# Questions on surfaces touching along a curve (why is it a curvature line?)

I have two questions on surfaces touching along a curve. I would really appreciate if you could help me.

i) Prove that if two surfaces touch along a curve and it is a curvature line on one of the surfaces, then it is a curvature line on the other surface as well.

ii) Show that if a plane touches a surface along a curve, then it is a curvature line and all of its points are parabolic.

Thank you very much in advance!

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I will only hint at the solution, leaving it to you to make things explicit. This way you will have to chase through the definitions, which, I claim, is a good exercise.

For simplicity I assume this is about 2-dimensional surfaces in $\mathbb{R}^3$. I also assume you know the 2nd fundamental form of a hypersurface can be expressed as the (covariant) derivative of the normal to the hypersurface along that hypersurface (up to sign).

If two surfaces $M_1, M_2$ touch along a curve, they share the normal along the curve (up to sign) (why?). The curve is a line of curvature iff it's tangent is an eigenvector of the 2nd fundamental form. This means: if $\phi$ parametrizes the curve along which the surfaces intersect and if $\nu$ denotes the normal you need to show that along $\phi$ $$\nabla^{M_1}_{\phi^'} \nu = \lambda \phi^' \Rightarrow \nabla^{M_2}_{\phi^'} \nu = \sigma \phi^'$$
where $\lambda, \sigma$ are scalar functions along $\phi$ and $\nabla^M$ denotes the derivative along the surface (i.e. the orthogonal projection of the derivative in the ambient space onto the tangent space of the surface). If you know how to compute this, the answer to i) should be quite obvious.

As for ii): convince yourself that the line of intersection is a curvature line in the plane. Since the surface touches the plane it is located in one of the halfspaces the plane defines. Write the surface locally as a graph of a nonnegative function $u$ over the plane, use the fact that $u$ has a minimum along the line of intersection. Check the implications of this observation for the second derivative and, in consequence, for the second fundamental form of the surface.

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Dear Thomas! Thank you very much for your quick answer. You helped me a lot. I chased through the definitions but I still have some questions. Unfortunately I couldn't find out how to compute that implication. (I am afraid I am not sure that I got all your notations.) As for ii) if I convince myself that the line of intersection is a curvature line in the plane then the first part of the statement follows from the previous problem, isn't it? All of its points are parabolic iff exactly one of the principal curvature is 0 for all points. I don't see how does it follow? I appreciate your help! – benny Dec 14 '11 at 11:51
Well it seems we are using different definitions. But with your definition of parabolic point, ii) is not true without some additional hypothesis. To see this look at the surface defined by $(x, y) \mapsto (x,y,y^4)$ and its intersection with $z=0$, which touches the surface along $y=0, z=0$, but according to your definition will not contain any parabolic point (the second fundamntal form vanishes along that line, all principal curvatures vanish). -- In differential geometry there are, unfortunately, many formalisms in use. Which notation causes you trouble and which are you using? – user20266 Dec 14 '11 at 16:28
First of all, I am very grateful for your help and patient. Secondly, you are right, my definition was not correct. A point is parabolic if LN-M^2=0 (where L, N, M are the coefficients of the second fundamental form). To tell the truth I don't really understand that implication containing nabla operator. (What does it mean? Why is it true? And why is it enough to show to solve i)?) Thanks for your help! – benny Dec 14 '11 at 17:44
There are several (equivalent) definitions of the 2nd fundamental form. You don't neet the nabla operator definition, but you should make sure you understand the 2nd ff may be looked at as i) a bilinear map $TM\times TM \rightarrow \mathbb{R}$ or ii) as a map (usually called Weingarten map) $TM \rightarrow TM$ - this is basic linear algebra. The $L N M$ notation you are using tends to hide this basic insight. What you need to do to prove i) is to show that the tangent to the curve of intersection is an eigenvector of the Weingarten map for both surfaces. This can be done in any formalism. – user20266 Dec 14 '11 at 19:48
One way to see this is to write one surface as a graph $(x,y, u(x,y)$ over the other one. Along the curve of intersection $u$ attains it's minimum, hence the gradient is zero. Now have a look at the formula for the normal for a graph. This works, of course, only for differentiable surfaces. – user20266 Dec 15 '11 at 6:12