Asymptotic curves and lines of curvature in a hyperbolic paraboloid

The explicit computation of the asymptotic curves and lines of a curvature in a given surface uses parametrizations. Thus their computations depend what is the chosen parametrization. We illustrate this problem with the hyperbolic paraboloid. The usual way to work with this surface is as $z=x^2-y^2$. Then $X(x,y)=(x,y,x^2-y^2)$ is a parametrization of the surface and the coefficients of $I$ are:

$$E=1+4x^2,\ F=-4xy,\ G=1+4y^2.$$

For the coefficients of the second fundamental form suffices to consider the numerators in the expressions of these coefficients, which are:

$$en=2,\ fn=0,\ gn=-2.$$

Then $\alpha(t)=X(x(t),y(t))$ is an asymptotic curve if $2x'^2-2y'^2=0$, that is, $y(t)=x(t)+c$ or $y(t)=-x(t)+c$, $c\in{\mathbb R}$. If $x(t)=t$, then the curves $X^{-1}\alpha$ are straight-lines of slope $1$ and $-1$. 

For the lines of curvature, $\alpha(t)=X(x(t),y(t))$ is line of curvature if 

$$\left|\begin{array}{ccc}y'^2&-x'y'&x'^2\\ 1+4x^2& -4xy& 1+4y^2\\ 2&0&-2\end{array}\right|=0.$$

If $x(t)=t$, then $$4ty(1-y'^2)=y'(2+4t^2+4y^2),$$ which it is very difficult to solve. 

We change of parametrization of the hyperbolic paraboloid. After a change of variables $u=x-y$, $v=x+y$ and a rotation about the $z$-axis, the surface is $z=xy$. Let $X(x,y)=(x,y,xy)$. Now we have $$E=1+y^2,\ F=xy,\ G=1+x^2$$ and $en=0,\ fn=1,\ gn=0$. Then $\alpha(t)=X(x(t),y(t))$ is asymptotic curve if $x'y'=0$, that is, $x(t)=c$ or $y(t)=c$, or in other words, they are the coordinate curves. For the lines of curvatures, the determinant to solve simplifies into $$(1+y^2)x'^2=(1+x^2)y'^2.$$ Letting $x(t)=t$, then 

$$\frac{y'}{1+ y^2}=\pm\frac{t}{1+t^2}\Rightarrow {\mbox arc}\sinh(y(t))={\mbox arc} \sinh(t)+c.$$

Then $$y(t)=\sinh(\pm {\mbox arc} \sinh(t)+c)=\pm\cosh(c)t+\sinh(c)\sqrt{1+t^2}.$$

Thus with this parametrization we have found the lines of curvature of the surfaces, in contrast to the initial parametrization.