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=0e(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(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−u3/3+uv2,v−v3/3+vu2,u2−v2).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+u2+v2)2,F=0E=G=(1+u^2+v^2)^2, F=0

e=2,g=−2,f=0e=2, g=-2, f=0

Part d. The lines of curvature are the coordinate curves.

Part e. The asymptototic curves are u+v=constu+v=const and u−v=constu-v=const.

My question:

Take d for instance,

when e=g=0e=g=0 and eg−f2<0eg-f^2<0, we only know u′v′=0u'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,∀tu'(t)v'(t)=0, \forall t implies that u′(t)=0,∀t or v′(t)=0,∀t?u'(t)=0, \forall t ~~\text{or}~~ v'(t)=0, \forall t?
Note: We only know that ∀t,u′(t)=0\forall t, u'(t)=0 or v′(t)=0v'(t)=0.
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1 Answer
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It's a continuity argument. Note that you cannot have any umbilic points, since K<0K<0, and so there's no way to switch continuously from uu-curve to vv-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.
Thank you! But what do you mean by "there's no way to switch continuously from u-curve to v-curve along a single curve"? How to prove this rigorously?
– Ti Wen
Oct 20 at 18:40
Since there are no umbilic points, we can number principal curvatures so that k1>k2k_1>k_2 everywhere. We parametrize so that the uu-curves are those with k1k_1 as principal curvature and the vv-curves are those with k2k_2 as principal curvature. This is globally well-defined since principal curvatures vary continuously along a line of curvature. Only when there are umbilic points (k1=k2k_1=k_2) can there be any possibility to have k1k_1 along part of the curve and k2k_2 along another part of the curve.

– Ted Shifrin

Oct 20 at 18:52

To make it slightly more succinct: Following up on my previous comment, consider the function k1−k2k_1-k_2 along any curve. Since this is a continuous function, it cannot ever change sign. In particular, on any line of curvature, it is either always positive or always negative.

– Ted Shifrin

Oct 20 at 19:28

Thank you for your explanation, but I am still seeking for a formal proof for this

– Ti Wen

Oct 20 at 21:04

I just gave you one, but … OK. You absolutely need to use continuity of k1k_1 and k2k_2.

– Ted Shifrin

Oct 20 at 21:57