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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