1
$\begingroup$

This can be regarded as a continuation of the question about focal surface posted in "Question about Focal surfaces".

More precisely, my question is part (b) of Exercise 3.5.9 of do Carmo's book "Differential geometry of curves and surfaces", while the above post is about part (a). Part(b) reads as follows.

Let $S$ be a regular surface without parabolic or umbilical points. Let $\boldsymbol{x}:U\to S$ be a parametrization of $S$ such that the coordinate curves are lines of curvature (if $U$ is small, this is no restriction). The parametrized surfaces

$$\boldsymbol{y}(u, v)=\boldsymbol{x}(u, v)+\rho_1N(u, v),$$

where $\rho_1=\frac{1}{\kappa_1}$ (I ignore the surface $\boldsymbol{z}$).

Part (b) asks: At the regular points, the directions on a focal surface corresponding to the principal directions on $\boldsymbol{x}(U)$ are conjugate. That means, for instance, that $\boldsymbol{y}_u$ and $\boldsymbol{y}_v$ are conjugate vectors in $\boldsymbol{y}(U)$ for all $(u, v)\in U$.

We have to prove the inner product of $N_*(\boldsymbol{y}_u)$ and $\boldsymbol{y}_v$ is zero, where $N$ is the unit normal vector field of $\boldsymbol{y}$.

But I don't know if there is an easy way to compute $N_*$. Of course we can find $N$ directly, but it seems to be not useful. The second way is to compute the coefficients of the first and the second fundamental forms, and make use of the matrix representation of $N_*$ in terms of $g_{ij}$ and $\ell_{ij}$ with respect to the basis $\boldsymbol{y}_u$ and $\boldsymbol{y}_v$. I computed that $\ell_{12}=0$, but then the $g_{ij}$'s and $\ell_{ii}$'s do not look nice. Perhaps there are other ways to the conjugacy of $\boldsymbol{y}_u$ and $\boldsymbol{y}_v$, but I don't know.

Any hint on this question will be appreciated. Thanks.

$\endgroup$
1
  • $\begingroup$ Well, I was so stupid that I didn't realize that $\ell_{12}=0$ implies the tangent vectors of the coordinate curves are conjugate. So this problem is solved. $\endgroup$ – GRR Aug 8 '14 at 17:35
1
$\begingroup$

The map $(u,v) \mapsto {\bf N}(u,v)$ is called the Gauss Map.

By ${\bf N}_*$ I assume you mean the differential of the Gauss Map.

The differential is, up to sign, the Shape Operator. It is a Self-Adjoint Operator.

For a tangent vector ${\bf v}$, the Shape Operator is defined by ${\bf v} \mapsto -D_{\bf v}{\bf N}$.

The Shape Operator is a linear map and we can take ${\bf Y}_u$ and ${\bf Y}_v$ as a basis for the tangent plane to get a $2\times 2$ matrix representation for the Shape Operator. If we do that, we have some nice facts:

  • The matrix is singular if, and only if, the point is a parabolic point.
  • The eigenvalues of the matrix are the principal curvatures.
  • The eigenvectors are in the principal directions.
  • The matrix is a multiple of the identity if, and only if, the point is an umbilic.
  • The determinant is the Gaussian Curvature.

Have a look at these slides for more details.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.