# If all points of a coordinate nbhood are parabolic, then the unique asymptotic curve is a straight line

Do Carmo's exercise: Let $$p$$ be a point of an oriented surface $$S$$ and assume that there is a neighborhood of $$p$$ in $$S$$ all points of which are parabolic. Prove that the (unique) asymptotic curve through $$p$$ is an open segment of a straight line.

At the end of the book the following hint is given:

Parametrize a neighborhood of $$p\in S$$ in such a way that the lines of curvature are the coordinate curves and that $$v= const.$$ are the asymptotic curves. It follows $$e_v=0$$, and from the Mainard-Codazzi equations, we conclude that $$E_v= 0$$. This implies that the geodesic curvature of $$v=const.$$ is zero.

Since all the points are not umbilical, I know that we may parametrize a neighborhood of $$p$$ in such a way that the coordinate curves are lines of curvature. I can't see though why we can make $$v=const.$$ be also an asymptotic curve. Assuming that we may do this, I also can't see why it implies $$e_v=0$$. The remaining affirmations I agree, after all the coordinate curves being lines of curvature implies $$f=F=0$$ and hence from the Mainard equations we have $$0 = e_v=E_v/2(e/E+g/G)$$ which implies $$E_v=0.$$ Since the geodesic curvature of the $$u$$-coordinate curve is $$-E_v/(2E\sqrt G),$$ it follows that the coordinate curve is geodesic and asymptotic simultaneously. Therefore, by another exercise in this same section, we conclude that it is a segment of a line.

Any help in clarifying these claims is much appreciated

• Note that the asymptotic curves are one of the families of lines of curvature. (The curve with principal curvature $k=0$ is the asymptotic curve.) Nov 2, 2019 at 4:31
• For an alternative argument, see Proposition 3.4 on p.61 of my differential geometry text. Nov 2, 2019 at 4:39
• @TedShifrin I think that I got what you said. Re-reading the proposition about the reparametrization, it states that the coordinate curves of the parametrization are $\textit{the}$ lines of curvature. As you said, the asymptotic curves are in particular lines of curvature, so we may suppose that the $v$-curve passing at $p$ is the asymptotic curve. But I still can't see how it implies $e_v=0$. Any hint? Nov 2, 2019 at 18:51
• I think doCarmo has his variables all confused. The rest of the argument seems to be following the assumption that the curves with principal curvature $0$ are the $u$-curves, not the $v$-curves. Look at his conclusion. At any rate, I still suggest you read my proof (which is quite similar but ends in a different way, without needing geodesic curvature). Nov 2, 2019 at 18:56

Let $$U$$ be the neighborhood of $$p$$ which exists a parametrization where the coordinate curves are lines of curvature and all points are parabolic.

Then, $$k_1 k_2 = \det (dN_q) = 0 \; (\forall q \in U)$$, where $$k_1, k_2$$ are principal curvatures. $$\forall q \in U$$, since $$dN_q \neq 0$$, one of $$k_1, k_2$$ is zero and the other is not.

Since the coordinate curves are lines of curvature in $$U$$, w.l.o.g., assume $$x_u, x_v$$ are principal directions corresponding to $$k_1, k_2$$, respectively. Note that principal directions are orthogonal.

W.l.o.g., assume $$k_1 = 0$$ at $$p$$, therefore $$x_u$$ is the only asymptotic direction of $$p$$ (since $$k_2 \neq 0$$). $$N_u = dN_p(x_u) = k_1 x_u = 0$$ $$N_v = dN_p(x_v) = k_2 x_v \neq 0$$ Since $$S$$ is $$C^\infty$$, $$N_v$$ is continous on $$U$$, therefore there exists a neighborhood $$V \subset U$$ such that $$N_v(q) \neq 0 \; (\forall q \in V)$$. Therefore $$N_u = 0 \; (\forall q \in V)$$.

Thus, $$v=const.$$ are the asymptotic curves in such neighborhood $$V$$ (In other words, all curves $$C \subset V$$ of the form $$v=const.$$ are asymptotic).

Moreover, $$e = - \langle N_u, x_u \rangle = 0 \; (\forall q \in V)$$. Hence $$e_v = 0$$.

Correct me if I am wrong, the following notations agree with those in Do Carmo's textbook.

Since $$u=const.$$ is line of curvature and $$v=const.$$ is asymptotic curve, it follows $$X_u$$ is asymptotic direction, and $$X_v$$ is principle direction, therefore $$N_v=\lambda X_v$$ for some real valued function $$\lambda$$, also $$II(X_u)=\langle -dG(X_u),X_u\rangle=\langle -N_u,X_u\rangle$$=0. By definition, $$e=\langle N,X_{uu}\rangle$$, thus \begin{align*} e_v&=\langle N_v,X_{uu}\rangle+\langle N,X_{uuv}\rangle=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_{uv}\rangle\\ &=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_u\rangle_v+\langle N_{uv},X_u\rangle\\ &=\lambda\langle X_v,X_{uu}\rangle-(0)_v+\langle (\lambda X_v)_u,X_u\rangle\\ &=\lambda\langle X_v,X_{uu}\rangle+\lambda\langle X_{uv},X_u\rangle+\lambda_u\langle X_v,X_u\rangle\\ &=\lambda(\langle X_v,X_{uu}\rangle+\langle X_{uv},X_u\rangle)+\lambda_uF=\lambda F_u+\lambda_u F=0 \end{align*} Then the result follows by the op's argument.