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I am reading this article on extrinsic curvature embedding diagrams in general relativity: it seems that these are used to visualize curved space. On page 2, it is stated that in the case of the constant Schwarzschild time hypersurface in a Schwarzschild spacetime, the extrinsic curvature embedding is a flat surface. Does that mean that if you take the $t=0$ hypersurface in Schwarzchild spacetime (ie. the spatial part of the Schwarzchild metric) given by

$g_{SC}= \Bigg(1 + \frac{m_0}{2 r} \Bigg) \delta$

that the extrinsic curvature $k_{ij}$ of this hypersurface is just the flat metric $\delta_{ij}$?

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No, it means that the extrinsic curvature is zero. Don't confuse the curvature with the metric! The metric of the hypersurface is in this case conformal to a flat metric, but that's not what we're saying. The extrinsic curvature is zero because the 4D metric is static, not because of any particular form of the 3D metric.

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  • $\begingroup$ I see, so even if take the 3D metric to be a small perturbation of the $t=0$ slice of the Schwarzchild spacetime, the extrinsic curvature will still vanish due the 4D metric? $\endgroup$
    – Tom
    Jun 14 '20 at 14:28
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    $\begingroup$ @Tom What do you mean by a small perturbation? The extrinsic curvature depends on how the hypersurface is embedded in the ambient spacetime; if you change it so that it isn't a $t = \text{const}$ surface anymore, the extrinsic curvature won't be zero. $\endgroup$
    – Javier
    Jun 14 '20 at 14:41
  • $\begingroup$ I guess what I am really asking is: take a 4D manifold where the metric is given by a small perturbation of the Schwarzchild metric ie. add some metric $h_{\alpha \beta}$ with components whose size is small in a suitable norm (in general I suppose this metric will neither be spherically symmettric or static). Now if we take a spacelike hypersurface of this 4D manifold, what would the induced metric and the extrinsic curvature be? I assume the extrinsic curvature is no longer going to vanish in this case $\endgroup$
    – Tom
    Jun 14 '20 at 16:56
  • $\begingroup$ @Tom yes, in general there would be no reason for the extrinsic curvature to vanish. That and the induced metric depend on the perturbation and on the way you choose the hypersurface, for which you have pretty much complete freedom. $\endgroup$
    – Javier
    Jun 14 '20 at 17:35
  • $\begingroup$ Could one just choose the hypersurface at $t=0$, as one wants an initial data anyway. In this case, what would the induced metric and the extrinsic curvature be for the hypersurface (or at least what would the general form be assuming any perturbation $h_{\alpha \beta}$? $\endgroup$
    – Tom
    Jun 14 '20 at 17:47

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