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Re: Can the Lobachevsky plane be embedded into R^3 ?
Posted:
Mar 28, 2007 4:03 PM
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In article <1173128819.567151.23640@c51g2000cwc.googlegroups.com>,
ksp4@msiu.ru 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.
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