There's a manifold where you take a hyper-spherinder and connect the ends. If you connect them the direct way, you get a S3 x S1 product manifold.
The S3 manifold is not the sphere but the glome. It's called S3 because the hypersurface itself is three-dimensional at each of its points. However, it can be embedded in R4. The circle is S1.
n topology, a torus is the product of circles. a 4-torus (T4) is S1 x S1 x S1 x S1. It is Euclidean. I call S(n-1) x S1 an "n-doughnut".
but still, what about polar euclidean space, aka 'counterspace' ? does it have anything to do with birational geometries ?
I don't know how many dimensions there are in five dimensions. I'm not sure if anyone does. The isotropic geometries (Sn, En, Hn) are easy, and so are the products. Projective and hyperbolic spaces can be real, complex, quaternion, or octonion. (Octonion dimensions are limited to 2, which has 16 real dimensions). The real n-projective spaces always have Sn geometry. Then there are geometries based on non-product fiber bundles, but questions arise as to how many different ways each bundle can "twist" or "squeeze".
Take something easier, like S2xE1, for example. How do the straight lines look there? Some will be the great circles of S2, some will be straight lines of E1, but what about the rest? The best I'm able to visualize them is as sort of spirals, with two coordinates going in a great circle while the third steadily moves upwards. How would look the lines, planes, hyperplanes, in the various non-isotropic geometries?
Is there a "parallel postulate" for counterspace? Say: Given a point and a line not through the point, there is exactly one point on the line that is not collinear with the given point... While at the same time you have a postulate saying that any two lines intersect?
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Yes there is !
In counterspace there are no parallel planes, but there are parallel points (which follows from the duality involved). So the postulate is:
"Given any line and a plane not containing it, there is exactly one line in that plane which is polar-parallel to the given line"
What this means is that there is only one line in the plane which contains a plane both through the given line and the counterspace absolute point (polar to the plane at infinity).
It is easily seen as the absolute point and the given line determine a plane which intersects the given plane in just one line.
What may be confusing if you are not used to counterspace is that polar-parallel lines meet!
However this is simply polar to the fact that in Beltrami's version of the postulate the two ordinarily parallel lines share a common plane.
I hope this helps.
Best wishes,
Nick Thomas
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