Talk:Clifford parallel

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@Steelpillow: : The context is given at elliptic geometry#Elliptic space where points are versors. The point 1 has angle 0 and r arbitrary. The variable r on the 2-sphere is the axis of a great circle where the versor lives. Parallel to which line? To the line through 1. Picturing the Clifford parallelism and other elliptic relations is challenging since the versors form a 3-sphere, something beyond our sketching. — Rgdboer (talk) 02:41, 11 November 2019 (UTC)[reply]

I wish I knew. Whether a line "is", or appears, metrically straight or curved can depend on its projection into the observer's space. The fact that these lines "are in fact curves" makes me wonder whether isocurves might be a better description, though they are somewhat different from ordinary isocurves. I am not sure if Clifford parallels can exist in a 2-sphere, I think they may need at least a 3-sphere or higher. The illustrations I recall once seeing were more like spirals around an axial line, but I cannot recall the context in which these spirals were "lines". My own interest in elliptic geometries comes from the broader axiomatic treatment, and especially the sense in which certain manifestations of projective geometry (where an isocurve to a line is a conic) can be understood as elliptic. This is very different from the analytic approach via quaternions, about which I know even less. I created this article at least in part in the hope that others would answer some of these questions. — Cheers, Steelpillow (Talk) 10:38, 11 November 2019 (UTC)[reply]

Right, the Clifford parallel occurs in the 3-sphere. An animation at Clifford torus is a related representation. The word line is used in this context in the sense of a geodesic in differential geometry. — Rgdboer (talk) 01:37, 12 November 2019 (UTC)[reply]

Thank you, I may try to add some of that to the article. As an aside, the description in the Clifford torus article that it is an example of Euclidean geometry appears to be wrong. The Euclidean plane is equivalent to

\mathbb {R} ^{2}

, but the Clifford torus is equivalent only to a finite subset. If you set up a Cartesian coordinate system with the origin at the centre of the fundamental square, it will bump into itself across the joining zippers, where finite coordinate values must suddenly change sign. Sure it is locally Euclidean, but then so is any smooth coordinate manifold. — Cheers, Steelpillow (Talk) 11:54, 12 November 2019 (UTC)[reply]