I'm researching about Hilbert's theorem which says that there isn't isometric immersion of a complete surface with constant negative Gaussian curvature in $\mathbb{R}^3$. I'm taking as a reference the book "Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo" and also "A Comprehensive Introduction to Differential Geometry, Vol. 3 by Michael Spivak". I have only two specific doubts:

In the attached document, the Spivak's proof is presented first in English and then the do Carmo's proof in Spanish: click here.

  • The first is that it must be shown that an asymptotic curve on a complete surface with constant negative Gaussian curvature can be defined in all $\mathbb{R}$, I have framed it in red in the attached document. The arguments mentioned in the documents I know are valid for compact surfaces but not for complete surfaces and I have not been able to come up with a clear proof of this result.
  • The second is about the injectivity of $X(s, t)$ in the first edition of do Carmo's book, he uses two lemmas within which he makes cases to arrive at the result, while in the current edition (2016) he mentions how to get there To that result of a faster one using coating applications but I haven't been able to find a clear proof, I haven't gotten stuck and I cann't get out of that, I have framed it in orange in the document.

I have been justifying the steps that both authors leave without proof, in order to fully understand this result but I cannot understand those two points that I mention, I hope they can help me, I thank you in advance.


1 Answer 1


This only answers your first confusion.

If the curve not defined on all of $\mathbb{R}$, then it can only be defined on some interval $I\subsetneq\mathbb{R}$. By symmetry, we may assume that either $I=[0,a)$ or $I=[0,a]$; but we assume that $a$ is maximal.

In the first case, we can take a sequence of times $\{t_j\}_j$ converging to $a$ from below. Since $c$ has unit derivative, it is $1$-Lipschitz; that is, since $\{t_j\}_j$ is Cauchy, so must the sequence $\{c(t_j)\}_j$ be. _Since $M$ is complete, we can extend $c$ to $a$ continuously. A little more work shows that the extension is in fact $C^1$, which reduces to the second case.

In the second case, note that $c(a)$ is some point. So we can create a new integral curve by solving our differential equation starting at $a$. But the right- and left-derivatives around $a$ are already prescribed by our ODE: they are precisely the vector field at $a$. So they must be equal, and we can extend $c$ to some larger interval.


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