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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. –  joriki Feb 26 '12 at 22:14
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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. –  Will Jagy Feb 26 '12 at 22:15
    
@joriki: I have edited it :) –  jason b Feb 26 '12 at 22:33
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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. –  Willie Wong Feb 26 '12 at 23:47
    
@WillieWong: Thank you! I will check it out. –  jason b Feb 27 '12 at 0:07

1 Answer 1

EDIT, 7:51 pm: the argument below, except for the quadratic form and the entries in the matrices $B,C,$ is exactly the argument that says, once you know that the equator is a geodesic on the unit sphere, then all the geodesics on the sphere are the great circles, based on the fact that the great circles are the intersections of the sphere with planes through the origin.

ORIGINAL: We are looking at the hyperboloid sheet $$ z = \sqrt{1 + x^2 + y^2 } $$ or $$ x^2 + y^2 - z^2 = -1, \; \; z > 0.$$

This is in Minkowski $(2+1)$ space, with quadratic form $x^2 + y^2 - z^2$ as a semi-Riemannian metric. All orientation preserving isometries of the space are the product, in whatever order you like, of the three matrices below:

$$ A \; = \; \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right) , $$

$$ B \; = \; \left( \begin{array}{ccc} \cosh u & 0 & \sinh u \\ 0 & 1 & 0 \\ \sinh u & 0 & \cosh u \end{array} \right) , $$

$$ C \; = \; \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cosh v & \sinh v \\ 0 & \sinh v & \cosh v \end{array} \right) , $$

Of course, these also preserve the sheet $ x^2 + y^2 - z^2 = -1, \; \; z > 0,$ and the subspace metric on the sheet (positive definite, needs separate proof) is preserved by definition.

Because the curve $$ (\sinh t, \; 0, \; \cosh t )$$ is invariant pointwise under the orientation reversing isometry $ y \rightarrow -y,$ it is a geodesic curve that lies in a plane $y=0$ through the origin. It can thus be carried to any other geodesic curve by the isometries. However, the isometries not only fix the origin, they take planes through the origin to to other planes through the origin.

That's about it.

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