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This is an exercises in A Course in Differential Geometry, which is 3.9.3 on page 68.

Needed definitions are in the image:enter image description here

I have trouble with (iii) in $\Rightarrow$ direction.

I think it is sufficient to show that the Gauss curvature is less than zero so I compute the curvature of $b_1$,

$b_{1,1}=f_{,1}+\frac{n_{,1}}{\kappa_1}-\frac{\kappa_{1,1}}{\kappa_1^2}n=f_{,1}-\frac{\kappa_1}{\kappa_1}f_{,1}-\frac{\kappa_{1,1}}{\kappa_1^2}n=-\frac{\kappa_{1,1}}{\kappa_1^2}n $

$b_{1,2}=(1-\frac{\kappa_2}{\kappa_1})f_{,2}-\frac{\kappa_{1,2}}{\kappa_1^2}n ,$

and the unit norm of $b_1$ is $n_{b_1}=\frac{f_{,1}}{|f_{,1}|}$, so

$n_{b_1,1}=\frac{f_{,11}}{|f_{,1}|}+\partial_1(\frac{1}{|f_{,1}|})f_{,1}$

$n_{b_1,2}=\frac{f_{,12}}{|f_{,1}|}+\partial_2(\frac{1}{|f_{,1}|})f_{,1} .$

Then, I compute the entries of the second fundamental form

$-L = n_{b_1,1} \cdot b_{1,1}=-\frac{\kappa_{1,1}}{\kappa_1^2|f_{,1}|}(n \cdot f_{,11}) $

$-M= n_{b_1,1} \cdot b_{1,2} = (1-\frac{\kappa_2}{\kappa_1})\frac{1}{|f_{,1}|}(f_{,11}\cdot f_{,2})-\frac{\kappa_{1,2}}{\kappa_1^2|f_{,1}|}(n\cdot f_{,11})$

$-M =n_{b_1,2} \cdot b_{1,1}=-\frac{\kappa_{1,1}}{\kappa_1^2|f_{,1}|}(n \cdot f_{,12}) $

$-N = n_{b_1,1} \cdot b_{1,2} = (1-\frac{\kappa_2}{\kappa_1})\frac{1}{|f_{,1}|}(f_{,12}\cdot f_{,2})-\frac{\kappa_{1,2}}{\kappa_1^2|f_{,1}|}(n\cdot f_{,12}) .$

I notice that $-M=n_{b_1,2} \cdot b_{1,1}=0$ since

$n \cdot f_{,12}=-n_{,1} \cdot f_{,2}=\kappa_1f_{,1}\cdot f_{,2}=0$

and $(n \cdot f_{,1}) = 0$, so

$n \cdot f_{,11} = (n \cdot f_{,1})_{,1} - n_{,1} \cdot f_{,1} = \kappa_{1} |f_{,1}|^{2} .$

Thus,

$LN-M^2=LN=-\frac{\kappa_{1,1}}{\kappa_1}(1-\frac{\kappa_2}{\kappa_1})(f_{,2}\cdot f_{,12}) .$

I expect that determinent of 2nd fundamental form is less than zero, but I don't know how to use the condition of being a Weingarten surface or the equality in (i).

Thanks in advance!

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