector fields, and $N$ a continuous unit normal field. Notation below follows the exam mentioned in the comments.

The first fundamental form is the quadratic form on $T_{p}S$ whose matrix with respect to the basis $\{f_{x}, f_{y}\}$ is $$ I = \left[\begin{array}{@{}cc@{}} f_{x} \cdot f_{x} & f_{x} \cdot f_{y} \\ f_{x} \cdot f_{y} & f_{y} \cdot f_{y} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} g_{11} & g_{12} \\ g_{12} & g_{22} \\ \end{array}\right]. $$ The second fundamental form is the quadratic form on $T_{p}S$ whose matrix with respect to the basis $\{f_{x}, f_{y}\}$ is $$ II = \left[\begin{array}{@{}cc@{}} N \cdot f_{xx} & N \cdot f_{xy} \\ N \cdot f_{xy} & N \cdot f_{yy} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} e & f \\ f & g \\ \end{array}\right]. $$

The shape operator is the endomorphism associated to $II$ by "raising an index" with respect to $I$. Its matrix in the basis $\{f_{x}, f_{y}\}$ is $$ \frac{1}{g_{11} g_{22} - g_{12}^{2}}\left[\begin{array}{@{}rr@{}} g_{22} & -g_{12} \\ -g_{12} & g_{11} \\ \end{array}\right]\left[\begin{array}{@{}cc@{}} e & f \\ f & g \\ \end{array}\right], $$ its eigenvalues are the principal curvatures, and its eigenvectors are the principal directions.

If the coordinate basis $\{f_{x}, f_{y}\}$ is orthonormal at $p$, then the matrix of the first fundamental form at $p$ is the identity, so the matrix of the shape operator coincides with the matrix of the second fundamental form, just as you've noticed; if the coordinate basis is not orthonormal, however, the matrices of the second fundamental form and the shape operator do not have the same entries.

Andrew D. Hwang
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