# Finding Parabolic and Planar Points

I am trying to solve this question, I understand the logic behind it which I will try to explain after the statement of the question. The question reads:

Let $S$ be an oriented regular surface that is tangent to the plane along a regular curve $\alpha$. Show that the points on $\alpha$ are parabolic or planar points of $S$.

Now considering the Normal Vector field $N$ along $\alpha$, we can see that this normal vector field is a normal vector field along the surface as well.

The plane has the same normal vector plane, and has a Gaussian Curvature of $0$. Thus because the surface has the same normal vector plane and is tangent to the plane there, shouldn't the Gaussian Curvature there also give $0$?

Is my logic sound or is there another way to prove this?

• I for one don't understand the question. Any regular curve on a regular surface has a plane that is tangent to it. So $\mathbb S^2$ seems like a counterexample. Is something missing from the question? Like some assumptions before the question? Think its the phrase "the plane" that mystifies me. – Martin Apr 18 '17 at 3:11
• There is no more to add to the question, I am assuming because the normal vector field to the plane is the same as the normal vector field to the surface along $\alpha$, and since the plane has 0 Gaussian Curvature, the surface would have 0 Gaussian Curvature there also? – Felicio Grande Apr 18 '17 at 3:18
• @Martin, there is a hint given, which says to consider the normal vector field of $\alpha$. – Felicio Grande Apr 18 '17 at 3:18
• But what plane? On any point on a curve there are infinetly many planes tangent to it- so the phrase "the plane" needs some clarification. Hmm - are all the regular surfaces in your textbook graphs of functions of two variables, or are they parimetrizations from $U \subseteq \mathbb R^2$ to $\mathbb R^3$? – Martin Apr 18 '17 at 3:23
• @Martin. This is from an exam, I'm assuming they just mean a general plane? Also I believe they are talking about a general surface as well. – Felicio Grande Apr 18 '17 at 3:25

You are right. Take $p \in S$ and let $\alpha \colon ]-\epsilon , \epsilon[ \to \mathbb R^3$ be a unit speed parametrization of $\Gamma$. Since $\Gamma$ is planar we know that $\alpha$'s binormal is constant (zero torsion). Hence we may pick a patch such that the binormal and the surface normals agree along $\alpha$. Now define $v = \alpha'(o)$ then we have argued that $dN_p(v)=0$.
Let $B(p,\delta)$ be so small that the graph of $\alpha$ partition $S \cap B(p,\delta)$ in two parts. If the surface lies to the same side of $\Pi$ on each part then all other sectional curvatures are either non-negative or non-positive (depending on the side of $\Pi$). Therefore $v$ is a principal direction and the gaussian curvature is 0. If $S$ lies on different sides of $\Pi$ in our small $S \cap B(p,\delta)$ neighbourhood. Then the sectional curvatures are 0. And again $v$ is a principal direction. Thus the gaussian curvature is 0.