The parameter curves are asymptotic curves

Question

I am looking at the following exercise:

Let $p$ be a hyperbolic point of a surface $S$.

Show that there is a patch of $S$ containing $p$ whose parameter curves are asymptotic curves.

Show that the second fundamental form of such a patch is of the form $2Mdudv$.

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I have done the following:

Let $\sigma (u,v)$ be a surface patch of $S$ that contains $p=\sigma (u_0, v_0)$.

Since $p$ is a hyperbolic point of $S$ we have that at $p$ the principal curvatures $\kappa_1$ and $\kappa_2$ are of opposite sign and non-zero.

Since the Gauss curvature is $K=\kappa_1\kappa_2$ we have that at $p$, $K<0$.

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How could we continue?

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P.S. A curve $\gamma$ on a surface $S$ is called asymptotic if its normal curvature is everywhere zero.

Answer 1

Since $p$ is a hyperbolic point there exists a neighborhood $U_p\subset S$ of $p$ such that $$K(q)<0,\ \forall q\in U_p.$$ Consider now a parametrization of $p$ in $U_p$, i.e. $$X\colon V\subset \mathbb{R}^2\longrightarrow U_p\subset S, \ \ (u,v)\mapsto X(u,v),\ \ \&\ p\in X(V),$$ where $V$ is an open subset of $\mathbb{R}^2$. Since every point of $X(V)$ is hyperbolic you have exactly two families of asymptotic curves, which you can find by solving the equation $$L(c(t))(u'(t))^2+2M(c(t))u'(t)v'(t)+N(c(t))(v'(t))^2=0,$$ where $$c\colon I\subset \mathbb{R}\rightarrow V,\ t\mapsto c(t)=(u(t),v(t)).$$ Next you change the parameters $(u,v)$ to $(\tilde{u},\tilde{v})$ in the following way: Let $$F\colon V\subset \mathbb{R}^2\rightarrow \mathbb{R}^2$$ the differentiable map that maps the asymptotic curves (of $V$) to parameter curves. By checking that the jacobian determinant of $F$ is $\neq 0$ you can apply the inverse function theorem and consider the parametrization $$Y:=X\circ F^{-1} \colon F(V)\subset \mathbb{R}^2\longrightarrow U_p\subset S,\ \ (\tilde{u},\tilde{v})\mapsto Y(\tilde{u},\tilde{v}).$$ Now, the parameter curves of $Y$ are asymptotic curves. Now, due to the parametrization $Y$, we obtain:

• $k_N(Y(\tilde{u},\tilde{v}),Y_\tilde{u}(\tilde{u},\tilde{v}))=0\Rightarrow II_{Y(\tilde{u},\tilde{v})}(Y_{\tilde{u}}(\tilde{u},\tilde{v}))=0\Rightarrow L(\tilde{u},\tilde{v})=0.$
• $k_N(Y(\tilde{u},\tilde{v}),Y_\tilde{v}(\tilde{u},\tilde{v}))=0\Rightarrow II_{Y(\tilde{u},\tilde{v})}(Y_{\tilde{v}}(\tilde{u},\tilde{v}))=0\Rightarrow N(\tilde{u},\tilde{v})=0.$

Thus the second fundamental form of $Y$ is of the desired form: $II=Mu'v'$.

PS : By $k_n(p,v)$ we denote the normal curvature at the point $p\in S$ in the direction $v\in T_pS$.