Topology of de Sitter and black hole

  1. Mar 18, 2006 #1
    What does it mean in the statement "Topologically, de Sitter space is R × S^n-1..."
    What is the topology of a Schwarzschild black hole?
     
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  3. Mar 18, 2006 #2

    George Jones

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    Every spacetime is a differentiable manifold, and every differentiable manifold is a topological space. The underlying topoogical space for de Sitter spacetime is the topological product R x S^3. Here, R is the set of real numbers, and is, roughly, a space of time coordinates. S^3, the compact 3-dimensional hypersurface of a 4-dimensional ball, is a space of spatial coordinates.

    R^2 x S^S. Here, S^2 is the 2-dimesional surface of a 3-dimensional ball, and represents the spherical symmetry of Schwarzschild spacetime. R^2 is the set of ordered pairs of real numbers, and is the space of t and r coordinates.

    Regards,
    George
     
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