In article <email@example.com>, firstname.lastname@example.org wrote:
> Can the Lobachevsky plane be embedded into R^3 ? > > (The Lobachevsky metric must be induced by the standard eucledean > metric in R^3). > > I have heard that the answer is "no". If so, anybody knows what are > the main ideas of the proof?
The following is a cut-and-paste from an email I wrote some time ago, but it should be fairly clear:
I don't know what the book meant by embedding, but usually they are referring to something like a C^2 embedding, which is not possible. The idea behind that goes as follows: consider the lines of curvature; they are two families of orthogonal curves. If you draw them on a disc, you will see that somewhere in the middle, you will run into n-prong singularities in these lines of curvatures. This is intuitively clear if the surface is very wavy near the edges. Basically you need to eliminate the case where one family of curvature lines foliates the disc.
Hilbert proved this way back when, but his proof was, I think, quite different. I seem to recall his proof was much more analytic. The sketch I gave is of a proof shown to me by Thurston (possibly he learned it from somewhere).
Now, the point is once you relax the C^2 condition, to say, C^1, the embedding is now possible. This was shown by Kuiper.