# Extrinsic Curvature in Schwarzchild Space

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 $$\delta_{ij}$$?

• 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? – Tom Jun 14 '20 at 14:28
• @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. – Javier Jun 14 '20 at 14:41
• 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 – Tom Jun 14 '20 at 16:56
• 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 $h_{\alpha \beta}$? – Tom Jun 14 '20 at 17:47