In Do Carmo's differential geometry of curves and surfaces, page 160, the differential equation of the asymptotic curves is
$$e(u')^2+2fu'v'+g(v')^2=0$$
and the differential equation of the lines of curvature is(page 161)
$$(fE-eF)(u')^2+(gE-eG)u'v'+(gF-fG)(v')^2=0$$
There is an exercise related to these equations on page 168(part d and part e):
Consider the parametrized surface
$$x(u,v)=(u-u^3/3+uv^2, v-v^3/3+vu^2, u^2-v^2).$$Some computations show that
$$E=G=(1+u^2+v^2)^2, F=0$$
$$e=2, g=-2, f=0$$Part d. The lines of curvature are the coordinate curves.
Part e. The asymptototic curves are $u+v=const$ and $u-v=const$.
My question:
Take d for instance,
when $e=g=0$ and $eg-f^2<0$, we only know $u'v'=0$. Clearly lines of curvature are the coordinate curves satisfy the equations. I wonder why the lines of curvature can only be the coordinate curves and no others?
Specifically, why does $u'(t)v'(t)=0, \forall t$ implies that $$u'(t)=0, \forall t ~~\text{or}~~ v'(t)=0, \forall t?$$
Note: We only know that $\forall t, u'(t)=0$ or $v'(t)=0$.
Best Answer
It's a continuity argument. Note that you cannot have any umbilic points, since $K<0$, and so there's no way to switch continuously from $u$-curve to $v$-curve along a single curve. When there are umbilic points, there are no longer everywhere precisely two distinct principal directions and all bets are off.