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I was reading the Wikipedia article on hyperbolic geometry and have come across the line

geodesic paths are described by intersections with planes through the origin

Why is this necessarily true? In other words, why must the geodesics on the hyperboloid necessarily of this form/generated in this way?

Thank you.

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    I wish people wouldn't refer to Wikipedia as "Wiki". A wiki is a far more general thing than Wikipedia.2012-02-26
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    It's a long story, so I won't make an answer. As such curves are geodesics, and pass through any given point with any prescribed direction, they are therefore all the geodesics. Uniqueness of ODE system. Equidistant curves are the intersection with planes parallel to those, horocycles from planes parallel to the edge of the cone, actual geodesic circles planes that intersect the hyperboloid sheet in a closed path.2012-02-26
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    @joriki: I have edited it :)2012-02-26
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    The statement, in fact, is true for any hyperquadric embedded in a semi-Riemannian scalar-product space. I don't have the time to write up a proof now (maybe later), so I'll give you a reference: see page 112, Proposition 4.28 of Barrett O'Neill's _Semi-Riemannian Geometry_.2012-02-26
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    @WillieWong: Thank you! I will check it out.2012-02-27
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    @joriki: I will remember that next time, when I am not that lazy. ;)2012-02-27

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