
Re: Can the Lobachevsky plane be embedded into R^3 ?
Posted:
Mar 15, 2007 10:55 AM


Hero wrote: > Denis Feldmann wrote: >> Hero a écrit : >> >>> Thomas Mautsch wrote: >>>> Hero wrote: >>>>> On 5 Mrz., 22:07, k...@msiu.ru wrote: >>>>>> Can the Lobachevsky plane be embedded into R^3 ? >>> ...... >>> ...... >>>>> The answer is no. >>>>> Lobachevsky is contradicting Euclid's fifth postulate. >>>> This is completely irrelevant nonsense! >>> You can read in Wiki: > >> Do you realize at what level the answer was posted ? Thanks for >> reminding us all basic truths, and insisting on your misconceptions... >> It was *still* irrelevant (nonsense, I would not have been so harsh on >> first posting, but now...) >> >>> ,,Lobachevsky would instead develop a geometry in which the fifth >>> postulate was not true >>> ... >>> Lobachevsky replaced Euclid's parallel postulate with the postulate >>> that there is more than one parallel line through any given point; a >>> famous consequence is that the sum of angles in a triangle must be >>> less than 180 degrees. ,, >>> http://en.wikipedia.org/wiki/Nicolai_Ivanovich_Lobachevsky >> Somehow, I had missed that. Who is that guy again? > > Here is that guy Lobachevsky: > Geometrical researches on the theory of parallels > http://www.hti.umich.edu/cgi/t/text/textidx?c=umhistmath;idno=AAN2339 > > Lower level of completely irrelevant nonsense... > >> Was Euclid wrong, >> after all? > > Lobachevsky says, that he was wrong. What is Your answer?
This is what I think: I suggest that you think about geometry on a spherical surface. Given points A and B on a sphere, if A !=B, then the shortest path on the sphere from A to B is an arc of circle whose center = center of sphere. So geodesics, "shortestroute" curves, are formed by great circles, those circles with center = center of sphere. In the Euclidean plane, every straight line (geodesic of the plane) has parallel companions, i.e. lines distinct from it, but which never intersect it.
Going back to a spherical surface, can two geodesics fail to intersect? The answer is no. Note that the geodesics on the sphere are circles in space, not straight lines.
As far as I know, what I have discussed with geodesics of a sphere is close to an example of an embedding.
Let's say: in an embedding of some Riemannian surface in R^n, the geodesics of the surface often are not made up of straight line segments of R^n.
David Bernier
[ warp sqrt(1) ahead]

