# asymptotic curve on a complete surface.

• Dec 26th 2011, 12:13 PM
Sogan
asymptotic curve on a complete surface.
Hello,

We have a 2-dimensional Riemannian Manifold immersed in $\displaystyle \mathbb{R}^3$, which is complete and has constant negative curvature.
Then the statement is, that for every point $\displaystyle p\in M$, there is an asymptotic curve $\displaystyle c:\mathbb{R}\rightarrow M$, parametrized by arclength, s.t. c(0)=p.

I don't know how i can prove this. I mean the Surface M is complete, that is we can extend every geodesic $\displaystyle c: (-a,a)\rightarrow M$ into a geodesic on the whole real line $\displaystyle c:\mathbb{R}\rightarrow M$.
And we know, locally there exists always a geodesic. But a geodesic is not an asymptotic curve. Since for an asymptotic curve we have that $\displaystyle c''(t)\in T_p M$. But for a geodesic we have that c''(t) is perpendicular to $\displaystyle T_p M$.

So we have to search for another idea:
I could show, that for any point p, we have locally an asymptotic curve, parametrized by arclength.

But i have no idea, why we can extend such an asymptotic curve to the real line.....
I hope you can help me with this problem.

Regards
• Dec 26th 2011, 06:31 PM
xxp9
Re: asymptotic curve on a complete surface.
Didn't look in detail yet but your first statement if false since according to Hilbert theorem, there is no complete regular surface of constant negative gaussian curvature immersed in R^3.
• Dec 26th 2011, 11:08 PM
Sogan
Re: asymptotic curve on a complete surface.
Yes you are right. I want to prove this theorem. And the author first assumes, that there is such an immersion of M, in the proof. (His strategy is proof by contradiction)
Then he says, that there must be such an asymptotic curve to each point.

Maybe we can do it like this:
We take an local asymptotic curve $\displaystyle c_1 : [-a,a]\rightarrow M$ with $\displaystyle c_1 (0)=p$. (as restriction of a regular curve on $\displaystyle (-a-\epsilon , a+\epsilon)$

Then we take the point $\displaystyle c(a)\in M$ and do the same process again with some function $\displaystyle c_2 : [-b,b]\rightarrow M$.
Now we define: $\displaystyle c : [-a,a+b]\rightarrow M$ by $\displaystyle c(x)=c_2 (x-a)$ if x>=b and $\displaystyle c(x):=c_1 (x)$ otherwise.

and so on.... Is this the right idea?

Regards
• Dec 27th 2011, 06:25 AM
xxp9
Re: asymptotic curve on a complete surface.
Since the Gaussian curvature is negative, the two principal curvatures k1 and k2, are of different signs. Let's say k1 > 0 and k2 < 0, then according to Euler's theorem, $\displaystyle k = k_1 cos^2\theta + k_2 sin^2\theta$, this is a continuous function of $\displaystyle \theta$ with $\displaystyle k(0)=k_1 > 0$ and $\displaystyle k(\pi / 2)=k_2 < 0$, so there must be a value of $\displaystyle \theta \in (0, \pi/2)$ so that k=0. This is the asymptotic direction.

So this defines a unit vector field X on the surface. At each point p, X(p) is the unit vector at the asymptotic direction. Since $\displaystyle k_1$, $\displaystyle k_2$ and the principal directions are smooth on the surface, the above construction of X is smooth on the surface. Then at each p, there is locally a unique integral curve of X. Since M is complete, we can extend this integral curve indefinitely.
• Dec 27th 2011, 10:05 AM
Sogan
Re: asymptotic curve on a complete surface.
Hello,

thank you so much for your help. But i do not unterstand your last point.

"Since M is complete, we can extend this integral curve indefinitely."

is every integral curve extendable to the real line, if the surface is complete?
I mean the definition of complete surfaces just says, that every geodesic is extendable in such a way.

Regards