Embedding Non-Euclidean Spaces in Euclidean Spaces

Is it possible to isometrically embed a non-Euclidean manifold in a
Euclidean manifold of higher dimension?  If we limit ourselves to
just ONE new dimension the answer is no.  This was proved around 
1901 by Hilbert, who showed that the original non-Euclidean space 
(the 2D hyperbolic plane of Lobachevski, Bolyai, et al) cannot be 
isometrically embedded in its entirety in 3D Euclidean space.  
However, it CAN be embedded in 6D Euclidean space, and I think even 
in 5D Euclidean space (see Gromov's "Partial Differential Relations).
Apparently the question of whether there exists a complete isometric 
embedding in 4D Euclidean space remains open.  In any case, we can 
always embed a smooth metrical non-Euclidean space in a higher-
dimensional Euclidean space, but it usually takes more than just 
one extra dimension.

Of course, if we consider semi-Riemannian manifolds it complicates 
the question a bit.  If we're willing to allow imaginary "distances"
(i.e., distances whose squares are negative), then it IS possible to 
embed the hyperbolic plane in 3D Euclidean space - as the surface of 
a sphere with imaginary radius.  In fact, as early as 1766 Lambert 
observed that if you assumed there are at least two distinct lines 
through a given point that don't intersect a given line, then the 
area of a triangle with angles a,b,c would be -R^2 (a+b+c-pi) for 
some constant R.  He knew that the area of a triangle on a real 
sphere of radius R was R^2 (a+b+c-pi), so he wrote "one could almost 
conclude that the new geometry would be true on a sphere of imaginary 
radius".  It turns out that if you just plug in iR as the radius in 
any formula for spherical geometry you get the corresponding formula 
for hyperbolic geometry.

On the other hand, it isn't clear that a formally Euclidean space 
with imaginary distances is any more intuitive than a curved space 
with strictly real distances.  In fact, I would argue that the most 
un-intuitive aspect of the relativistic "metric" of spacetime is not 
it's curvature, but the fact that it isn't really a metric at all.  

A metric space, in the strict sense of the term, is a manifold that 
satisfies the triangle inequality, which is the property that leads 
to our intuitive impressions of "locality".  In particular, locality 
is transitive in a metric space, meaning that if A is close to B, 
and B is close to C, then A can't be too far from C.  Spacetime 
doesn't satisfy this condition.  There exist points A,B,C such that 
the absolute distances AB and BC are both less than the Planck 
length (10^-35 meters), and yet the distance AC is the radius of 
the observeable universe.  This has nothing to do with curvature; 
it's strictly a consequence of the fact that spacetime is not a 
metrical space and does not satisfy the triangle inequality.  In 
view of this, a little bit of curvature should be the least of our

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