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 worries.

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